Chemical Kinetics and Nuclear Chemistry
211 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Morning Shift
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.
(Given : ln 2 = 0.693)
Correct Answer: 200
Explanation:
$\mathrm{k}_{\mathrm{A}}=\frac{\ln 2}{100} ; \mathrm{k}_{\mathrm{B}}=\frac{\ln 2}{50}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$
$\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$
$\mathrm{A}_0=\mathrm{B}_0$
$\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}$
$\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$
$\mathrm{t}=200 ~ \mathrm{sec}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{-\mathrm{k}_{\mathrm{A}} \mathrm{t}}$
$\mathrm{A}_{\mathrm{t}}=\mathrm{A}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{100} \times \mathrm{t}\right)}$
$\mathrm{B}_{\mathrm{t}}=\mathrm{B}_0 \times \mathrm{e}^{\left(\frac{-\ln 2}{50} \times \mathrm{t}\right)}$
$\mathrm{A}_0=\mathrm{B}_0$
$\& \mathrm{~A}_{\mathrm{t}}=4 \mathrm{~B}_{\mathrm{t}}$
$\mathrm{e}^{-\frac{\ln 2}{100} \times \mathrm{t}}=4 \times \mathrm{e}^{-\frac{\ln 2}{50} \times \mathrm{t}}$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\mathrm{e}^{\frac{\ln 2}{100} \times \mathrm{t}}=4$
$\frac{\ln 2}{100} \times \mathrm{t}=\ln 4=2 \ln 2$
$\mathrm{t}=200 ~ \mathrm{sec}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Evening Shift
At 345 K, the half life for the decomposition of a sample of a gaseous compound initially at 55.5 kPa was 340 s. When the pressure was 27.8 kPa, the half life was found to be 170 s. The order of the reaction is ____________. [integer answer]
Correct Answer: 0
Explanation:
$t_{1 / 2} \propto \frac{1}{\left[P_{0}\right]^{n-1}}$
$ \begin{aligned} &\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\frac{\left[\mathrm{P}_{0}\right]_{2}^{\mathrm{n}-1}}{\left[\mathrm{P}_{0}\right]_{1}^{\mathrm{n}-1}} \\\\ &\frac{340}{170}=\left(\frac{27.8}{55.5}\right)^{\mathrm{n}-1} \\\\ &2=\left(\frac{1}{2}\right)^{\mathrm{n}-1} \\\\ &2=(2)^{1-n} \\\\ &1-\mathrm{n}=1 \\\\ &\mathrm{n}=0 \end{aligned} $
$ \begin{aligned} &\frac{\left(\mathrm{t}_{1 / 2}\right)_{1}}{\left(\mathrm{t}_{1 / 2}\right)_{2}}=\frac{\left[\mathrm{P}_{0}\right]_{2}^{\mathrm{n}-1}}{\left[\mathrm{P}_{0}\right]_{1}^{\mathrm{n}-1}} \\\\ &\frac{340}{170}=\left(\frac{27.8}{55.5}\right)^{\mathrm{n}-1} \\\\ &2=\left(\frac{1}{2}\right)^{\mathrm{n}-1} \\\\ &2=(2)^{1-n} \\\\ &1-\mathrm{n}=1 \\\\ &\mathrm{n}=0 \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
For a given chemical reaction
$\gamma$1A + $\gamma$2B $\to$ $\gamma$3C + $\gamma$4D
Concentration of C changes from 10 mmol dm$-$3 to 20 mmol dm$-$3 in 10 seconds. Rate of appearance of D is 1.5 times the rate of disappearance of B which is twice the rate of disappearance A. The rate of appearance of D has been experimentally determined to be 9 mmol dm$-$3 s$-$1. Therefore, the rate of reaction is _____________ mmol dm$-$3 s$-$1. (Nearest Integer)
Correct Answer: 1
Explanation:
Rate $=\frac{1}{r_{1}}\left(\frac{-d[A]}{d t}\right)=\frac{1}{r_{2}}\left(\frac{-d[B]}{d t}\right)=\frac{1}{r_{3}}\left(\frac{-d[C]}{d t}\right)$
$ =\frac{1}{r_{4}}\left(\frac{d[D]}{d t}\right) $
$\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{\mathrm{r}_{2}}\left(\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)$
$ \frac{r_{4}}{r_{2}}=\frac{3}{2} $
$\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\left(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) \Rightarrow \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=2$
$r_{2}=2 r_{1}$
$ \begin{aligned} &\mathrm{r}_{4}=1.5 \mathrm{r}_{2}=3 \mathrm{r}_{1} \\\\ &\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=1 \mathrm{~m} \cdot \mathrm{mol} ~\mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{r_{3}} \cdot \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{r}_{4}}{\mathrm{r}_{3}}=9 \\\\ &\mathrm{r}_{4}=9 \mathrm{r}_{3}=3 \mathrm{r}_{1} \\\\ &\Rightarrow \mathrm{r}_{1}=3 \mathrm{r}_{3} \\\\ &\begin{array}{rr} 3 \mathrm{r}_{3} \mathrm{~A}+6 \mathrm{r}_{3} \mathrm{~B} \rightarrow \mathrm{r}_{3} \mathrm{C}+9 \mathrm{r}_{3} \mathrm{D} \end{array} \\\\ &\begin{aligned} \therefore \text { rate of reaction } &=\frac{1}{9} \times 9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &=1 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \end{aligned} \end{aligned} $
$ =\frac{1}{r_{4}}\left(\frac{d[D]}{d t}\right) $
$\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{\mathrm{r}_{2}}\left(\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}\right)$
$ \frac{r_{4}}{r_{2}}=\frac{3}{2} $
$\frac{-\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\left(\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}\right) \Rightarrow \frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}=2$
$r_{2}=2 r_{1}$
$ \begin{aligned} &\mathrm{r}_{4}=1.5 \mathrm{r}_{2}=3 \mathrm{r}_{1} \\\\ &\frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}}=1 \mathrm{~m} \cdot \mathrm{mol} ~\mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &\frac{\mathrm{d}[\mathrm{D}]}{\mathrm{dt}}=\frac{\mathrm{r}_{4}}{r_{3}} \cdot \frac{\mathrm{d}[\mathrm{C}]}{\mathrm{dt}} \Rightarrow \frac{\mathrm{r}_{4}}{\mathrm{r}_{3}}=9 \\\\ &\mathrm{r}_{4}=9 \mathrm{r}_{3}=3 \mathrm{r}_{1} \\\\ &\Rightarrow \mathrm{r}_{1}=3 \mathrm{r}_{3} \\\\ &\begin{array}{rr} 3 \mathrm{r}_{3} \mathrm{~A}+6 \mathrm{r}_{3} \mathrm{~B} \rightarrow \mathrm{r}_{3} \mathrm{C}+9 \mathrm{r}_{3} \mathrm{D} \end{array} \\\\ &\begin{aligned} \therefore \text { rate of reaction } &=\frac{1}{9} \times 9 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \\\\ &=1 \mathrm{~m} \cdot \mathrm{mol}~ \mathrm{dm}^{-3} \mathrm{sec}^{-1} \end{aligned} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Morning Shift
The rate constants for decomposition of acetaldehyde have been measured over the temperature range 700 - 1000 K. The data has been analysed by plotting ln k vs ${{{{10}^3}} \over T}$ graph. The value of activation energy for the reaction is ___________ kJ mol$-$1. (Nearest integer)
(Given : R = 8.31 J K$-$1 mol$-$1)
Correct Answer: 154
Explanation:
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}$
$\therefore$ Slope of the graph $=-\frac{E_{a}}{R \times 10^{3}}=-18.5$
$\therefore E_{a}=18.5 \times 8.31 \times 1000 \simeq 154 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 28th July Morning Shift
For kinetic study of the reaction of iodide ion with $\mathrm{H}_{2} \mathrm{O}_{2}$ at room temperature :
(A) Always use freshly prepared starch solution.
