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|---|---|---|---|---|---|---|---|---|---|---|---|
luz2uxam
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>The total number of species from the following in which one unpaired electron is present, is _______.</p>
<p>$$\mathrm{N}_2, \mathrm{O}_2, \mathrm{C}_2^{-}, \mathrm{O}_2^{-}, \mathrm{O}_2^{2-}, \mathrm{H}_2^{+}, \mathrm{CN}^{-}, \mathrm{He}_2^{+}$$</p>
|
[]
| null |
4
|
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-baqh{text-align:center;vertical-align:top}
.tg .tg-amwm{font-weight:bold;text-align:center;vertical-align:top}
</style>
<table class="tg" style="undefined;table-layout: fixed; width: 351px">
<colgroup>
<col style="width: 175px">
<col style="width: 176px">
</colgroup>
<thead>
<tr>
<th class="tg-amwm">Species</th>
<th class="tg-amwm">Unpaired e</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-baqh">$$<br>\mathrm{N}_2<br>$$</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{O}_2<br>$$</td>
<td class="tg-baqh">2</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{C}_2^{-}<br>$$</td>
<td class="tg-baqh">1</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{O}_2^{-}<br>$$</td>
<td class="tg-baqh">1</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{O}_2^{2-}<br>$$</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{H}_2^{+}<br>$$</td>
<td class="tg-baqh">1</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{CN}^{-}<br>$$</td>
<td class="tg-baqh">0</td>
</tr>
<tr>
<td class="tg-baqh">$$<br>\mathrm{He}_2^{+}<br>$$</td>
<td class="tg-baqh">1</td>
</tr>
</tbody>
</table></p>
|
integer
|
jee-main-2024-online-9th-april-morning-shift
| 988
|
lv0vytmr
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Number of molecules/species from the following having one unpaired electron is ________.</p>
<p>$$\mathrm{O}_2, \mathrm{O}_2^{-1}, \mathrm{NO}, \mathrm{CN}^{-1}, \mathrm{O}_2^{2-}$$</p>
|
[]
| null |
2
|
<p>$$\mathrm{O}_2^{-} \text {and } \mathrm{NO} \text { have } 1 \text { unpaired electron. }$$</p>
|
integer
|
jee-main-2024-online-4th-april-morning-shift
| 989
|
lv3xm5q9
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>When $$\psi_{\mathrm{A}}$$ and $$\psi_{\mathrm{B}}$$ are the wave functions of atomic orbitals, then $$\sigma^*$$ is represented by :</p>
|
[{"identifier": "A", "content": "$$\\psi_A+2 \\psi_B$$\n"}, {"identifier": "B", "content": "$$\\psi_{\\mathrm{A}}-\\psi_{\\mathrm{B}}$$\n"}, {"identifier": "C", "content": "$$\\psi_A-2 \\psi_B$$\n"}, {"identifier": "D", "content": "$$\\psi_{\\mathrm{A}}+\\psi_{\\mathrm{B}}$$"}]
|
["B"]
| null |
<p>In Molecular Orbital Theory, molecular orbitals are formed by the linear combination of atomic orbitals (LCAO). The wave functions of atomic orbitals $\psi_A$ and $\psi_B$ can combine in two ways:</p>
<ol>
<li><strong>Constructive Interference</strong>: This leads to the formation of a bonding molecular orbital ($\sigma$):</li>
</ol>
<p>$ \sigma = \psi_A + \psi_B $</p>
<ol>
<li><strong>Destructive Interference</strong>: This leads to the formation of an antibonding molecular orbital ($\sigma^*$):</li>
</ol>
<p>$ \sigma^* = \psi_A - \psi_B $</p>
<h3>Conclusion</h3>
<p>The antibonding molecular orbital $\sigma^*$ is represented by:</p>
<p>$ \sigma^* = \psi_A - \psi_B $</p>
<p>Therefore, the correct answer is:</p>
<p>Option B: $\psi_{\mathrm{A}} - \psi_{\mathrm{B}}$</p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
| 990
|
lv40v9yu
|
chemistry
|
chemical-bonding-and-molecular-structure
|
molecular-orbital-theory
|
<p>Number of molecules having bond order 2 from the following molecules is _________.</p> <p>$$\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$$</p>
|
[]
| null |
2
|
<p>To determine the number of molecules with a bond order of 2 from the given molecules, we need to first calculate the bond order for each molecule. The bond order can be determined using Molecular Orbital Theory (MOT). The bond order is given by the formula:</p>
<p>
<p>$$\text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of anti-bonding electrons}}{2}$$</p>
</p>
<p>Let's evaluate each molecule individually:</p>
<p><strong>1. $$\mathrm{C}_2$$:</strong></p>
<p>For $$\mathrm{C}_2$$, the total number of electrons is 12. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$$. There are 8 bonding electrons and 4 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{8 - 4}{2} = 2$$</p>
</p>
<p><strong>2. $$\mathrm{O}_2$$:</strong></p>
<p>For $$\mathrm{O}_2$$, the total number of electrons is 16. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^1 \left( \pi_{2p_y}^* \right)^1$$. There are 10 bonding electrons and 6 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{10 - 6}{2} = 2$$</p>
</p>
<p><strong>3. $$\mathrm{Be}_2$$:</strong></p>
<p>For $$\mathrm{Be}_2$$, the total number of electrons is 8. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2$$. There are 4 bonding electrons and 4 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{4 - 4}{2} = 0$$</p>
</p>
<p><strong>4. $$\mathrm{Li}_2$$:</strong></p>
<p>For $$\mathrm{Li}_2$$, the total number of electrons is 6. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2$$. There are 4 bonding electrons and 2 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{4 - 2}{2} = 1$$</p>
</p>
<p><strong>5. $$\mathrm{Ne}_2$$:</strong></p>
<p>For $$\mathrm{Ne}_2$$, the total number of electrons is 20. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2 \left( \pi_{2p_x}^* \right)^2 \left( \pi_{2p_y}^* \right)^2 \left( \sigma_{2p_z}^* \right)^2$$. There are 10 bonding electrons and 10 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{10 - 10}{2} = 0$$</p>
</p>
<p><strong>6. $$\mathrm{N}_2$$:</strong></p>
<p>For $$\mathrm{N}_2$$, the total number of electrons is 14. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2 \left( \sigma_{2s} \right)^2 \left( \sigma_{2s}^* \right)^2 \left( \sigma_{2p_z} \right)^2 \left( \pi_{2p_x} \right)^2 \left( \pi_{2p_y} \right)^2$$. There are 10 bonding electrons and 4 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{10 - 4}{2} = 3$$</p>
</p>
<p><strong>7. $$\mathrm{He}_2$$:</strong></p>
<p>For $$\mathrm{He}_2$$, the total number of electrons is 4. The molecular orbital configuration will be $$\left( \sigma_{1s} \right)^2 \left( \sigma_{1s}^* \right)^2$$. There are 2 bonding electrons and 2 anti-bonding electrons:</p>
<p>
<p>$$\text{Bond Order} = \frac{2 - 2}{2} = 0$$</p>
</p>
<p>Thus, the molecules with a bond order of 2 from the given list are $$\mathrm{C}_2$$ and $$\mathrm{O}_2$$. Therefore, the number of molecules having bond order 2 is <strong>2</strong>.</p>
|
integer
|
jee-main-2024-online-8th-april-evening-shift
| 991
|
1krx8h5dq
|
chemistry
|
chemical-bonding-and-molecular-structure
|
resonance
|
Identify the species having one $$\pi$$-bond and maximum number of canonical forms from the following :
|
[{"identifier": "A", "content": "SO<sub>3</sub>"}, {"identifier": "B", "content": "O<sub>2</sub>"}, {"identifier": "C", "content": "SO<sub>2</sub>"}, {"identifier": "D", "content": "CO$$_3^{2 - }$$"}]
|
["D"]
| null |
Among SO<sub>3</sub>, O<sub>2</sub>, SO<sub>2</sub> and CO$$_3^{2 - }$$, only O<sub>2</sub> and CO$$_3^{2 - }$$ has only one $$\pi$$-bond.<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264623/exam_images/ta7fcu4vvzhmhpraq0w4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Evening Shift Chemistry - Chemical Bonding & Molecular Structure Question 124 English Explanation">
|
mcq
|
jee-main-2021-online-25th-july-evening-shift
| 992
|
jaoe38c1lsfjlsrq
|
chemistry
|
chemical-bonding-and-molecular-structure
|
resonance
|
<p>The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is</p>
|
[{"identifier": "A", "content": "electromeric energy\n"}, {"identifier": "B", "content": "resonance energy\n"}, {"identifier": "C", "content": "ionization energy\n"}, {"identifier": "D", "content": "hyperconjugation energy"}]
|
["B"]
| null |
<p>The difference in energy between the actual structure and the lowest energy resonance structure for the given compound is known as resonance energy.</p>
|
mcq
|
jee-main-2024-online-29th-january-morning-shift
| 993
|
SH9c4822qbHkKs4MGODP9
|
chemistry
|
chemical-bonding-and-molecular-structure
|
valence-bond-theory
|
The pair that contains two P – H bonds in each of the oxoacids is
|
[{"identifier": "A", "content": "H<sub>3</sub>PO<sub>2</sub> and H<sub>4</sub>P<sub>2</sub>O<sub>5</sub>"}, {"identifier": "B", "content": "H<sub>4</sub>P<sub>2</sub>O<sub>5</sub> and H<sub>4</sub>P<sub>2</sub>O<sub>6</sub>"}, {"identifier": "C", "content": "H<sub>4</sub>P<sub>2</sub>O<sub>5</sub> and H<sub>3</sub>PO<sub>3</sub>"}, {"identifier": "D", "content": "H<sub>3</sub>PO<sub>3</sub> and H<sub>3</sub>PO<sub>2</sub>"}]
|
["A"]
| null |
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263401/exam_images/a7dwpsa6n7njmos7lcxb.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263999/exam_images/xrv5xps3c0fvcjfmqcev.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 169 English Explanation 1"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265378/exam_images/sm33nsrvszcyeivjhrcg.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264626/exam_images/zhofsp7zcvgqhmzovays.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 169 English Explanation 2"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267507/exam_images/n9rsmyclopql45lr67s1.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266548/exam_images/uwwerstvtftk5tuzpkru.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 169 English Explanation 3"></picture>
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267628/exam_images/fswywbubdxaxq5r4ncxk.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267312/exam_images/dvvhhtk6jrakbgytptip.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Chemistry - Chemical Bonding & Molecular Structure Question 169 English Explanation 4"></picture>
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
| 994
|
lsbn3ilu
|
chemistry
|
chemical-bonding-and-molecular-structure
|
valence-bond-theory
|
The lowest oxidation number of an atom in a compound $\mathrm{A}_2 \mathrm{B}$ is -2 . The number of electrons in its valence shell is _______.
|
[]
| null |
6
|
$\mathrm{A}_2 \mathrm{~B} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{B}^{-2}$
<p>When an atom has the lowest oxidation number of -2 in a compound, it means the atom has gained two electrons beyond its neutral state to achieve this oxidation state. This is because gaining electrons makes the oxidation number more negative. In a neutral atom, the electrons in the outermost shell are known as valence electrons, which play a crucial role in chemical reactions and bonding.</p><p>In the given compound $\mathrm{A}_2\mathrm{B}$, atom B has an oxidation state of -2, indicating it has gained two electrons. For an atom to gain two electrons to complete its octet implies that its neutral state had six valence electrons. Therefore, before gaining electrons, the atom B would have had six electrons in its valence shell to begin with. As it gains two more electrons, it reaches an oxidation number of -2, implying it now has a complete octet or eight electrons in its valence shell, but the original count of valence electrons in its neutral state was six.</p>
|
integer
|
jee-main-2024-online-1st-february-morning-shift
| 995
|
VLonTGvaaXvC2Rn2Px7k9k2k5dz61w9
|
chemistry
|
chemical-bonding-and-molecular-structure
|
van-der-walls-forces
|
The relative strength of interionic/intermolecular forces in decreasing order is :
|
[{"identifier": "A", "content": "dipole-dipole $$>$$ ion-dipole $$>$$ ion-ion"}, {"identifier": "B", "content": "ion-dipole $$>$$ dipole-dipole $$>$$ ion-ion"}, {"identifier": "C", "content": "ion-dipole $$>$$ ion-ion $$>$$ dipole-dipole"}, {"identifier": "D", "content": "ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole"}]
|
["D"]
| null |
Ionic interactions are stronger as compared to
van der waal interactions.
<br><br>So, correct order is
<br><br>ion-ion $$>$$ ion-dipole $$>$$ dipole-dipole
|
mcq
|
jee-main-2020-online-7th-january-morning-slot
| 996
|
zIzC0tNqd3aPhBvL
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction CO (g) + (1/2) O<sub>2</sub> (g) $$\leftrightharpoons$$ CO<sub>2</sub> (g), K<sub>p</sub>/K<sub>c</sub> is :
|
[{"identifier": "A", "content": "RT"}, {"identifier": "B", "content": "(RT)<sup>-1</sup>"}, {"identifier": "C", "content": "(RT)<sup>-1/2</sup>"}, {"identifier": "D", "content": "(RT)<sup>1/2</sup>"}]
|
["C"]
| null |
$${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$$
<br><br>$$\Delta n = 1 - \left( {1 + {1 \over 2}} \right)$$
<br><br>$$ = 1 - {3 \over 2} = - {1 \over 2}.$$
<br><br>$$\therefore$$ $$\,\,\,\,{{{K_p}} \over {{K_c}}} = {\left( {RT} \right)^{ - 1/2}}$$
|
mcq
|
aieee-2002
| 997
|
xnMc4pIlDWh0AoYp
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction, CO(g) + Cl<sub>2</sub>(g) $$\leftrightharpoons$$ COCl<sub>2</sub>(g) the $${{{K_p}} \over {{K_c}}}$$ is equal to :
|
[{"identifier": "A", "content": "$$\\sqrt {RT} $$"}, {"identifier": "B", "content": "RT"}, {"identifier": "C", "content": "1/RT"}, {"identifier": "D", "content": "1.0"}]
|
["C"]
| null |
$${K_p} = {K_c}{\left( {RT} \right)^{\Delta n}};$$
<br><br>Here $$\Delta n = 1 - 2 = - 1$$
<br><br>$$\therefore$$ $${{{k_p}} \over {{K_c}}} = {1 \over {RT}}$$
|
mcq
|
aieee-2004
| 999
|
FOPtW9AewX3ZEoEr
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
What is the equilibrium expression for the reaction <br/>
P<sub>4</sub> (s) + 5O<sub>2</sub> $$\leftrightharpoons$$ P<sub>4</sub>O<sub>10</sub> (s)?
