question_id stringlengths 8 35 | subject stringclasses 3
values | chapter stringclasses 90
values | topic stringclasses 459
values | question stringlengths 17 24.5k | options stringlengths 2 4.26k | correct_option stringclasses 6
values | answer stringclasses 460
values | explanation stringlengths 1 10.6k | question_type stringclasses 3
values | paper_id stringclasses 154
values | __index_level_0__ int64 2 13.4k |
|---|---|---|---|---|---|---|---|---|---|---|---|
3q0krlWokhcZPlqUCoIQV | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | If a reaction follows the Arrhenius equation, the plot ln k vs $${1 \over {\left( {RT} \right)}}$$ gives straight line with a gradient ($$-$$ y) unit.
<br/><br/>The energy required to active the reactant is : | [{"identifier": "A", "content": "y unit"}, {"identifier": "B", "content": "y/R unit"}, {"identifier": "C", "content": "yR unit"}, {"identifier": "D", "content": "$$-$$y unit"}] | ["A"] | null | According to Arrhenius equation,
<br>k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>or ln k = ln A - $${{{{E_a}} \over {RT}}}$$
<br><br>Comparing the above equation
with straight line equation,
<br><br>y = mx + c,
<br><br>we get, slope or gradient (m) = –E<sub>a</sub>
<br>and Intercept (c) = ln A
<br><br>Also given tha... | mcq | jee-main-2019-online-11th-january-morning-slot | 1,103 |
BX0f3gu7YGoPxKiu1m8Kn | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Consider the given plot of enthalpy of the
following reaction between A and B.<br/>
A+ B $$ \to $$ C + D<br/>
Identify the incorrect statement.
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266787/exam_images/jb7muupvpdrqunp3kbqg.webp"/><img src="data:image/p... | [{"identifier": "A", "content": "Formation of A and B from C has highest\nenthalpy of activation."}, {"identifier": "B", "content": "D is kinetically stable product."}, {"identifier": "C", "content": "C is the thermodynamically stable product"}, {"identifier": "D", "content": "Activation enthalpy to form C is 5kJ mol<s... | ["D"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263418/exam_images/qgwtvctnzwier557jg2q.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265202/exam_images/xzqg3ows3yturz8x6vtv.webp" style="max-width: 100%;height: auto;display: block;margi... | mcq | jee-main-2019-online-9th-april-evening-slot | 1,104 |
IVPcWimfkzGXc9GmeQ3rsa0w2w9jx0vyjfc | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | For the reaction of H<sub>2</sub> with I<sub>2</sub>, the rate constant is 2.5 × 10<sup>–4</sup> dm<sup>3</sup>
mol<sup>–1</sup>s<sup>–1</sup>
at 327°C and 1.0 dm<sup>3</sup>
mol<sup>–1</sup>
at
527°C. The activation energy for the reaction, in kJ mole<sup>–1</sup>
is : (R = 8.314 JK<sup>–1</sup>
mol<sup>–1</sup>... | [{"identifier": "A", "content": "59"}, {"identifier": "B", "content": "166"}, {"identifier": "C", "content": "72"}, {"identifier": "D", "content": "150"}] | ["B"] | null | From Arrhenius equation, we get
<br><br>$$\log {{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
<br><br>$$ \Rightarrow $$ $$\log {1 \over {2.5 \times {{10}^{ - 4}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {600}} - {1 \over {800}}} \right)$$
<br><... | mcq | jee-main-2019-online-10th-april-evening-slot | 1,105 |
yVktXwXyziBaymmJTG7k9k2k5h5sfuo | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate of a certain biochemical reaction at physiological temperature (T) occurs 10<sup>6</sup> times faster with
enzyme than without. The change in the activation energy upon adding enzyme is : | [{"identifier": "A", "content": "+ 6RT"}, {"identifier": "B", "content": "\u2013 6 (2.303)RT"}, {"identifier": "C", "content": "\u2013 6RT"}, {"identifier": "D", "content": "+ 6(2.303)RT"}] | ["B"] | null | The rate constant of a reaction without catalyst is
<br><br>$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>The rate constant in presence of catalyst is
given by
<br><br>$$k' = A{e^{ - {{E{'_a}} \over {RT}}}}$$
<br><br>$$ \therefore $$ $${{k'} \over k} = {e^{ - {{\left( {E{'_a} - {E_a}} \right)} \over {RT}}}}$$
<br><br>... | mcq | jee-main-2020-online-8th-january-morning-slot | 1,106 |
WwY6fIA8HXnjgAQAtgjgy2xukg4o13gz | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate of a reaction decreased by 3.555
times when the temperature was changed
from 40<sup>o</sup>C to 30<sup>o</sup>C. The activation energy
(in kJ mol<sup>–1</sup>) of the reaction is _______.
<br/><br/>Take;
R = 8.314 J mol<sup>–1</sup> K<sup>–1</sup> ln 3.555 = 1.268 | [] | null | 100 | k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>$$ \therefore $$ $$\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$
<br><br>$$ \Rightarrow $$ ln (3.555) = $${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$$
<br><br>$$ \Rightarrow $$ E<sub>a... | integer | jee-main-2020-online-6th-september-evening-slot | 1,107 |
lX2NTCfJLnRUqPdoJijgy2xukfjbf57i | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | A flask contains a mixture of compounds A and
B. Both compounds decompose by first-order
kinetics. The half-lives for A and B are 300 s
and 180 s, respectively. If the concentrations
of A and B are equal initially, the time required
for the concentration of A to be four times that
of B(in s) :
<br/>(Use ln 2 = 0.693) | [{"identifier": "A", "content": "180"}, {"identifier": "B", "content": "120"}, {"identifier": "C", "content": "300"}, {"identifier": "D", "content": "900"}] | ["D"] | null | A<sub>t</sub> = A<sub>0</sub>.e<sup>-k<sub>1</sub>t</sup>
<br><br>B<sub>t</sub> = B<sub>0</sub>.e<sup>-k<sub>2</sub>t</sup>
<br><br>k<sub>1</sub> = $${{\ln 2} \over {300}}$$
<br><br>k<sub>2</sub> = $${{\ln 2} \over {180}}$$
<br><br>Given, A<sub>0</sub> = B<sub>0</sub>
<br><br>and A<sub>t</sub>
and B<sub>t</sub>
are r... | mcq | jee-main-2020-online-5th-september-morning-slot | 1,109 |
7X0pxzOenCqz3yDKKljgy2xukfchqtep | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The number of molecules with energy greater
than the threshold energy for a reaction
increases five fold by a rise of temperature
from 27<sup>o</sup>C to 42<sup>o</sup>C. Its energy of activation in
J/mol is _____.
<br/>(Take ln 5 = 1.6094; R = 8.314 J mol<sup>–1</sup> K<sup>–1</sup>)
| [] | null | 84297.47to84297.48 | $$ \because $$ k = A$${e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>T<sub>1</sub> = 300K, T<sub>2</sub> = 315K
<br><br>As per question K<sub>T<sub>2</sub></sub> = 5K<sub>T<sub>2</sub></sub> as molecules activated are increased five times so K will increases five time.
<br><br>$$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}... | integer | jee-main-2020-online-4th-september-evening-slot | 1,110 |
eIOYvmNDljtJn4Xrf17k9k2k5llve3e | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | A sample of milk splits after 60 min. at 300 K
and after 40 min. at 400 K when the population
of lactobacillus acidophilus in it doubles. The
activa tion energy (in kJ/ mol) for this process
is closest to__________.<br/><br/>
(Given, R = 8.3 J mol<sup>–1</sup> K<sup>–1</sup>, $$\ln \left( {{3 \over 2}} \right) = 0.4$$,... | [] | null | 3.98TO3.99 | Using Arrehenius equation
<br><br>K = A$${e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>$$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$$
<br><br>$$ \Rightarrow $$ $$\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 ... | integer | jee-main-2020-online-9th-january-evening-slot | 1,111 |
39ebrqeMsCvjMWY1ET7k9k2k5ibxkiy | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | For the following reactions<br/>
$$A\buildrel {700K} \over
\longrightarrow {\mathop{\rm Product}\nolimits} $$<br/>
$$A\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits} $$<br/>
it was found that E<sub>a</sub> is decreased by 30 kJ/mol
in the presence of catalyst.<br/>
... | [{"identifier": "A", "content": "198 kJ/mol"}, {"identifier": "B", "content": "135 kJ/mol"}, {"identifier": "C", "content": "105 kJ/mol"}, {"identifier": "D", "content": "75 kJ/mol"}] | ["D"] | null | K<sub>1</sub> = A$${e^{ - {{{E_a}} \over {R \times 700}}}}$$
<br><br>K<sub>2</sub> = A$${e^{ - {{\left( {{E_a} - 30} \right)} \over {R \times 500}}}}$$
<br><br>$$ \because $$Rate is same
<br><br>$$ \therefore $$ Rate constant will also be same
<br><br>K<sub>1</sub> = K<sub>2</sub>
<br><br>A$${e^{ - {{{E_a}} \over {R \t... | mcq | jee-main-2020-online-9th-january-morning-slot | 1,112 |
RwXPf82Y5CjdeEq3H27k9k2k5hkz5pa | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Consider the following plots of rate constant
versus $${1 \over T}$$ for four different reactions. Which
of the following orders is correct for the
activation energies of these reactions?<br/><br/>
<img src="data:image/png;base64,UklGRnINAABXRUJQVlA4IGYNAABwWgCdASqPAQ0BPm00l0kkIqIhIdN6CIANiWlu62AzT+lYzs69P07/k3bL/f+h59... | [{"identifier": "A", "content": "E<sub>c</sub> > E<sub>a</sub> > E<sub>d</sub> > E<sub>b</sub>"}, {"identifier": "B", "content": "E<sub>a</sub> > E<sub>c</sub> > E<sub>d</sub> > E<sub>b</sub>"}, {"identifier": "C", "content": "E<sub>b</sub> > E<sub>a</sub> > E<sub>d</sub> > E<sub>c</sub>"}, {... | ["A"] | null | We know
<br><br>$$k = A{e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>$$ \Rightarrow $$ $${\log _e}k = {\log _e}A - {{{E_a}} \over {2.303RT}}$$
<br><br>According to Arhenius equation plot of $${\log _e}k$$ versus $${1 \over T}$$ is linear.