(B) Always keep the concentration of sodium thiosulphate solution less than that of KI solution.
(C) Record the time immediately after the appearance of blue colour.
(D) Record the time immediately before the appearance of blue colour.
(E) Always keep the concentration of sodium thiosulphate solution more than that of KI solution.
Choose the correct answer from the options given below :
A.
$(\mathrm{A}),(\mathrm{B}),(\mathrm{C})$ only
B.
$(\mathrm{A}),(\mathrm{D}),(\mathrm{E})$ only
C.
$(\mathrm{D}),(\mathrm{E})$ only
D.
$(\mathrm{A}),(\mathrm{B}),(\mathrm{E})$ only
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 26th July Evening Shift
At $30^{\circ} \mathrm{C}$, the half life for the decomposition of $\mathrm{AB}_{2}$ is $200 \mathrm{~s}$ and is independent of the initial concentration of $\mathrm{AB}_{2}$. The time required for $80 \%$ of the $\mathrm{AB}_{2}$ to decompose is
Given: $\log 2=0.30$ $\quad \log 3=0.48$
A.
200 s
B.
323 s
C.
467 s
D.
532 s
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 24th June Evening Shift
For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is
(Given : ln 10 = 2.303 and log 2 = 0.3010)
A.
1.12
B.
2.43
C.
3.32
D.
33.31
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Evening Shift
For the reaction A $\to$ B, the rate constant k(in s$-$1) is given by
${\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}$
The energy of activation in kJ mol$-$1 is ____________. (Nearest integer) [Given : R = 8.314 J K$-$1 mol$-$1]
${\log _{10}}k = 20.35 - {{(2.47 \times {{10}^3})} \over T}$
The energy of activation in kJ mol$-$1 is ____________. (Nearest integer) [Given : R = 8.314 J K$-$1 mol$-$1]
Correct Answer: 47
Explanation:
Given,
$\log K = 20.35 - {{2.47 \times {{10}^3}} \over T}$
We know
$\log K = \log A - {{{E_a}} \over {2.303RT}}$
$ \Rightarrow {{{E_a}} \over {2.303RT}} = 2.47 \times {10^3}$
${E_a} = 2.47 \times {10^3} \times 2.303 \times {{8.314} \over {1000}}$ KJ/mole
= 47.29 = 47 (Nearest integer)
$\log K = 20.35 - {{2.47 \times {{10}^3}} \over T}$
We know
$\log K = \log A - {{{E_a}} \over {2.303RT}}$
$ \Rightarrow {{{E_a}} \over {2.303RT}} = 2.47 \times {10^3}$
${E_a} = 2.47 \times {10^3} \times 2.303 \times {{8.314} \over {1000}}$ KJ/mole
= 47.29 = 47 (Nearest integer)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
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)
Correct Answer: 2
Explanation:
$k = {{2.303} \over t}\log {a \over {a - x}}$
${{2.303} \over {{t_{50\% }}}}\log {{100} \over {100 - 50}} = {{2.303} \over {{t_{75\% }}}}\log {{100} \over {100 - 75}}$
$ \Rightarrow $ ${t_{75\% }} = 2{t_{50\% }}$
${{2.303} \over {{t_{50\% }}}}\log {{100} \over {100 - 50}} = {{2.303} \over {{t_{75\% }}}}\log {{100} \over {100 - 75}}$
$ \Rightarrow $ ${t_{75\% }} = 2{t_{50\% }}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
According to the following figure, the magnitude of the enthalpy change of the reaction
A + B $\to$ M + N in kJ mol$-$ is equal to ___________. (Integer answer)
A + B $\to$ M + N in kJ mol$-$ is equal to ___________. (Integer answer)
Correct Answer: 45
Explanation:
$\Delta$H = EProduct $-$ EReactant
= 15 $-$ (15 + 45)
= $-$45 KJ/mol
| $\Delta$H | = 45 KJ/mol
= 15 $-$ (15 + 45)
= $-$45 KJ/mol
| $\Delta$H | = 45 KJ/mol
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
The first order rate constant for the decomposition of CaCO3 at 700 K is 6.36 $\times$ 10$-$3s$-$1 and activation energy is 209 kJ mol$-$1. Its rate constant (in s$-$1) at 600 K is x $\times$ 10$-$6. The value of x is ___________. (Nearest integer)
[Given R = 8.31 J K$-$ mol$-$1; log 6.36 $\times$ 10$-$3 = $-$2.19, 10$-$4.79 = 1.62 $\times$ 10$-$5]