|
[{"identifier": "A", "content": "K<sub>c</sub> = [P<sub>4</sub>O<sub>10</sub>] / 5[P<sub>4</sub>] [O<sub>2</sub>] "}, {"identifier": "B", "content": "K<sub>c</sub> = 1/[O<sub>2</sub>]<sup>5</sup>"}, {"identifier": "C", "content": "K<sub>c</sub> = [P<sub>4</sub>O<sub>10</sub>] / [P<sub>4</sub>] [O<sub>2</sub>]<sup>5</sup>"}, {"identifier": "D", "content": "K<sub>c</sub> = [O<sub>2</sub>]<sup>5</sup>"}]
|
["B"]
| null |
For $${P_4}\left( s \right) + 5{O_2}\left( g \right)\,\rightleftharpoons\,{P_4}{O_{10}}\left( 8 \right)$$
<br><br>$${K_c} = {1 \over {{{\left( {{O_2}} \right)}^5}}}.$$ The solids have concentration unity
|
mcq
|
aieee-2004
| 1,000
|
jBYItI7R5Faztuq9
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant for the reaction N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\leftrightharpoons$$ 2NO(g) at temperature T is
4 $$\times$$ 10<sup>-4</sup>. The value of K<sub>c</sub> for the reaction NO(g) $$\leftrightharpoons$$ $$1 \over 2$$N<sub>2</sub> (g) + $$1 \over 2$$O<sub>2</sub> (g) at the same temperature is :
|
[{"identifier": "A", "content": "2.5 $$\\times$$ 10<sup>2</sup>"}, {"identifier": "B", "content": "4 $$\\times$$ 10<sup>-4</sup>"}, {"identifier": "C", "content": "50"}, {"identifier": "D", "content": "0.02"}]
|
["C"]
| null |
$${K_c} = {{{{\left[ {NO} \right]}^2}} \over {\left[ {{N_2}} \right]\left[ {{O_2}} \right]}} = 4 \times {10^{ - 4}}$$
<br><br>$$K{'_c} = {{{{\left[ {{N_2}} \right]}^{1/2}}{{\left[ {{Q_2}} \right]}^{1/2}}} \over {\left[ {NO} \right]}}$$
<br><br>$$ = {1 \over {\sqrt {{K_c}} }}$$
<br><br>$$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$$
<br><br>$$ = 50$$
|
mcq
|
aieee-2004
| 1,001
|
es64H9nSDhNRSnA4
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction
2NO<sub>2</sub> (g) $$\leftrightharpoons$$ 2NO (g) + O<sub>2</sub> (g), (K<sub>c</sub> = 1.8 $$\times$$ 10<sup>-6</sup> at 184<sup>o</sup>C) (R = 0.0831 kJ/(mol. K))<br/>
When K<sub>p</sub> and K<sub>c</sub> are compared at 184<sup>o</sup>C , it is found that :
|
[{"identifier": "A", "content": "K<sub>p</sub> is greater than K<sub>c</sub>"}, {"identifier": "B", "content": "K<sub>p</sub> is less than K<sub>c</sub>"}, {"identifier": "C", "content": "K<sub>p</sub> = K<sub>c</sub>"}, {"identifier": "D", "content": "Whether K<sub>p</sub> is greater than, less than or equal to K<sub>c</sub> depends upon the total gas\npressure "}]
|
["A"]
| null |
For the reaction : -
<br><br>$$2N{O_2}\left( g \right)\rightleftharpoons2NO\left( g \right) + {O_2}\left( g \right)$$
<br><br>Given $${K_c} = 1.8 \times {10^{ - 6}}\,\,$$ at $$\,\,{184^ \circ }C$$
<br><br>$$R=0.0831$$ $$\,\,kJ/mol.k$$
<br><br>$${K_p} = 1.8 \times {10^{ - 6}} \times 0.0831 \times 457$$
<br><br>$$ = 6.836 \times {10^{ - 6}}$$
<br><br>$$\left[ {} \right.$$ as $$\,\,\,\,\,{184^ \circ }C = \left( {273 + 184} \right) = 457\,k,\,\,$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. {\Delta n = \left( {2 + 1, - 1} \right) = 1} \right]$$
<br><br>Hence it is clear that $${K_p} > {K_c}$$
|
mcq
|
aieee-2005
| 1,002
|
7T1O7dmxQQdrgE6B
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Phosphorus pentachloride dissociates as follows, in a closed reaction vessel<br/>
PCl<sub>5</sub> (g) $$\leftrightharpoons$$ PCl<sub>3</sub> (g) + Cl<sub>2</sub> (g) <br/>
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl<sub>5</sub> is x, the
partial pressure of PCl<sub>3</sub> will be
|
[{"identifier": "A", "content": "$$\\left( {{x \\over {x + 1}}} \\right)P$$ "}, {"identifier": "B", "content": "$$\\left( {{2x \\over {1 - x}}} \\right)P$$ "}, {"identifier": "C", "content": "$$\\left( {{x \\over {x - 1}}} \\right)P$$ "}, {"identifier": "D", "content": "$$\\left( {{x \\over {1 - x}}} \\right)P$$ "}]
|
["A"]
| null |
$$\mathop {PC{l_5}\left( g \right)}\limits_{1 - x} \mathop {\,\rightleftharpoons\,PC{l_3}\left( g \right)}\limits_x \,\, + \,\,\mathop {C{l_2}\left( g \right)}\limits_x $$
<br><br>Total moles after dissociation
<br><br>$$1 - x + x + x = 1 + x$$
<br><br>$${P_{PC{l_3}}} = $$ mole fraction of $$PCl{}_3 \times $$ Total pressure
<br><br>$$ = \left( {{x \over {1 + x}}} \right)P$$
|
mcq
|
aieee-2006
| 1,004
|
BN4EdeIy0Gczcmki
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant for the reaction <br/>
SO<sub>3</sub> (g) $$\leftrightharpoons$$ SO<sub>2</sub> (g) + $$1 \over 2$$ O<sub>2</sub> (g)<br/>
is K<sub>c</sub> = 4.9 $$\times$$ 10<sup>–2</sup>. The value of K<sub>c</sub> for the reaction<br/>
2SO<sub>2</sub> (g) + O<sub>2</sub> (g) $$\leftrightharpoons$$ 2SO<sub>3</sub> (g) will be :
|
[{"identifier": "A", "content": "416"}, {"identifier": "B", "content": "9.8 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "C", "content": "4.9 $$\\times$$ 10<sup>-2</sup>"}, {"identifier": "D", "content": "2.40 $$\\times$$ 10<sup>-3</sup>"}]
|
["A"]
| null |
$$S{O_3}\left( g \right)\,\rightleftharpoons\,S{O_2}\left( g \right)\,\, + \,\,{1 \over 2}{O_2}\left( g \right)$$
<br><br>$${K_c} = {{\left[ {S{O_2}} \right]{{\left[ {{O_2}} \right]}^{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left[ {S{O_3}} \right]}}$$
<br><br>$$ = 4.9 \times {10^{ - 2}};$$
<br><br>On taking the square of the above reaction
<br><br>$${{{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]} \over {{{\left[ {S{O_3}} \right]}^2}}}$$
<br><br>$$ = 24.01 \times {10^{ - 4}}$$
<br><br>now $$K{'_C}$$
<br><br>for $$\,\,2S{O_2}\left( g \right)\,\, + \,\,{O_2}\left( g \right)\,\rightleftharpoons\,2S{O_3}$$
<br><br>$$ = {{{{\left[ {S{O_3}} \right]}^2}} \over {{{\left[ {S{O_2}} \right]}^2}\left[ {{O_2}} \right]}}$$
<br><br>$$ = {1 \over {24.01 \times {{10}^{ - 4}}}}$$
<br><br>$$ = 416$$
|
mcq
|
aieee-2006
| 1,005
|
3yX4MhgoPTGVHeB5
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constants K<sub>P1</sub> and K<sub>P2</sub> for the reactions X $$\leftrightharpoons$$ 2Y and Z $$\leftrightharpoons$$ P + Q, respectively are in
the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressure at
these equilibria is :
|
[{"identifier": "A", "content": "1 : 36"}, {"identifier": "B", "content": "1 : 1"}, {"identifier": "C", "content": "1 : 3"}, {"identifier": "D", "content": "1 : 9"}]
|
["A"]
| null |
Let the initial moles of $$X$$ be $$'a'$$
<br><br>and that of $$Z$$ be $$'b'$$ the for the given reactions,
<br><br>we have $$X\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,2Y$$
<br><br>$$\eqalign{
& Initial\,\,\,\,\,\,\,\,\,\,\,\,a\,\,moles\,\,\,\,\,\,\,0 \cr
& At\,\,equi.\,\,\,\,\,\,\,a\left( {1 - \alpha } \right)\,\,\,\,\,\,\,2a\alpha \cr
& (moles) \cr} $$
<br><br>Total no. of moles $$ = a\left( {1 - \alpha } \right) + 2a\alpha $$
<br><br>$$ = a - a\alpha + 2a\alpha $$ $$ = a\left( {1 + \alpha } \right)$$
<br><br>Now, $$\,\,\,{K_{{p_1}}} = {{{{\left( {{n_y}} \right)}^2}} \over {{n_x}}} \times {\left( {{{{P_{{T_1}}}} \over {\sum n }}} \right)^{\Delta n}}$$
<br><br>or, $$\,\,\,{K_{{p_1}}} = {{{{\left( {2a\alpha } \right)}^2}.{P_{{T_1}}}} \over {\left[ {a\left( {1 - \alpha } \right)} \right]\left[ {a\left( {1 + \alpha } \right)} \right]}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$Z\,\,\,\,\,\,\,\,\,\rightleftharpoons\,\,\,\,\,\,\,\,\,P+Q$$
<br><br>$$\eqalign{
& Initial\,\,\,\,\,\,\,\,\,\,\,b\,\,moles\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,0 \cr
& Ar\,\,equi.\,\,\,\,\,\,b\left( {1 - \alpha } \right)\,\,\,\,\,\,\,\,b\alpha \,\,\,\,\,\,\,b\alpha \cr
& (moles) \cr} $$
<br><br>Total no. of moles
<br><br>$$ = b\left( {1 - \alpha } \right) + b\alpha + b\alpha $$
<br><br>$$ = b - b\alpha + b\alpha + b\alpha $$
<br><br>$$ = b\left( {1 + \alpha } \right)$$
<br><br>Now $$\,\,\,{K_{{P_2}}} = {{{n_Q} \times {n_P}} \over {{n_z}}} \times {\left[ {{{{P_{{T_2}}}} \over {\sum\nolimits_n \, }}} \right]^{\Delta n}}$$
<br><br>or$$\,\,\,{K_{{P_2}}} = {{\left( {b\alpha } \right)\left( {b\alpha } \right).{P_{{T_2}}}} \over {\left[ {b\left( {1 - \alpha } \right)} \right]\left[ {b\left( {1 + \alpha } \right)} \right]}}$$
<br><br>or$$\,\,\,$$ $$\,\,\,{{{K_{{P_1}}}} \over {{K_{P2}}}} = {{4{\alpha ^2}.{P_{{T_1}}}} \over {\left( {1 - {\alpha ^2}} \right)}} \times {{{{\left( {1 - \alpha } \right)}^2}} \over {{P_{{T_2}}}.{\alpha ^2}}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$ = {{4{P_{{T_1}}}} \over {{P_{{T_2}}}}}$$
<br>or$$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over 9}\,\,\,\,\,$$ [ as $$\,\,\,{{{K_{{P_1}}}} \over {{K_{{P_1}}}}} = {1 \over 9}\,\,$$ given ]
<br><br>or$$\,\,\,{{{P_{{T_1}}}} \over {{P_{T2}}}} = {1 \over {36}}$$
<br><br>or$$\,\,\,1:36$$
|
mcq
|
aieee-2008
| 1,006
|
07XXSkPNgWGKLYw1
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
A vessel at 1000 K contains CO<sub>2</sub> with a pressure of 0.5 atm. Some of the CO<sub>2</sub> is converted into CO on
the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
|
[{"identifier": "A", "content": "3 atm "}, {"identifier": "B", "content": "0.3 atm "}, {"identifier": "C", "content": "0.18 atm"}, {"identifier": "D", "content": "1.8 atm "}]
|
["D"]
| null |
$$C{O_2} + {C_{\left( {grapnite} \right)}}\,\rightleftharpoons\,2CO$$
<br><br>$${P_{initial}}\,\,0.5atm\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$$
<br><br>$${P_{final}}\,\,\,\left( {0.5 - x} \right)atm\,\,\,\,\,\,\,2x\,atm$$
<br><br>Total $$P$$ at equilibrium
<br><br>$$ = 0.5 - x + 2x = 0.5 + x\,atm$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0.8 = 0.5 + x$$
<br><br>$$\therefore$$ $$\,\,\,x = 0.8 - 0.5 = 0.3\,atm$$
<br><br>Now $$\,\,\,{k_p} = {\left( {{P_{CO}}} \right)^2}/{P_{C{O_2}}}$$
<br><br>$$ = {{{{\left( {2 \times 0.3} \right)}^2}} \over {\left( {0.5 - 0.3} \right)}} = {{{{\left( {0.6} \right)}^2}} \over {\left( {0.2} \right)}}$$
<br><br>$$ = 1.8\,atm$$
|
mcq
|
aieee-2011
| 1,009
|
OZsFoYVKT9unygaB
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant (K<sub>C</sub>) for the reaction N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\to$$ 2NO(g) at temperature T is 4 $$\times$$ 10<sup>–4</sup>. The value of K<sub>C</sub> for the reaction, NO(g) $$\to$$ 1/2N<sub>2</sub>(g) + 1/2O<sub>2</sub>(g) at the same temperature is :
|
[{"identifier": "A", "content": "0.02"}, {"identifier": "B", "content": "2.5 $$\\times$$ 10<sup>2</sup>"}, {"identifier": "C", "content": "4 $$\\times$$ 10<sup>-4</sup>"}, {"identifier": "D", "content": "50.0"}]
|
["D"]
| null |
For the reaction
<br><br>$${N_2} + {O_2} \to 2NO$$
<br><br>$$\,K = 4 \times {10^{ - 4}}$$
<br><br>Hence for the reaction
<br><br>$$NO \to {1 \over 2}{N_2} + {1 \over 2}{O_2}$$
<br><br>$$K' = {1 \over {\sqrt K }}$$
<br><br>$$ = {1 \over {\sqrt {4 \times {{10}^{ - 4}}} }}$$
<br><br>$$ = 50$$
|
mcq
|
aieee-2012
| 1,010
|
I2FfcEHEOHxjLXgs
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction SO<sub>2</sub> (g) + $${1 \over 2} O_2(g) \leftrightharpoons$$ SO<sub>3</sub>(g). <br/>
if K<sub>P</sub> = K<sub>C</sub>(RT)<sup>x</sup> where the symbols have usual meaning then
the value of x is: (assuming ideality)
|
[{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "-1/2"}, {"identifier": "C", "content": "1/2"}, {"identifier": "D", "content": "1"}]
|
["B"]
| null |
$$S{O_2}\left( g \right) + {1 \over 2}{O_2}\left( g \right)\,\rightleftharpoons\,S{O_3}\left( g \right)$$
<br><br>$${K_p} = {K_C}{\left( {RT} \right)^x}$$
<br><br>where $$x = \Delta {n_g} = $$ number of gaseous moles in product
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$-$$ number of gaseous moles in reactant
<br><br>$$ = 1 - \left( {1 + {1 \over 2}} \right)$$
<br><br>$$ = 1 - {3 \over 2} = - {1 \over 2}$$
|
mcq
|
jee-main-2014-offline
| 1,011
|
slcAxWaeZqwRmZga
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The standard Gibbs energy change at 300 K for the reaction 2A $$\leftrightharpoons$$ B + C is 2494.2 J. At a given time,
the composition of the reaction mixture is [A] = 1/2, [B] = 2 and [C] = 1/2. The reaction proceeds in the: [R = 8.314 J/K/mol, e = 2.718]
|
[{"identifier": "A", "content": "reverse direction because Q > K<sub>c</sub>"}, {"identifier": "B", "content": "forward direction because Q < K<sub>c</sub>"}, {"identifier": "C", "content": "reverse direction because Q < K<sub>c</sub>"}, {"identifier": "D", "content": "forward direction because Q > K<sub>c</sub>"}]
|
["A"]
| null |
$$\Delta {G^ \circ } = 2494.2J$$
<br><br>$$2A\,\rightleftharpoons\,B + C$$
<br><br>$$R = 8.314\,J/K/mol.$$
<br><br>$$e = 2.718$$
<br><br>$$\left[ A \right] = {1 \over 2},\left[ B \right] = 2,$$
<br><br>$$\left[ C \right] = {1 \over 2};Q = {{\left[ B \right]\left[ C \right]} \over {{{\left[ A \right]}^2}}}$$
<br><br>$$ = {{2 \times 1/2} \over {{{\left( {{1 \over 2}} \right)}^2}}} = 4$$
<br><br>$$\Delta {G^ \circ } = - 2.303\,\,RT\,\log \,{K_c}.$$
<br><br>$$2494.2J = - 2.303 \times \left( {8.314J/K/mol} \right)$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \times \left( {300K} \right)\log {K_c}$$
<br><br>$$ \Rightarrow \log \,{K_c}$$
<br><br>$$ = - {{2494.2\,J} \over {2.303 \times 8.314\,J/K/mol \times 300\,K}}$$
<br><br>$$ \Rightarrow \log \,{K_c} = - 0.4341;\,\,{K_c} = 0.37;\,\,Q > {K_c}.$$
|
mcq
|
jee-main-2015-offline
| 1,012
|
iBWkn6XhSK3d12VG
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant at 298 K for a reaction A + B $$\leftrightharpoons$$ C + D is 100. If the initial concentration of
all the four species were 1M each, then equilibrium concentration of D (in mol L<sup>–1</sup>) will be:
|
[{"identifier": "A", "content": "0.818"}, {"identifier": "B", "content": "1.818"}, {"identifier": "C", "content": "1.182"}, {"identifier": "D", "content": "0.182"}]
|
["B"]
| null |
<b>Given, </b>
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267730/exam_images/gpdkhpydwpbnbn6omhdy.webp" loading="lazy" alt="JEE Main 2016 (Offline) Chemistry - Chemical Equilibrium Question 78 English Explanation">
<br><br>$$\therefore$$ $$\,\,\,{K_c} = {\left( {{{1 + a} \over {1 - a}}} \right)^2} = 100$$
<br><br>$$\therefore$$ $$\,\,\,{{1 + a} \over {1 - a}} = 10$$
<br><br>On solving
<br><br>$$a=0.81$$
<br><br>$${\left[ D \right]_{At\,eq}} = 1 + a = 1 + 0.81 = 1.81$$
|
mcq
|
jee-main-2016-offline
| 1,013
|
ax46xkqWgYsItJK5jVP6o
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. K<sub>p</sub> for this reaction is :
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "100"}]
|
["C"]
| null |
<p>To determine the equilibrium constant, $K_p$, for the decomposition reaction of the solid compound XY into its gaseous components X and Y, we need to consider the following reaction:</p>
<p>
<p>$$ \text{XY (s)} \rightleftharpoons \text{X (g)} + \text{Y (g)} $$</p>
</p>
<p>For a reaction where a solid decomposes into gases, the equilibrium constant $K_p$ is defined in terms of the partial pressures of the gaseous products. Since XY is a solid, its activity is considered to be 1 and does not appear in the equilibrium expression. Hence the expression for $K_p$ is:</p>
<p>
<p>$$ K_p = P_X \cdot P_Y $$</p>
</p>
<p>Where $P_X$ and $P_Y$ are the partial pressures of the gases X and Y, respectively.</p>
<p>Given that the total pressure at equilibrium is 10 bar, and assuming that X and Y are present in equal amounts due to the stoichiometry of the decomposition reaction (i.e., 1:1), we can write:</p>
<p>
<p>$$ P_X = P_Y = \frac{10}{2} = 5 \, \text{bar} $$</p>
</p>
<p>Substituting these partial pressures into the expression for $K_p$ gives:</p>
<p>
<p>$$ K_p = P_X \cdot P_Y = 5 \, \text{bar} \cdot 5 \, \text{bar} = 25 \, \text{bar}^2 $$</p>
</p>
<p>Therefore, the equilibrium constant $K_p$ for this reaction is 25. Thus, the correct answer is:</p>
<p>Option C - 25</p>
|
mcq
|
jee-main-2016-online-10th-april-morning-slot
| 1,014
|
CprqaIhBf2B6XthNgHb7s
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
At a certain temperature in a $$5$$ $$L$$ vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction,
<br/> CO + Cl<sub>2</sub> $$\rightleftharpoons$$ COCl<sub>2</sub>
<br/>At equilibrium, if one mole of CO is present then equilibrium constant (K<sub>c</sub>) for the reaction is :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "2.5"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}]
|
["B"]
| null |
<table class="tg">
<tbody><tr>
<th class="tg-p7ly"></th>
<th class="tg-p7ly">CO</th>
<th class="tg-p7ly">+</th>
<th class="tg-fbrz">Cl<sub>2</sub></th>
<th class="tg-fbrz">$$\rightleftharpoons$$</th>
<th class="tg-fbrz">COCl<sub>2</sub></th>
</tr>
<tr>
<td class="tg-13k7">Initially number of moles</td>
<td class="tg-p7ly">2</td>
<td class="tg-p7ly"></td>
<td class="tg-fbrz">3</td>
<td class="tg-fbrz"></td>
<td class="tg-fbrz">0</td>
</tr>
<tr>
<td class="tg-60hs">At equilibrium number of moles</td>
<td class="tg-fbrz">1</td>
<td class="tg-fbrz"></td>
<td class="tg-fbrz">2</td>
<td class="tg-fbrz"></td>
<td class="tg-fbrz">1</td>
</tr>
</tbody></table>
<br><br>The equilibrium constant,
<br><br>K<sub>c</sub> = $${{\left[ {COC{l_2}} \right]} \over {\left[ {CO} \right]\left[ {C{l_2}} \right]}}$$
<br><br>= $${{\left( {{1 \over 5}} \right)} \over {\left( {{1 \over 5}} \right) \times \left( {{2 \over 5}} \right)}}$$
<br><br>= $${5 \over 2}$$
<br><br>= 2.5
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
| 1,015
|
5DyBjOOhCKpSEcMoeo3rsa0w2w9jx97n6gj
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
In which one of the following equilibria, K<sub>p</sub> $$ \ne $$ K<sub>C</sub> ?