<br><br>Slope = $$ - {{{E_a}} \over {2.303R}}$$
<br><br>From plot we conclude :
<br><br... | mcq | jee-main-2020-online-8th-january-evening-slot | 1,113 |
iR2uWj0H5Y0qjsKbby1klsd39wx | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | For the reaction, aA + bB $$ \to $$ cC + dD, the plot of log k vs $${1 \over T}$$ is given below :<br/><br/><img src="data:image/png;base64,UklGRo4HAABXRUJQVlA4IIIHAACwMwCdASr4ALEAPm02lkkkIyIhIRVZ+IANiWlu/HyZGutQxv0l/kH43eB/9c8KfDz5P9m/RZrIv03t+fdP4v/P/9760f5Twj90moF6n/yv8Z/arzpdpQAD6S/5v+T/2Lxtv3b+LeoH0y/vX5Ge5n/sf5j7... | [] | null | 526 | $${\log _{10}}K = {\log _{10}}A - {{{E_a}} \over {2.303RT}}$$<br><br>Slope $$ = {{{E_a}} \over {2.303R}} = - 10000$$<br><br>$${\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$<br><br>$${\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 100... | integer | jee-main-2021-online-25th-february-morning-slot | 1,114 |
FihxpHl17L5XyJRPhj1kltchaxj | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The rate constant of a reaction increases by five times on increase in temperature from 27$$^\circ$$C to 52$$^\circ$$C. The value of activation energy in kJ mol<sup>$$-$$1</sup> is _________. (Rounded off to the nearest integer)<br/><br/>[R = 8.314 J K<sup>$$-$$1</sup>mol<sup>$$-$$1</sup>] | [] | null | 52 | T<sub>1</sub> = (273 + 27) = 300 K, T<sub>2</sub> = (273 + 52) = 325 K<br/><br/>Given, temperature coefficient of the reaction,<br/><br/>$${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$$<br/><br/>$$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}... | integer | jee-main-2021-online-25th-february-evening-slot | 1,115 |
vFMPcCRiZfR8ujzA9L1kmhv8djf | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 $$\times$$ 10<sup>$$-$$3</sup> s<sup>$$-$$1</sup> and the activation energy E<sub>a</sub> = 11.488 kJ mol<sup>$$-$$1</sup>, the rate constant at 200 K is ____________ $$\times$$ 10<sup>$$-$$5</sup> s<sup>... | [] | null | 10 | $$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$<br><br>k<sub>1</sub> (at 200 K) = ?<br><br>k<sub>2</sub> (at 300 K) = $$1 \times {10^{ - 3}}{s^{ - 1}}$$<br><br>$$\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \time... | integer | jee-main-2021-online-16th-march-morning-shift | 1,117 |
SDpnDNbGhPZ8CFh1LS1kmiuxl8n | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is _________ min. (Round off to the Nearest Integer). | [] | null | 108 | Initially : $$\left[ {{A_0}} \right] = \left[ {{B_0}} \right] = a$$<br><br>After time 't' min : $$\left[ A \right] = 16\left[ B \right]$$<br><br>$$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$$<br><br>$$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$$<br><br>$$ \Rightarrow a\,.\,{e^{ - {k_A}t}} = ... | integer | jee-main-2021-online-16th-march-evening-shift | 1,118 |
1ktftdmjy | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | The first order rate constant for the decomposition of CaCO<sub>3</sub> at 700 K is 6.36 $$\times$$ 10<sup>$$-$$3</sup>s<sup>$$-$$1</sup> and activation energy is 209 kJ mol<sup>$$-$$1</sup>. Its rate constant (in s<sup>$$-$$1</sup>) at 600 K is x $$\times$$ 10<sup>$$-$$6</sup>. The value of x is ___________. (Nearest ... | [] | null | 16 | K<sub>700</sub> = 6.36 $$\times$$ 10<sup>$$-$$3</sup>s<sup>$$-$$1</sup>;<br><br>K<sub>600</sub> = x $$\times$$ 10<sup>$$-$$6</sup>s<sup>$$-$$1</sup><br><br>E<sub>a</sub> = 209 kJ/mol<br><br>Applying;<br><br>$$\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}... | integer | jee-main-2021-online-27th-august-evening-shift | 1,119 |
1ktjz0okt | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | For the reaction A $$\to$$ B, the rate constant k(in s<sup>$$-$$1</sup>) is given by<br/><br/>$${\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}$$<br/><br/>The energy of activation in kJ mol<sup>$$-$$1</sup> is ____________. (Nearest integer) [Given : R = 8.314 J K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup>] | [] | null | 47 | Given, <br><br>$$\log K = 20.35 - {{2.47 \times {{10}^3}} \over T}$$<br><br>We know<br><br>$$\log K = \log A - {{{E_a}} \over {2.303RT}}$$<br><br>$$ \Rightarrow {{{E_a}} \over {2.303RT}} = 2.47 \times {10^3}$$<br><br>$${E_a} = 2.47 \times {10^3} \times 2.303 \times {{8.314} \over {1000}}$$ KJ/mole<br><br>= 47.29 = 47 (... | integer | jee-main-2021-online-31st-august-evening-shift | 1,120 |
1ktn05dzc | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | Which one of the following given graphs represents the variation of rate constant (k) with temperature (T) for an endothermic reaction? | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1kwj5wez0/dcb59aaf-53dc-418c-af13-f1a02937ec16/b7e2c7d0-503e-11ec-87d6-07ff3bfe6155/file-1kwj5wez1.jpeg?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1kwj5wez0/dcb59aaf-53dc-418c-af13-f1a02937ec16/b7... | ["A"] | null | According to Arrhenius equation
<br><br>$$K = A{e^{ - {{{E_a}} \over {RT}}}}$$
<br><br>The graph will varies as
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwj5wez0/dcb59aaf-53dc-418c-af13-f1a02937ec16/b7e2c7d0-503e-11ec-87d6-07ff3bfe6155/file-1kwj5wez1.jpeg?format=png" data-orsrc="https://... | mcq | jee-main-2021-online-1st-september-evening-shift | 1,121 |
1l54a49uc | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>The activation energy of one of the reactions in a biochemical process is 532611 J mol<sup>$$-$$1</sup>. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k<sub>300</sub> = x $$\times$$ 10<sup>$$-$$3</sup> k<sub>310</sub>. The value of x is ____________.</p>
<p>[Given : $$\ln 10... | [] | null | 1 | $$
\begin{aligned}
& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\\\
& \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=\frac{532611}{8.3} \times\left(\frac{10}{310 \times 300}\right)
\end{aligned}
$$<br/>... | integer | jee-main-2022-online-29th-june-morning-shift | 1,122 |
1l54zbevw | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>The equation <br/><br/>k = (6.5 $$\times$$ 10<sup>12</sup>s<sup>$$-$$1</sup>)e<sup>$$-$$26000K/T</sup> <br/><br/>is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol<sup>$$-$$1</sup>. [nearest integer]</p>
<p>(Given : R = 8.314 J K<sup>$$-$$1</sup> mol<sup>$$-$$1... | [] | null | 216 | $k=A e^{\frac{-E_{a}}{R T}}$
<br/><br/>
$$
\begin{aligned}
\frac{E_{a}}{R T} &=\frac{26000}{T} \\\\
E_{a} &=26000 \times 8.314 \\\\
&=216164 \mathrm{~J} \\\\
&=216 \mathrm{~kJ}
\end{aligned}
$$ | integer | jee-main-2022-online-29th-june-evening-shift | 1,123 |
1l56yzq2n | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>It has been found that for a chemical reaction with rise in temperature by 9 K the rate constant gets doubled. Assuming a reaction to be occurring at 300 K, the value of activation energy is found to be ____________ kJ mol<sup>$$-$$1</sup>. [nearest integer]</p>
<p>(Given ln10 = 2.3, R = 8.3 J K<sup>$$-$$1</sup> mol... | [] | null | 59 | $T_{1}=300 \mathrm{~K}$
<br/><br/>
(Rate constant)
<br/><br/>
$\mathrm{K}_{2}=2 \mathrm{~K}_{1}$, on increase temperature by $9 \mathrm{~K}$
<br/><br/>
$\mathrm{T}_{2}=309 \mathrm{~K}$
<br/><br/>
$\mathrm{Ea}=?$
<br/><br/>
$\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathr... | integer | jee-main-2022-online-27th-june-evening-shift | 1,124 |
1l57t32id | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>The rate constant for a first order reaction is given by the following equation:</p>
<p>$$\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T}$$</p>
<p>The activation energy for the reaction is given by ____________ kJ mol<sup>$$-$$1</sup>. (In nearest integer) (Given : R = 8.3 J K<sup>$$-$$1</sup> mol<sup>$$-$$1</sup... | [] | null | 166 | $\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}$<br/><br/>
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$<br/><br/>
$$
\begin{aligned}
&\therefore \text { on comparing } \frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{R}}=2.0 \times 10^4 \\\\
&\therefore \mathrm{E}_{\mathrm{A}}=2.0 \times 10... | integer | jee-main-2022-online-27th-june-morning-shift | 1,125 |
1l58ejomt | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>A flask is filled with equal moles of A and B. The half lives of A and B are 100 s and 50 s respectively and are independent of the initial concentration. The time required for the concentration of A to be four times that of B is ___________ s.</p>
<p>(Given : ln 2 = 0.693)</p> | [] | null | 200 | $\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}$<br/><br/>
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$<br/><br/>
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$<br/><br/>... | integer | jee-main-2022-online-26th-june-morning-shift | 1,126 |
1l58kgm1l | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>Catalyst A reduces the activation energy for a reaction by 10 kJ mol<sup>$$-$$1</sup> at 300 K. The ratio of rate constants, $${{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}}$$ is e<sup>x</sup>. The value of x is ___________. [nearest integer]</p>
<p>[Assume that the pre-exponential factor is same in both the case... | [] | null | 4 | Activation Energy : It is the minimum amount of energy required to activate the molecules/ atoms so that they can undergo the chemical reaction.