[Given R = 8.31 J K$-$ mol$-$1; log 6.36 $\times$ 10$-$3 = $-$2.19, 10$-$4.79 = 1.62 $\times$ 10$-$5]
Correct Answer: 16
Explanation:
K700 = 6.36 $\times$ 10$-$3s$-$1;
K600 = x $\times$ 10$-$6s$-$1
Ea = 209 kJ/mol
Applying;
$\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$
$\log \left( {{{{K_{700}}} \over {{K_{600}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {700}} - {1 \over {600}}} \right)$
$\log \left( {{{6.36 \times {{10}^{ - 3}}} \over {{K_{600}}}}} \right) = {{ + 209 \times 1000} \over {2.303 \times 8.31}}\left( {{{100} \over {700 \times 600}}} \right)$
log(6.36 $\times$ 10$-$3) $-$ logK600 = 2.6
$\Rightarrow$ logK600 = $-$2.19 $-$ 2.6 = $-$4.79
$\Rightarrow$ K600 = 10$-$4.79 = 1.62 $\times$ 10$-$5
= 1.62 $\times$ 10$-$6
= x $\times$ 10$-$6
$\Rightarrow$ x = 16
K600 = x $\times$ 10$-$6s$-$1
Ea = 209 kJ/mol
Applying;
$\log \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right)$
$\log \left( {{{{K_{700}}} \over {{K_{600}}}}} \right) = {{ - {E_a}} \over {2.303R}}\left( {{1 \over {700}} - {1 \over {600}}} \right)$
$\log \left( {{{6.36 \times {{10}^{ - 3}}} \over {{K_{600}}}}} \right) = {{ + 209 \times 1000} \over {2.303 \times 8.31}}\left( {{{100} \over {700 \times 600}}} \right)$
log(6.36 $\times$ 10$-$3) $-$ logK600 = 2.6
$\Rightarrow$ logK600 = $-$2.19 $-$ 2.6 = $-$4.79
$\Rightarrow$ K600 = 10$-$4.79 = 1.62 $\times$ 10$-$5
= 1.62 $\times$ 10$-$6
= x $\times$ 10$-$6
$\Rightarrow$ x = 16
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Morning Shift
The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is
$2{K_2}C{r_2}{O_7} + 8{H_2}S{O_4} + 3{C_2}{H_6}O \to 2C{r_2}{(S{O_4})_3} + 3{C_2}{H_4}{O_2} + 2{K_2}S{O_4} + 11{H_2}O$
If the rate of appearance of Cr2(SO4)3 is 2.67 mol min$-$1 at a particular time, the rate of disappearance of C2H6O at the same time is _____________ mol min$-$1. (Nearest integer)
$2{K_2}C{r_2}{O_7} + 8{H_2}S{O_4} + 3{C_2}{H_6}O \to 2C{r_2}{(S{O_4})_3} + 3{C_2}{H_4}{O_2} + 2{K_2}S{O_4} + 11{H_2}O$
If the rate of appearance of Cr2(SO4)3 is 2.67 mol min$-$1 at a particular time, the rate of disappearance of C2H6O at the same time is _____________ mol min$-$1. (Nearest integer)
Correct Answer: 4
Explanation:
$\left( {{{Rate\,of\,disappearance\,of\,{C_2}{H_6}O} \over 3}} \right)$
$ = \left( {{{Rate\,of\,disappearance\,of\,C{r_2}{{(S{O_4})}_3}} \over 2}} \right)$
$ \Rightarrow \left( {{{2.67\,mol/\min \, \times 3} \over 2}} \right) = $ rate of disappearance of C2H6O.
$\Rightarrow$ Rate of disappearance of C2H6O = 4.005 mol/min.
$ = \left( {{{Rate\,of\,disappearance\,of\,C{r_2}{{(S{O_4})}_3}} \over 2}} \right)$
$ \Rightarrow \left( {{{2.67\,mol/\min \, \times 3} \over 2}} \right) = $ rate of disappearance of C2H6O.
$\Rightarrow$ Rate of disappearance of C2H6O = 4.005 mol/min.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
The following data was obtained for chemical reaction given below at 975 K.
2NO(g) + 2H2(g) $\to$ N2(g) + 2H2O(g)
The order of the reaction with respect to NO is ___________. [Integer answer]
2NO(g) + 2H2(g) $\to$ N2(g) + 2H2O(g)
The order of the reaction with respect to NO is ___________. [Integer answer]
Correct Answer: 1
Explanation:
7 $\times$ 10$-$9 = K $\times$ (8 $\times$ 10$-$5)x (8 $\times$ 10$-$5)y ......... (1)
2.1 $\times$ 10$-$8 = K $\times$ (24 $\times$ 10$-$5)x (8 $\times$ 10$-$5)y ....... (2)
${1 \over 3} = {\left( {{1 \over 3}} \right)^x} \Rightarrow x = 1$
2.1 $\times$ 10$-$8 = K $\times$ (24 $\times$ 10$-$5)x (8 $\times$ 10$-$5)y ....... (2)
${1 \over 3} = {\left( {{1 \over 3}} \right)^x} \Rightarrow x = 1$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
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).