|
[{"identifier": "A", "content": "2NO(g) \u21cb N<sub>2</sub>(g) + O<sub>2</sub>(g)"}, {"identifier": "B", "content": "2C(s) + O<sub>2</sub>(g) \u21cb 2CO(g)"}, {"identifier": "C", "content": "2HI(g) \u21cb H<sub>2</sub>(g) + I<sub>2</sub>(g)"}, {"identifier": "D", "content": "NO<sub>2</sub>(g) + SO<sub>2</sub>(g) \u21cb NO(g) + SO<sub>3</sub>(g)"}]
|
["B"]
| null |
We know,
<br>K<sub>p</sub> = K<sub>C</sub>(RT)<sup>$$\Delta $$n<sub>g</sub></sup>
<br><br>K<sub>p</sub> = K<sub>C</sub> when $$\Delta $$n<sub>g</sub> = 0
<br>K<sub>p</sub> $$ \ne $$ K<sub>C</sub> when $$\Delta $$n<sub>g</sub> $$ \ne $$ 0 and T $$ \ne $$ 12 K
<br><br>In this reaction, 2C(s) + O<sub>2</sub>(g) ⇋ 2CO(g)
<br>$$\Delta $$n<sub>g</sub> = 1, so K<sub>p</sub> $$ \ne $$ K<sub>C</sub>
|
mcq
|
jee-main-2019-online-12th-april-evening-slot
| 1,016
|
WTrxdDGwVVa5QCJNsLbQe
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the following reactions, equilibrium
constants are given :<br/>
<br/>S(s) + O<sub>2</sub>(g) ⇋ SO<sub>2</sub>(g); K<sub>1</sub> = 10<sup>52</sup><br/>
<br/>2S(s) + 3O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g); K<sub>2</sub> = 10<sup>129</sup><br/>
<br/>The equilibrium constant for the reaction,<br/>
<br/>2SO<sub>2</sub>(g) + O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g) is :
|
[{"identifier": "A", "content": "10<sup>181</sup>"}, {"identifier": "B", "content": "10<sup>25</sup>"}, {"identifier": "C", "content": "10<sup>77</sup>"}, {"identifier": "D", "content": "10<sup>154</sup>"}]
|
["B"]
| null |
S(s) + O<sub>2</sub>(g) ⇋ SO<sub>2</sub>(g); K<sub>1</sub> = 10<sup>52</sup>
<br><br>By reversing the equation, we get
<br><br>SO<sub>2</sub>(g); ⇋ S(s) + O<sub>2</sub>(g) ; $${1 \over {{K_1}}} = {1 \over {{{10}^{52}}}}$$ ..............(1)
<br><br>2S(s) + 3O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g); K<sub>2</sub> = 10<sup>129</sup> ....................(2)
<br><br>Performing (2) - 2 $$ \times $$ (1), we get
<br><br>2SO<sub>2</sub>(g) + O<sub>2</sub>(g) ⇋ 2SO<sub>3</sub>(g)
<br><br>$$ \therefore $$ Equilibrium constant of this reaction is
<br><br>= $${{{K_2}} \over {K_1^2}}$$ = $${{{{10}^{129}}} \over {{{\left( {{{10}^{52}}} \right)}^2}}}$$ = 10<sup>25</sup>
|
mcq
|
jee-main-2019-online-8th-april-evening-slot
| 1,017
|
FvoCSr01x8ku0onxV2jEr
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Two solids dissociate as follows –
<br/><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266268/exam_images/uqjldhdlltvid4j6irjs.webp"/><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266103/exam_images/vtezzbhjgelh6illgc2g.webp"/><img src="data:image/png;base64,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"/></picture>
<br/>The total pressure when both the solids dissociated simultaneously is -
|
[{"identifier": "A", "content": "(x + y) atm "}, {"identifier": "B", "content": "$$\\left( {\\sqrt {x + y} } \\right)$$ atm"}, {"identifier": "C", "content": "2$$\\left( {\\sqrt {x + y} } \\right)$$ atm"}, {"identifier": "D", "content": "x<sup>2</sup> + y<sup>2</sup> atm"}]
|
["C"]
| null |
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263582/exam_images/l0unbj43efmzlu6ogqyh.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266285/exam_images/g3hpfyrd6hvz7dcm9qmy.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263735/exam_images/gcbrwwnktne3josz5m3p.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Chemical Equilibrium Question 67 English Explanation"></picture>
<br><br>K<sub>p<sub>1</sub></sub> = P<sub>1</sub>(P<sub>1</sub> + P<sub>2</sub>)
<br><br>K<sub>p<sub>2</sub></sub> = P<sub>2</sub>(P<sub>1</sub> + P<sub>2</sub>)
<br><br>$$ \therefore $$ K<sub>p<sub>1</sub></sub> + K<sub>p<sub>2</sub></sub> = (P<sub>1</sub> + P<sub>2</sub>)<sup>2</sup>
<br><br>$$ \Rightarrow $$ x + y = (P<sub>1</sub> + P<sub>2</sub>)<sup>2</sup>
<br><br>$$ \Rightarrow $$ (P<sub>1</sub> + P<sub>2</sub>) = $$\sqrt {x + y} $$ .....(1)
<br><br>Now total pressure
<br><br>P<sub>T</sub>
= P<sub>C</sub> + P<sub>B</sub> + P<sub>E</sub>
<br><br>= (P<sub>1</sub>
+ P<sub>2</sub>) + P<sub>1</sub> + P<sub>2</sub>
<br><br>= 2(P<sub>1</sub> + P<sub>2</sub>)
<br><br>= $$2\sqrt {x + y} $$ [From equation (1)]
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
| 1,018
|
JX0ZZZbwalKbgS6Jp2vRQ
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
5.1 g NH<sub>4</sub>SH is introduced in 3.0 L evacuated flask at 327ºC. 30% of the solid NH<sub>4</sub>SH decomposed to NH<sub>3</sub> and H<sub>2</sub>S as gases . The Kp of the reaction at 327<sup>o</sup>C is (R = 0.082 L atm mol<sup>–1</sup> K<sup>–1</sup>, Molar mass of S = 32 g mol<sup>–1</sup> molar mass of N = 14 g mol<sup>–1</sup>)
|
[{"identifier": "A", "content": "0.242 $$ \\times $$ 10<sup>$$-$$4</sup> atm<sup>2</sup>"}, {"identifier": "B", "content": "1 $$ \\times $$ 10<sup>\u20134</sup> atm<sup>2</sup>\n"}, {"identifier": "C", "content": "4.9 $$ \\times $$ 10<sup>$$-$$3</sup> atm<sup>2</sup>"}, {"identifier": "D", "content": "0.242 atm<sup>2</sup>"}]
|
["D"]
| null |
NH<sub>4</sub>SH(s) $$\rightleftharpoons$$ NH<sub>3</sub>(g) + H<sub>2</sub>S(g)
<br><br>$$n = {{5.1} \over {51}} = .1\,mole\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,$$
<br><br>$$.1\left( { - 1 - \alpha } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.1\alpha $$
<br><br>$$\alpha \,\, = \,\,30\% = .3$$
<br><br>so number of moles at equilibrium
<br><br>$$\,\,\,\,\,\,\,.1\,(1 - .3)\,\,\,\,\,.1\,\, \times \,\,.3\,\,\,\,\,\,\,\,\,.1\,\, \times \,.3$$
<br><br>$$ = \,\,\,\,.07\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = .03\;\,\,\,\,\,\,\,\,\, = .03$$
<br><br>Now use PV = nRT at equilibrium
<br><br>P<sub>total</sub> $$ \times $$ 3 lit = (.03 + .03) $$ \times $$ .082 $$ \times $$ 600
<br><br>P<sub>total</sub> = .984 atm
<br><br>At equilibrium
<br><br><sup>P<sub>NH<sub>3</sub></sub> = P<sub>H<sub>2</sub>S</sub> = $${{{P_{total}}} \over 2}$$ = .492
<br><br>So k<sub>p</sub> = P<sub>NH<sub>3</sub></sub> . P<sub>H<sub>2</sub>S</sub> = (.492) (.492)
<br><br>k<sub>p</sub> = .242 atm<sup>2</sup></sup>
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
| 1,020
|
sCuCIyzQctkzU3Yg4cnVF
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The values of K<sub>p</sub>/K<sub>c</sub> for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm<sup>3</sup> atm mol<sup>–1</sup>)
<br/><br/>N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightleftharpoons$$ 2 NO(g)
<br/><br/>N<sub>2</sub>O<sub>4</sub>(g) $$\rightleftharpoons$$ 2 NO(g)
<br/><br/>N<sub>2</sub>(g) + 3H<sub>2</sub>(g) $$\rightleftharpoons$$ 2 NH<sub>3</sub>(g)
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[{"identifier": "A", "content": "24.62 dm<sup>3</sup> atm mol<sup>\u20131</sup>, 606.0 dm<sup>6</sup> atm<sup>2</sup> mol<sup>\u20132</sup> 1.65 $$ \\times $$ 10<sup>\u20133</sup> dm<sup>\u20136</sup> atm<sup>\u20132</sup> mol<sup>2</sup>"}, {"identifier": "B", "content": "1,4.1 $$ \\times $$ 10<sup>\u20132</sup> dm<sup>\u20133</sup> atm<sup>\u20131</sup> mol, 606 dm<sup>6</sup> atm<sup>2</sup> mol<sup>\u20132</sup>"}, {"identifier": "C", "content": "1,24.62 dm<sup>3</sup> atm mol<sup>\u20131</sup> , 1.65 $$ \\times $$ 10<sup>\u20133</sup> dm<sup>\u20136</sup> atm <sup>\u20132</sup> mol<sup>2</sup>"}, {"identifier": "D", "content": "1, 24.62 dm<sup>3</sup> atm mol<sup>\u20131</sup>, 606.0 dm<sup>6</sup> atm<sup>2</sup> mol<sup>\u20132</sup>\n"}]
|
["C"]
| null |
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267661/exam_images/nflpizlvl6pwqgygmlb2.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Chemistry - Chemical Equilibrium Question 71 English Explanation">
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
| 1,021
|
OXleJcmajnlCRgGiZwAFa
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Consider the following reversible chemical reactions :
<br/><br/><img src="data:image/png;base64,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"/>
<br/><br/>The relation between K<sub>1</sub> and K<sub>2</sub> is :
|
[{"identifier": "A", "content": "K<sub>1</sub>K<sub>2</sub> = $${1 \\over 3}$$"}, {"identifier": "B", "content": "K<sub>2</sub> = K<sub>1</sub><sup>3</sup>"}, {"identifier": "C", "content": "K<sub>2</sub> = K<sub>1</sub><sup>$$-$$3</sup>"}, {"identifier": "D", "content": "K<sub>1</sub>K<sub>2</sub> = 3"}]
|
["C"]
| null |
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267683/exam_images/j4h1nyops5hm9wtmf5en.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Chemistry - Chemical Equilibrium Question 72 English Explanation">
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
| 1,022
|
fvvAe6DDxfHLuSybPhg5E
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
In a chemical reaction,
<br/><br/><img src="data:image/png;base64,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"/>
<br/>the initial concentration of B was 1.5 times of the
concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is -
|
[{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "1/4"}, {"identifier": "D", "content": "4"}]
|
["D"]
| null |
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264739/exam_images/q9upzfam6hyhsoh6m2o7.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265850/exam_images/bms988tdon2t2diysvau.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Chemical Equilibrium Question 68 English Explanation"></picture>
<br><br>At equilibrium [A] = [B]
<br><br>$$ \Rightarrow $$ a - x = 1.5a - 2x
<br><br>$$ \Rightarrow $$ x = 0.5a
<br><br>K<sub>c</sub> = $${{{{\left[ C \right]}^2}\left[ D \right]} \over {\left[ A \right]{{\left[ B \right]}^2}}}$$
<br><br>= $${{{{\left( a \right)}^2}\left( {0.5a} \right)} \over {\left( {0.5a} \right){{\left( {0.5a} \right)}^2}}}$$ = 4
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
| 1,023
|
FBD5i44Aa7d4LIF4CCjgy2xukfq9iah8
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For a reaction X + Y ⇌ 2Z , 1.0 mol of X, 1.5 mol <br/>of Y and 0.5 mol of Z were taken in a 1 L
vessel and<br/> allowed to react. At equilibrium, the concentration<br/> of Z was 1.0 mol L<sup>–1</sup>. The equilibrium constant of reaction<br/> is $${x \over {15}}$$. The value of x is _________.
|
[]
| null |
16
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266846/exam_images/epachbfsh2pmgpid9cpi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 5th September Evening Slot Chemistry - Chemical Equilibrium Question 57 English Explanation">
<br><br>K<sub>eq</sub> = $${{{{\left( 1 \right)}^2}} \over {{3 \over 4} \times {5 \over 4}}}$$ = $${{16} \over {15}}$$
<br><br>$$ \therefore $$ x = 16
|
integer
|
jee-main-2020-online-5th-september-evening-slot
| 1,024
|
dfT78WspfSOKCkAFl9jgy2xukg3fgqra
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The value of K<sub>C</sub> is 64 at 800 K for the reaction
<br/><br/>N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g)
<br/><br/>The value of K<sub>C</sub> for the following reaction is :
<br/><br/>NH<sub>3</sub>(g) ⇌ $${1 \over 2}$$N<sub>2</sub>(g) + $${3 \over 2}$$H<sub>2</sub>(g)
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "$${1 \\over 8}$$"}, {"identifier": "C", "content": "$${1 \\over 4}$$"}, {"identifier": "D", "content": "$${1 \\over {64}}$$"}]
|
["B"]
| null |
N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ⇌ 2NH<sub>3</sub>(g) ; K<sub>C</sub>
<br><br>2NH<sub>3</sub>(g) ⇌ N<sub>2</sub>(g) + 3H<sub>2</sub>(g) ; $${1 \over {{K_C}}}$$
<br><br>Multiplying by $${1 \over 2}$$, reaction becomes
<br><br>NH<sub>3</sub>(g) ⇌ $${1 \over 2}$$N<sub>2</sub>(g) + $${3 \over 2}$$H<sub>2</sub>(g) ;
<br><br>$$ \therefore $$ New K<sub>C</sub> = $${\left( {{1 \over {{K_C}}}} \right)^{{1 \over 2}}}$$ = $${\left( {{1 \over {64}}} \right)^{{1 \over 2}}}$$ = $${1 \over 8}$$
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
| 1,025
|
tGqFP5AUBgiAasKCDRjgy2xukfuptcel
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The variation of equilibrium constant with
temperature is given below :
<br/><br/><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;}
.tg .tg-0lax{text-align:left;vertical-align:top}
</style>
<table class="tg">
<thead>
<tr>
<th class="tg-0lax">Temperature</th>
<th class="tg-0lax">Equilibrium Constant</th>
</tr>
</thead>
<tbody>
<tr>
<td class="tg-0lax">T<sub>1</sub> = 25<sup>o</sup>C</td>
<td class="tg-0lax">K<sub>1</sub> = 10</td>
</tr>
<tr>
<td class="tg-0lax">T<sub>2</sub> = 100<sup>o</sup>C</td>
<td class="tg-0lax">K<sub>2</sub> = 100</td>
</tr>
</tbody>
</table>
<br/>The values of $$\Delta $$H<sup>o</sup>, $$\Delta $$G<sup>o</sup> at <br/>T<sub>1</sub> and $$\Delta $$G<sup>o</sup> at T<sub>2</sub>
(in kJ mol<sup>–1</sup>) respectively, are close to :<br/>[Use
R = 8.314 J K<sup>–1</sup> mol<sup>–1</sup>]
|
[{"identifier": "A", "content": "28.4, \u20135.71 and \u201314.29"}, {"identifier": "B", "content": "0.64, \u20137.14 and \u20135.71"}, {"identifier": "C", "content": "28.4, \u20137.14 and \u20135.71"}, {"identifier": "D", "content": "0.64, \u20135.71 and \u201314.29"}]
|
["A"]
| null |
ln $$\left[ {{{{k_2}} \over {{k_1}}}} \right]$$ = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right\}$$
<br><br>$$ \Rightarrow $$ ln(10) = $${{\Delta H^\circ } \over R}\left\{ {{1 \over {298}} - {1 \over {373}}} \right\}$$
<br><br>$$ \Rightarrow $$ $${\Delta H^\circ }$$ = 28.37 kJ/mol
<br><br>$$\Delta $$G<sup>o</sup> = –RT ln K
<br><br>T<sub>1</sub> = 25<sup>o</sup>C K<sub>1</sub> = 10
<br><br>$$\Delta $$G<sup>o</sup> at T<sub>1</sub> = –8.314 × 298 × 2.303 × log 10
<br><br>= –5.71 kJ/mol
<br><br>$$\Delta $$G<sup>o</sup> at T<sub>2</sub> = –8.314 × 373 × 2.303 × log(100)
<br><br>= –14.29 kJ/mol
|
mcq
|
jee-main-2020-online-6th-september-morning-slot
| 1,026
|
Wymv2TIWJaheArgnEdjgy2xukfi6jvls
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Consider the following reaction:
<br/><br/>N<sub>2</sub>O<sub>4</sub>(g) ⇌ 2NO<sub>2</sub>(g); $$\Delta $$H<sup>o</sup> = +58 kJ
<br/><br/>For each of the following cases (a, b), the direction in which the equilibrium shifts is :
<br/>(a) Temperature is decreased.
<br/>(b) Pressure is increased by adding N<sub>2</sub>
at constant T.
|
[{"identifier": "A", "content": "(a) towards reactant, (b) towards product"}, {"identifier": "B", "content": "(a) towards reactant, (b) no change"}, {"identifier": "C", "content": "(a) towards product, (b) towards reactant"}, {"identifier": "D", "content": "(a) towards product, (b) no change"}]
|
["B"]
| null |
$$ \because $$ Given reaction is endothermic.
<br><br>$$ \therefore $$ On decreasing temperature backward
reaction will be favoured.
<br><br>On adding N<sub>2</sub>, pressure is increased at
constant T, and volume would also be constant
so no change is observed.
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
| 1,028
|
TIkZn72VO1Voziu4yojgy2xukfc91f9y
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
If the equilibrium constant for<br/> A ⇌ B + C is $$K_{eq}^{(1)}$$ and that of
<br/>B + C ⇌ P is $$K_{eq}^{(2)}$$, the equilibrium
<br/>constant for A ⇌ P is :
|
[{"identifier": "A", "content": "$${{K_{eq}^{(1)}} \\over {K_{eq}^{(2)}}}$$"}, {"identifier": "B", "content": "$${K_{eq}^{(1)}}$$ + $${K_{eq}^{(2)}}$$"}, {"identifier": "C", "content": "$${K_{eq}^{(2)}}$$ - $${K_{eq}^{(1)}}$$"}, {"identifier": "D", "content": "$${K_{eq}^{(1)}}$$ $${K_{eq}^{(2)}}$$"}]
|
["D"]
| null |
A ⇌ B + C $$K_{eq}^{(1)}$$ = $${{\left[ B \right]\left[ C \right]} \over {\left[ A \right]}}$$ ....(1)
<br><br>B + C ⇌ P $$K_{eq}^{(2)}$$ = $${{\left[ P \right]} \over {\left[ B \right]\left[ C \right]}}$$ ....(2)
<br><br>For A ⇌ P K<sub>eq</sub> = $${{\left[ P \right]} \over {\left[ A \right]}}$$
<br><br>Multiplying equation (1) & (2),
<br><br>$$K_{eq}^{(1)}$$ $$ \times $$ $$K_{eq}^{(2)}$$ = $${{\left[ P \right]} \over {\left[ A \right]}}$$ = K<sub>eq</sub>
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
| 1,029
|
be9mFEh0uaOIM9Asfgjgy2xukf92vfgc
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the equilibrium A ⇌ B , the variation of the
rate of the forward (a) and reverse (b) reaction
with time is given by :
|
[{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263534/exam_images/cpqestv7mxua9v68f391.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Chemical Equilibrium Question 60 English Option 1\">"}, {"identifier": "B", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265895/exam_images/urb1tbjeqq3k27wmyzpy.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Chemical Equilibrium Question 60 English Option 2\">"}, {"identifier": "C", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267735/exam_images/pqjarfgibpeyiutnhvf5.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Chemical Equilibrium Question 60 English Option 3\">"}, {"identifier": "D", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263861/exam_images/ndlq71kmqwb77gilzkfw.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Chemistry - Chemical Equilibrium Question 60 English Option 4\">"}]
|
["C"]
| null |
At equilibrium,
<br><br>rate of forward reaction = Rate of backward
reaction
|
mcq
|
jee-main-2020-online-4th-september-morning-slot
| 1,030
|
yEdUCUZPSH3TjZAnJAjgy2xukevgnseq
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
An open beaker of water in equilibrium with
water vapour is in a sealed container. When a
few grams of glucose are added to the beaker
of water, the rate at which water molecules :
|
[{"identifier": "A", "content": "leaves the solution increases"}, {"identifier": "B", "content": "leaves the vapour increases"}, {"identifier": "C", "content": "leaves the vapour decreases"}, {"identifier": "D", "content": "leaves the solution decreases"}]
|
["D"]
| null |
With addition of solute in solvent, surface area for vapourisation decreases causes lowering in vapour
pressure.
|
mcq
|
jee-main-2020-online-2nd-september-morning-slot
| 1,031
|
UZbTpLs7LFbcmlQe4b7k9k2k5ll44t0
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
In the figure shown below reactant A
(represented by square) is in equilibrium with
product B (represented by circle). The
equilibrium constant is :
<img src="data:image/png;base64,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"/>
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "4"}]
|
["B"]
| null |
In the figure, Reactant A is represented by square.