<br/><br/>A catalyst increases the rate of a reaction by lowering the activation energy so that more reactant molecules collide with enough energy to surmount the smaller ene... | integer | jee-main-2022-online-26th-june-evening-shift | 1,127 |
1l5w6s143 | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>For the reaction P $$\to$$ B, the values of frequency factor A and activation energy E<sub>A</sub> are 4 $$\times$$ 10<sup>13</sup> s<sup>$$-$$1</sup> and 8.3 kJ mol<sup>$$-$$1</sup> respectively. If the reaction is of first order, the temperature at which the rate constant is 2 $$\times$$ 10<sup>$$-$$6</sup> s<sup>... | [] | null | 225 | $$
\begin{aligned}
&k=A \mathrm{e}^{\frac{-E_{a}}{R T}} \\\\
&\ln k=\ln A-\frac{E_{a}}{R T} \\\\
&\Rightarrow \log \left(2 \times 10^{-6}\right)=\log \left(4 \times 10^{13}\right)-\frac{8.3 \times 10^{3}}{8.3 \times T \times 2.3} \\\\
&\Rightarrow \log (2)-6=2 \times \log (2)+13-\frac{8.3 \times 10^{3}}{8.3 \times T \t... | integer | jee-main-2022-online-30th-june-morning-shift | 1,129 |
1l6f7sxp2 | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>For the decomposition of azomethane.</p>
<p>CH<sub>3</sub>N<sub>2</sub>CH<sub>3</sub>(g) $$\to$$ CH<sub>3</sub>CH<sub>3</sub>(g) + N<sub>2</sub>(g), a first order reaction, the variation in partial pressure with time at 600 K is given as</p>
<p><img src="data:image/png;base64,UklGRpYJAABXRUJQVlA4IIoJAABwsgCdASoAA0gC... | [] | null | 2 | For first order reaction,
<br/><br/>
$$
\begin{aligned}
&\ln A=\ln A_{0}-k t \\
&\text { Hence Slope }=-k \\
&-k=-3.465 \times 10^{4} \\
&k=\frac{0.693}{t_{1 / 2}} \\
&3.465 \times 10^{4}=\frac{0.693}{t_{1 / 2}} \\
&t_{1 / 2}=2 \times 10^{-5} \mathrm{~s}
\end{aligned}
$$ | integer | jee-main-2022-online-25th-july-evening-shift | 1,130 |
1l6ny40pq | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>For a reaction, given below is the graph of $$\ln k$$ vs $${1 \over T}$$. The activation energy for the reaction is equal to ____________ $$\mathrm{cal} \,\mathrm{mol}^{-1}$$. (nearest integer)</p>
<p>(Given : $$\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}$$ )</p>
<p><img src="data:image/png;ba... | [] | null | 8 | $\begin{aligned}
&\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}} \\\\
&\ln \mathrm{k}=\frac{-\mathrm{Ea}}{\mathrm{RT}}+\ln \mathrm{A} \\\\
&\text { Slope }=\frac{\mathrm{Ea}}{\mathrm{R}}=\frac{20}{5} \\\\
&\mathrm{E}_{\mathrm{a}}=4 \mathrm{R}=8 \,\mathrm{Cal} / \mathrm{mol}
\end{aligned}$ | integer | jee-main-2022-online-28th-july-evening-shift | 1,131 |
1lh04l8va | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>The number of given statement/s which is/are correct is __________.</p>
<p>(A) The stronger the temperature dependence of the rate constant, the higher is the activation energy.</p>
<p>(B) If a reaction has zero activation energy, its rate is independent of temperature.</p>
<p>(C) The stronger the temperature depend... | [] | null | 2 | <p>Clearly, if $\mathrm{E}_a=0, \mathrm{~K}$ is temperature independent if $\mathrm{E}_a>0, \mathrm{~K}$ increase with increase in temperature if $\mathrm{E}_a<0, \mathrm{~K}$ decrease with increase in temperature</p>
<ul>
<li>Rate constant increases with increase in temperature. This is due to a greater number o... | integer | jee-main-2023-online-8th-april-morning-shift | 1,133 |
jaoe38c1lsfk61gm | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>For a reaction taking place in three steps at same temperature, overall rate constant $$\mathrm{K}=\frac{\mathrm{K}_1 \mathrm{~K}_2}{\mathrm{~K}_3}$$. If $$\mathrm{Ea}_1, \mathrm{Ea}_2$$ and $$\mathrm{Ea}_3$$ are 40, 50 and $$60 \mathrm{~kJ} / \mathrm{mol}$$ respectively, the overall $$\mathrm{Ea}$$ is ________ $$\m... | [] | null | 30 | <p>$$\begin{aligned} & \mathrm{K}=\frac{\mathrm{K}_1 \cdot \mathrm{K}_2}{\mathrm{~K}_3}=\frac{\mathrm{A}_1 \cdot \mathrm{A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}} \\ & \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm... | integer | jee-main-2024-online-29th-january-morning-shift | 1,134 |
lv0vyu96 | chemistry | chemical-kinetics-and-nuclear-chemistry | arrhenius-equation | <p>Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.</p>
<p><img src="data:image/png;base64,UklGRpoQAABXRUJQVlA4II4QAADwYQCdASoAA4EAPm00l0gkIqIhI3IqgIANiWlu+FGoamM3IDux8kfLP9t7bv8L+VPoj5S/a3t5ymesPM3+Z/cj9L/dvbD/Gf8Lwf4Av5R/Nv9FvfIAPzH+0... | [] | null | 100 | <p>$$\begin{aligned}
& k=\frac{k_1 k_2}{k_3} \\
& E_{a_{e f f}}=E_{a_1}+E_{a_2}-E_{a_3}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& 400=300+200-E_{\mathrm{a}_3} \\
& 400=500-E_{\mathrm{a}_3} \\
& E_{\mathrm{a}_3}=100 \mathrm{~kJ} \mathrm{~mole}^{-1}
\end{aligned}$$</p> | integer | jee-main-2024-online-4th-april-morning-shift | 1,135 |
fkD6oNW2wDzvFnrU | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | Consider the reaction, 2A + B $$\to$$ products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A
alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is | [{"identifier": "A", "content": "L mol<sup>-1</sup> s<sup>-1</sup> "}, {"identifier": "B", "content": "no unit"}, {"identifier": "C", "content": "mol L<sup>-1</sup> s<sup>-1</sup> "}, {"identifier": "D", "content": "s<sup>-1</sup> "}] | ["A"] | null | Rate = k [A]<sup>x</sup>[B]<sup>y</sup>
<br><br>When [B] is doubled, keeping [A] constant half-life of the reaction
does not change.
<br><br>For a first order reaction $${t_{1/2}} = {{0.693} \over K}\,\,$$ i.e. for a first order reaction $${t_{1/2}}$$ does not depend up on the concentration. Hence the reaction is firs... | mcq | aieee-2007 | 1,136 |
E3S8gyHmFrA7wN2W | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | At 518<sup>o</sup>C the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr,
was 1.00 Torr s<sup>–1</sup> when 5% had reacted and 0.5 Torr s<sup>–1</sup> when 33% had reacted. The order of the reaction is | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}] | ["B"] | null | For a n<sup>th</sup> order reaction, the rate of reaction at time t ,
<br><br>Rate = K [P<sub>t</sub>] <sup>n</sup>
<br><br>Here P<sub>t</sub> = pressure at time t, k = constant.
<br><br><b><u>Note :</u></b>
<br><br>Here instead of concentration of product, pressure of product is given.
<br><br>When 5% is reacted at... | mcq | jee-main-2018-offline | 1,137 |
mhzhGRed4mor9bPOHgjgy2xukeyfg6n8 | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | The results given in the below table were
obtained during kinetic studies of the following
reaction
<br/><br/>2A + B $$ \to $$ C + D
<img src="data:image/png;base64,UklGRgYhAABXRUJQVlA4IPogAABQqgCdASr0ASoBPm0wlEckIqIhI7D+UIANiWlu+BAbvXgaSJ48AGA3u0SYx3gDpyUkfFf8i/JfzK/qX9Y/uP7O/3T1L/FPmf7H+U39e9mH+46UD+K9GP5Z9Zft/8s/pn+... | [{"identifier": "A", "content": "0.3, 0.4"}, {"identifier": "B", "content": "0.4, 0.3"}, {"identifier": "C", "content": "0.4, 0.4"}, {"identifier": "D", "content": "0.3, 0.3"}] | ["A"] | null | Rate = k[A]<sup>a</sup>[B]<sup>b</sup>
<br><br>6 × 10<sup>–3</sup> = k(0.1)<sup>a</sup>(0.1)<sup>b</sup> ...(1)
<br><br>2.4 × 10<sup>–3</sup> = k(0.1)<sup>a</sup>(0.2)<sup>b</sup> ...(2)
<br><br>1.2 × 10<sup>–3</sup> = k(0.2)<sup>a</sup> (0.1)<sup>b</sup> ...(3)
<br><br>(3) ÷ (1) $$ \Rightarrow $$ x = 1
<br><br>(2) ÷ (... | mcq | jee-main-2020-online-2nd-september-evening-slot | 1,139 |
NIE7Ke9JOo1hWLwkZejgy2xukfun274k | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | Consider the following reactions
<br/><br/>A $$ \to $$ P<sub>1</sub> ; B $$ \to $$ P<sub>2</sub> ; C $$ \to $$ P<sub>3</sub> ; D $$ \to $$ P<sub>4</sub>,
<br/><br/>The order of the above reactions are a, b, c,
and d, respectively. The following graph is
obtained when log[rate] vs. log[conc.] are
plotted
<img src="data:... | [{"identifier": "A", "content": "d > b > a > c"}, {"identifier": "B", "content": "d > a > b > c"}, {"identifier": "C", "content": "a > b > c > d"}, {"identifier": "D", "content": "c > a > b > d"}] | ["A"] | null | Rate = k[A]<sup>n</sup>
<br><br>log[Rate] = log k + n log [A]
<br><br>slope = n [n is order of the reaction]
<br><br>So, Which one has higher slope will have higher order.
<br><br>$$ \therefore $$ Correct sequence for the order of the
reaction is
<br>d > b > a > c | mcq | jee-main-2020-online-6th-september-morning-slot | 1,140 |
1ktb72496 | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | The following data was obtained for chemical reaction given below at 975 K.<br/><br/>2NO<sub>(g)</sub> + 2H<sub>2(g)</sub> $$\to$$ N<sub>2(g)</sub> + 2H<sub>2</sub>O<sub>(g)</sub><br/><br/><img src="data:image/png;base64,UklGRrQcAABXRUJQVlA4IKgcAACwiwCdASo0AqkAPm00lkekIyIhJPL66IANiWlu/HRYPF9ZMo3PwR/JPyE8yP6j/Xf7b+0H959... | [] | null | 1 | 7 $$\times$$ 10<sup>$$-$$9</sup> = K $$\times$$ (8 $$\times$$ 10<sup>$$-$$5</sup>)<sup>x</sup> (8 $$\times$$ 10<sup>$$-$$5</sup>)<sup>y</sup> ......... (1)<br><br>2.1 $$\times$$ 10<sup>$$-$$8</sup> = K $$\times$$ (24 $$\times$$ 10<sup>$$-$$5</sup>)<sup>x</sup> (8 $$\times$$ 10<sup>$$-$$5</sup>)<sup>y</sup> ....... (2)<... | integer | jee-main-2021-online-26th-august-morning-shift | 1,142 |
1l6e29xmv | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>The half life for the decomposition of gaseous compound $$\mathrm{A}$$ is $$240 \mathrm{~s}$$ when the gaseous pressure was 500 Torr initially. When the pressure was 250 Torr, the half life was found to be $$4.0$$ min. The order of the reaction is ______________. (Nearest integer)</p> | [] | null | 1 | $$\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{A}}=240 \mathrm{~s}$$ when $$\mathrm{P}=500$$ torr
<br/><br/>
$$\left(t_{1 / 2}\right)_{A}=4 \min =4 \times 60=240$$ sec when $$P=250$$ torr
<br/><br/>
If means half-life is independent of concentration of reactant present.