[Use : ln 2 = 0.69, ln 10 = 2.3]
Properties of logarithms : ln xy = y ln x;
$\ln \left( {{x \over y}} \right) = \ln x - \ln y$
(Round off to the nearest integer)
[Use : ln 2 = 0.69, ln 10 = 2.3]
Properties of logarithms : ln xy = y ln x;
$\ln \left( {{x \over y}} \right) = \ln x - \ln y$
(Round off to the nearest integer)
Correct Answer: 600to700
Explanation:
Now, $t = {{{t_{1/2}}} \over {\ln 2}} \times {{[{A_0}]} \over {[{A_t}]}}$
$100 = {{{t_{1/2}}} \over {\ln 2}} \times \ln {1 \over {0.9}} \Rightarrow {t_{1/2}} = 690$ min (taking ln 3 = 1.11)
Ans. 600 to 700
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Evening Shift
For a chemical reaction A $\to$ B, it was found that concentration of B is increased by 0.2 mol L$-$ in 30 min. The average rate of the reaction is ____________ $\times$ 10$-$1 mol L$-$1 h$-$1. (in nearest integer)
Correct Answer: 4
Explanation:
A $\to$ B
$\matrix{ {t = 0} & 0 \cr {t = 30\,\min } & {0.2M} \cr } $
Av. rate of reaction = $ - {{\Delta [A]} \over {\Delta t}} = {{\Delta [B]} \over {\Delta t}} = {{(0.2 - 0)} \over {{1 \over 2}}}$
$ = 0.4 = 4 \times {10^{ - 1}}$ mol / L $\times$ hr
$\matrix{ {t = 0} & 0 \cr {t = 30\,\min } & {0.2M} \cr } $
Av. rate of reaction = $ - {{\Delta [A]} \over {\Delta t}} = {{\Delta [B]} \over {\Delta t}} = {{(0.2 - 0)} \over {{1 \over 2}}}$
$ = 0.4 = 4 \times {10^{ - 1}}$ mol / L $\times$ hr
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
${N_2}{O_{5(g)}} \to 2N{O_{2(g)}} + {1 \over 2}{O_{2(g)}}$
In the above first order reaction the initial concentration of N2O5 is 2.40 $\times$ 10$-$2 mol L$-$1 at 318 K. The concentration of N2O5 after 1 hour was 1.60 $\times$ 10$-$2 mol L$-$1. The rate constant of the reaction at 318 K is ______________ $\times$ 10$-$3 min$-$1. (Nearest integer)
[Given : log 3 = 0.477, log 5 = 0.699]
In the above first order reaction the initial concentration of N2O5 is 2.40 $\times$ 10$-$2 mol L$-$1 at 318 K. The concentration of N2O5 after 1 hour was 1.60 $\times$ 10$-$2 mol L$-$1. The rate constant of the reaction at 318 K is ______________ $\times$ 10$-$3 min$-$1. (Nearest integer)
[Given : log 3 = 0.477, log 5 = 0.699]
Correct Answer: 7
Explanation:
$K = {{2.303} \over t}\log {{{{[{N_2}{O_5}]}_0}} \over {{{[{N_2}{O_5}]}_t}}}$
$ = {{2.303} \over {60}}\log {{2.4} \over {1.6}} = 6.76 \times {10^{ - 3}}$ min$-$1 $ \approx $ 7 $\times$ 10$-$3 min$-$1
$ = {{2.303} \over {60}}\log {{2.4} \over {1.6}} = 6.76 \times {10^{ - 3}}$ min$-$1 $ \approx $ 7 $\times$ 10$-$3 min$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
$PC{l_5}(g) \to PC{l_3}(g) + C{l_2}(g)$
In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L$-$1 to 10 mol L$-$1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x $\times$ 10$-$2 min$-$1. The value of x is __________. [Given log5 = 0.6989]
In the above first order reaction the concentration of PCl5 reduces from initial concentration 50 mol L$-$1 to 10 mol L$-$1 in 120 minutes at 300 K. The rate constant for the reaction at 300 K is x $\times$ 10$-$2 min$-$1. The value of x is __________. [Given log5 = 0.6989]
Correct Answer: 1
Explanation:
$PC{l_5}(g)\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{300K}^{I\,order}} PC{l_3}(g) + C{l_2}(g)$
t = 0
50M
t = 120min
10 M
$ \Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$
$ \Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$
$ \Rightarrow K = {{2.303} \over {120}} \times 0.6989 = 0.013413$ min$-$1
$ = 1.3413 \times {10^{ - 2}}$ min$-$1
1.34 $\Rightarrow$ Nearest integer = 1
t = 0
50M
t = 120min
10 M
$ \Rightarrow K = {{2.303} \over t}\log {{[{A_0}]} \over {[{A_t}]}}$
$ \Rightarrow K = {{2.303} \over {120}}\log {{50} \over {10}}$
$ \Rightarrow K = {{2.303} \over {120}} \times 0.6989 = 0.013413$ min$-$1
$ = 1.3413 \times {10^{ - 2}}$ min$-$1
1.34 $\Rightarrow$ Nearest integer = 1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Morning Shift
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$-$3 min$-$1. (Nearest integer)
[Use : ln 10 = 2.303; log10 3 = 0.477; property
of logarithm : log xy = y log x]
[Use : ln 10 = 2.303; log10 3 = 0.477; property
of logarithm : log xy = y log x]
Correct Answer: 106
Explanation:
Unit of rate constant is min$-$1, so it must be a first order reaction. For first order reaction,
k $\times$ t = 2.303 log${{{A_0}} \over {{A_t}}}$
k is the rate constant
t is the time
A0 is the initial conc.
At is the conc. at time, t
Using formula,
A0 = 100, At = 90 min 1 min
So, K $\times$ 1 = 2.303 $\times$ log${{100} \over {90}}$
= 2.303 $\times$ (log 10 $-$ 2 log 3)
= 2.303 $\times$ (1 $-$ 2 $\times$ 0.477)
= 0.10593
= 105.93 $\times$ 10$-$3
$\approx$ 106
Hence, the rate constant for viral inactivation is 106.
k $\times$ t = 2.303 log${{{A_0}} \over {{A_t}}}$
k is the rate constant
t is the time
A0 is the initial conc.
At is the conc. at time, t
Using formula,
A0 = 100, At = 90 min 1 min
So, K $\times$ 1 = 2.303 $\times$ log${{100} \over {90}}$
= 2.303 $\times$ (log 10 $-$ 2 log 3)
= 2.303 $\times$ (1 $-$ 2 $\times$ 0.477)
= 0.10593
= 105.93 $\times$ 10$-$3
$\approx$ 106
Hence, the rate constant for viral inactivation is 106.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
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]
Correct Answer: 10
Explanation:
Given t1/2 = 1 min
$ \therefore $ $K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$
From formula we know,
$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$
$ \Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$
$ \Rightarrow 0.3 = {1 \over t}\ln \left( {{{10}^3}} \right)$
$ \Rightarrow 0.3 = {1 \over t} \times 3$
$ \Rightarrow t = 10$ min
$ \therefore $ $K = {{\ln 2} \over {{t_{1/2}}}} = {{\ln 2} \over 1}$
From formula we know,
$Kt = \ln \left( {{{100} \over {100 - x}}} \right)$
$ \Rightarrow {{\ln 2} \over 1} = {1 \over t}\ln \left( {{{100} \over {0.1}}} \right)$
$ \Rightarrow 0.3 = {1 \over t}\ln \left( {{{10}^3}} \right)$
$ \Rightarrow 0.3 = {1 \over t} \times 3$
$ \Rightarrow t = 10$ min
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
2NO(g) + Cl2(g) $\rightleftharpoons $ 2NOCl(s)
This reaction was studied at $-$10$^\circ$ and the following data was obtained
${[NO]_0}$ and ${[C{l_2}]_0}$ are the initial concentrations and r0 is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).
This reaction was studied at $-$10$^\circ$ and the following data was obtained
| Run | ${[NO]_0}$ | ${[C{l_2}]_0}$ | ${r_0}$ |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.18 |
| 2 | 0.10 | 0.20 | 0.35 |
| 3 | 0.20 | 0.20 | 1.40 |
${[NO]_0}$ and ${[C{l_2}]_0}$ are the initial concentrations and r0 is the initial reaction rate.