<br>And there are 6 squares. So at equilibrium
<br>[A]<sub>eq</sub> = 6
<br><br>In the figure, Product B is represented by circle.
<br>And there are 11 circles. So at equilibrium
<br>[B]<sub>eq</sub> = 11
<br><br>For reaction, A ⇌ B
<br><br>K<sub>eq</sub> = $${{\left[ B \right]} \over {\left[ A \right]}}$$ = $${{11} \over 6}$$ $$ \approx $$ 2
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
| 1,032
|
aYXsHNIdb8o9MrAUiR1klrgxhot
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
At 1990 K and 1 atm pressure, there are equal number of Cl<sub>2</sub>, molecules and Cl atoms in the
reaction mixture. The value of K<sub>p</sub> for the reaction Cl<sub>2 (g)</sub> $$ \rightleftharpoons $$ 2Cl<sub>(g)</sub> under the above conditions
is x $$ \times $$ 10<sup>-1</sup>.<br/>
The value of x is _______. (Rounded off to the nearest integer)
|
[]
| null |
5
|
Cl<sub>2</sub>(g) $$\rightleftharpoons$$ 2Cl(g)<br/><br/>Let moles of both of Cl<sub>2</sub> and Cl molecule be x.<br/><br/>Partial pressure of Cl is, $${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$$<br/><br/>Partial pressure of Cl<sub>2</sub> is, $${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$$<br/><br/>Now, $${K_p} = {{{{({p_{Cl}})}^2}} \over {{p_{C{l_2}}}}} $$
<br/><br/>$$\Rightarrow {K_p} = {{{{(1/2)}^2}} \over {1/2}} = {1 \over 2} = 0.5$$<br/><br/>= 5 $$\times$$ 10<sup>$$-$$1</sup><br/><br/>Hence, x $$\times$$ 10<sup>$$-$$1</sup><br/><br/>x = 5
|
integer
|
jee-main-2021-online-24th-february-morning-slot
| 1,034
|
nfBK2z0vo8pk8hKlIF1klueivgm
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
A homogeneous ideal gaseous reaction $$A{B_{2(g)}} \rightleftharpoons {A_{(g)}} + 2{B_{(g)}}$$ is carried out in a 25 litre flask at 27$$^\circ$$C. The initial amount of AB<sub>2</sub> was 1 mole and the equilibrium pressure was 1.9 atm. The value of K<sub>p</sub> is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is _________. (Integer answer)<br/><br/>[R = 0.08206 dm<sup>3</sup>atm K<sup>$$-$$1</sup>mol<sup>$$-$$1</sup>]
|
[]
| null |
72TO75
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l3l0tjjc/ca37916d-b8bf-49b7-9aee-4573c6e9abc7/ef4a1a80-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0tjjd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l3l0tjjc/ca37916d-b8bf-49b7-9aee-4573c6e9abc7/ef4a1a80-dbd9-11ec-adf6-01b3c1852b0a/file-1l3l0tjjd.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 26th February Morning Shift Chemistry - Chemical Equilibrium Question 50 English Explanation"></p>
<p>[p = Total pressure at equilibrium = 1.9 atm]</p>
<p>Now, at equilibrium pV = (1 + 2x)RT</p>
<p>$$ \Rightarrow 1 + 2x = {{pV} \over {RT}} = {{1.9 \times 25} \over {0.082 \times 300}} = 1.93$$</p>
<p>[V = 25 L, R = 0.082 L atm mol<sup>$$-$$1</sup> K<sup>$$-$$1</sup> T = 300 K]</p>
<p>$$ \Rightarrow x = {{1.93 - 1} \over 2} = 0.465$$</p>
<p>$$ \Rightarrow {K_p} = {{{p_A} \times p_B^2} \over {{p_{A{B_2}}}}} \Rightarrow {{\left( {{x \over {1 + 2x}}p} \right) \times {{\left( {{{2x} \over {1 + 2x}}p} \right)}^2}} \over {\left( {{{1 - x} \over {1 + 2x}}p} \right)}}$$</p>
<p>$$ = {{4{x^3} \times {p^3}} \over {{{(1 + 2x)}^3}}} \times {{(1 + 2x)} \over {(1 - x) \times p}} = {{4{x^3} \times {p^2}} \over {{{(1 + 2x)}^2} \times (1 - x)}}$$</p>
<p>$$ = {{4 \times {{(0.465)}^3} \times {{(1.9)}^2}} \over {{{(1 + 2 \times 0.465)}^2} \times (1 - 0.465)}} = 0.7285$$ atm</p>
<p>$$ = 72.85 \times {10^{ - 2}}$$ atm $$ \simeq 73 \times {10^{ - 2}} = x \times {10^{ - 2}}$$</p>
<p>$$\therefore$$ $$x = 73$$</p>
|
integer
|
jee-main-2021-online-26th-february-morning-slot
| 1,036
|
9T536VrN9LRuXuLDCM1kmhuu6ix
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction $$A(g) \rightleftharpoons B(g)$$ at 495 K, $$\Delta$$<sub>r</sub>G$$^\circ$$ = $$-$$9.478 kJ mol<sup>$$-$$1</sup>.<br/>If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.<br/> (Round off to the Nearest Integer). [R = 8.314 J mol<sup>$$-$$1</sup> K<sup>$$-$$1</sup>; ln 10 = 2.303]
|
[]
| null |
20
|
$$\Delta$$G<sup>o</sup> = $$-$$RT ln Keq<br><br>$$-$$9.478 $$\times$$ 10<sup>3</sup> = $$-$$495 $$\times$$ 8.314 ln Keq<br><br>ln Keq = 2.303 = ln 10<br><br>So, Keq = 10<br><br>Now, A(g) $$\rightleftharpoons$$ B(g)<br><br>$$\matrix{
{t = 0} & {22} & 0 \cr
{t = t} & {22 - x} & x \cr
} $$<br><br>$$Keq = {{[B]} \over {[A]}} = {x \over {(22 - x)}} = 10$$<br><br>x = 20<br><br>So, millimoles of B = 20
|
integer
|
jee-main-2021-online-16th-march-morning-shift
| 1,037
|
AALr4KCta5vwmOHeFi1kmj8rwvb
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
0.01 moles of a weak acid HA (K<sub>a</sub> = 2.0 $$\times$$ 10<sup>$$-$$6</sup>) is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is __________ $$\times$$ 10<sup>$$-$$5</sup> (Round off to the Nearest Integer). <br/><br/>[Neglect volume change on adding HA. Assume degree of dissociation <<1 ]
|
[]
| null |
2
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263732/exam_images/apw95glwmk7vsc8xzgye.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Morning Shift Chemistry - Chemical Equilibrium Question 48 English Explanation"><br><br>Now, <br><br>$${K_a} = {{[{H^ + }][{A^ - }]} \over {[HA]}}$$<br><br>$$ \Rightarrow 2 \times {10^{ - 6}} = {{(0.1)({{10}^{ - 2}}\alpha )} \over {{{10}^{ - 2}}}}$$<br><br>$$ \Rightarrow \alpha = 2 \times {10^{ - 5}}$$
|
integer
|
jee-main-2021-online-17th-march-morning-shift
| 1,038
|
5FwMvVp0Oixhc7DsGj1kmkjmcj4
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Consider the reaction<br/><br/> $$N_{2}O_{4}\left( g\right) \rightleftharpoons 2NO_{2}\left( g\right) $$<br/><br/>The temperature at which K<sub>C</sub> = 20.4 and K<sub>P</sub> = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>]
|
[]
| null |
354
|
N<sub>2</sub>O<sub>4</sub>(g) $$\rightleftharpoons$$ 2NO<sub>2</sub>(g)<br><br>$$\Delta$$n<sub>g</sub> = 2 $$-$$ 1 = 1<br><br>K<sub>P</sub> = K<sub>C</sub>(RT)<sup>$$\Delta$$n<sub>g</sub></sup><br><br>600.1 = 20.4 (0.0831 $$\times$$ T)<sup>1</sup><br><br>$$ \Rightarrow $$ T = $${{600.1} \over {20.4 \times 0.0831}}$$ = 354 K
|
integer
|
jee-main-2021-online-17th-march-evening-shift
| 1,039
|
3c3jXS1kKN0TMCSAPU1kmm24oyg
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The gas phase reaction $$2A(g) \rightleftharpoons {A_2}(g)$$ at 400 K has $$\Delta$$G<sup>o</sup> = + 25.2 kJ mol<sup>-1</sup>.<br/><br/>The equilibrium constant K<sub>C</sub> for this reaction is ________ $$\times$$ 10<sup>$$-$$2</sup>. (Round off to the Nearest Integer).<br/><br/>[Use : R = 8.3 J mol<sup>$$-$$1</sup> K<sup>$$-$$1</sup>, ln 10 = 2.3 log<sub>10</sub> 2 = 0.30, 1 atm = 1 bar]<br/><br/>[antilog ($$-$$0.3) = 0.501]
|
[]
| null |
2
|
Using formula,<br><br>$$\Delta$$G$$^\circ$$ = $$-$$ RTln(K<sub>p</sub>)<br><br>$$ \Rightarrow $$ 25.2 $$\times$$ 10<sup>3</sup> = $$-$$8.3 $$\times$$ 400 $$\times$$ 2.3 log (K<sub>p</sub>)<br><br>$$ \Rightarrow $$ K<sub>p</sub> = 10<sup>$$-$$3.3</sup><br><br>= 10<sup>$$-$$3</sup> $$\times$$ 0.501<br><br>= 5.01 $$\times$$ 10<sup>$$-$$4</sup> Bar<sup>$$-$$1</sup><br><br>Also, <br><br>$${{{K_p}} \over {{K_c}}} = {(RT)^{\Delta {n_g}}}$$<br><br>$$ \Rightarrow {{{K_p}} \over {{K_c}}} = {(RT)^{ - 1}}$$<br><br>$$ \Rightarrow {K_c} = {K_p}(RT)$$<br><br>$$ = 5.01 \times {10^{ - 4}} \times 8.3 \times 400$$<br><br>$$ = 1.66 \times {10^{ - 5}}$$ m<sup>3</sup>/mole<br><br>$$ = 1.66 \times {10^{ - 2}}$$ L/mol
|
integer
|
jee-main-2021-online-18th-march-evening-shift
| 1,040
|
1krq6szsp
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
2SO<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightleftharpoons$$ 2SO<sub>3</sub>(g)<br/><br/>In an equilibrium mixture, the partial pressures are <br/><br/>P<sub>SO<sub>3</sub></sub> = 43 kPa; P<sub>O<sub>2</sub></sub> = 530 Pa and P<sub>SO<sub>2</sub></sub> = 45 kPa. The equilibrium constant K<sub>P</sub> = ___________ $$\times$$ 10<sup>$$-$$2</sup>. (Nearest integer)
|
[]
| null |
172
|
On reaction, 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) $$\rightarrow$$ 2SO<sub>3</sub>(g)<br/><br/>Given values are : p<sub>SO<sub>3</sub></sub> = 45kPa, p<sub>SO<sub>2</sub></sub> = 530 Pa = 0.53 kPa<br/><br/>p<sub>SO<sub>2</sub></sub> = 43 kPa<br/><br/>Now, $${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}$$<br/><br/>On putting given values, we get<br/><br/>$$ \Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}$$<br/><br/>$$ = {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}$$<br/><br/>$$ = 1.7228$$<br/><br/>$$ = {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172$$<br/><br/>Hence, the equilibrium constant, K<sub>p</sub> = 172.
|
integer
|
jee-main-2021-online-20th-july-morning-shift
| 1,041
|
1krt6vg2f
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Value of K<sub>P</sub> for the equilibrium reaction<br/><br/>N<sub>2</sub>O<sub>4(g)</sub> $$\rightleftharpoons$$ 2NO<sub>2(g)</sub> at 288 K is 47.9. The K<sub>C</sub> for this reaction at same temperature is ____________. (Nearest integer)<br/><br/>(R = 0.083 L bar K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>)
|
[]
| null |
2
|
$${K_C} = {{{K_P}} \over {RT}} = {{47.9} \over {0.083 \times 288}} = 2$$
|
integer
|
jee-main-2021-online-22th-july-evening-shift
| 1,042
|
1krutyzlo
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
For the reaction<br/><br/>A + B $$\rightleftharpoons$$ 2C<br/><br/>the value of equilibrium constant is 100 at 298 K. If the initial concentration of all the three species is 1 M each, then the equilibrium concentration of C is x $$\times$$ 10<sup>$$-$$1</sup> M. The value of x is ____________. (Nearest integer)
|
[]
| null |
25
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265358/exam_images/wpk0kq9fjtjueiqpsv61.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Chemistry - Chemical Equilibrium Question 43 English Explanation"> <br><br>$$K = {{[C]_{eq}^2} \over {{{[A]}_{eq}}{{[B]}_{eq}}}} = {{{{(1 + 2x)}^2}} \over {(1 - x)(1 - x)}}$$<br><br>$$100 = {\left( {{{1 + 2x} \over {1 - x}}} \right)^2}$$<br><br>$$\left( {{{1 + 2x} \over {1 - x}}} \right) = 10$$<br><br>$$x = {3 \over 4}$$<br><br>$$[C]{e_{q.}} = 1 + 2x$$<br><br>$$ = 1 + 2\left( {{3 \over 4}} \right)$$<br><br>= 2.5 M<br><br>= 25 $$\times$$ 10<sup>-1</sup> M
|
integer
|
jee-main-2021-online-25th-july-morning-shift
| 1,043
|
1krxcezww
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
Assuming that Ba(OH)<sub>2</sub> is completely ionised in aqueous solution under the given conditions the concentration of H<sub>3</sub>O<sup>+</sup> ions in 0.005 M aqueous solution of Ba(OH)<sup>2</sup> at 298 K is ______________ $$\times$$ 10<sup>$$-$$12</sup> mol L<sup>$$-$$1</sup>. (Nearest integer)
|
[]
| null |
1
|
$$Ba{(OH)_2} \to B{a^{ + 2}} + 2O{H^ - } \downarrow 2 \times 0.005 = 0.01 = {10^{ - 2}}$$<br><br>At 298 K : in aq. solution $$[{H_3}{O^ + }][O{H^ - }] = {10^{ - 14}}$$<br><br>$$[{H_3}{O^ + }] = {{{{10}^{ - 14}}} \over {{{10}^{ - 2}}}} = {10^{ - 12}}$$
|
integer
|
jee-main-2021-online-25th-july-evening-shift
| 1,044
|
1krz31m2x
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
PCl<sub>5</sub> $$\rightleftharpoons$$ PCl<sub>3</sub> + Cl<sub>2</sub><br/><br/>K<sub>c</sub> = 1.844<br/><br/>3.0 moles of PCl<sub>5</sub> is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl<sub>5</sub> at equilibrium is ______________ $$\times$$ 10<sup>$$-$$3</sup>. (Round off to the Nearest Integer)
|
[]
| null |
1400
|
PCl<sub>5(g)</sub> $$\rightleftharpoons$$ PCl<sub>3(g)</sub> + Cl<sub>2(g)</sub> K<sub>2</sub> = 1.844<br><br>t = 0 3moles<br><br>t = $$\infty$$ x x<br><br>$$ \Rightarrow {{[PC{l_3}][C{l_2}]} \over {[PC{l_5}]}} = {{{x^2}} \over {3 - x}} = 1.844$$<br><br>$$ \Rightarrow {x^2} + 1.844 - 5.532 = 0$$<br><br>$$ \Rightarrow x = {{ - 1.844 + \sqrt {{{(1.844)}^2} + 4 \times 5.532} } \over 2}$$<br><br>$$ \cong 1.604$$<br><br>$$\Rightarrow$$ Moles of PCl<sub>5</sub> = 3 $$-$$ 1.604 $$\cong$$ 1.396
|
integer
|
jee-main-2021-online-27th-july-morning-shift
| 1,045
|
1ks1ixqw8
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant for the reaction<br/><br/>A(s) $$\rightleftharpoons$$ M(s) + $${1 \over 2}$$O<sub>2</sub>(g)<br/><br/>is K<sub>p</sub> = 4. At equilibrium, the partial pressure of O<sub>2</sub> is _________ atm. (Round off to the nearest integer)
|
[]
| null |
16
|
k<sub>p</sub> = Po$$_2^{1/2}$$ = 4<br><br>$$\therefore$$ Po<sub>2</sub> = 16 bar = 16 atm
|
integer
|
jee-main-2021-online-27th-july-evening-shift
| 1,046
|
1ktb5zysh
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The OH<sup>$$-$$</sup> concentration in a mixture of 5.0 mL of 0.0504 M NH<sub>4</sub>Cl and 2 mL of 0.0210 M NH<sub>3</sub> solution is x $$\times$$ 10<sup>$$-$$6</sup> M. The value of x is ___________. (Nearest integer)<br/><br/>[Given K<sub>w</sub> = 1 $$\times$$ 10<sup>$$-$$14</sup> and K<sub>b</sub> = 1.8 $$\times$$ 10<sup>$$-$$5</sup>]
|
[]
| null |
3
|
$$\left[ {NH_4^ + } \right]$$ = 0.0504 & [NH<sub>3</sub>] = 0.0210<br><br>So, $${K_b} = {{[NH_4^ + ][H{O^ - }]} \over {[N{H_3}]}}$$<br><br>$$[H{O^ - }] = {{{K_b} \times [N{H_3}]} \over {[NH_4^ + ]}} = 1.8 \times {10^{ - 5}} \times {2 \over 5} \times {{210} \over {504}} = 3 \times {10^{ - 6}}$$
|
integer
|
jee-main-2021-online-26th-august-morning-shift
| 1,047
|
1ktcry9cn
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The equilibrium constant K<sub>c</sub> at 298 K for the reaction A + B $$\rightleftharpoons$$ C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1M, the equilibrium concentration of D is ___________ $$\times$$ 10<sup>$$-$$2</sup> M. (Nearest integer)
|
[]
| null |
182
|
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ky4hs1vt/7e63d559-006a-4e38-b17c-173400ecb985/d91ab790-6fc5-11ec-8887-d75613c1bf3a/file-1ky4hs1vu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ky4hs1vt/7e63d559-006a-4e38-b17c-173400ecb985/d91ab790-6fc5-11ec-8887-d75613c1bf3a/file-1ky4hs1vu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 26th August Evening Shift Chemistry - Chemical Equilibrium Question 38 English Explanation"></p>
<p>$$\therefore$$ $${K_C} = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$</p>
<p>$$100 = {\left( {{{1 + x} \over {1 - x}}} \right)^2}$$</p>
<p>$${{1 + x} \over {1 - x}} = 10$$</p>
<p>$$x = {9 \over {11}}$$</p>
<p>Moles of D = 1 + x</p>
<p>$$ = 1 + {9 \over {11}} = {{20} \over {11}}$$</p>
<p>$$ = 1.818 = 181.8 \times {10^{ - 2}} = 181.8 \times {10^{ - 2}}$$</p>
<p>$$ \cong 182 \times {10^{ - 2}}$$ M</p>
|
integer
|
jee-main-2021-online-26th-august-evening-shift
| 1,048
|
1ktctc6nr
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The reaction rate for the reaction<br/><br/>[PtCl<sub>4</sub>]<sup>2$$-$$</sup> + H<sub>2</sub>O $$\rightleftharpoons$$ [Pt(H<sub>2</sub>O)Cl<sub>3</sub>]<sup>$$-$$</sup> + Cl<sup>$$-$$</sup><br/><br/>was measured as a function of concentrations of different species. It was observed that $${{ - d\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right]} \over {dt}} = 4.8 \times {10^{ - 5}}\left[ {{{\left[ {PtC{l_4}} \right]}^{2 - }}} \right] - 2.4 \times {10^{ - 3}}\left[ {{{\left[ {Pt({H_2}O)C{l_3}} \right]}^ - }} \right]\left[ {C{l^ - }} \right]$$.<br/><br/>where square brackets are used to denote molar concentrations. The equilibrium constant K<sub>c</sub> = ____________ . (Nearest integer)
|
[]
| null |
0
|
The rate equation provided in your question describes the rates of the forward and reverse reactions for this chemical system. The equilibrium constant, $K_c$, is the ratio of the forward rate constant ($k_f$) to the reverse rate constant ($k_r$). At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, so the rate equation simplifies to:
<br/><br/>$$k_f[\text{{PtCl}}_4]^{2-} = k_r[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$$
<br/><br/>Given the rate equation:
<br/><br/>$$-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$$
<br/><br/>At equilibrium, $-\frac{d[\text{{PtCl}}_4]^{2-}}{dt} = 0$, so:
<br/><br/>$$0 = 4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} - 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$$
<br/><br/>Rearranging terms, we find:
<br/><br/>$$4.8 \times 10^{-5} [\text{{PtCl}}_4]^{2-} = 2.4 \times 10^{-3} [\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]$$
<br/><br/>Now, the equilibrium constant $K_c$ is defined as the ratio of the concentrations of the products to the reactants, each raised to the power of their stoichiometric coefficients. For the reaction in question, we have:
<br/><br/>$$K_c = \frac{[\text{{Pt(H}}_2\text{{O)Cl}}_3]^-[\text{{Cl}}^-]}{[\text{{PtCl}}_4]^{2-}}$$
<br/><br/>Dividing both sides of our rate equation by $[\text{{PtCl}}_4]^{2-}$, we find that:
<br/><br/>$$K_c = \frac{4.8 \times 10^{-5}}{2.4 \times 10^{-3}} = \frac{1}{50}$$ = 0.02
<br/><br/>So, the equilibrium constant $K_c$ for this reaction is approximately 0, when rounded to the nearest integer.