<br/><br/>
$\therefore$ Order of reaction $=1$ | integer | jee-main-2022-online-25th-july-morning-shift | 1,144 |
1l6jmrjx0 | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>$$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$</p>
<p>The above reaction has been studied at $$800^{\circ} \mathrm{C}$$. The related data are given in the table below</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;... | [] | null | 2 | Let the rate of reaction ( $r$ ) is as
<br/><br/>
$$
\mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}}
$$
<br/><br/>
From $1^{\text {st }}$ data
<br/><br/>
$$
0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1)
$$
<br/><br/>
From $2^{\text {nd }}$ data
<br/><br/>
$$
0... | integer | jee-main-2022-online-27th-july-morning-shift | 1,145 |
1l6mdte0d | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>For kinetic study of the reaction of iodide ion with $$\mathrm{H}_{2} \mathrm{O}_{2}$$ at room temperature :</p>
<p>(A) Always use freshly prepared starch solution.</p>
<p>(B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution.</p>
<p>(C) Record the time immediately after the... | [{"identifier": "A", "content": "$$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{C})$$ only"}, {"identifier": "B", "content": "$$(\\mathrm{A}),(\\mathrm{D}),(\\mathrm{E})$$ only"}, {"identifier": "C", "content": "$$(\\mathrm{D}),(\\mathrm{E})$$ only"}, {"identifier": "D", "content": "$$(\\mathrm{A}),(\\mathrm{B}),(\\mathrm{E})... | ["A"] | null | To minimize contamination, use freshly prepared starch solution to determine end point. As $\mathrm{KI}$ is used in excess to consume all the $\mathrm{H}_{2} \mathrm{O}_{2}$ the concentration of sodium thiosulphate solution is less than $\mathrm{KI}$ solution. After appearance of blue colour record the time immediately... | mcq | jee-main-2022-online-28th-july-morning-shift | 1,146 |
1l6p92cbn | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>The reaction between X and Y is first order with respect to X and zero order with respect to Y.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5... | [] | null | 40 | $$r=k[X][Y]^{0}=k[X]$$
<br/><br/>
Using I & II
<br/><br/>
$\frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\left(\frac{L}{0.1}\right) \quad \Rightarrow \quad \mathrm{L}=0.2$
<br/><br/>
Using I & III
<br/><br/>
$\frac{M \times 10^{-3}}{2 \times 10^{-3}}=\frac{0.4}{0.1} \quad \Rightarrow \quad \mathrm{M}=8$
<br/><br/>
$\frac{M... | integer | jee-main-2022-online-29th-july-morning-shift | 1,147 |
1ldst1xtp | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>For certain chemical reaction $$X\to Y$$, the rate of formation of product is plotted against the time as shown in the figure. The number of $$\mathrm{\underline {correct} }$$ statement/s from the following is ___________.</p>
<p><img src="data:image/png;base64,UklGRqwPAABXRUJQVlA4IKAPAADwuACdASoAA5cBP4G+1mW2LywnIZN... | [] | null | 1 | <p>In region I and II, slope of the graph is positive so reaction is nagative order.</p>
<p>In region III, slope of the graph is zero so the reaction is of zero order.</p>
<p>$$ \therefore $$ Order of this reaction can't be determined. So only (B) is correct.</p> | integer | jee-main-2023-online-29th-january-morning-shift | 1,149 |
1lgrlo6dq | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>The reaction $$2 \mathrm{NO}+\mathrm{Br}_{2} \rightarrow 2 \mathrm{NOBr}$$</p>
<p>takes places through the mechanism given below:</p>
<p>$$\mathrm{NO}+\mathrm{Br}_{2} \Leftrightarrow \mathrm{NOBr}_{2}$$ (fast)</p>
<p>$$\mathrm{NOBr}_{2}+\mathrm{NO} \rightarrow 2 \mathrm{NOBr}$$ (slow)</p>
<p>The overall order of the... | [] | null | 3 | <p>The overall order of a reaction is determined by the slow (rate-determining) step. </p>
<p>Here, the slow step is:
$ \text{NOBr}_2 + \text{NO} \rightarrow 2 \text{NOBr} $</p>
<p>This is a second-order reaction: first-order with respect to $\text{NOBr}_2$ and first-order with respect to NO.</p>
<p>However, $\text{NOB... | integer | jee-main-2023-online-12th-april-morning-shift | 1,151 |
1lgsyyzmb | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>For a chemical reaction $$\mathrm{A}+\mathrm{B} \rightarrow$$ Product, the order is 1 with respect to $$\mathrm{A}$$ and $$\mathrm{B}$$.</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-si... | [{"identifier": "A", "content": "80 and 4"}, {"identifier": "B", "content": "160 and 4"}, {"identifier": "C", "content": "80 and 2"}, {"identifier": "D", "content": "40 and 4"}] | ["C"] | null | $$
\begin{aligned}
& \mathrm{r}=\mathrm{K}[\mathrm{A}]^1[\mathrm{~B}]^1 \\\\
& 0.1=\mathrm{K}(20)^1(0.5)^1 ........(i) \\\\
& 0.40=\mathrm{K}(\mathrm{x})^1(0.5)^1 ........(ii) \\\\
& 0.80=\mathrm{K}(40)^1(\mathrm{y})^1 ........(iii)
\end{aligned}
$$
<br/><br/>From (i) and (ii)
<br/><br/>$$
x=80
$$
<br/><br/>From (i) an... | mcq | jee-main-2023-online-11th-april-evening-shift | 1,152 |
jaoe38c1lsc6egnz | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>Consider the following data for the given reaction</p>
<p>$$2 \mathrm{HI}_{(\mathrm{g})} \rightarrow \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})}$$</p>
<p><img src="data:image/png;base64,UklGRlgUAABXRUJQVlA4IEwUAABQcACdASoAA6sAPm02mEikIyKhI9LZoIANiWlu/HyZyOtQyv0+/xnan/k/ya8/fIL7Hzwcp/YlqX/Hfs3+t/uX7qfHv+W... | [] | null | 2 | <p>Let, $$\mathrm{R}=\mathrm{k}[\mathrm{HI}]^{\mathrm{n}}$$</p>
<p>using any two of given data,</p>
<p>$$\begin{aligned}
& \frac{3 \times 10^{-3}}{7.5 \times 10^{-4}}=\left(\frac{0.01}{0.005}\right)^{\mathrm{n}} \\\\
& \mathrm{n}=2
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift | 1,153 |
luy1mwyx | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>Consider the following first order gas phase reaction at constant temperature $$
\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$$</p>
<p>If the total pressure of the gases is found to be 200 torr after 23 $$\mathrm{sec}$$. and 300 torr upon the complete decomposition of A after a... | [] | null | 3 | <p>To find the rate constant of the first-order reaction indicated, let us first understand the reaction dynamics based on the total pressure change over time. The reaction is:</p>
<p>$$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$$</p>
<p>Initially, only A is present. As the reaction progresses, A decrea... | integer | jee-main-2024-online-9th-april-evening-shift | 1,154 |
lv5gsy3l | chemistry | chemical-kinetics-and-nuclear-chemistry | different-methods-to-determine-order-of-reaction | <p>Consider the following reaction</p>
<p>$$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$$</p>
<p>The time taken for A to become $$1 / 4^{\text {th }}$$ of its initial concentration is twice the time taken to become $$1 / 2$$ of the same. Also, when the change of concentration of B is plotted against time, the resultin... | [] | null | 1 | <p>To determine the overall order of the reaction given by $$\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$$, we can derive the information based on the given details about the kinetics of reactant A and the graphical behavior of reactant B.</p>
<p>The first piece of information tells us that the time for the concent... | integer | jee-main-2024-online-8th-april-morning-shift | 1,156 |
0pAS0aK9pMdrCa6s | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is | [{"identifier": "A", "content": "16 grams"}, {"identifier": "B", "content": "2 grams"}, {"identifier": "C", "content": "32 grams"}, {"identifier": "D", "content": "8 grams"}] | ["D"] | null | $${t_{1/2}} = 5\,\,$$ years, $$T=15$$ years hence -
<br><br>total number of half life periods $$ = {{15} \over 3} = 3.$$
<br><br>$$\therefore$$ Amount left $$ = {{64} \over {{{\left( 2 \right)}^3}}} = 8g$$ | mcq | aieee-2002 | 1,158 |
eHXEeCKSb75YwkZC | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively | [{"identifier": "A", "content": "sec<sup>-1</sup>, Msec<sup>-1</sup>"}, {"identifier": "B", "content": "sec<sup>-1</sup>, M"}, {"identifier": "C", "content": "Msec<sup>-1</sup>, sec<sup>-1</sup>"}, {"identifier": "D", "content": "M, sec<sup>-1</sup>"}] | ["A"] | null | For a zero order reaction.
<br><br>rate $$ = k{\left[ A \right]^ \circ }\,\,\,$$ i.e. rate $$=k$$
<br><br>hence unit of $$k = M.{\sec ^{ - 1}}$$
<br><br>For a first order reaction.
<br><br>rate $$ = k\left[ A \right]$$
<br><br>$$k=M.$$ $${\sec ^{ - 1}}/M = {\sec ^{ - 1}}$$ | mcq | aieee-2002 | 1,160 |
3zlLJSuwq89PodyV | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For the reaction system:
<br/><br/>2NO(g) + O<sub>2</sub>(g) $$\to$$ 2NO<sub>2</sub>(g) volume is suddenly reduce to half its value by increasing the pressure on it. If the reaction is of first order with respect to O<sub>2</sub> and second order with respect to NO, the rate of reaction will | [{"identifier": "A", "content": "diminish to one-eighth of its initial value"}, {"identifier": "B", "content": "increase to eight times of its initial value"}, {"identifier": "C", "content": "increase to four times of its initial value"}, {"identifier": "D", "content": "diminish to one-fourth of its initial value"}] | ["B"] | null | $$r = k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}.$$
<br><br>When the volume is reduced to $$1/2,$$ the conc. will double
<br><br>$$\therefore$$ $$\,\,\,\,\,$$ New rate $$ = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8k\left[ {{O^2}} \right]{\left[ {NO} \right]^2}$$
<br><br>The new rate increases to eight ... | mcq | aieee-2003 | 1,161 |
rQQOGpPKH7jhMLFx | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | A reaction involving two different reactants can never be | [{"identifier": "A", "content": "Unimolecular reaction "}, {"identifier": "B", "content": "First order reaction"}, {"identifier": "C", "content": "second order reaction"}, {"identifier": "D", "content": "Bimolecular reaction "}] | ["A"] | null | The molecularity of reaction is the number of reactant molecules taking part in a single step of the reaction.