The overall order of the reaction is __________. (Round off to the Nearest Integer).
Correct Answer: 3
Explanation:
We know,
$Rate = K{\left[ A \right]^x}{\left[ B \right]^y}$
Here from,
Exp 1 : $0.18 = K{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}$ ..... (1)
Exp. 2 : $0.35 = K{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}$ .... (2)
Exp. 3 : $1.40 = K{\left[ {0.2} \right]^x}{\left[ {0.2} \right]^y}$ .... (3)
(2) $\div$ (3)
${{0.35} \over {1.40}} = {{K \times {{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}} \over {K \times {{\left[ {0.2} \right]}^x}{{\left[ {0.2} \right]}^y}}}$
$ \Rightarrow {1 \over 4} = {\left( {{1 \over 2}} \right)^x}$
$ \Rightarrow x = 2$
(1) $\div$ (2)
$\left( {{1 \over 2}} \right) = {\left( {{1 \over 2}} \right)^y}$
$ \Rightarrow y = 1$
$ \therefore $ x + y = 3
$Rate = K{\left[ A \right]^x}{\left[ B \right]^y}$
Here from,
Exp 1 : $0.18 = K{\left[ {0.1} \right]^x}{\left[ {0.1} \right]^y}$ ..... (1)
Exp. 2 : $0.35 = K{\left[ {0.1} \right]^x}{\left[ {0.2} \right]^y}$ .... (2)
Exp. 3 : $1.40 = K{\left[ {0.2} \right]^x}{\left[ {0.2} \right]^y}$ .... (3)
(2) $\div$ (3)
${{0.35} \over {1.40}} = {{K \times {{\left[ {0.1} \right]}^x}{{\left[ {0.2} \right]}^y}} \over {K \times {{\left[ {0.2} \right]}^x}{{\left[ {0.2} \right]}^y}}}$
$ \Rightarrow {1 \over 4} = {\left( {{1 \over 2}} \right)^x}$
$ \Rightarrow x = 2$
(1) $\div$ (2)
$\left( {{1 \over 2}} \right) = {\left( {{1 \over 2}} \right)^y}$
$ \Rightarrow y = 1$
$ \therefore $ x + y = 3
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
The reaction 2A + B2 $ \to $ 2AB is an elementary reaction.
For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____________. (Round off to the Nearest Integer).
For a certain quantity of reactants, if the volume of the reaction vessel is reduced by a factor of 3, the rate of the reaction increases by a factor of ____________. (Round off to the Nearest Integer).
Correct Answer: 27
Explanation:
Let initial concentration of A & B is a & b
Rate = $K{[A]^2}{[{B_2}]^1}$
${r_1} = K{\left( {{a \over V}} \right)^2}{\left( {{b \over V}} \right)^1}$
${r_2} = K{\left( {{{3a} \over V}} \right)^2}{\left( {{{3b} \over V}} \right)^1}$
$\left( {{{{r_2}} \over {{r_1}}}} \right) = 27$
Rate = $K{[A]^2}{[{B_2}]^1}$
${r_1} = K{\left( {{a \over V}} \right)^2}{\left( {{b \over V}} \right)^1}$
${r_2} = K{\left( {{{3a} \over V}} \right)^2}{\left( {{{3b} \over V}} \right)^1}$
$\left( {{{{r_2}} \over {{r_1}}}} \right) = 27$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Morning Shift
For a certain first order reaction 32% of the reactant is left after 570s. The rate constant of this reaction is _________ $\times$ 10$-$3 s$-$1. (Round off to the Nearest Integer). [Given : log102 = 0.301, ln10 = 2.303]
Correct Answer: 2
Explanation:
$k = {1 \over t}\ln \left[ {{a \over {a - x}}} \right]$
$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$
$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$
$k = {{2.303} \over {570}} \times 0.5$
$k = 2 \times {10^{ - 3}}{s^{ - 1}}$
$k = {{2.303} \over {570}}\log \left( {{{100} \over {32}}} \right)$
$k = {{2.303} \over {570}}\left[ {\log ({{10}^2}) - \log {2^5}} \right]$
$k = {{2.303} \over {570}} \times 0.5$
$k = 2 \times {10^{ - 3}}{s^{ - 1}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
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).
Correct Answer: 108
Explanation:
Initially : $\left[ {{A_0}} \right] = \left[ {{B_0}} \right] = a$
After time 't' min : $\left[ A \right] = 16\left[ B \right]$
$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$
$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$
$ \Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$
$ \Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16$
$ \Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16$
$ \Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2$
$ \Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108$ min
After time 't' min : $\left[ A \right] = 16\left[ B \right]$
$\left[ A \right] = \left[ {{A_0}} \right]{e^{ - {k_A}t}}$
$\left[ B \right] = \left[ {{B_0}} \right]{e^{ - {k_B}t}}$
$ \Rightarrow a\,.\,{e^{ - {k_A}t}} = 16a{e^{ - {k_B}t}}$
$ \Rightarrow {e^{ - \left( {{k_A} - {k_B}} \right)t}} = 16$
$ \Rightarrow \left( {{k_B} - {k_A}} \right)t = \ln 16$
$ \Rightarrow \ln 2\left( {{1 \over {18}} - {1 \over {54}}} \right)t = 4\ln 2$
$ \Rightarrow t = {{54 \times 18 \times 4} \over {36}} = 108$ min
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Morning Shift
The decomposition of formic acid on gold surface follows first order kinetics. If the rate constant at 300 K is 1.0 $\times$ 10$-$3 s$-$1 and the activation energy Ea = 11.488 kJ mol$-$1, the rate constant at 200 K is ____________ $\times$ 10$-$5 s$-$1. (Round off to the Nearest Integer). (Given : R = 8.314 J mol$-$1 K$-$1)
Correct Answer: 10
Explanation:
$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
k1 (at 200 K) = ?
k2 (at 300 K) = $1 \times {10^{ - 3}}{s^{ - 1}}$
$\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \times 8.314}}\left[ {{1 \over {600}}} \right] = 1$
$ \Rightarrow $ ${{1 \times {{10}^{ - 3}}} \over {{k_1}}} = 10$
$ \Rightarrow $ ${k_1} = 10 \times {10^{ - 5}}{s^{ - 1}}$
k1 (at 200 K) = ?