|
integer
|
jee-main-2021-online-26th-august-evening-shift
| 1,049
|
1kteel4k2
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
The number of moles of NH<sub>3</sub>, that must be added to 2L of 0.80 M AgNO<sub>3</sub> in order to reduce the concentration of Ag<sup>+</sup> ions to 5.0 $$\times$$ 10<sup>$$-$$8</sup> M (K<sub>formation</sub> for [Ag(NH<sub>3</sub>)<sub>2</sub>]<sup>+</sup> = 1.0 $$\times$$ 10<sup>8</sup>) is ____________. (Nearest integer)<br/><br/>[Assume no volume change on adding NH<sub>3</sub>]
|
[]
| null |
4
|
Let moles added = a<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267050/exam_images/zpyygyyimoipd1vht27c.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Morning Shift Chemistry - Chemical Equilibrium Question 36 English Explanation"><br>$${{0.8} \over {(5 \times {{10}^{ - 8}})\left( {{a \over 2} - 1.6} \right)}} = {10^8}$$<br><br>$$\Rightarrow$$ $${{a \over 2}}$$ $$-$$ 1.6 = 0.4 $$\Rightarrow$$ a = 4
|
integer
|
jee-main-2021-online-27th-august-morning-shift
| 1,050
|
1ktfupc62
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
When 5.1 g of solid NH<sub>4</sub>HS is introduced into a two litre evacuated flask at 27$$^\circ$$C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K<sub>p</sub> for the reaction at 27$$^\circ$$C is x $$\times$$ 10<sup>$$-$$2</sup>. The value of x is _____________. (Integer answer) [Given R = 0.082 L atm K<sup>$$-$$1</sup> mol<sup>$$-$$</sup>]
|
[]
| null |
6
|
moles of NH<sub>4</sub>HS initially taken = $${{5.1g} \over {51g/mol}}$$ = 0.1 mol<br><br>volume of vessel = 2 litre<br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l0p3p902/e913b719-7346-4ca3-8c6e-3f5922dcc83d/afc32410-a2b3-11ec-a4c5-57354ee38743/file-1l0p3p903.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l0p3p902/e913b719-7346-4ca3-8c6e-3f5922dcc83d/afc32410-a2b3-11ec-a4c5-57354ee38743/file-1l0p3p903.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 27th August Evening Shift Chemistry - Chemical Equilibrium Question 35 English Explanation"> <br><br>$$\Rightarrow$$ partial pressure of each component<br><br>$$P = {{nRT} \over V} = {{0.1 \times 0.2 \times 0.082 \times 300} \over 2}$$<br><br>= 0.246 atm<br><br>$$\Rightarrow$$ k<sub>P</sub> = P<sub>$$N{H_3}$$</sub> $$\times$$ P<sub>$${H_2}S$$</sub> = (0.246)<sup>2</sup> = 0.060516<br><br>= 6.05 $$\times$$ 10<sup>$$-$$2</sup><br><br>$$ \therefore $$ x = 6
|
integer
|
jee-main-2021-online-27th-august-evening-shift
| 1,051
|
1l54y48r6
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>4.0 moles of argon and 5.0 moles of PCl<sub>5</sub> are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K<sub>p</sub> for the reaction is :</p>
<p>[Given : R = 0.082 L atm K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>]</p>
|
[{"identifier": "A", "content": "2.25"}, {"identifier": "B", "content": "6.24"}, {"identifier": "C", "content": "12.13"}, {"identifier": "D", "content": "15.24"}]
|
["A"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mkagsh/cdd0b8a6-eb19-44ed-9d57-1b85785143ab/4ac8df10-044b-11ed-b6d5-555c254d75e0/file-1l5mkagsi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mkagsh/cdd0b8a6-eb19-44ed-9d57-1b85785143ab/4ac8df10-044b-11ed-b6d5-555c254d75e0/file-1l5mkagsi.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2022 (Online) 29th June Evening Shift Chemistry - Chemical Equilibrium Question 34 English Explanation"></p>
<p>Here 4 moles of inert gas argon also present.</p>
<p>$$\therefore$$ Total moles of mixture present at equilibrium,</p>
<p>n<sub>T</sub> = 5 + x + 4</p>
<p>= 9 + x</p>
<p>At equilibrium, total pressure (p<sub>T</sub>) = 6 atm</p>
<p>Volume (v) = 100 L</p>
<p>Temperature = 610 K</p>
<p>$$\therefore$$ Using ideal gas equation,</p>
<p>$${P_T}V = {n_T}RT$$</p>
<p>$$ \Rightarrow 6 \times 100 = (9 + x) \times 0.082 \times 610$$</p>
<p>$$ \Rightarrow x = 3$$</p>
<p>Now,</p>
<p>$${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$$</p>
<p>$$ = {{\left[ {{3 \over {9 + 3}} \times 6} \right] \times \left[ {{3 \over {9 + 3}} \times 6} \right]} \over {\left[ {{{5 - 3} \over {9 + 3}} \times 6} \right]}}$$</p>
<p>$$ = {{27} \over {12}}$$</p>
<p>$$ = {9 \over 4}$$</p>
<p>= 2.25 atm</p>
<p>Note :</p>
<p>Inert gas always contribute to total mole and pressure calculation.</p>
|
mcq
|
jee-main-2022-online-29th-june-evening-shift
| 1,052
|
1l54z1uy4
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>A box contains 0.90 g of liquid water in equilibrium with water vapour at 27$$^\circ$$C. The equilibrium vapour pressure of water at 27$$^\circ$$C is 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be __________ litre. [nearest integer]</p>
<p>(Given : R = 0.082 L atm K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>)</p>
<p>(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)</p>
|
[]
| null |
29
|
<p>We know, 760 Torr = 1 atm</p>
<p>$$\therefore$$ 32 Torr = $${{32} \over {760}}$$ atm</p>
<p>As all the liquid water evaporates so entire water is in gaseous state.</p>
<p>$$\therefore$$ Weight of water vapour = 0.9 g</p>
<p>$$\therefore$$ Moles of water vapour (n) = $${{0.9} \over {18}}$$</p>
<p>Pressure (P) = $${{32} \over {760}}$$ atm</p>
<p>Temperature (T) = (27 + 273) K = 300 K</p>
<p>R = 0.082 L atm K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup></p>
<p>Given water vapour act as an ideal gas, so we can apply ideal gas equation.</p>
<p>From ideal gas equation,</p>
<p>PV = nRT</p>
<p>$$ \Rightarrow {{32} \over {760}} \times v = {{0.9} \over {18}} \times 0.082 \times 300$$</p>
<p>$$ \Rightarrow v = 29$$ L</p>
|
integer
|
jee-main-2022-online-29th-june-evening-shift
| 1,053
|
1l57szlay
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>2NOCl(g) $$\rightleftharpoons$$ 2NO(g) + Cl<sub>2</sub>(g)</p>
<p>In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 mol/L. The equilibrium constant at 30$$^\circ$$C is ______________ $$\times$$ 10<sup>$$-$$4</sup>.</p>
|
[]
| null |
125
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mje0dm/e9d41c32-a58b-4de7-8d83-428404631f67/c436d3b0-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mje0dn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mje0dm/e9d41c32-a58b-4de7-8d83-428404631f67/c436d3b0-0447-11ed-9d9a-21bdd0f00aa4/file-1l5mje0dn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Chemistry - Chemical Equilibrium Question 32 English Explanation"></p>
<p>Given that at equilibrium, concentration of NO = 0.4 mol/L</p>
<p>$$\therefore$$ 2x = 0.4</p>
<p>$$\Rightarrow$$ x = 0.2</p>
<p>$$\therefore$$ Concentration of NOCl at equilibrium,</p>
<p>[NOCl]<sub>eq</sub> = 2 $$-$$ 2 $$\times$$ 0.2 = 1.6</p>
<p>and [NO]<sub>eq</sub> = 0.4</p>
<p>and [Cl<sub>2</sub>]<sub>eq</sub> = 0.2</p>
<p>We know,</p>
<p>$${K_C} = {{{{[NO]}^2}[C{l_2}]} \over {{{[NOCl]}^2}}}$$</p>
<p>$$ = {{{{[0.4]}^2}[0.2]} \over {{{[1.6]}^2}}}$$</p>
<p>$$ \Rightarrow {K_C} = 12.5 \times {10^{ - 3}}$$</p>
<p>$$ \Rightarrow {K_C} = 125 \times {10^{ - 4}}$$</p>
|
integer
|
jee-main-2022-online-27th-june-morning-shift
| 1,054
|
1l5ammdi1
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>The standard free energy change ($$\Delta$$G$$^\circ$$) for 50% dissociation of N<sub>2</sub>O<sub>4</sub> into NO<sub>2</sub> at 27$$^\circ$$C and 1 atm pressure is $$-$$ x J mol<sup>$$-$$1</sup>. The value of x is ___________. (Nearest Integer)</p>
<p>[Given : R = 8.31 J K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>, log 1.33 = 0.1239 ln 10 = 2.3]</p>
|
[]
| null |
710
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l8mdj5pa/8120212a-92fd-4e3f-937a-70f4e4ec93ae/82bbdbe0-3f95-11ed-8fcd-e94b6c00f3b4/file-1l8mdj5pb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l8mdj5pa/8120212a-92fd-4e3f-937a-70f4e4ec93ae/82bbdbe0-3f95-11ed-8fcd-e94b6c00f3b4/file-1l8mdj5pb.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th June Morning Shift Chemistry - Chemical Equilibrium Question 30 English Explanation"><br>
$\mathrm{k}_{\mathrm{P}}=\frac{\left(\frac{1}{1.5} \times 1\right)^2}{\left(\frac{0.5}{1.5} \times 1\right)}=\frac{1}{0.75}=\frac{100}{75}$<br><br>
$=1.33$<br><br>
$\Delta \mathrm{G}^0=-\mathrm{RT} \ell \mathrm{nk}_{\mathrm{P}}$<br><br>
$=-8.31 \times 300 \times \ln (1.33)=-710.45 \mathrm{~J} / \mathrm{mol}$<br><br>
$=-710 \mathrm{~J} / \mathrm{mol}$
|
integer
|
jee-main-2022-online-25th-june-morning-shift
| 1,055
|
1l5bdwvkz
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>PCl<sub>5</sub> dissociates as</p>
<p>PCl<sub>5</sub>(g) $$\rightleftharpoons$$ PCl<sub>3</sub>(g) + Cl<sub>2</sub>(g)</p>
<p>5 moles of PCl<sub>5</sub> are placed in a 200 litre vessel which contains 2 moles of N<sub>2</sub> and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant K<sub>p</sub> for the dissociation of PCl<sub>5</sub> is __________ $$\times$$ 10<sup>$$-$$3</sup>. (nearest integer)</p>
<p>(Given : R = 0.082 L atm K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>; Assume ideal gas behaviour)</p>
|
[]
| null |
1107
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mkc6w5/9bcd3b0f-bd25-4002-a7df-7f91aec78977/7ac1c150-044b-11ed-b6d5-555c254d75e0/file-1l5mkc6w6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mkc6w5/9bcd3b0f-bd25-4002-a7df-7f91aec78977/7ac1c150-044b-11ed-b6d5-555c254d75e0/file-1l5mkc6w6.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Chemistry - Chemical Equilibrium Question 29 English Explanation"></p>
<p>Here 2 moles of N<sub>2</sub> also present that is why 2 moles always have to add in total mole calculation.</p>
<p>At equilibrium,</p>
<p>Pressure (P) = 2.46 atm</p>
<p>Volume (V) = 200 L</p>
<p>Temperature (T) = 600 K</p>
<p>$$\therefore$$ Applying ideal gas equation,</p>
<p>PV = nRT</p>
<p>$$\Rightarrow$$ 2.46 $$\times$$ 200 = (7 + x) $$\times$$ 0.082 $$\times$$ 600</p>
<p>$$\Rightarrow$$ x = 3</p>
<p>Now,</p>
<p>$${K_P} = {{{P_{PC{l_3}}} \times {P_{C{l_2}}}} \over {{P_{PC{l_5}}}}}$$</p>
<p>$$ = {{\left[ {{3 \over {7 + 3}} \times 2.46} \right]\left[ {{3 \over {7 + 3}} \times 2.46} \right]} \over {\left[ {{{5 - 3} \over {7 + 3}} \times 2.46} \right]}}$$</p>
<p>$$ = {{{3 \over {10}} \times {3 \over {10}} \times {{(2.46)}^2}} \over {{2 \over {10}} \times 2.46}}$$</p>
<p>$$ = {9 \over {20}} \times 2.46$$</p>
<p>$$ = 1107 \times {10^{ - 3}}$$ atm</p>
|
integer
|
jee-main-2022-online-24th-june-evening-shift
| 1,056
|
1l5c689au
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>For a reaction at equilibrium</p>
<p>A(g) $$\rightleftharpoons$$ B(g) + $${1 \over 2}$$ C(g)</p>
<p>the relation between dissociation constant (K), degree of dissociation ($$\alpha$$) and equilibrium pressure (p) is given by :</p>
|
[{"identifier": "A", "content": "$$K = {{{\\alpha ^{{1 \\over 2}}}{p^{{3 \\over 2}}}} \\over {{{\\left( {1 + {3 \\over 2}\\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$"}, {"identifier": "B", "content": "$$K = {{{\\alpha ^{{3 \\over 2}}}{p^{{1 \\over 2}}}} \\over {{{\\left( {2 + \\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$"}, {"identifier": "C", "content": "$$K = {{{{(\\alpha \\,p)}^{{3 \\over 2}}}} \\over {{{\\left( {1 + {3 \\over 2}\\alpha } \\right)}^{{1 \\over 2}}}(1 - \\alpha )}}$$"}, {"identifier": "D", "content": "$$K = {{{{(\\alpha \\,p)}^{{3 \\over 2}}}} \\over {{{\\left( {1 + \\alpha } \\right)}}{{(1 - \\alpha )}^{{1 \\over 2}}}}}$$"}]
|
["B"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5mjgrml/bddc4d35-1e47-45de-af09-7895683e36e4/10e198d0-0448-11ed-9d9a-21bdd0f00aa4/file-1l5mjgrmm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5mjgrml/bddc4d35-1e47-45de-af09-7895683e36e4/10e198d0-0448-11ed-9d9a-21bdd0f00aa4/file-1l5mjgrmm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Morning Shift Chemistry - Chemical Equilibrium Question 27 English Explanation"></p>
<p>Now,</p>
<p>$${K_P}$$ or $$K = {{{P_B} \times {{\left( {{P_C}} \right)}^{{1 \over 2}}}} \over {{P_A}}}$$</p>
<p>$$ = {{\left( {{\alpha \over {1 + {\alpha \over 2}}}} \right)P \times {{\left[ {\left( {{{{\alpha \over 2}} \over {1 + {\alpha \over 2}}}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{1 - \alpha } \over {1 + {\alpha \over 2}}}} \right)P}}$$</p>
<p>$$ = {{\left( {{{2\alpha } \over {2 + \alpha }}} \right)P \times {{\left[ {\left( {{\alpha \over {2 + \alpha }}} \right)P} \right]}^{{1 \over 2}}}} \over {\left( {{{2(1 - \alpha )} \over {2 + \alpha }}} \right)P}}$$</p>
<p>$$= {\alpha \over {1 - \alpha }} \times {\left( {{{\alpha P} \over {2 + \alpha }}} \right)^{{1 \over 2}}}$$</p>
<p>$$ = {{{\alpha ^{{3 \over 2}}}\,.\,{P^{{1 \over 2}}}} \over {(1 - \alpha ){{(2 + \alpha )}^{{1 \over 2}}}}}$$</p>
|
mcq
|
jee-main-2022-online-24th-june-morning-shift
| 1,057
|
1l5c7268y
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>2O<sub>3</sub>(g) $$\rightleftharpoons$$ 3O<sub>2</sub>(g)</p>
<p>At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is ($$-$$) ____________ J mol<sup>$$-$$1</sup>. (Nearest integer)</p>
<p>[Given : ln 1.35 = 0.3 and R = 8.3 J K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>]</p>
|
[]
| null |
747
|
$\underset{1-x}{2 \mathrm{O}_3(\mathrm{~g})} \rightleftharpoons \underset{\frac{3 \mathrm{x}}{2}}{3 \mathrm{O}_2(\mathrm{~g})}$<br/><br/>
Given, $x=0.5$
<br/><br/>
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{[3(0.5)]^{3} \times 1}{[2]^{3} \times(0.5)^{2} \times 1.25}$
<br/><br/>
$\therefore \mathrm{k}_{\mathrm{p}}=\frac{27}{8} \times \frac{0.5}{1.25}=1.35$
<br/><br/>
$$
\begin{aligned}
\Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\\\
&=-2.303 \times 8.3 \times 300 \log 1.35 \\\\
&=-8.3 \times 300 \ln (1.35) \\\\
&=-747 \mathrm{~J} \mathrm{~mol}^{-1}
\end{aligned}
$$
|
integer
|
jee-main-2022-online-24th-june-morning-shift
| 1,058
|
1l5w5kyld
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>The equilibrium constant for the reversible reaction</p>
<p>2A(g) $$\rightleftharpoons$$ 2B(g) + C(g) is K<sub>1</sub></p>
<p>$${3 \over 2}$$A(g) $$\rightleftharpoons$$ $${3 \over 2}$$B(g) + $${3 \over 4}$$C(g) is K<sub>2</sub>.</p>
<p>K<sub>1</sub> and K<sub>2</sub> are related as :</p>
|
[{"identifier": "A", "content": "$${K_1} = \\sqrt {{K_2}} $$"}, {"identifier": "B", "content": "$${K_2} = \\sqrt {{K_1}} $$"}, {"identifier": "C", "content": "$${K_2} = K_1^{3/4}$$"}, {"identifier": "D", "content": "$${K_1} = K_2^{3/4}$$"}]
|
["C"]
| null |
$2 \mathrm{A}(\mathrm{g})=2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g}) \quad K_{1} \quad\quad ...(i)$
<br/><br/>
$$
\frac{3}{2} \mathrm{A}(\mathrm{g})=\frac{3}{2} \mathrm{B}(\mathrm{g})+\frac{3}{4} \mathrm{C}(\mathrm{g}) \quad K_{2}\quad\quad...(ii)
$$
<br/><br/>
eq. (ii) is $\frac{3}{4}$ times of eq. (i), hence,
<br/><br/>
$K_{2}=\left(K_{1}\right)^{\frac{3}{4}}$
|
mcq
|
jee-main-2022-online-30th-june-morning-shift
| 1,059
|
1l6nxtp6c
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>At $$600 \mathrm{~K}, 2 \mathrm{~mol}$$ of $$\mathrm{NO}$$ are mixed with $$1 \mathrm{~mol}$$ of $$\mathrm{O}_{2}$$.</p>