<br><br><b>NOTE :</b> The reaction involving two different reactant can never be unimolecular. | mcq | aieee-2005 | 1,163 |
y3iZdaYIT0PcqWtg | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | t<sub>1/4</sub> can be taken as the time taken for the concentration of a reactant to drop to $$3 \over 4$$ of its initial value. If the rate constant for a first order reaction is K, the t<sub>1/4</sub> can be written as | [{"identifier": "A", "content": "0.10 / K"}, {"identifier": "B", "content": "0.29 / K"}, {"identifier": "C", "content": "0.69 / K "}, {"identifier": "D", "content": "0.75 / K"}] | ["B"] | null | $${t_{1/4}} = {{2.303} \over K}\log {1 \over {3/4}}$$
<br><br>$$ = {{2.303} \over K}\log {4 \over 3}$$
<br><br>$$ = {{2.303} \over K}\left( {\log \,4 - \log 3} \right)$$
<br><br>$$ = {{2.303} \over K}\left( {2{{\log }^2} - \log 3} \right)$$
<br><br>$$ = {{2.303} \over K}\left( {2 \times 0.301 - 0.4771} \right)$$
<br><b... | mcq | aieee-2005 | 1,164 |
IHfYxx3h2h00in84 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The following mechanism has been proposed for the reaction of NO with Br<sub>2</sub> to form NOBr: <br/>
NO(g) + Br<sub>2</sub> (g) $$\leftrightharpoons$$ NOBr<sub>2</sub> (g) <br/>
NOBr<sub>2</sub> (g) + NO (g) $$\to$$ 2NOBr (g)<br/>
If the second step is the rate determining step, the order of the reaction with resp... | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["D"] | null | $$\left( i \right)\,\,\,\,\,\,\,\,\,NO\left( g \right) + B{r_2}\left( g \right)\,\,\,\rightleftharpoons\,NOB{r_2}\left( g \right)$$
<br><br>$$\left( {ii} \right)\,\,\,\,\,\,\,\,\,NOB{r_2}\left( g \right) + NO\left( g \right)\,\buildrel \, \over
\longrightarrow 2NOBr\left( g \right)$$
<br><br>Rate law equation $$ = k... | mcq | aieee-2006 | 1,165 |
COrQxrpmaLWvDtVr | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The half life period of a first order chemical reaction is 6.93 minutes. The time required for the
completion of 99% of the chemical reaction will be (log 2=0.301) : | [{"identifier": "A", "content": "230.3 minutes"}, {"identifier": "B", "content": "23.03 minutes "}, {"identifier": "C", "content": "46.06 minutes "}, {"identifier": "D", "content": "460.6 minutes "}] | ["C"] | null | For first order reaction
<br><br>$$k = {{2.303} \over t}\log {{100} \over {100 - 99}}$$
<br><br>$${{0.693} \over {6.93}} = {{2.303} \over t}\log {{100} \over 1}$$
<br><br>$${{0.693} \over {6.93}} = {{2.303 \times 2} \over t}$$
<br><br>$$ \Rightarrow t = 46.06\min $$ | mcq | aieee-2009 | 1,166 |
hLktTOPDljWkQjIX | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For a first order reaction, (A) $$\to$$ products, the concentration of A changes from 0.1 M to 0.025 M in 40
minutes. The rate of reaction when the concentration of A is 0.01 M is : | [{"identifier": "A", "content": "1.73 x 10<sup>\u20135</sup> M/ min"}, {"identifier": "B", "content": "3.47 x 10<sup>\u20134</sup> M/min"}, {"identifier": "C", "content": "3.47 x 10<sup>\u20135 </sup> M/min"}, {"identifier": "D", "content": "1.73 x 10<sup>\u20134</sup> M/min"}] | ["B"] | null | For a first order reaction
<br><br>$$k = {{2.0303} \over t}\,\log \,{a \over {a - x}}$$
<br><br>$$ = {{2.303} \over {40}}\log {{0.1} \over {0.025}}$$
<br><br>$$ = {{2.303} \over {40}}\log 4$$
<br><br>$$ = {{2.303 \times 0.6020} \over {40}}$$
<br><br>$$ = 3.47 \times {10^{ - 2}}$$
<br><br>$$R = K{\left( A \right)^1} = 3... | mcq | aieee-2012 | 1,168 |
tGD5YikwpgBoGz2p | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | Higher order (>3) reactions are rare due to | [{"identifier": "A", "content": "increase in entropy and activation energy as more molecules are involved"}, {"identifier": "B", "content": "shifting of equilibrium towards reactants due to elastic collisions"}, {"identifier": "C", "content": "loss of active species on collision"}, {"identifier": "D", "content": "low p... | ["D"] | null | Reactions of higher order $$\left( { > 3} \right)$$ are very rate due to very less chances of many molecules to undergo effective collisions. | mcq | jee-main-2015-offline | 1,169 |
VI1ZCGKHXMmHURKiRYM1q | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | N<sub>2</sub>O<sub>5</sub> decomposes to NO<sub>2</sub> and O<sub>2</sub> and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be : | [{"identifier": "A", "content": "175.0 mmHg"}, {"identifier": "B", "content": "116.25 mmHg"}, {"identifier": "C", "content": "136.25 mmHg"}, {"identifier": "D", "content": "106.25 mmHg"}] | ["D"] | null | <table class="tg">
<tbody><tr>
<th class="tg-4kyz"></th>
<th class="tg-4kyz">N<sub>2</sub>O<sub>5</sub></th>
<th class="tg-4kyz">$$\rightleftharpoons$$</th>
<th class="tg-4kyz">2NO<sub>2</sub></th>
<th class="tg-bzci">+ $${1 \over 2}$$O<sub>2</sub></th>
</tr>
<tr>
<td class="tg-4kyz">At t... | mcq | jee-main-2018-online-15th-april-morning-slot | 1,171 |
YlCdJY31C8nFgTVGaXo1t | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For a first order reaction, A $$ \to $$ P, t<sub>1/2</sub> (half-life) is 10 days The time required for $${1 \over 4}$$<sup>th</sup> conversion of A (in days) is : (ln 2 = 0.693, ln 3 = 1.1) | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "3.2"}, {"identifier": "C", "content": "4.1"}, {"identifier": "D", "content": "2.5"}] | ["C"] | null | For first order reaction,
<br><br>The half life, t<sub>$${1 \over 2}$$</sub> = $${{0.693} \over k}$$
<br><br>Here given t<sub>$${1 \over 2}$$</sub> = 10 days
<br><br>$$\therefore\,\,\,$$ k = $${{0.693} \over {10}}$$ = 0.0693 days<sup>$$-$$1</sup>
<br><br>Now, the time required for $${1 \over 4}$$th conversion of A is ... | mcq | jee-main-2018-online-15th-april-evening-slot | 1,172 |
t0FqjpWS7TTNgAE8QXJGS | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | If 50% of a reaction occurs in 100 second and 75% of the reaction occurs in 200 secod, the order of this reaction is : | [{"identifier": "A", "content": "Zero "}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Assume initial concentration of the reactant = 1 M
<br><br>After 100 second, concentration becomes of the reactant
<br><br>= 1 $$ \times $$ $${{50} \over {100}}$$ = 0.5 M
<br><br>After 200 second, concentration of the reactant becomes
<br><br>= 1 $$ \times $$ $${{25} \over {100}}$$ = 0.25 M.
<br><br>So, in first 10... | mcq | jee-main-2018-online-16th-april-morning-slot | 1,173 |
Ft8kB9UjMuylZgY1AeiLm | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The reaction 2X $$ \to $$ B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will
be:
| [{"identifier": "A", "content": "18.0 h"}, {"identifier": "B", "content": "9.0 h"}, {"identifier": "C", "content": "7.2 h"}, {"identifier": "D", "content": "12.0 h"}] | ["A"] | null | For zero order reaction,
<br><br>t<sub>1/2</sub> = $${{{a_0}} \over {2k}}$$
<br><br>$$ \Rightarrow $$ k = $${{{a_0}} \over {2{t_{1/2}}}}$$ = $${{0.2} \over {2 \times 6}}$$ = 1.67 $$ \times $$ 10<sup>-2</sup> mol L<sup>–1</sup>h<sup>–1</sup>
<br><br>For zero order reaction,
<br><br>A<sub>0</sub> - A<sub>t</sub> = kt
<br... | mcq | jee-main-2019-online-11th-january-evening-slot | 1,174 |
6t6QNoFnJSEbjAZcG3jgy2xukf25i9vq | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | It is true that : | [{"identifier": "A", "content": "A first order reaction is always a single step reaction"}, {"identifier": "B", "content": "A zero order reaction is a multistep reaction"}, {"identifier": "C", "content": "A zero order reaction is a single step reaction"}, {"identifier": "D", "content": "A second order reaction is alway... | ["B"] | null | Zero order reaction has complex mechanism.