k2 (at 300 K) = $1 \times {10^{ - 3}}{s^{ - 1}}$
$\log {{1 \times {{10}^{ - 3}}} \over {{k_1}}} = {{11.488 \times {{10}^3}} \over {2.303 \times 8.314}}\left[ {{1 \over {600}}} \right] = 1$
$ \Rightarrow $ ${{1 \times {{10}^{ - 3}}} \over {{k_1}}} = 10$
$ \Rightarrow $ ${k_1} = 10 \times {10^{ - 5}}{s^{ - 1}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
If the activation energy of a reaction is 80.9 kJ mol$-$1, the fraction of molecules at 700 K, having enough energy to react to form
products is e$-$x. The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K$-$1 mol$-$1]
products is e$-$x. The value of x is __________. (Rounded off to the nearest integer) [Use R = 8.31 J K$-$1 mol$-$1]
Correct Answer: 14
Explanation:
Ea = 80.9 kJ/mol
Fraction of molecules able to cross energy barrier = e$-$Ea/RT = e$-$x
x = ${{{E_a}} \over {RT}} = {{80.9 \times 1000} \over {8.31 \times 700}} = 13.91$
$ \Rightarrow $ x $ \simeq $ 14
Fraction of molecules able to cross energy barrier = e$-$Ea/RT = e$-$x
x = ${{{E_a}} \over {RT}} = {{80.9 \times 1000} \over {8.31 \times 700}} = 13.91$
$ \Rightarrow $ x $ \simeq $ 14
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
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$-$1 is _________. (Rounded off to the nearest integer)
[R = 8.314 J K$-$1mol$-$1]
[R = 8.314 J K$-$1mol$-$1]
Correct Answer: 52
Explanation:
T1 = (273 + 27) = 300 K, T2 = (273 + 52) = 325 K
Given, temperature coefficient of the reaction,
${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$
$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$
$\log {{{K_{325}}} \over {{K_{300}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{{325 - 300} \over {300 \times 325}}} \right)$
$\log 5 = {{{E_a}} \over {2.303 \times 8.319}} \times {{25} \over {300 \times 325}}$
Ea = 52194.78 J mol$-$
= 52.194 kJ mol$-$1
$\simeq$ 52 kJ mol$-$1
Given, temperature coefficient of the reaction,
${\alpha _T} = {{{K_{325}}} \over {{K_{300}}}} = 5$
$\log {{{K_{{T_2}}}} \over {{K_{{T_1}}}}} = {{{E_a}} \over {2.303R}} \times \left( {{{{T_2} - {T_1}} \over {{T_1}{T_2}}}} \right)$
$\log {{{K_{325}}} \over {{K_{300}}}} = {{{E_a}} \over {2.303 \times 8.314}}\left( {{{325 - 300} \over {300 \times 325}}} \right)$
$\log 5 = {{{E_a}} \over {2.303 \times 8.319}} \times {{25} \over {300 \times 325}}$
Ea = 52194.78 J mol$-$
= 52.194 kJ mol$-$1
$\simeq$ 52 kJ mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Morning Shift
For the reaction, aA + bB $ \to $ cC + dD, the plot of log k vs ${1 \over T}$ is given below :
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
The temperature at which the rate constant of the reaction is 10-4 s-1 is _________ K. (Rounded off to the nearest integer)
[Given : The rate constant of the reaction is 10-5 s-1 at 500 K.]
Correct Answer: 526
Explanation:
${\log _{10}}K = {\log _{10}}A - {{{E_a}} \over {2.303RT}}$
Slope $ = {{{E_a}} \over {2.303R}} = - 10000$
${\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
${\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$
$ \Rightarrow $ $1 = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$
$ \Rightarrow $ ${1 \over {10000}} = {1 \over {500}} - {1 \over T}$
$ \Rightarrow $ ${1 \over T} = {1 \over {500}} - {1 \over {10000}}$
$ \Rightarrow $${1 \over T} = {{20 - 1} \over {10000}} = {{19} \over {10000}}$
$ \Rightarrow $ $T = {{10000} \over {19}} = 526$ K
Slope $ = {{{E_a}} \over {2.303R}} = - 10000$
${\log _{10}}{{{K_2}} \over {{K_1}}} = {{{E_a}} \over {2.303R}} \times \left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$
${\log _{10}}{{{{10}^{ - 4}}} \over {{{10}^{ - 5}}}} = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$
$ \Rightarrow $ $1 = 10000 \times \left[ {{1 \over {500}} - {1 \over T}} \right]$
$ \Rightarrow $ ${1 \over {10000}} = {1 \over {500}} - {1 \over T}$
$ \Rightarrow $ ${1 \over T} = {1 \over {500}} - {1 \over {10000}}$
$ \Rightarrow $${1 \over T} = {{20 - 1} \over {10000}} = {{19} \over {10000}}$
$ \Rightarrow $ $T = {{10000} \over {19}} = 526$ K
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
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$-$2. (Rounded off to the nearest integer)
[Assume : ln 10 = 2.303, ln 2 = 0.693]
[Assume : ln 10 = 2.303, ln 2 = 0.693]
Correct Answer: 81
Explanation:
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} $
${t_{1/2}} = {{10} \over 3}h$
At t = 0, a = [A]0 (initial conc.)
t = 9h, a $-$ x = [A]t [conc. at time t]
For using 1st order equation,
$K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}} $
$\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}$
${{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124$ ($\because$ $k = {{\ln 2} \over {{t_{1/2}}}}$)
$\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}$
x = 81.24 or x $\approx$ 81
${t_{1/2}} = {{10} \over 3}h$
At t = 0, a = [A]0 (initial conc.)
t = 9h, a $-$ x = [A]t [conc. at time t]
For using 1st order equation,
$K = {{2.303} \over t}\log {{{{[A]}_0}} \over {{{[A]}_t}}} $
$\Rightarrow {{K \times t} \over {2.303}} = \log {{{{[A]}_0}} \over {{{[A]}_t}}}$
${{\ln 2 \times 9} \over {10/3 \times 2.303}} = \log \left( {{1 \over F}} \right) \Rightarrow \log \left( {{1 \over F}} \right) = 0.8124$ ($\because$ $k = {{\ln 2} \over {{t_{1/2}}}}$)
$\log \left( {{1 \over F}} \right) = 81.24 \times {10^{ - 2}}$
x = 81.24 or x $\approx$ 81
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Morning Shift
Gaseous cyclobutene isomerises to butadiene in a first order process which has a 'k' value of 3.3 $ \times $ 10-4 s-1 at 153°C. The time in minutes it takes for the isomerization to proceed 40% to completion at this temperature is ______.