<p>$$2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}_{2}(\mathrm{g})$$</p>
<p>The reaction occurring as above comes to equilibrium under a total pressure of 1 atm. Analysis of the system shows that $$0.6 \mathrm{~mol}$$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is ________. (Nearest integer)</p>
|
[]
| null |
2
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7vyiyym/b3d0ff82-278a-44ca-8925-29649ed4cb48/73b469e0-310e-11ed-9c98-ad39d53b642b/file-1l7vyiyyn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7vyiyym/b3d0ff82-278a-44ca-8925-29649ed4cb48/73b469e0-310e-11ed-9c98-ad39d53b642b/file-1l7vyiyyn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Evening Shift Chemistry - Chemical Equilibrium Question 24 English Explanation"><br>Partial pressure of $\mathrm{NO}(\mathrm{g})=\frac{1.2}{2.6} \times 1$
<br><br>
Partial pressure of $\mathrm{O}_{2}(\mathrm{~g})=\frac{0.6}{2.6}$
<br><br>
Partial pressure of $\mathrm{NO}_{2}(\mathrm{~g})=\frac{0.8}{2.6}$
<br><br>
$$
\begin{aligned}
\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{NO}_{2}}\right)^{2}}{\left(\mathrm{P}_{\mathrm{NO}}\right)^{2}\left(\mathrm{P}_{\mathrm{O}_{2}}\right)} &=\frac{0.8 \times 0.8 \times 2.6}{1.2 \times 1.2 \times 0.6} \\\\
&=1.925 \\\\
& \approx 2
\end{aligned}
$$
|
integer
|
jee-main-2022-online-28th-july-evening-shift
| 1,061
|
1ldoltihj
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>(i) $$\mathrm{X}(\mathrm{g}) \rightleftharpoons \mathrm{Y}(\mathrm{g})+\mathrm{Z}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 1}=3$$</p>
<p>(ii) $$\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{g}) \quad \mathrm{K}_{\mathrm{p} 2}=1$$</p>
<p>If the degree of dissociation and initial concentration of both the reactants $$\mathrm{X}(\mathrm{g})$$ and $$\mathrm{A}(\mathrm{g})$$ are equal, then the ratio of the total pressure at equilibrium $$\left(\frac{p_{1}}{p_{2}}\right)$$ is equal to $$\mathrm{x}: 1$$. The value of $$\mathrm{x}$$ is _____________ (Nearest integer)</p>
|
[]
| null |
12
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le7i662y/833b3985-b140-440c-8deb-848011ab2b75/31afaaa0-ae31-11ed-b364-6dadf34ccd07/file-1le7i662z.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le7i662y/833b3985-b140-440c-8deb-848011ab2b75/31afaaa0-ae31-11ed-b364-6dadf34ccd07/file-1le7i662z.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Chemistry - Chemical Equilibrium Question 22 English Explanation 1">
<br><br>$$
\begin{aligned}
& \mathrm{k}_{\mathrm{p}_1}=\frac{\left(\frac{\alpha}{1+\alpha} \times \mathrm{p}_1\right)^2}{\frac{1-\alpha}{1+\alpha} \mathrm{p}_1} \\\\
& 3=\frac{\alpha^2 \times \mathrm{p}_1}{1-\alpha^2}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le7i87dz/9403a96c-5169-4e41-86de-96a0455c6e96/6a504770-ae31-11ed-b364-6dadf34ccd07/file-1le7i87e0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le7i87dz/9403a96c-5169-4e41-86de-96a0455c6e96/6a504770-ae31-11ed-b364-6dadf34ccd07/file-1le7i87e0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Chemistry - Chemical Equilibrium Question 22 English Explanation 2">
<br><br>$$
\begin{aligned}
& \mathrm{k}_{\mathrm{p}_2}=\frac{\left(\frac{2 \alpha}{1+\alpha} \times \mathrm{p}_2\right)^2}{\frac{1-\alpha}{1+\alpha} \times \mathrm{p}_2} \\\\
& 1=\frac{4 \alpha^2 \times \mathrm{p}_2}{1-\alpha^2} \\\\
& \frac{\mathrm{k}_{\mathrm{p}_1}}{\mathrm{k}_{\mathrm{p}_2}}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\
& \frac{3}{1}=\frac{\mathrm{p}_1}{4 \mathrm{p}_2} \\\\
& {\mathrm{p}_1} : {\mathrm{p}_2} = 12 : 1 \\\\
& \therefore \mathrm{x}=12
\end{aligned}
$$
|
integer
|
jee-main-2023-online-1st-february-morning-shift
| 1,062
|
1ldpqfe1q
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>For reaction : $$\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})$$</p>
<p>$$\mathrm{K}_{\mathrm{p}}=2 \times 10^{12}$$ at $$27^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure. The $$\mathrm{K}_{\mathrm{c}}$$ for the same reaction is ____________ $$\times 10^{13}$$. (Nearest integer)</p>
<p>(Given $$\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$)</p>
|
[]
| null |
1
|
$$
\begin{aligned}
& \mathrm{SO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{SO}_{3(\mathrm{~g})} \\\\
& \mathrm{K}_{\mathrm{P}}=2 \times 10^{12} \text { at } 300 \mathrm{~K} \\\\
& \mathrm{~K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \times(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\\\
& \Rightarrow 2 \times 10^{12}=\mathrm{K}_{\mathrm{C}} \times(0.082 \times 300)^{-1 / 2} \\\\
& \Rightarrow \mathrm{~K}_{\mathrm{C}}=9.92 \times 10^{12} \\\\
& \Rightarrow \mathrm{~K}_{\mathrm{C}}=0.992 \times 10^{13}
\end{aligned}
$$
|
integer
|
jee-main-2023-online-31st-january-morning-shift
| 1,063
|
1lgsza2h5
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>4.5 moles each of hydrogen and iodine is heated in a sealed ten litre vessel. At equilibrium, 3 moles of $$\mathrm{HI}$$ were found. The equilibrium constant for $$\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$ is _________.</p>
|
[]
| null |
1
|
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1liclq95i/c8c19de4-06d3-45eb-a68d-4a6e0ce940bd/e4643760-002f-11ee-a99f-43454a56aaa3/file-1liclq95j.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1liclq95i/c8c19de4-06d3-45eb-a68d-4a6e0ce940bd/e4643760-002f-11ee-a99f-43454a56aaa3/file-1liclq95j.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Chemistry - Chemical Equilibrium Question 16 English Explanation">
<br><br>$$
\mathrm{K}_{\mathrm{c}}=\frac{[\mathrm{HI}]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=\frac{(3)^2}{3 \times 3}=\frac{9}{9}=1
$$
|
integer
|
jee-main-2023-online-11th-april-evening-shift
| 1,066
|
1lgv10uu5
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>A mixture of 1 mole of $$\mathrm{H}_{2} \mathrm{O}$$ and 1 mole of $$\mathrm{CO}$$ is taken in a 10 litre container and heated to $$725 \mathrm{~K}$$. At equilibrium $$40 \%$$ of water by mass reacts with carbon monoxide according to the equation :
<br/><br/>$$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})$$.
<br/><br/> The equilibrium constant $$\mathrm{K}_{\mathrm{c}} \times 10^{2}$$ for the reaction is ____________. (Nearest integer)</p>
|
[]
| null |
44
|
The given reaction is :
<br/><br/>$$\mathrm{CO(g)} + \mathrm{H_{2}O(g)} \rightleftharpoons \mathrm{CO_{2}(g)} + \mathrm{H_{2}(g)} $$
<br/><br/>We are given that $40\%$ of water by mass reacts with carbon monoxide.
<br/><br/>The molecular weight of water (H<sub>2</sub>O) is approximately $18 \, \mathrm{g/mol}$, so $1 \, \mathrm{mole}$ of water weighs $18 \, \mathrm{g}$. Therefore, the mass of the reacted water is $0.40 \times 18 \, \mathrm{g} = 7.2 \, \mathrm{g}$.
<br/><br/>Since from the stoichiometry of the reaction we can see that $1 \, \mathrm{mole}$ of CO reacts with $1 \, \mathrm{mole}$ of H<sub>2</sub>O to form $1 \, \mathrm{mole}$ of CO<sub>2</sub> and $1 \, \mathrm{mole}$ of H<sub>2</sub>, this means that $7.2 \, \mathrm{g}$ of H<sub>2</sub>O is equivalent to $7.2 \, \mathrm{g} / 18 \, \mathrm{g/mol} = 0.4 \, \mathrm{mole}$.
<br/><br/>We start with $1 \, \mathrm{mole}$ of CO and $1 \, \mathrm{mole}$ of H2O. At equilibrium, we have :
<br/><br/>- CO: $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$
<br/><br/>- H<sub>2</sub>O : $1 \, \mathrm{mole} - 0.4 \, \mathrm{mole} = 0.6 \, \mathrm{mole}$
<br/><br/>- CO<sub>2</sub> : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$
<br/><br/>- H<sub>2</sub> : $0 \, \mathrm{mole} + 0.4 \, \mathrm{mole} = 0.4 \, \mathrm{mole}$
<br/><br/>The volume of the container is $10 \, \mathrm{litres}$. Therefore, we can convert the moles to concentrations (in M = moles/L) as :
<br/><br/>- [CO] = $0.6 \, \mathrm{M}$
<br/><br/>- [H<sub>2</sub>O] = $0.6 \, \mathrm{M}$
<br/><br/>- [CO<sub>2</sub>] = $0.4 \, \mathrm{M}$
<br/><br/>- [H<sub>2</sub>] = $0.4 \, \mathrm{M}$
<br/><br/>The equilibrium constant $K_{c}$ for the reaction can be calculated as :
<br/><br/>$$K_{c} = \frac{[\mathrm{CO_{2}}][\mathrm{H_{2}}]}{[\mathrm{CO}][\mathrm{H_{2}O}]}$$
<br/><br/>So, substituting the values we get :
<br/><br/>$$K_{c} = \frac{(0.4 \times 0.4)}{(0.6 \times 0.6)} = 0.44$$
<br/><br/>As per the question, we need to calculate the value of $K_{c} \times 10^{2}$ :
<br/><br/>$$K_{c} \times 10^{2} = 0.44 \times 10^{2} = 44 \, (\text{rounded to the nearest
integer})$$
<br/><br/>Therefore, $K_{c} \times 10^{2} = 44$.
|
integer
|
jee-main-2023-online-11th-april-morning-shift
| 1,067
|
1lgvveuca
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>$$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g)$$</p>
<p>For the given reaction, if the initial pressure is $$450 \mathrm{~mm} ~\mathrm{Hg}$$ and the pressure at time $$\mathrm{t}$$ is $$720 \mathrm{~mm} ~\mathrm{Hg}$$ at a constant temperature $$\mathrm{T}$$ and constant volume $$\mathrm{V}$$. The fraction of $$\mathrm{A}(\mathrm{g})$$ decomposed under these conditions is $$x \times 10^{-1}$$. The value of $$x$$ is ___________ (nearest integer)</p>
|
[]
| null |
3
|
<p>Given the reaction:</p>
<p>$$\mathrm{A}_{(\mathrm{g})} \rightleftharpoons 2 \mathrm{~B}_{(\mathrm{g})}+\mathrm{C}_{(\mathrm{g})}$$</p>
<p>At the beginning of the reaction (at time = 0), the total pressure is solely due to A, hence it is 450 mmHg.</p>
<p>As the reaction progresses, let's denote 'x' as the pressure decrease due to the decomposition of A. Correspondingly, the pressure increases by '2x' and 'x' for B and C respectively, following the stoichiometry of the reaction.</p>
<p>At time 't', the total pressure (P(T)) is the sum of the pressures of A, B, and C, which is given to be 720 mmHg.</p>
<p>This gives us the equation:</p>
<p>$$450 \text{ mmHg} - x \text{ mmHg (decrease in A's pressure)} + 2x \text{ mmHg (increase in B's pressure)} + x \text{ mmHg (increase in C's pressure)} = 720 \text{ mmHg (total pressure at time t)}$$</p>
<p>Solving for 'x' gives:</p>
<p>$$x = 135 \text{ mmHg}$$</p>
<p>The fraction of A decomposed would then be this change in pressure divided by the initial pressure:</p>
<p>$$\text{Fraction decomposed} = \frac{x}{\text{Initial pressure}} = \frac{135 \text{ mmHg}}{450 \text{ mmHg}} = 0.3 = 3 \times 10^{-1}$$</p>
|
integer
|
jee-main-2023-online-10th-april-evening-shift
| 1,068
|
1lh27t8xf
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p> For a concentrated solution of a weak electrolyte ($$\mathrm{K}_{\text {eq }}=$$ equilibrium constant) $$\mathrm{A}_{2} \mathrm{B}_{3}$$ of concentration '$$c$$', the degree of dissociation '$$\alpha$$' is :</p>
|
[{"identifier": "A", "content": "$$\\left(\\frac{K_{e q}}{25 c^{2}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "B", "content": "$$\\left(\\frac{K_{e q}}{108 c^{4}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "C", "content": "$$\\left(\\frac{K_{e q}}{5 c^{4}}\\right)^{\\frac{1}{5}}$$"}, {"identifier": "D", "content": "$$\\left(\\frac{K_{e q}}{6 c^{5}}\\right)^{\\frac{1}{5}}$$"}]
|
["B"]
| null |
$$
\begin{aligned}
& \mathrm{A}_2 \mathrm{~B}_3(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{~A}_{\text {(aq.) }}^{3+}+3 \mathrm{~B}_{\text {(aq) }}^{2-} \\\\
& \mathrm{c}(1-\alpha) \quad\quad\quad 2 \mathrm{c} \alpha\quad\quad\quad 3 \mathrm{c}\alpha\\\\
& \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{~B}^{2-}\right]^3}{\left[\mathrm{~A}_2 \mathrm{~B}_3\right]}=\frac{4 \mathrm{c}^2 \alpha^2 \times 27 \mathrm{c}^3 \alpha^3}{\mathrm{c}(1-\alpha)} \\\\
& \mathrm{K}_{\mathrm{eq}}=\frac{108 \mathrm{c}^5 \alpha^5}{\mathrm{c}} \alpha=\left(\frac{\mathrm{K}_{\mathrm{eq}}}{108 \mathrm{c}^4}\right)^{\frac{1}{5}}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
| 1,069
|
1lh332lfs
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>The equilibrium composition for the reaction $$\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}$$ at $$298 \mathrm{~K}$$ is given below:</p>
<p>$$\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.40 \mathrm{~mol} \mathrm{~L}^{-1}$$</p>
<p>If $$0.2 \mathrm{~mol}$$ of $$\mathrm{Cl}_{2}$$ is added at the same temperature, the equilibrium concentrations of $$\mathrm{PCl}_{5}$$ is __________ $$\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$$</p>
<p>Given : $$\mathrm{K}_{\mathrm{c}}$$ for the reaction at $$298 \mathrm{~K}$$ is 20</p>
|
[]
| null |
49
|
<p>The initial equilibrium concentrations are given as:<br/><br/>
$[\mathrm{PCl}_{3}]_{\text{eq}} = 0.2 \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}]_{\text{eq}} = 0.1 \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}]_{\text{eq}} = 0.4 \, \mathrm{mol/L}$.</p>
<p>After adding $0.2 \, \mathrm{mol}$ of $\mathrm{Cl_2}$, the new concentration of $\mathrm{Cl_2}$ becomes $0.1 + 0.2 = 0.3 \, \mathrm{mol/L}$.</p>
<p>Let the change in concentration be $x$, so at the new equilibrium, the concentrations are:<br/><br/>
$[\mathrm{PCl}_{3}] = 0.2 - x \, \mathrm{mol/L}$, $[\mathrm{Cl}_{2}] = 0.3 - x \, \mathrm{mol/L}$, $[\mathrm{PCl}_{5}] = 0.4 + x \, \mathrm{mol/L}$.</p>
<p>Using the equilibrium constant expression $K_c = 20$, we get:<br/><br/>
$20 = \frac{0.4+x}{(0.2-x)(0.3-x)}$.</p>
<p>Solving for $x$, we find $x \approx 0.086$.</p>
<p>Therefore, the new equilibrium concentration of $\mathrm{PCl}_5$ is:<br/><br/>
$[\mathrm{PCl}_5]_{\text{new eq}} = 0.4 + 0.086 = 0.486 \, \mathrm{mol/L} = 48.6 \times 10^{-2} \, \mathrm{mol/L}$.</p>
|
integer
|
jee-main-2023-online-6th-april-evening-shift
| 1,070
|
jaoe38c1lse7hf3x
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>For the given reaction, choose the correct expression of $$\mathrm{K}_{\mathrm{C}}$$ from the following :-</p>
<p>$$\mathrm{Fe}_{(\mathrm{aq})}^{3+}+\mathrm{SCN}_{(\mathrm{aq})}^{-} \rightleftharpoons(\mathrm{FeSCN})_{(\mathrm{aq})}^{2+}$$</p>
|
[{"identifier": "A", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}{\\left[\\mathrm{FeSCN}^{2+}\\right]}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n"}, {"identifier": "C", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]^2}{\\left[\\mathrm{Fe}^{3+}\\right]\\left[\\mathrm{SCN}^{-}\\right]}$$\n"}, {"identifier": "D", "content": "$$\\mathrm{K}_{\\mathrm{C}}=\\frac{\\left[\\mathrm{FeSCN}^{2+}\\right]}{\\left[\\mathrm{Fe}^{3+}\\right]^2\\left[\\mathrm{SCN}^{-}\\right]^2}$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \mathrm{K}_{\mathrm{C}}=\frac{\text { Products ion conc. }}{\text { Reactants ion conc. }} \\
& \mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{FeSCN}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]\left[\mathrm{SCN}^{-}\right]}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-31st-january-morning-shift
| 1,072
|
jaoe38c1lsfk477k
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>For the reaction $$\mathrm{N}_2 \mathrm{O}_{4(\mathrm{~g})} \rightleftarrows 2 \mathrm{NO}_{2(\mathrm{~g})}, \mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}$$ at $$300 \mathrm{~K} . \mathrm{K}_{\mathrm{c}}$$ for the reaction at same temperature is _________ $$\times 10^{-2}$$.</p>