<br>Zero order reaction is a multistep reaction. | mcq | jee-main-2020-online-3rd-september-morning-slot | 1,177 |
aV4tfk8ZVnss57JCmIjgy2xukf9ksjxc | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | If 75% of a first order reaction was completed
in 90 minutes, 60% of the same reaction would
be completed in approximately (in minutes)
_______.
<br/><br/>(Take : log 2 = 0.30; log 2.5 = 0.40) | [] | null | 60 | t<sub>75%</sub> = 90 min = 2 × t<sub>1/2</sub>
<br><br>$$ \Rightarrow $$ t<sub>1/2</sub> = 45 min
<br><br>Rate constant, K = $${{0.693} \over {45}}$$ min<sup>-1</sup>
<br><br>Time for completion of 60% of the reaction,
<br><br>t<sub>60%</sub> = $${{2.303} \over K}\log {{10} \over 4}$$
<br><br>= $${{2.303 \times 45} \ov... | integer | jee-main-2020-online-4th-september-morning-slot | 1,178 |
fYTdCPGXwGKwJN0tV61klrheni9 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 $$ \times $$ 10<sup>-4</sup> s<sup>-1</sup> at 153°C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is ______.<br/>
(Rounded off to the nearest integer) | [] | null | 26 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxfqpwtb/35daed4c-f81a-4b73-8e5a-e633f7c9ecbc/377f3ef0-6229-11ec-95de-59bb36457aaa/file-1kxfqpwtc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxfqpwtb/35daed4c-f81a-4b73-8e5a-e633f7c9ecbc/377f3ef0-6229-11ec-95de-59bb36457aaa/fi... | integer | jee-main-2021-online-24th-february-morning-slot | 1,179 |
Sht7PowVRG3RGD0fCO1klrvkkn4 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | Sucrose hydrolyses in acid solution into glucose and fructose following first order rate law with a half-life of 3.33 h at 25$$^\circ$$C. After 9 h, the fraction of sucrose remaining is f. The value of $${\log _{10}}\left( {{1 \over f}} \right)$$ is ________ $$\times$$ 10<sup>$$-$$2</sup>. (Rounded off to the nearest i... | [] | null | 81 | Given, $$\mathop {{C_{12}}{H_{22}}{O_{11}}}\limits_{Sucrose} + {H_2}O\buildrel {1st\,order} \over
\longrightarrow \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose} $$<br/><br/>$${t_{1/2}} = {{10} \over 3}h$$<br/><br/>At t = 0, a = [A]<sub>0</sub>  ... | integer | jee-main-2021-online-24th-february-evening-slot | 1,180 |
ggbQtu3VJuar2itwcr1kmj8plke | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ $$\times$$ 10<sup>$$-$$3</sup> s<sup>$$-$$1</sup>. (Round off to the Nearest Integer). [Given : log<sub>10</sub>2 = 0.301, ln10 = 2.303] | [] | null | 2 | $$k = {1 \over t}\ln \left[ {{a \over {a - x}}} \right]$$<br><br>$$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$$<br><br>$$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$$<br><br>$$k = {{2.303} \over {570}} \times 0.5$$<br><br>$$k = 2 \times {10^{ - 3}}{s^{ - 1}}$$ | integer | jee-main-2021-online-17th-march-morning-shift | 1,181 |
EWgArk15drIBMzFdtx1kmm25u10 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is _________ min. (Round off to the Nearest Integer). [Use : ln 2 = 0.69; ln 10 = 2.3] | [] | null | 10 | Given t<sub>1/2</sub> = 1 min<br><br>$$ \therefore $$ $$K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$$<br><br>From formula we know,<br><br>$$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$$<br><br>$$ \Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$$<br><br>$$ \Rightarrow 0.... | integer | jee-main-2021-online-18th-march-evening-shift | 1,182 |
1krq6wwao | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The inactivation rate of a viral preparation is proportional to the amount of virus. In the first minute after preparation, 10% of the virus is inactivated. The rate constant for viral inactivation is ___________ $$\times$$ 10<sup>$$-$$3</sup> min<sup>$$-$$1</sup>. (Nearest integer)<br/><br/>[Use : ln 10 = 2.303; log<s... | [] | null | 106 | Unit of rate constant is min<sup>$$-$$1</sup>, so it must be a first order reaction. For first order reaction,<br/><br/>k $$\times$$ t = 2.303 log$${{{A_0}} \over {{A_t}}}$$<br/><br/>k is the rate constant<br/><br/> t is the time<br/><br/> A<sub>0</sub> is the initial conc. <br/><br/>A<sub>t</sub> is the conc. at time,... | integer | jee-main-2021-online-20th-july-morning-shift | 1,183 |
1krrlfd0r | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | $$PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)$$<br/><br/>In the above first order reaction the concentration of PCl<sub>5</sub> reduces from initial concentration 50 mol L<sup>$$-$$1</sup> to 10 mol L<sup>$$-$$1</sup> in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x $$\times$$ 10<sup>$$-$$2</sup> min... | [] | null | 1 | $$PC{l_5}(g)\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{300K}^{I\,order}} PC{l_3}(g) + C{l_2}(g)$$<br><br>t = 0<br><br>50M<br><br>t = 120min<br><br>10 M<br><br>$$ \Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$$<br><br>$$ \Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$$<br><br>$$ ... | integer | jee-main-2021-online-20th-july-evening-shift | 1,184 |
1ks1k9mag | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For the first order reaction A $$\to$$ 2B, 1 mole of reactant A gives 0.2 moles of B after 100 minutes. The half life of the reaction is __________ min. (Round off to the nearest integer).<br/><br/>[Use : ln 2 = 0.69, ln 10 = 2.3]<br/><br/>Properties of logarithms : ln x<sup>y</sup> = y ln x;<br/><br/>$$\ln \left( {{x ... | [] | null | 600to700 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265294/exam_images/swrl81evlrkmvs0qzurq.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Chemistry - Chemical Kinetics and Nuclear Chemistry Question 75 English Ex... | integer | jee-main-2021-online-27th-july-evening-shift | 1,187 |
1ktih92er | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ____________. (Integer answer) | [] | null | 2 | $$k = {{2.303} \over t}\log {a \over {a - x}}$$<br><br>$${{2.303} \over {{t_{50\% }}}}\log {{100} \over {100 - 50}} = {{2.303} \over {{t_{75\% }}}}\log {{100} \over {100 - 75}}$$<br><br>$$ \Rightarrow $$ $${t_{75\% }} = 2{t_{50\% }}$$ | integer | jee-main-2021-online-31st-august-morning-shift | 1,188 |
1ktihd4xf | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | According to the following figure, the magnitude of the enthalpy change of the reaction<br/><br/>A + B $$\to$$ M + N in kJ mol<sup>$$-$$</sup> is equal to ___________. (Integer answer)<br/><br/><img src="data:image/png;base64,UklGRrQcAABXRUJQVlA4IKgcAAAwigCdASqfAQEBPm0ylkikIqIhIxBrSIANiWlu/HyXRh5nZ12fpX/Hfxw8xn6L/Tv61+... | [] | null | 45 | $$\Delta$$H = E<sub>Product</sub> $$-$$ E<sub>Reactant</sub><br><br>= 15 $$-$$ (15 + 45)<br><br>= $$-$$45 KJ/mol<br><br>| $$\Delta$$H | = 45 KJ/mol | integer | jee-main-2021-online-31st-august-morning-shift | 1,189 |
1l55o22wj | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)</p>
<p>(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)</p> | [] | null | 75 | $\lambda=\frac{2.303}{t} \log \frac{A_{0}}{A}$
<br/><br/>
$$
\begin{aligned}
&\frac{0.693}{200}=\frac{2.303}{83} \log \frac{A_{0}}{A} \\\\
&\frac{A}{A_{0}}=0.75
\end{aligned}
$$
<br/><br/>
Hence, percentage of original activity remaining after 83 days is $75 \%$ | integer | jee-main-2022-online-28th-june-evening-shift | 1,190 |
1l56bto6n | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>For a first order reaction A $$\to$$ B, the rate constant, k = 5.5 $$\times$$ 10<sup>$$-$$14</sup> s<sup>$$-$$1</sup>. The time required for 67% completion of reaction is x $$\times$$ 10<sup>$$-$$1</sup> times the half life of reaction. The value of x is _____________ (Nearest integer)</p>
<p>(Given : log 3 = 0.4771... | [] | null | 16 | $$
\begin{aligned}
&\because \mathrm{kt}=\ln \frac{\mathrm{A}_{0}}{\mathrm{~A}} \\\\
&\frac{\ln 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\ln \frac{\mathrm{A}_{0}}{0.33 \mathrm{~A}_{0}} \\\\
&\frac{\log 2}{\mathrm{t}_{\frac{1}{2}}} \mathrm{t}_{67 \%}=\log \frac{1}{0.33} \\\\
&\mathrm{t}_{67 \%}=1.566 \mathrm{t}_{... | integer | jee-main-2022-online-28th-june-morning-shift | 1,191 |
1l6gsm8wx | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>For a reaction $$\mathrm{A} \rightarrow 2 \mathrm{~B}+\mathrm{C}$$ the half lives are $$100 \mathrm{~s}$$ and $$50 \mathrm{~s}$$ when the concentration of reactant $$\mathrm{A}$$ is $$0.5$$ and $$1.0 \mathrm{~mol} \mathrm{~L}^{-1}$$ respectively. The order of the reaction is ______________ . (Nearest Integer)</p> | [] | null | 2 | $t_{1 / 2} \propto \frac{1}{\left(a_{0}\right)^{n-1}}$
<br/><br/>
$$
\begin{array}{ll}
\mathrm{t}_{1 / 2}=100 \,\mathrm{sec} & \mathrm{a}_{0}=0.5 \\
\mathrm{t}_{1 / 2}=50 \,\mathrm{sec} & \mathrm{a}_{0}=1
\end{array}
$$
<br/><br/>
$$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$$
<br/><br/>
$$(2)=(2)^{n-1}$$
<b... | integer | jee-main-2022-online-26th-july-morning-shift | 1,193 |
1l6kurge0 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>$$\matrix{
{[A]} & \to & {[B]} \cr
{{\mathop{\rm Reactant}\nolimits} } & {} & {{\mathop{\rm Product}\nolimits} } \cr
} $$</p>
<p>If formation of compound $$[\mathrm{B}]$$ follows the first order of kinetics and after 70 minutes the concentration of $$[\mathrm{A}]$$ was found to be half ... | [] | null | 165 | $\mathrm{K}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{70 \times 60}$<br/><br/>
$=\frac{6930}{7 \times 6} \times 10^{-6}$<br/><br/>
$=165 \times 10^{-6} \mathrm{~s}^{-1}$ | integer | jee-main-2022-online-27th-july-evening-shift | 1,195 |
1l6meac34 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>For the given first order reaction</p>
<p>$$\mathrm{A} \rightarrow \mathrm{B}$$</p>
<p>the half life of the reaction is $$0.3010 \mathrm{~min}$$. The ratio of the initial concentration of reactant to the concentration of reactant at time $$2.0 \mathrm{~min}$$ will be equal to ___________. (Nearest integer)</p> | [] | null | 100 | $\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \quad\quad \mathrm{t}_{1 / 2}$ given $=0.3010$
<br/><br/>
$$
\begin{aligned}
&K=\frac{0.693}{0.3010} \\\\
&K=2.30
\end{aligned}
$$
<br/><br/>
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
<br/><br/>
$A_{0... | integer | jee-main-2022-online-28th-july-morning-shift | 1,196 |
1l6rlrh7w | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>Assuming $$1 \,\mu \mathrm{g}$$ of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ______ $$\times\, 10^{-1} \mu \mathrm{g}$$.</p>
<p>[Given : ln 10 = 2.303; log 2 = 0.30]</p> | [] | null | 1 | $$t=\frac{1}{\lambda} \ln \left(\frac{a}{a-x}\right)$$
<br/><br/>
$$
\begin{aligned}
&\Rightarrow100=\left(\frac{30}{\ln 2}\right)\left[\ln \left(\frac{1}{w}\right)\right] \\
&\Rightarrow{\left[\frac{100 \times \log 2}{30}\right]=\log \left(\frac{1}{w}\right)} \\
&\Rightarrow1=\log \left(\frac{1}{w}\right) \\
&\Rightar... | integer | jee-main-2022-online-29th-july-evening-shift | 1,197 |
1ldo4awir | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>$$\mathrm{A}$$ $$\rightarrow \mathrm{B}$$</p>
<p>The above reaction is of zero order. Half life of this reaction is $$50 \mathrm{~min}$$. The time taken for the concentration of $$\mathrm{A}$$ to reduce to one-fourth of its initial value is ____________ min. (Nearest integer)</p> | [] | null | 75 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4ep7dn/80539e04-2b31-4fad-91c3-e23e318096d9/4498afa0-ac7d-11ed-85a4-571c003beb83/file-1le4ep7do.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le4ep7dn/80539e04-2b31-4fad-91c3-e23e318096d9/4498afa0-ac7d-11ed-85a4-571c003beb83/fi... | integer | jee-main-2023-online-1st-february-evening-shift | 1,198 |
ldoa4gw5 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The rate constant for a first order reaction is $20 \mathrm{~min}^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is _______ $\times 10^{-2} \mathrm{~min}$. (Nearest integer)<br/><br/>
(Given : $\ln 10=2.303$ and $$ \log 2=0.3010 \text { )}$$ | [] | null | 17 | $K=20 \mathrm{~min}^{-1}$
<br/><br/>$$
\mathrm{t}_{1 / 2}=\frac{0.6932}{20}=\frac{\ln 2}{20}
$$
<br/><br/>Required time $=\mathrm{n} \times \mathrm{t}_{1 / 2}$
<br/><br/>$$C = {{{C_0}} \over {{2^n}}} = {{{C_0}} \over {32}}$$
<br/><br/>$$ \Rightarrow $$ $${2^n} = 32$$ = $${2^5}$$
<br/><br/>$$ \Rightarrow $$ n = 5
<b... | integer | jee-main-2023-online-31st-january-evening-shift | 1,199 |
ldqy0fq0 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | An organic compound undergoes first-order decomposition. If the time taken for the $60 \%$ decomposition is $540 \mathrm{~s}$, then the time required for $90 \%$ decomposition will be ________ s. (Nearest integer).