(Rounded off to the nearest integer)
(Rounded off to the nearest integer)
Correct Answer: 26
Explanation:
It is a first order isomerisation reaction. Integrated rate law for 1st order reaction is
$kt = \ln {{{{[A]}_0}} \over {{{[A]}_t}}}$ ...(i)
Here,
k = rate constant
${[A]_0}$ = initial concentration
${[A]_t}$ = concentration at time 't'
Given, k = 3.3 $\times$ 10$-$4 s$-$1
${[A]_0} = 100 \Rightarrow {[A]_t} = 100 - 40 = 60$
Put values in Eq. (i), we get
$3.3 \times {10^{ - 4}}{s^{ - 1}} \times t = \ln {{100} \over {60}}$
t = 1547.95 s = 25.79 min (1 min = 60 s or 1s = ${1 \over {60}}$ min) = 26 minutes
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 1st September Evening Shift
Which one of the following given graphs represents the variation of rate constant (k) with temperature (T) for an endothermic reaction?
A.
B.
C.
D.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 27th July Morning Shift
For a reaction of order n, the unit of the rate constant is :
A.
mol1$-$n L1$-$n s
B.
mol1$-$n L2n s$-$1
C.
mol1$-$n Ln$-$1 s$-$1
D.
mol1$-$n L1$-$n s$-$1
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 25th July Morning Shift
For the following graphs,
Choose from the options given below, the correct one regarding order of reaction is :
Choose from the options given below, the correct one regarding order of reaction is :
A.
(b) Zero order (c) and (e) First order
B.
(a) and (b) Zero order (e) First order
C.
(b) and (d) Zero order (e) First order
D.
(a) and (b) Zero order (c) and (e) First order
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 22th July Evening Shift
Isotope(s) of hydrogen which emits low energy $\beta$$-$ particles with t1/2 value > 12 years is/are
A.
Protium
B.
Tritium
C.
Deuterium
D.
Deuterium and Tritium
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Evening Slot
The rate of a reaction decreased by 3.555
times when the temperature was changed
from 40oC to 30oC. The activation energy
(in kJ mol–1) of the reaction is _______.
Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.268
Take; R = 8.314 J mol–1 K–1 ln 3.555 = 1.268
Correct Answer: 100
Explanation:
k = A${e^{ - {{{E_a}} \over {RT}}}}$
$ \therefore $ $\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
$ \Rightarrow $ ln (3.555) = ${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$
$ \Rightarrow $ Ea = ${{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}$
= 99980.7 = 99.98 kJ/mol $ \simeq $ 100 kJ/mol
$ \therefore $ $\ln {{{k_2}} \over {{k_1}}} = {{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
$ \Rightarrow $ ln (3.555) = ${{{E_a}} \over {8.314}}\left( {{1 \over {303}} - {1 \over {313}}} \right)$
$ \Rightarrow $ Ea = ${{1.268 \times 8.314 \times 3.3 \times 313} \over {10}}$
= 99980.7 = 99.98 kJ/mol $ \simeq $ 100 kJ/mol
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
The number of molecules with energy greater
than the threshold energy for a reaction
increases five fold by a rise of temperature
from 27oC to 42oC. Its energy of activation in
J/mol is _____.
(Take ln 5 = 1.6094; R = 8.314 J mol–1 K–1)
(Take ln 5 = 1.6094; R = 8.314 J mol–1 K–1)
Correct Answer: 84297.47to84297.48
Explanation:
$ \because $ k = A${e^{ - {{{E_a}} \over {RT}}}}$
T1 = 300K, T2 = 315K
As per question KT2 = 5KT2 as molecules activated are increased five times so K will increases five time.
$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right)$ = ${{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
$ \Rightarrow $ ln 5 = ${{{E_a}} \over R}\left( {{{15} \over {300 \times 315}}} \right)$
$ \Rightarrow $ Ea = ${{1.6094 \times 8.314 \times 300 \times 315} \over {15}}$
= 84297.47 Joules/mole
T1 = 300K, T2 = 315K
As per question KT2 = 5KT2 as molecules activated are increased five times so K will increases five time.
$\ln \left( {{{{K_{{T_2}}}} \over {{K_{{T_1}}}}}} \right)$ = ${{{E_a}} \over R}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$
$ \Rightarrow $ ln 5 = ${{{E_a}} \over R}\left( {{{15} \over {300 \times 315}}} \right)$
$ \Rightarrow $ Ea = ${{1.6094 \times 8.314 \times 300 \times 315} \over {15}}$
= 84297.47 Joules/mole
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Morning Slot
If 75% of a first order reaction was completed
in 90 minutes, 60% of the same reaction would
be completed in approximately (in minutes)
_______.
(Take : log 2 = 0.30; log 2.5 = 0.40)
(Take : log 2 = 0.30; log 2.5 = 0.40)
Correct Answer: 60
Explanation:
t75% = 90 min = 2 × t1/2
$ \Rightarrow $ t1/2 = 45 min
Rate constant, K = ${{0.693} \over {45}}$ min-1
Time for completion of 60% of the reaction,
t60% = ${{2.303} \over K}\log {{10} \over 4}$
= ${{2.303 \times 45} \over {0.693}}\log 2.5$
= 60 min
$ \Rightarrow $ t1/2 = 45 min
Rate constant, K = ${{0.693} \over {45}}$ min-1
Time for completion of 60% of the reaction,
t60% = ${{2.303} \over K}\log {{10} \over 4}$
= ${{2.303 \times 45} \over {0.693}}\log 2.5$
= 60 min
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Evening Slot
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__________.