<p>(Given : $$\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$$)</p>
|
[]
| null |
2
|
<p>$$\begin{aligned}
& \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\
& \Delta \mathrm{n}_{\mathrm{g}}=1 \\
& \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{0.492}{0.082 \times 300}=2 \times 10^{-2}
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-29th-january-morning-shift
| 1,073
|
lv2er9pz
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>The equilibrium constant for the reaction</p>
<p>$$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$$</p>
<p>is $$\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$$. The value of $$\mathrm{K}_{\mathrm{c}}$$ for the reaction given below is $$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$$ is :</p>
|
[{"identifier": "A", "content": "49"}, {"identifier": "B", "content": "416"}, {"identifier": "C", "content": "41.6"}, {"identifier": "D", "content": "4.9"}]
|
["B"]
| null |
<p>The reaction</p>
<p>$$2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$$</p>
<p>can be formed from the given reaction by reverting it and multiplying coefficients by 2.</p>
<p>Thus,</p>
<p>$$\begin{aligned}
\mathrm{K}_c^{\prime} & =\mathrm{K}_{\mathrm{c}}^{-2}=\frac{1}{\mathrm{~K}_{\mathrm{c}}^2}=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\
& =416
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
| 1,075
|
lv5gspnf
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>For the given hypothetical reactions, the equilibrium constants are as follows :</p>
<p>$$\begin{aligned}
& \mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0 \\
& \mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0 \\
& \mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0
\end{aligned}$$</p>
<p>The equilibrium constant for the reaction $$\mathrm{X} \rightleftharpoons \mathrm{W}$$ is</p>
|
[{"identifier": "A", "content": "12.0"}, {"identifier": "B", "content": "8.0"}, {"identifier": "C", "content": "6.0"}, {"identifier": "D", "content": "7.0"}]
|
["B"]
| null |
<p>To find the equilibrium constant for the overall reaction $$\mathrm{X} \rightleftharpoons \mathrm{W}$$, we need to combine the equilibrium constants for the individual reactions given. Let's analyze this step-by-step:</p>
<p>The given reactions and their equilibrium constants are:</p>
<p>$$\begin{aligned} & \mathrm{X} \rightleftharpoons \mathrm{Y} \quad K_1 = 1.0 \\ & \mathrm{Y} \rightleftharpoons \mathrm{Z} \quad K_2 = 2.0 \\ & \mathrm{Z} \rightleftharpoons \mathrm{W} \quad K_3 = 4.0 \end{aligned}$$</p>
<p>The equilibrium constant for the overall reaction $$\mathrm{X} \rightleftharpoons \mathrm{W}$$ is the product of the equilibrium constants for the individual steps. This is because the equilibrium constant for a composite reaction is the product of the equilibrium constants for the sequential reactions that lead to the composite reaction.</p>
<p>So, we have:</p>
<p>$$K_{\text{overall}} = K_1 \cdot K_2 \cdot K_3$$</p>
<p>Now, substituting the given values:</p>
<p>$$K_{\text{overall}} = 1.0 \cdot 2.0 \cdot 4.0 = 8.0$$</p>
<p>Therefore, the equilibrium constant for the reaction $$\mathrm{X} \rightleftharpoons \mathrm{W}$$ is 8.0, which corresponds to Option B.</p>
|
mcq
|
jee-main-2024-online-8th-april-morning-shift
| 1,076
|
lv9s2805
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>Given below are two statements :</p>
<p>Statement I : On passing $$\mathrm{HCl}_{(\mathrm{g})}$$ through a saturated solution of $$\mathrm{BaCl}_2$$, at room temperature white turbidity appears.</p>
<p>Statement II : When $$\mathrm{HCl}$$ gas is passed through a saturated solution of $$\mathrm{NaCl}$$, sodium chloride is precipitated due to common ion effect.</p>
<p>In the light of the above statements, choose the most appropriate answer from the options given below :</p>
|
[{"identifier": "A", "content": "Both Statement I and Statement II are correct\n"}, {"identifier": "B", "content": "Statement I is incorrect but Statement II is correct\n"}, {"identifier": "C", "content": "Both Statement I and Statement II are incorrect\n"}, {"identifier": "D", "content": "Statement I is correct but Statement II is incorrect"}]
|
["D"]
| null |
<p>To determine the correctness of the statements, we'll analyze each one individually by applying principles of solubility, common ion effect, and chemical equilibria.</p>
<hr />
<h3><strong>Statement I:</strong></h3>
<p><strong>On passing $\mathrm{HCl}_{(g)}$ through a saturated solution of $\mathrm{BaCl}_2$ at room temperature, white turbidity appears.</strong></p>
<p><strong>Analysis:</strong></p>
<p><p><strong>Dissolution of HCl Gas:</strong></p></p>
<p><p>When $\mathrm{HCl}_{(g)}$ is bubbled through water, it dissolves and dissociates completely:</p>
<p>$ \mathrm{HCl}_{(aq)} \rightarrow \mathrm{H}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $</p></p>
<p><p>This increases the concentration of $\mathrm{Cl}^-$ ions in the solution.</p></p>
<p><p><strong>Effect on BaCl₂ Solubility:</strong></p></p>
<p><p>The solubility equilibrium of $\mathrm{BaCl}_2$ in water is:</p>
<p>$ \mathrm{BaCl}_2 \leftrightarrow \mathrm{Ba}^{2+}_{(aq)} + 2\mathrm{Cl}^-_{(aq)} $</p></p>
<p><p>Adding more $\mathrm{Cl}^-$ shifts the equilibrium to the left (Le Chatelier's Principle), causing $\mathrm{BaCl}_2$ to precipitate.</p></p>
<p><p>The precipitation of $\mathrm{BaCl}_2$ manifests as a <strong>white turbidity</strong>.</p></p>
<p><strong>Conclusion:</strong></p>
<p><strong>Statement I is correct.</strong></p>
<hr />
<h3><strong>Statement II:</strong></h3>
<p><strong>When $\mathrm{HCl}$ gas is passed through a saturated solution of $\mathrm{NaCl}$, sodium chloride is precipitated due to common ion effect.</strong></p>
<p><strong>Analysis:</strong></p>
<p><p><strong>Dissolution of HCl Gas:</strong></p></p>
<p><p>Similar to before, $\mathrm{HCl}$ increases $\mathrm{Cl}^-$ ion concentration.</p></p>
<p><p><strong>Effect on NaCl Solubility:</strong></p></p>
<p><p>The solubility equilibrium of $\mathrm{NaCl}$ is:</p>
<p>$ \mathrm{NaCl} \leftrightarrow \mathrm{Na}^+_{(aq)} + \mathrm{Cl}^-_{(aq)} $</p></p>
<p><p><strong>However, $\mathrm{NaCl}$ is highly soluble in water, and its solubility is not significantly affected by the common ion effect from $\mathrm{Cl}^-$.</strong></p></p>
<p><p>The solubility product ($K_{sp}$) of $\mathrm{NaCl}$ is large, and the addition of $\mathrm{Cl}^-$ ions does not cause $\mathrm{NaCl}$ to precipitate under normal conditions.</p></p>
<p><p><strong>No precipitation occurs; the solution remains clear.</strong></p></p>
<p><strong>Conclusion:</strong></p>
<p><strong>Statement II is incorrect.</strong></p>
<hr />
<h3><strong>Final Answer:</strong></h3>
<p><strong>Statement I is correct but Statement II is incorrect.</strong></p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
| 1,077
|
lvb2a7x3
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>The ratio $$\frac{K_P}{K_C}$$ for the reaction :</p>
<p>$$\mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})}$$ is :</p>
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$$\n(\\mathrm{RT})^{1 / 2}\n$$"}, {"identifier": "C", "content": "RT"}, {"identifier": "D", "content": "$$\\mathrm{\n\\frac{1}{\\sqrt{R T}}}\n$$"}]
|
["D"]
| null |
<p>To solve this problem, we need to understand the relationship between the equilibrium constant in terms of pressure, $$K_P$$, and the equilibrium constant in terms of concentration, $$K_C$$. The relationship between these two constants for a general reaction is given by:
<p>$$ K_P = K_C (RT)^{\Delta n} $$</p>
<p>where:</p>
</p>
<p>
<p>$$\Delta n$$ is the change in the number of moles of gas (moles of gaseous products minus moles of gaseous reactants),</p>
</p>
<p>
<p>$$R$$ is the ideal gas constant, and</p>
</p>
<p>
<p>$$T$$ is the temperature in Kelvin.</p>
</p>
<p>For the given reaction:
<p>$$ \mathrm{CO}_{(\mathrm{g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightleftharpoons \mathrm{CO}_{2(\mathrm{~g})} $$</p>
<p>we need to calculate $$\Delta n$$.</p></p>
<p>On the reactants side, we have:</p>
<ul>
<li>1 mole of $$\mathrm{CO}$$ (g)</li>
<li>0.5 moles of $$\mathrm{O}_2$$ (g)</li>
</ul>
<p>So, the total number of moles of reactants is:</p>
<p>$$ 1 + \frac{1}{2} = \frac{3}{2} = 1.5 $$</p>
<p>On the products side, we have:</p>
<ul>
<li>1 mole of $$\mathrm{CO}_2$$ (g)</li>
</ul>
<p>The total number of moles of products is:</p>
<p>$$ 1 $$</p>
<p>Therefore, $$\Delta n$$ is:</p>
<p>$$ \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 1.5 = -0.5 $$</p>
<p>Now, we can use the relationship between $$K_P$$ and $$K_C$$:</p>
<p>$$ K_P = K_C (RT)^{\Delta n} $$</p>
<p>Substituting $$\Delta n = -0.5$$, we get:</p>
<p>$$ K_P = K_C (RT)^{-0.5} $$</p>
<p>or equivalently:</p>
<p>$$ \frac{K_P}{K_C} = (RT)^{-0.5} $$</p>
<p>Therefore, the correct option is:</p>
<p><strong>Option D</strong>: $$\frac{1}{\sqrt{R T}}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-evening-shift
| 1,078
|
lvc587ob
|
chemistry
|
chemical-equilibrium
|
chemical-equilibrium,-law-of-mass-action-and-equilibrium-constant
|
<p>At $$-20^{\circ} \mathrm{C}$$ and $$1 \mathrm{~atm}$$ pressure, a cylinder is filled with equal number of $$\mathrm{H}_2, \mathrm{I}_2$$ and $$\mathrm{HI}$$ molecules for the reaction
$$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$, the $$\mathrm{K}_{\mathrm{p}}$$ for the process is $$x \times 10^{-1}$$.</p>
<p>$$\mathrm{x}=$$ __________.</p>
<p>[Given : $$\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$$]</p>
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "0.01"}]
|
["C"]
| null |
<p>At $$-20^{\circ} \mathrm{C}$$ and a pressure of $$1 \mathrm{~atm}$$, a cylinder contains equal quantities of $$\mathrm{H}_2, \mathrm{I}_2,$$ and $$\mathrm{HI}$$ molecules. The equilibrium constant for the reaction</p>
<p>$$\mathrm{H}_2(\mathrm{~g})+\mathrm{I}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})$$</p>
<p>denoted as $$\mathrm{K}_{\mathrm{p}}$$, is given by the expression $$x \times 10^{-1}$$.</p>
<p>To solve for $$x$$, use the provided equilibrium expression:</p>
<p>$$\mathrm{K_{P} = \mathrm{K_C}} =\frac{\left[\mathrm{HI}\right]^2}{\left[\mathrm{H}_2\right]\left[\mathrm{I}_2\right]}=1 = 10 \times 10^{-1}$$</p>
<p>Therefore, $$x = 10$$.</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
| 1,079
|
lAbL1dOH9madyBTF
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
Change in volume of the system does not alter which of the following equilibria?
|
[{"identifier": "A", "content": "N<sub>2</sub>(g) + O<sub>2</sub>(g) $$\\leftrightharpoons$$ 2NO (g)"}, {"identifier": "B", "content": "PCl<sub>5</sub>(g) $$\\leftrightharpoons$$ PCl<sub>3</sub> (g) + Cl<sub>2</sub> (g)"}, {"identifier": "C", "content": "N<sub>2</sub>(g) + 3H<sub>2</sub>(g) $$\\leftrightharpoons$$ 2NH<sub>3</sub> (g)"}, {"identifier": "D", "content": "SO<sub>2</sub>Cl<sub>2</sub> (g) $$\\leftrightharpoons$$ SO<sub>2</sub> (g) + Cl<sub>2</sub> (g)"}]
|
["A"]
| null |
In this reaction the ratio of number of moles of reactants to products is same $$i.e.\,\,2:2,$$ hence change in volume will not alter the number of moles.
|
mcq
|
aieee-2002
| 1,080
|
3F69yMnkhThKnTAU
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
Consider the reaction equilibrium<br/>
2 SO<sub>2</sub> (g) + O<sub>2</sub> (g) $$\leftrightharpoons$$ 2 SO<sub>3</sub> (g); $$\Delta H^o$$ = -198 kJ<br/>
One the basis of Le Chatelier's principle, the condition favourable for the forward reaction is :
|
[{"identifier": "A", "content": "increasing temperature as well as pressure"}, {"identifier": "B", "content": "lowering the temperature and increasing the pressure"}, {"identifier": "C", "content": "any value of temperature and pressure"}, {"identifier": "D", "content": "lowering temperature as well as pressure"}]
|
["B"]
| null |
Due to exothermicity of reaction low or optimum temperature will be required. Since $$3$$ moles are changing to $$2$$ moles.
<br><br>$$\therefore$$ High pressure will be required.
|
mcq
|
aieee-2003
| 1,081
|
iCB432yoEFUORD9G
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
The exothermic formation of ClF<sub>3</sub> is represented by the equation: <br/>
Cl<sub>2</sub> (g) + 3F<sub>2</sub> (g) $$\leftrightharpoons$$ 2ClF<sub>3</sub> (g); $$\Delta H$$ = -329 kJ<br/>
Which of the following will increase the quantity of ClF<sub>3</sub> in an equilibrium mixture of
Cl<sub>2</sub>, F<sub>2</sub> and ClF<sub>3</sub>?
|
[{"identifier": "A", "content": "Increasing the temperature"}, {"identifier": "B", "content": "Removing Cl<sub>2</sub>"}, {"identifier": "C", "content": "Increasing the volume of the container "}, {"identifier": "D", "content": "Adding F<sub>2</sub>"}]
|
["D"]
| null |
The reaction given is an exothermic reaction thus accordingly to Lechatalier's principle lowering of temperature, addition of $${F_2}$$ and or $$C{l_2}$$ favour the for ward direction and hence the production of $$Cl{F_3}.$$
|
mcq
|
aieee-2005
| 1,082
|
6vmuqmKCuf8FOH8rAjOWz
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
The following reaction occurs in the Blast Furnace where iron ore is reduced to iron
metal :
<br/><br/>Fe<sub>2</sub>O<sub>3</sub>(s) + 3 CO(g) $$\rightleftharpoons$$ 2 Fe(1) + 3 CO<sub>2</sub>(g)
<br/><br/>Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
|
[{"identifier": "A", "content": "Removal of CO "}, {"identifier": "B", "content": "Removal of CO<sub>2</sub> "}, {"identifier": "C", "content": "Addition of CO<sub>2</sub>"}, {"identifier": "D", "content": "Addition of Fe<sub>2</sub>O<sub>3</sub>"}]
|
["D"]
| null |
<p>Addition of a solid component to a system at constant
pressure has no effect on the equilibrium. Therefore,
addition of Fe<sub>2</sub>O<sub>3</sub> will not disturb the equilibrium.</p>
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
| 1,083
|
O2mqgIpwasZAg7L4e5sLG
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
In which of the following reactions, an increase in the volume of the container will favour the formation of products?
|
[{"identifier": "A", "content": "2NO<sub>2</sub>(g) $$\\rightleftharpoons$$ 2NO(g) + O<sub>2</sub>(g)"}, {"identifier": "B", "content": "H<sub>2</sub>(g) + I<sub>2</sub>(g) $$\\rightleftharpoons$$ 2HI(g)"}, {"identifier": "C", "content": "4NH<sub>3</sub>(g) + 5O<sub>2</sub>(g) $$\\rightleftharpoons$$ 4NO(g) + 6H<sub>2</sub>O(1)"}, {"identifier": "D", "content": "3O<sub>2</sub> (g) $$\\rightleftharpoons$$ 2O<sub>3</sub>(g)"}]
|
["A"]
| null |
From Boyle's law, we know when volume increase then pressure decreases, and to keep the pressure on original value the reaction should proceeds in the direction where the number of moles of gases increases.
<br><br>In this reaction, only satisfy this.