<br/><br/>
Given: $\ln 10=2.3 ; \log 2=0.3$ | [] | null | 1350 | For the first order reaction,
<br/><br/>$$
k=\frac{1}{t} \ln \frac{a}{a-x} $$
<br/><br/>$$ \Rightarrow $$ $$ t=\frac{1}{k} \ln \frac{a}{a-x}
$$
<br/><br/>When reaction is $60 \%$ completed,
<br/><br/>$$
x=\frac{60}{100} a=0.6 a, t=540 \text { seconds } ; $$
<br/><br/>$k= \frac{1}{t_1} \ln \frac{a}{a-0.6 a}$
<br/><br/>... | integer | jee-main-2023-online-30th-january-evening-shift | 1,201 |
1ldr4uo5p | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>If compound A reacts with B following first order kinetics with rate constant $$2.011 \times 10^{-3} \mathrm{~s}^{-1}$$. The time taken by $$\mathrm{A}$$ (in seconds) to reduce from $$7 \mathrm{~g}$$ to $$2 \mathrm{~g}$$ will be ___________. (Nearest Integer)</p>
<p>$$[\log 5=0.698, \log 7=0.845, \log 2=0.301]$$</p> | [] | null | 623 | <p>$$t = {{2.303} \over k}\log {{{C_0}} \over {{C_t}}}$$</p>
<p>$$ = {{2.303} \over {2.011 \times {{10}^{ - 3}}}}\log {7 \over 2}$$</p>
<p>$$ = {{2.303 \times {{10}^3}} \over {2.011}}(.845 - .301)$$</p>
<p>$$ = 622.99$$</p>
<p>$$ \approx 623$$ sec.</p> | integer | jee-main-2023-online-30th-january-morning-shift | 1,202 |
1ldu3leo3 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>A first order reaction has the rate constant, $$\mathrm{k=4.6\times10^{-3}~s^{-1}}$$. The number of correct statement/s from the following is/are __________</p>
<p>Given : $$\mathrm{\log3=0.48}$$</p>
<p>A. Reaction completes in 1000 s.</p>
<p>B. The reaction has a half-life of 500 s.</p>
<p>C. The time required for ... | [] | null | 1 | (A) $\underset{1-\alpha}{\mathrm{A}} \longrightarrow$ Products
<br/><br/>
$\mathrm{k}=4.6 \times 10^{-3} \mathrm{~s}^{-1}$
<br/><br/>
$\mathrm{kt}=\ln \frac{1}{1-\alpha}$
<br/><br/>
$\alpha=1-\mathrm{e}^{-\mathrm{kt}}$
<br/><br/>
Reaction completes at infinite time.
<br/><br/>(B) For first order reaction,
<br/><br/>
Ha... | integer | jee-main-2023-online-25th-january-evening-shift | 1,203 |
1ldv10608 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>For the first order reaction A $$\to$$ B, the half life is 30 min. The time taken for 75% completion of the reaction is _________ min. (Nearest integer)</p>
<p>Given : log 2 = 0.3010</p>
<p>log 3 = 0.4771</p>
<p>log 5 = 0.6989</p> | [] | null | 60 | Time taken for $75 \%$ completion
<br/><br/>
$=2 \times \mathrm{t}_{1 / 2}$
<br/><br/>
$=2 \times 30$
<br/><br/>
$=60 \mathrm{~min}$ | integer | jee-main-2023-online-25th-january-morning-shift | 1,204 |
1lgp3cfsl | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>A(g) $$\to$$ 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800 mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will be _________ mm Hg. (Nearest integer)</p> | [] | null | 2200 | At 10 minutes, the pressure increased by 2x:
<br/><br/>
$$800 + 2x = 1600$$<br/><br/>
$$2x = 800$$<br/><br/>
$$x = 400$$
<br/><br/>
The rate constant (k) can be found as:
<br/><br/>
$$k = \frac{2.303}{10} \log \frac{800}{400} = \frac{2.303 \times \log 2}{10}$$
<br/><br/>
For 30 minutes, we can set up the equation:
<br/... | integer | jee-main-2023-online-13th-april-evening-shift | 1,205 |
1lgvv9smd | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>The number of <b>incorrect</b> statement/s from the following is ___________</p>
<p>A. The successive half lives of zero order reactions decreases with time.</p>
<p>B. A substance appearing as reactant in the chemical equation may not affect the rate of reaction</p>
<p>C. Order and molecularity of a chemical reactio... | [] | null | 1 | <p>A. The successive half lives of zero order reactions decreases with time. This statement is <strong>correct</strong>. The half-life of a zero-order reaction is given by the formula $t_{1/2} = [A]_0/2k$, where $[A]_0$ is the initial concentration and k is the rate constant. As the concentration decreases, the half-li... | integer | jee-main-2023-online-10th-april-evening-shift | 1,207 |
1lh31x80m | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>Consider the following reaction that goes from A to B in three steps as shown below:</p>
<p><img src="data:image/png;base64,UklGRlgOAABXRUJQVlA4IEwOAAAwswCdASoAA3UBP4HA2WS2MKynItMpksAwCWlu4XPT0mNwur6qtDbPHuF/h/Cm8zKGn3ep/epM3sB9vAHG+3gDjfaocBnZ2ITNJCkJmkhSEzSQll35y0T9KBqqLLMwC789DueM+a5W8tR3niU86utlVYoWyqsULZVT5ycvV... | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lkep71ou/00f8d8be-c626-4913-9620-52b7c2e1505c/72d376e0-28ef-11ee-8dcd-3152cee9aeb3/file-6y3zli1lkep71ov.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/6y3zli1lkep71ou/00f8d8be-c626-4913-962... | ["B"] | null | Step with highest activation energy is RDS, so step II is RDS<br><br>
No. of activated complex $=3$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmjv8hbn/6d0e0cb7-c184-429d-9efa-8da538f85f81/6b7dd030-535f-11ee-a6c7-77f5d3c4aba9/file-6y3zli1lmjv8hbo.png?format=png" data-orsrc="https://a... | mcq | jee-main-2023-online-6th-april-evening-shift | 1,209 |
lsapa057 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | The following data were obtained during the first order thermal decomposition of a gas A at constant volume :<br/><br/>
$\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ <br/><br/>
<style>
table {
border-collapse: collapse;
width: 100%;
}
th, td {
border: 1px solid... | [] | null | 2 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsc7sq84/87ceba2b-60db-485b-be19-a6b831dacb88/932da130-c5f3-11ee-942d-ad97c2c39cb5/file-6y3zli1lsc7sq85.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsc7sq84/87ceba2b-60db-485b-be19-a6b831dacb88/932da130-c5f3-11ee-94... | integer | jee-main-2024-online-1st-february-evening-shift | 1,210 |
jaoe38c1lsdagmj2 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>$$\mathrm{r}=\mathrm{k}[\mathrm{A}]$$ for a reaction, $$50 \%$$ of $$\mathrm{A}$$ is decomposed in 120 minutes. The time taken for $$90 \%$$ decomposition of $$\mathrm{A}$$ is _________ minutes.</p> | [] | null | 399 | <p>$$\mathrm{r}=\mathrm{k}[\mathrm{A}]$$</p>
<p>So, order of reaction $$=1$$</p>
<p>$$\mathrm{t}_{1 / 2}=120 \mathrm{~min}$$</p>
<p>For $$90 \%$$ completion of reaction</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right) \\
& \Rightarrow ... | integer | jee-main-2024-online-31st-january-evening-shift | 1,212 |
1lsgz859k | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>The rate of First order reaction is $$0.04 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$ at 10 minutes and $$0.03 \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}$$ at 20 minutes after initiation. Half life of the reaction is _______ minutes.<br/><br/> (Given $$\log 2=0.3010, \log 3=0.4771$$)</p> | [] | null | 24 | <p>$$\begin{aligned}
& 0.04=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 10 \times 60} \quad \text{..... (1)}\\
& 0.03=\mathrm{k}[\mathrm{A}]_0 \mathrm{e}^{-\mathrm{k} \times 20 \times 60} \quad \text{..... (2)}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\frac{4}{3} & =\mathrm{e}^{600 \mathrm{k}(2-1)} /(2) \\
... | integer | jee-main-2024-online-30th-january-morning-shift | 1,214 |
lv2es2g1 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>Consider the following reaction, the rate expression of which is given below</p>
<p>$$\begin{aligned}
& \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \\
& \text { rate }=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2}
\end{aligned}$$</p>
<p>The reaction is initiated by taking $$1 \mathrm{~M}$$ concentrati... | [] | null | 50 | <p>$$\begin{aligned}
& A+B \rightarrow C \\
& \frac{-d[A]}{d t}=k[A]^{1 / 2}[B]^{1 / 2}
\end{aligned}$$</p>
<p>Since, $$[A]=[B]$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \frac{-d[A]}{d t}=k[A] \\
& \Rightarrow \quad k t=\ln \frac{[A]_0}{[A]} \\
& \Rightarrow \quad t=\frac{1}{4.6 \times 10^{-2}} \times \ln \left(\f... | integer | jee-main-2024-online-4th-april-evening-shift | 1,215 |
lvb2abd4 | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>Consider the two different first order reactions given below</p>
<p>$$\begin{aligned}
& \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \text { (Reaction 1) } \\
& \mathrm{P} \rightarrow \mathrm{Q} \text { (Reaction 2) }
\end{aligned}$$</p>
<p>The ratio of the half life of Reaction 1 : Reaction 2 is $$5: 2$$ I... | [] | null | 17 | <p>$$\mathrm{A+B \rightarrow C}$$ Reaction .... (1)</p>