(Given, R = 8.3 J mol–1 K–1, $\ln \left( {{3 \over 2}} \right) = 0.4$, e–3 = 4.0)
(Given, R = 8.3 J mol–1 K–1, $\ln \left( {{3 \over 2}} \right) = 0.4$, e–3 = 4.0)
Correct Answer: 3.98TO3.99
Explanation:
Using Arrehenius equation
K = A${e^{ - {{{E_a}} \over {RT}}}}$
$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$
$ \Rightarrow $ $\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$
$ \Rightarrow $ $\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$
$ \therefore $ Ea = 0.4 × 1200 × 8.3 = 3984 J/mol
Ea = 3.984 kJ/mol = 3.98 kJ/mol
K = A${e^{ - {{{E_a}} \over {RT}}}}$
$\ln \left( {{{{k_{400}}} \over {{k_{300}}}}} \right) = {{{E_a}} \over R}\left( {{1 \over {300}} - {1 \over {400}}} \right)$
$ \Rightarrow $ $\ln \left( {{{60} \over {40}}} \right) = {{{E_a}} \over R}\left( {{{100} \over {300 \times 400}}} \right)$
$ \Rightarrow $ $\ln \left( {{3 \over 2}} \right) = {{{E_a}} \over {1200R}}$
$ \therefore $ Ea = 0.4 × 1200 × 8.3 = 3984 J/mol
Ea = 3.984 kJ/mol = 3.98 kJ/mol
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 7th January Morning Slot
During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 $\mu $ g of 90Sr was absorbed in the bones of newly born baby in placed of Ca, how much time, in years, is required to reduce much time, in year, is required to reduce it by 90% if it not lost metabolically.
Correct Answer: 23to23.03
Explanation:
All nuclear decays follow first order kinetics
t = ${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$
= ${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$
= 10 × 2.303 × 1
= 23.03 years
Shortcut Method :
t90% = ${{10} \over 3} \times {t_{50\% }}$
= ${{10} \over 3} \times 6.93$ = 23.1
t = ${1 \over k}\ln {{\left[ {{A_0}} \right]} \over {\left[ A \right]}}$
= ${{\left( {{t_{1/2}}} \right)} \over {0.693}} \times 2.303{\log _{10}}10$
= 10 × 2.303 × 1
= 23.03 years
Shortcut Method :
t90% = ${{10} \over 3} \times {t_{50\% }}$
= ${{10} \over 3} \times 6.93$ = 23.1
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Consider the following reactions
A $ \to $ P1 ; B $ \to $ P2 ; C $ \to $ P3 ; D $ \to $ P4,
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
Among the following, the correct sequence for the order of the reactions is :
A $ \to $ P1 ; B $ \to $ P2 ; C $ \to $ P3 ; D $ \to $ P4,
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
Among the following, the correct sequence for the order of the reactions is :
A.
d > b > a > c
B.
d > a > b > c
C.
a > b > c > d
D.
c > a > b > d
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The rate constant (k) of a reaction is measured at differenct temperatures (T), and the data are
plotted in the given figure. The activation energy of the reaction in kJ mol–1 is :
(R is gas constant)
(R is gas constant)
A.
R
B.
2R
C.
${1 \over R}$
D.
${1 \over {2R}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Morning Slot
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) :
(Use ln 2 = 0.693)
(Use ln 2 = 0.693)
A.
180
B.
120
C.
300
D.
900
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Evening Slot
For the reaction
2A + 3B + ${3 \over 2}$C $ \to $ 3P, which statement is correct ?
2A + 3B + ${3 \over 2}$C $ \to $ 3P, which statement is correct ?
A.
${{d{n_A}} \over {dt}} = {{d{n_B}} \over {dt}} = {{d{n_C}} \over {dt}}$
B.
${{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}$
C.
${{d{n_A}} \over {dt}} = {3 \over 2}{{d{n_B}} \over {dt}} = {3 \over 4}{{d{n_C}} \over {dt}}$
D.
${{d{n_A}} \over {dt}} = {2 \over 3}{{d{n_B}} \over {dt}} = {4 \over 3}{{d{n_C}} \over {dt}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
It is true that :
A.
A first order reaction is always a single step reaction
B.
A zero order reaction is a multistep reaction
C.
A zero order reaction is a single step reaction
D.
A second order reaction is always a multistep reaction
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
The results given in the below table were
obtained during kinetic studies of the following
reaction
2A + B $ \to $ C + D
X and Y in the given table are respectively :
2A + B $ \to $ C + D
X and Y in the given table are respectively :
A.
0.3, 0.4
B.
0.4, 0.3
C.
0.4, 0.4
D.
0.3, 0.3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
For the following reactions
$A\buildrel {700K} \over \longrightarrow {\mathop{\rm Product}\nolimits} $
$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits} $
it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):
$A\buildrel {700K} \over \longrightarrow {\mathop{\rm Product}\nolimits} $
$A\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{catalyst}^{500K}} {\mathop{\rm Product}\nolimits} $
it was found that Ea is decreased by 30 kJ/mol in the presence of catalyst.
If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same):
A.
198 kJ/mol
B.
135 kJ/mol
C.
105 kJ/mol
D.
75 kJ/mol
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Evening Slot
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?
A.
Ec > Ea > Ed > Eb
B.
Ea > Ec > Ed > Eb
C.
Eb > Ea > Ed > Ec
D.
Eb > Ed > Ec > Ea
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with
enzyme than without. The change in the activation energy upon adding enzyme is :
A.
+ 6RT
B.
– 6 (2.303)RT
C.
– 6RT
D.
+ 6(2.303)RT
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
For the reaction
2H2(g) + 2NO(g) $ \to $ N2(g) + 2H2O(g)
the observed rate expression is, rate = Kf[NO]2[H2]. The rate expression for the reverse reaction is :
2H2(g) + 2NO(g) $ \to $ N2(g) + 2H2O(g)
the observed rate expression is, rate = Kf[NO]2[H2]. The rate expression for the reverse reaction is :
A.
Kb[N2][H2O]
B.
Kb[N2][H2O]2/[H2]
C.
Kb[N2][H2O]2/[NO]
D.
Kb[N2][H2O]2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation,
2N2O5(g) $ \to $ 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :
2N2O5(g) $ \to $ 4NO2(g) + O2(g).
The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is :
A.
2.083 × 10–3
mol L–1
min–1
B.
8.333 × 10–3
mol L–1
min–1
C.
4.167 × 10–3
mol L–1
min–1
D.
1.667 × 10–2
mol L–1
min–1
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
In the following reaction; xA $ \to $ yB
${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010$
'A' and 'B' respectively can be :
${\log _{10}}\left[ { - {{d\left[ A \right]} \over {dt}}} \right] = {\log _{10}}\left[ {{{d\left[ B \right]} \over {dt}}} \right] + 0.3010$
'A' and 'B' respectively can be :
A.
n-Butane and Iso-butane
B.
C2H4 and C4H8
C.
C2H4 and C6H6
D.
N2O4 and NO2