<br><br>2 NO<sub>2</sub>(g) $$\rightleftharpoons$$ 2 NO(g) + O<sub>2</sub> (g)
<br><br>$$\therefore\,\,\,\,$$ $$\Delta $$n<sub>g</sub> = (2 + 1) $$-$$ 2 = 1
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
| 1,084
|
ujFoukI0ZKA9R7QeWXVur
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
The gas phase reaction 2NO<sub>2</sub>(g) $$ \to $$ N<sub>2</sub>O<sub>4</sub>(g) is an exothermic reaction. The decomposition of N<sub>2</sub>O<sub>4</sub>, in equilibrium mixture of NO<sub>2</sub>(g) and N<sub>2</sub>O<sub>4</sub>(g), can be increased by :
|
[{"identifier": "A", "content": "lowering the temperature."}, {"identifier": "B", "content": "increasing the pressure."}, {"identifier": "C", "content": "addition of an inert gas at constant volume. "}, {"identifier": "D", "content": "addition of an inert gas at constant pressure. "}]
|
["D"]
| null |
2NO<sub>2</sub> (g) $$\buildrel \, \over
\longrightarrow $$ N<sub>2</sub>O<sub>4</sub> (g); $$\Delta $$H = $$-$$ ve
<br><br>At equilibrium, N<sub>2</sub>O<sub>4</sub> $$\rightleftharpoons$$ 2NO<sub>2</sub> ; $$\Delta $$H = + ve
<br><br> On adding the inert gas at constant pressure, the number of moles per unit volume of various reactants and products will decrease. Hence, the equilibrium will shift towards the direction in which there is increase in number of moles of gases. Therefore, in above case, reaction will move towards forward direction which will lead to the decomposition of dinitrogen tetraoxide (N<sub>2</sub>O<sub>4</sub>).
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
| 1,085
|
d02h5fIw8PMD42WGT93rsa0w2w9jx0x1gkb
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
For the reaction,
<br/> 2SO<sub>2</sub>(g) + O<sub>2</sub>(g) = 2SO<sub>3</sub>(g), $$\Delta $$H = –57.2 kJ mol<sup>–1</sup>
and K<sub>C</sub> = 1.7 × 10<sup>16</sup>
<br/>Which of the following statement is incorrect ?
|
[{"identifier": "A", "content": "The equilibrium will shift in forward direction as the pressure increase."}, {"identifier": "B", "content": "The addition of inert gas at constant volume will be not affect the equilibrium constant."}, {"identifier": "C", "content": "The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is\nrequired."}, {"identifier": "D", "content": "The equilibrium constant decreases as the temperature increase."}]
|
["C"]
| null |
Equilibrium constant has no relation with
catalyst.
<br><br>Catalyst only affects the rate with
which a reaction proceeds.
<br><br>Here we use catalyst V<sub>2</sub>O<sub>5</sub> to
speed up the reaction.
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
| 1,086
|
bkjyti2pc4gu9E5DFQ3rsa0w2w9jx83y3v4
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
The INCORRECT match in the following is :
|
[{"identifier": "A", "content": "$$\\Delta $$G<sup>o</sup>\n = 0, K = 1"}, {"identifier": "B", "content": "$$\\Delta $$G<sup>o</sup>\n < 0, K < 1"}, {"identifier": "C", "content": "$$\\Delta $$G<sup>o</sup>\n > 0, K < 1"}, {"identifier": "D", "content": "$$\\Delta $$G<sup>o</sup>\n < 0, K > 1"}]
|
["B"]
| null |
We know, $$\Delta {G^o} = - RT\ln K$$
<br>Case 1 :
<br>If $$\Delta $$G<sup>o</sup> < 0
<br>$$ \Rightarrow $$ $$- RT\ln K$$ < 0
<br>$$ \Rightarrow $$ $$\ln K$$ > 0
<br>$$ \Rightarrow $$ K > 1
<br><br>Case 2 :
<br>If $$\Delta $$G<sup>o</sup> > 0
<br>$$ \Rightarrow $$ $$- RT\ln K$$ > 0
<br>$$ \Rightarrow $$ $$\ln K$$ < 0
<br>$$ \Rightarrow $$ K < 1
<br><br>Case 2 :
<br>If $$\Delta $$G<sup>o</sup> = 0
<br>$$ \Rightarrow $$ $$- RT\ln K$$ = 0
<br>$$ \Rightarrow $$ $$\ln K$$ = 0
<br>$$ \Rightarrow $$ K = 1
|
mcq
|
jee-main-2019-online-12th-april-evening-slot
| 1,087
|
ldqy3up7
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
Consider the following equation:
<br/><br/>
$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \Delta H=-190 \mathrm{~kJ}$
<br/><br/>
The number of factors which will increase the yield of $\mathrm{SO}_{3}$ at equilibrium from the following is _______.<br/><br/>
A. Increasing temperature<br/><br/>
B. Increasing pressure<br/><br/>
C. Adding more $\mathrm{SO}_{2}$<br/><br/>
D. Adding more $\mathrm{O}_{2}$<br/><br/>
E. Addition of catalyst
|
[]
| null |
3
|
<p>$$\mathrm{2SO_2+O_2\rightleftharpoons 2SO_3~~\Delta H=-190~kJ}$$</p>
<p>It is an exothermic reaction</p>
<p>$$\therefore$$ factor B, C, D will increase the amount of SO$$_3$$.</p>
|
integer
|
jee-main-2023-online-30th-january-evening-shift
| 1,090
|
1ldsdsiwh
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
<p>At 298 K</p>
<p>$$\mathrm{N_2~(g)+3H_2~(g)\rightleftharpoons~2NH_3~(g),~K_1=4\times10^5}$$</p>
<p>$$\mathrm{N_2~(g)+O_2~(g)\rightleftharpoons~2NO~(g),~K_2=1.6\times10^{12}}$$</p>
<p>$$\mathrm{H_2~(g)+\frac{1}{2}O_2~(g)\rightleftharpoons~H_2O~(g),~K_3=1.0\times10^{-13}}$$</p>
<p>Based on above equilibria, then equilibrium constant of the reaction, $$\mathrm{2NH_3(g)+\frac{5}{2}O_2~(g)\rightleftharpoons~2NO~(g)+3H_2O~(g)}$$ is ____________ $$\times10^{-33}$$ (Nearest integer).</p>
|
[]
| null |
4
|
<p>$$\mathrm{2NH_3(g)\frac{5}{2}O_2(g)\rightleftharpoons 2NO(g)+3H_2O(g)}$$</p>
<p>Clearly, $$\mathrm{{K_{eq}} = {1 \over {{k_1}}} \times {k_2} \times k_3^3}$$</p>
<p>$$ = {{1.6 \times {{10}^{12}} \times {{10}^{ - 39}}} \over {4 \times {{10}^5}}}$$</p>
<p>$$ = 0.4 \times {10^{ - 32}}$$</p>
<p>$$ = 4 \times {10^{ - 33}}$$</p>
|
integer
|
jee-main-2023-online-29th-january-evening-shift
| 1,091
|
1lgyhl1dn
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
<p>The number of correct statement/s involving equilibria in physical processes from the following is ________</p>
<p>(A) Equilibrium is possible only in a closed system at a given temperature.</p>
<p>(B) Both the opposing processes occur at the same rate.</p>
<p>(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.</p>
<p>(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.</p>
|
[]
| null |
3
|
<p><b>(A) Equilibrium is possible only in a closed system at a given temperature.</b></p>
<ul>
<li>This statement is true. In a closed system, there is no exchange of matter with the surroundings. This allows the reaction to reach equilibrium at a constant temperature.</li>
</ul>
<p><b>(B) Both the opposing processes occur at the same rate.</b></p>
<ul>
<li>This statement is true. At equilibrium, the forward and reverse reactions occur at the same rate, meaning the concentrations of reactants and products remain constant.</li>
</ul>
<p><b>(C) When equilibrium is attained at a given temperature, the value of all its parameters became equal.</b></p>
<ul>
<li>This statement is false. When a system reaches equilibrium, it doesn't mean that the values of all parameters (such as the concentrations of reactants and products) are equal. It means that these values are constant, i.e., they are not changing with time because the rates of the forward and reverse reactions are equal.</li>
</ul>
<p><b>(D) For dissolution of solids in liquids, the solubility is constant at a given temperature.</b></p>
<ul>
<li>This statement is true. The solubility of a solid in a liquid (in a saturated solution) is indeed constant at a given temperature and pressure.</li>
</ul>
<p>Therefore, there are 3 correct statements: (A), (B), and (D).</p>
|
integer
|
jee-main-2023-online-10th-april-morning-shift
| 1,092
|
lv7v4nyl
|
chemistry
|
chemical-equilibrium
|
le-chatelier's-principle-and-factors-affecting-chemical-equilibrium
|
<p>The following reaction occurs in the Blast furnance where iron ore is reduced to iron metal</p>
<p>$$\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightleftharpoons \mathrm{Fe}_{(j)}+3 \mathrm{CO}_{2(g)}$$</p>
<p>Using the Le-chatelier's principle, predict which one of the following will not disturb the equilibrium.</p>
|
[{"identifier": "A", "content": "Addition of $$\\mathrm{CO}_2$$\n"}, {"identifier": "B", "content": "Removal of $$\\mathrm{CO}$$\n"}, {"identifier": "C", "content": "Addition of $$\\mathrm{Fe}_2 \\mathrm{O}_3$$\n"}, {"identifier": "D", "content": "Removal of $$\\mathrm{CO}_2$$"}]
|
["C"]
| null |
<p>For the reaction :</p>
<p>$$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}+3 \mathrm{CO}_{(\mathrm{g})} \rightleftharpoons \mathrm{Fe}_{(\mathrm{l})}+3 \mathrm{CO}_{2(\mathrm{~g})}$$</p>
<p>Addition or removal of $$\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{s})}$$ and/or $$\mathrm{Fe}_{\text {(l) }}$$ will not affect the equilibrium quotient and the equilibrium.</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
| 1,093
|
VcCAgyN3fplIli6M
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
In respect of the equation k = Ae<sup>-E<sub>a</sub>/RT</sup> in chemical kinetics, which one of the following statements is correct?
|
[{"identifier": "A", "content": "A is adsorption factor"}, {"identifier": "B", "content": "E<sub>a</sub> is energy of activation"}, {"identifier": "C", "content": "R is Rydberg\u2019s constant"}, {"identifier": "D", "content": "k is equilibrium constant"}]
|
["B"]
| null |
In equation $$K = A{e^{ - {E_a}/RT}};A = $$ Frequency factor
<br><br>$$K = $$ velocity constant, $$R=$$ gas constant and
<br><br>$${E_b} = $$ energy of activation
|
mcq
|
aieee-2003
| 1,094
|
kJAXWKzFjLzd8NnZ
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
A schematic plot of $$ln$$ $${K_{eq}}$$ versus inverse of temperature for a reaction is shown below
<br/><br/><img src="data:image/png;base64,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"/>
<br/>The reaction must be
|
[{"identifier": "A", "content": "highly spontaneous at ordinary temperature"}, {"identifier": "B", "content": "one with negligible enthalpy change "}, {"identifier": "C", "content": "endothermic "}, {"identifier": "D", "content": "exothermic "}]
|
["D"]
| null |
The graph show that reaction is exothermic.
<br><br>$$\log \,k = {{ - \Delta H} \over {RT}} + 1$$
<br><br>For exothermic reaction $$\Delta H < 0$$
<br><br>$$\therefore$$ $$\,\,\,\,\,log\,\,k\,\,Vs{1 \over T}\,\,$$ would be negative straight line with positive slope.
|
mcq
|
aieee-2005
| 1,095
|
bFM8KL83corCGEsg
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
Rate of a reaction can be expressed by Arrhenius equation as: <br/>
$$$k = A\,{e^{ - E/RT}}$$$
In this equation, E represents
|
[{"identifier": "A", "content": "the energy above which all the colliding molecules will react"}, {"identifier": "B", "content": "the energy below which colliding molecules will not react "}, {"identifier": "C", "content": "the total energy of the reacting molecules at a temperature, T "}, {"identifier": "D", "content": "the fraction of molecules with energy greater than the activation energy of the reaction"}]
|
["A"]
| null |
In Arrhenius equation $$K = A\,{e^{ - E/RT}},\,\,E$$ is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products.
|
mcq
|
aieee-2006
| 1,096
|
HCILgcwKXnYVtRGp
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
The energies of activation for forward and reverse reactions for A<sub>2</sub> + B<sub>2</sub> $$\leftrightharpoons$$ 2AB are 180 kJ mol<sup>−1</sup>
and 200 kJ mol<sup>−1</sup> respectively. The presence of catalyst lowers the activation energy of both (forward and reverse) reactions by 100 kJ mol<sup>−1</sup>. The enthalpy change of the reaction ( A<sub>2</sub> + B<sub>2</sub> $$\to$$ 2AB) in the presence of catalyst will be (in kJ mol<sup>−1</sup>)
|
[{"identifier": "A", "content": "300"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "200"}, {"identifier": "D", "content": "20"}]
|
["D"]
| null |
$$\Delta {H_R} = {E_f} - {E_b} = 180 - 200 = - 20kJ/mol$$
<br><br>The nearest correct answer given in choices may be obtained by neglecting sign.
|
mcq
|
aieee-2007
| 1,097
|
hfgD93hASiXKWbBb
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such
a reaction will be:(R = 8.314 JK<sup>–1</sup> mol<sup>–1</sup> and log 2 = 0.301)
|
[{"identifier": "A", "content": "48.6 kJ mol<sup>\u20131</sup> "}, {"identifier": "B", "content": "58.5 kJ mol<sup>\u20131</sup>"}, {"identifier": "C", "content": "60.5 kJ mol<sup>\u20131</sup> "}, {"identifier": "D", "content": "53.6 kJ mol<sup>\u20131</sup>"}]
|
["D"]
| null |
Activation energy can be calculated from the equation
<br><br>$${{\log \,{k_2}} \over {\log \,{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
<br><br>given $${{{k_2}} \over {{k_1}}} = 2\,\,{T_2} = 310\,K\,\,{T_1} = 300\,K$$
<br><br>$$ = \log 2 = {{ - {E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {310}} - {1 \over {300}}} \right)$$
<br><br>$${E_a} = 53598.6J/mol$$
<br><br>$$ = 53.6kJ/mol.$$
|
mcq
|
jee-main-2013-offline
| 1,098
|
dwGTF5etVJOvFtl9
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
Two reactions R<sub>1</sub> and R<sub>2</sub> have identical pre-exponential factors. Activation energy of R<sub>1</sub> exceeds that of R<sub>2</sub> by 10 kJ mol<sup>–1</sup>. If k<sub>1</sub> and k<sub>2</sub> are rate constants for reactions R<sub>1</sub> and R<sub>2</sub> respectively at 300 K, then ln(k<sub>2</sub>/k<sub>1</sub>) is equal to : <br/>
(R = 8.314 J mol<sup>–1</sup> K<sup>–1</sup>)
|
[{"identifier": "A", "content": "12"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}]
|
["C"]
| null |
We know, from arrhenius equation,
<br><br>k = A.$${e^{{{ - {E_a}} \over {RT}}}}$$
<br><br>$$ \therefore $$ k<sub>1</sub> = A.$${e^{{{ - {E_{{a_1}}}} \over {RT}}}}$$ ......(1)
<br><br>k<sub>2</sub> = A.$${e^{{{ - {E_{{a_2}}}} \over {RT}}}}$$ ......(2)
<br><br>On dividing equation (2) by (1), we get
<br><br>$${{{k_2}} \over {{k_1}}} = {e^{{{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}}}$$
<br><br>$$ \Rightarrow $$ $$\ln \left( {{{{k_2}} \over {{k_1}}}} \right) = {{\left( {{E_{{a_1}}} - {E_{{a_2}}}} \right)} \over {RT}}$$ = $${{10,000} \over {8.314 \times 300}}$$ = 4
|
mcq
|
jee-main-2017-offline
| 1,099
|
GXzTlAJFalSuW7k6TcPVS
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be Increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.
|
[{"identifier": "A", "content": "9.84 K"}, {"identifier": "B", "content": "4.92 K"}, {"identifier": "C", "content": "2.45 K"}, {"identifier": "D", "content": "19.67 K"}]
|
["B"]
| null |
For reaction A, T<sub>1</sub> = 300 K, T<sub>2</sub> = 310 K, k<sub>2</sub> = 2 k<sub>1</sub>
<br><br>$$\log {{{k_2}} \over {{k_1}}} = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$
<br><br>$$ \therefore $$ $$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$ ...(1)
<br><br>For reaction B, T<sub>1</sub> = 300 K, T<sub>2</sub> = ?, k<sub>2</sub> = 2k<sub>1</sub>, E<sub>a<sub>2</sub></sub> = 2E<sub>a<sub>1</sub></sub>
<br><br>$$\log {{2{k_1}} \over {{k_1}}} = \log 2 = {{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right]$$
<br><br>From eq. (i) and (ii), we get
<br><br>$${{2{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = {{{E_{{a_1}}}} \over {2.303R}}\left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
<br><br>$$ \Rightarrow $$ $$2\left[ {{1 \over {300}} - {1 \over {{T_2}}}} \right] = \left[ {{1 \over {300}} - {1 \over {310}}} \right]$$
<br><br>$$ \Rightarrow $$ T<sub>2</sub> = $${{{300 \times 310} \over {610}}}$$ $$ \times $$ 2 = 304.92 K
<br><br>$$ \therefore $$ Increased temperature = (304.92 – 300) = 4.92 K
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
| 1,100
|
OU0GIizMN5uNzlPOBbG31
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :
<br/><br/>(Assume activation energy and preexponential factor are independent of
temperature; ln 2 = 0.693; R = 8.314 J mol<sup>−1</sup> K<sup>−1</sup>)
|
[{"identifier": "A", "content": "107.2 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "53.6 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "26.8 kJ mol<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "214.4 kJ mol<sup>$$-$$1</sup>"}]
|
["A"]
| null |
<p>According to Arrhenius equation,</p>
<p>$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$$</p>
<p>Substituting the given values in the equation, we get</p>
<p>$$\log 4 = {{{E_a}} \over {2.303 \times 8.314\,J\,{K^{ - 1}}mo{l^{ - 1}}}}\left( {{{310 - 300} \over {300 \times 310}}} \right)$$</p>
<p>$${E_a} = {{0.602 \times 2.303 \times 8.314 \times 300 \times 310} \over {10}}$$</p>
<p>= 107197.12 J/mol<sup>$$-$$1</sup> or 107.2 kJ/mol<sup>$$-$$1</sup></p>
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
| 1,101
|
BuRyLYEbdumiFeMieMLSG
|
chemistry
|
chemical-kinetics-and-nuclear-chemistry
|
arrhenius-equation
|
Consider the given plots for a reaction obeying Arrhenius equation (0<sup>o</sup>C < T < 300<sup>o</sup>C) : (K and Ea are rate constant and activation energy, respectively)
<br/><br/><img src="data:image/png;base64,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"/>
<br/><br/>Choose the correct option :
|
[{"identifier": "A", "content": "I is right but II is wrong "}, {"identifier": "B", "content": "Both I and II are correct "}, {"identifier": "C", "content": "Both I and II are wrong "}, {"identifier": "D", "content": "I is wrong but II is right "}]
|
["B"]
| null |
Arrhenius Equation -
<br><br>
Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.
<br><br>
- wherein<br><br>
$$k = A{e^{ - {E_a}/RT}}$$<br><br>
$${\mathop{\rm lnk}\nolimits} = lnA - {{{E_a}} \over {RT}}$$<br><br>
k = Rate constant
<br><br>
as we have learned in Arrhenius equation when we increase Ea, K should decrease
As T increases, power of exponential increases so k also increase.
<br><br>
Hence, both the plots are correct.
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
| 1,102
|
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