<p>$$\mathrm{P} \rightarrow \mathrm{Q} \quad$$ Reaction .... (2)</p>
<p>$$\frac{\left(t_{\frac{1}{2}}\right)_1}{\left(t_{\frac{1}{2}}\right)_2}=\frac{k}{k_1}= \frac{5}{2}$$</p>
<p>$$\begin{aligned}
& \frac{\frac{t_2}{3}}{t_{\frac{4}{5}}^5}=\frac{k_2}{k_1} \frac{\l... | integer | jee-main-2024-online-6th-april-evening-shift | 1,216 |
lvc58dzo | chemistry | chemical-kinetics-and-nuclear-chemistry | integrated-rate-law-equations | <p>Time required for $$99.9 \%$$ completion of a first order reaction is _________ times the time required for completion of $$90 \%$$ reaction.(nearest integer)</p> | [] | null | 3 | <p>To determine the time required for a certain level of completion ($x\%$) of a first-order reaction, we can use the formula that relates the time $ t $, the rate constant $ k $, and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ ... | integer | jee-main-2024-online-6th-april-morning-shift | 1,217 |
CZAJ4QtIBawaIF5v | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | $$\beta$$ - particle is emitted in radioactivity by | [{"identifier": "A", "content": "conversion of proton to neutron"}, {"identifier": "B", "content": "from outermost orbit"}, {"identifier": "C", "content": "conversion of neutron to proton"}, {"identifier": "D", "content": "$$\\beta$$ -particle is not emitted"}] | ["C"] | null | $${}_0{n^1} \to {}_{ + 1}{p^1} + {}_{ - 1}{e^{}}$$ | mcq | aieee-2002 | 1,218 |
T6sg5c4kC0Snw7S3 | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | The radionucleide $${}_{90}^{234}Th$$ undergoes two successive $$\beta$$ -decays followed by one $$\alpha$$-decay. The atomic number
and the mass number respectively of the resulting radionucleide are | [{"identifier": "A", "content": "94 and 230"}, {"identifier": "B", "content": "90 and 230"}, {"identifier": "C", "content": "92 and 230"}, {"identifier": "D", "content": "92 and 234"}] | ["B"] | null | $${}_{90}^{234}Th\buildrel { - \beta } \over
\longrightarrow {}_{91}^{234}X\buildrel {\, - \beta } \over
\longrightarrow {}_{92}^{234}Th\buildrel { - \alpha } \over
\longrightarrow {}_{90}^{230}Th$$ | mcq | aieee-2003 | 1,219 |
4OgYvm3kGvpGgflZ | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | The half-life of a radioactive isotope is three hours. If the initial mass of the isotope were 256 g, the mass of
it remaining undecayed after 18 hours would be | [{"identifier": "A", "content": "8.0 g"}, {"identifier": "B", "content": "12.0 g"}, {"identifier": "C", "content": "16.0 g"}, {"identifier": "D", "content": "4.0 g"}] | ["D"] | null | $${t_{1/2}} = 3\,$$ hrs. $$T=18$$ hours
<br><br>as $$\,\,\,\,T = n \times {t_{1/2}}$$
<br><br>$$\therefore$$ $$\,\,\,\,n = {{18} \over 3} = 6$$
<br><br>Initial mass $$\left( {{C_0}} \right) = 256\,g$$
<br><br>$$\therefore$$ $$\,\,\,\,{C_n} = {{{C_0}} \over {{2^n}}} = {{256} \over {{{\left( 2 \right)}^6}}} = {{256} \... | mcq | aieee-2003 | 1,220 |
kUkCNXWLkUBsUDI0 | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | In the transformation of $${}_{92}^{238}U$$ to $${}_{92}^{234}U$$, if one emission is an α-particle, what should be the other
emission(s)? | [{"identifier": "A", "content": "Two $$\\beta^-$$"}, {"identifier": "B", "content": "Two $$\\beta^-$$ and one $$\\beta^+$$"}, {"identifier": "C", "content": "One $$\\beta^-$$ and one $$\\gamma$$"}, {"identifier": "D", "content": "One $$\\beta^+$$ and One $$\\beta^-$$"}] | ["A"] | null | $${}_{92}^{238}U\buildrel { - \alpha } \over
\longrightarrow {}_{90}^{234}Th\buildrel {\, - 2\beta } \over
\longrightarrow {}_{92}^{234}U$$ | mcq | aieee-2006 | 1,225 |
rQtJEhNZohnohoAz | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial
activity is ten times the permissible value, after how many days will it be safe to enter the room? | [{"identifier": "A", "content": "1000 days"}, {"identifier": "B", "content": "300 days "}, {"identifier": "C", "content": "10 days"}, {"identifier": "D", "content": "100 days"}] | ["D"] | null | Suppose activity of safe working $$=A$$
<br><br>Given $${A_0} = 10A$$
<br><br>$$\lambda = {{0.693} \over {{t_{1/2}}}} = {{0.693} \over {30}}$$
<br><br>$${t_{{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}}} = {{2.303} \over \lambda }\log {{{A_0}} \over A}$$
<br><br>$$ ... | mcq | aieee-2007 | 1,226 |
FQS6yLzHkmhyTZ0M | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | Which of the following nuclear reactions will generate an isotope? | [{"identifier": "A", "content": "neutron particle emission"}, {"identifier": "B", "content": "positron emission "}, {"identifier": "C", "content": "$$\\alpha$$-particle emission"}, {"identifier": "D", "content": "$$\\beta$$-particle emission"}] | ["A"] | null | <b>NOTE :</b> Isotopes are atoms of same element having same atomic number but different atomic masses. Neutron has atomic number $$0$$ and atomic mass $$1.$$ So loss of neutron will generate isotope. $$e.g.,$$
<br>$$${}_{92}{U^{238}} + {}_0{n^1} \to {}_{92}{U^{239}}$$$ | mcq | aieee-2007 | 1,227 |
X58IrT2rNmlJDvYAZn3rsa0w2w9jwvcemf3 | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | A bacterial infection in an internal wound grows as N'(t) = N<sub>0</sub> exp(t), where the time t is in hours. A does of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as $${{dN} \over {dt}} = - 5{N^2}$$.
What will be the plot of $${{{N_0}} \over ... | [{"identifier": "A", "content": "<picture><source media=\"(max-width: 320px)\" srcset=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266126/exam_images/p18civrhfx5y2lorpwya.webp\"><img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264212/exam_images/th0zrbux20wch4hvbge6.webp\" style=\"max-width... | ["D"] | null | $${{dN} \over {dt}} = - 5{N^2}$$ for t > 1 hr
<br><br>'N' increases upto 1 hr and then start decreasing after 1 hr.
<br><br>$$ \therefore $$ $${{{N_0}} \over N}$$ will increase after 1 hr.
<br><br>Therefore option (D) is correct. | mcq | jee-main-2019-online-10th-april-morning-slot | 1,228 |
1krt3yw5x | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | Isotope(s) of hydrogen which emits low energy $$\beta$$<sup>$$-$$</sup> particles with t<sub>1/2</sub> value > 12 years is/are | [{"identifier": "A", "content": "Protium"}, {"identifier": "B", "content": "Tritium"}, {"identifier": "C", "content": "Deuterium"}, {"identifier": "D", "content": "Deuterium and Tritium"}] | ["B"] | null | $$_1^1$$H and $$_1^2$$H are stable while $$_1^3$$H is radioactive. | mcq | jee-main-2021-online-22th-july-evening-shift | 1,230 |
lsbn7mjs | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | The ratio of $\frac{{ }^{14} \mathrm{C}}{{ }^{12} \mathrm{C}}$ in a piece of wood is $\frac{1}{8}$ part that of atmosphere. If half life of ${ }^{14} \mathrm{C}$ is 5730 years, the age of wood sample is ________ years. | [] | null | 17190 | <p>The given problem involves calculating the age of a wood sample using the carbon-14 dating method. Carbon-14 ($^{14}C$) is a radioactive isotope of carbon that decays over time, and its ratio compared to carbon-12 ($^{12}C$) can be used to date organic materials. The ratio of $\frac{^{14}C}{^{12}C}$ in the wood is $... | integer | jee-main-2024-online-1st-february-morning-shift | 1,231 |
jaoe38c1lsfpmc1s | chemistry | chemical-kinetics-and-nuclear-chemistry | nuclear-chemistry | <p>The half-life of radioisotope bromine - 82 is 36 hours. The fraction which remains after one day is ________ $$\times 10^{-2}$$.</p>
<p>(Given antilog $$0.2006=1.587$$)</p> | [] | null | 63 | <p>Half life of bromine $$-82=36$$ hours</p>
<p>$$\begin{aligned}
& \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\
& \mathrm{~K}=\frac{0.693}{36}=0.01925 \mathrm{~hr}^{-1} \\
& 1^{\text {st }} \text { order rxn kinetic equation } \\
& \mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \... | integer | jee-main-2024-online-29th-january-evening-shift | 1,232 |
A4qBbd9hlrVScvAw | chemistry | chemical-kinetics-and-nuclear-chemistry | rate-laws-and-rate-constant | For the reaction A + 2B $$\to$$ C, rate is given by R = [A] [B]<sup>2</sup> then the order of the reaction is | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "7"}] | ["A"] | null | <b>NOTE :</b> Order is the sum of the power of the concentrations terms in rate law expression.
<br><br>Hence the order of reaction is $$=1+2=3$$ | mcq | aieee-2002 | 1,233 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.