Chemical Kinetics and Nuclear Chemistry
262 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th July Evening Shift
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)
(Given : $\mathrm{R}=2 \,\mathrm{cal} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}$ )
Correct Answer: 8
Explanation:
$\begin{aligned}
&\mathrm{K}=\mathrm{Ae}^{-\mathrm{Ea} / \mathrm{RT}} \\\\
&\ln \mathrm{k}=\frac{-\mathrm{Ea}}{\mathrm{RT}}+\ln \mathrm{A} \\\\
&\text { Slope }=\frac{\mathrm{Ea}}{\mathrm{R}}=\frac{20}{5} \\\\
&\mathrm{E}_{\mathrm{a}}=4 \mathrm{R}=8 \,\mathrm{Cal} / \mathrm{mol}
\end{aligned}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th July Morning Shift
For the given first order reaction
$\mathrm{A} \rightarrow \mathrm{B}$
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)
Correct Answer: 100
Explanation:
$\mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \quad\quad \mathrm{t}_{1 / 2}$ given $=0.3010$
$ \begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned} $
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
$A_{0} \rightarrow$ initial concentration of reactant
$A_{t} \rightarrow$ concentration of reactant at time $t$
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$100=\frac{A_{0}}{A_{t}}$
$ \begin{aligned} &K=\frac{0.693}{0.3010} \\\\ &K=2.30 \end{aligned} $
$\mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\left(\mathrm{A}_{0}\right)}{\left(\mathrm{A}_{\mathrm{t}}\right)}$
$A_{0} \rightarrow$ initial concentration of reactant
$A_{t} \rightarrow$ concentration of reactant at time $t$
$2.303=\frac{2.303}{2} \log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$2=\log \frac{\left(A_{0}\right)}{\left(A_{t}\right)}$
$100=\frac{A_{0}}{A_{t}}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th July Evening Shift
$\matrix{ {[A]} & \to & {[B]} \cr {{\mathop{\rm Reactant}\nolimits} } & {} & {{\mathop{\rm Product}\nolimits} } \cr } $
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 of its initial concentration. Then the rate constant of the reaction is $x \times 10^{-6} \mathrm{~s}^{-1}$. The value of $x$ is ______________. (Nearest Integer)
Correct Answer: 165
Explanation:
$\mathrm{K}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{70 \times 60}$
$=\frac{6930}{7 \times 6} \times 10^{-6}$
$=165 \times 10^{-6} \mathrm{~s}^{-1}$
$=\frac{6930}{7 \times 6} \times 10^{-6}$
$=165 \times 10^{-6} \mathrm{~s}^{-1}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th July Morning Shift
$2 \mathrm{NO}+2 \mathrm{H}_{2} \rightarrow \mathrm{N}_{2}+2 \mathrm{H}_{2} \mathrm{O}$
The above reaction has been studied at $800^{\circ} \mathrm{C}$. The related data are given in the table below
| Reaction serial number | Initial Pressure of ${H_2}/kPa$ | Initial Pressure of $NO/kPa$ | Initial rate $\left( {{{ - dp} \over {dt}}} \right)/(kPa/s)$ |
|---|---|---|---|
| 1 | 65.6 | 40.0 | 0.135 |
| 2 | 65.6 | 20.1 | 0.033 |
| 3 | 38.6 | 65.6 | 0.214 |
| 4 | 19.2 | 65.6 | 0.106 |
The order of the reaction with respect to NO is ___________.
Correct Answer: 2
Explanation:
Let the rate of reaction ( $r$ ) is as
$ \mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}} $
From $1^{\text {st }}$ data
$ 0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1) $
From $2^{\text {nd }}$ data
$ 0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2) $
On dividing equation (1) by equation (2)
$ \begin{aligned} &\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\ &4=(2)^{n} \\\\ &\therefore n=2 \\\\ &\therefore \text { Order of reaction w.r.t. NO is } 2 . \end{aligned} $
$ \mathrm{r}=\mathrm{K}[\mathrm{NO}]^{n}\left[\mathrm{H}_{2}\right]^{\mathrm{m}} $
From $1^{\text {st }}$ data
$ 0.135=\mathrm{K}[40]^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(1) $
From $2^{\text {nd }}$ data
$ 0.033=\mathrm{K}(20.1)^{\mathrm{n}} \cdot(65.6)^{\mathrm{m}}\quad\quad...(2) $
On dividing equation (1) by equation (2)
$ \begin{aligned} &\frac{0.135}{0.033}=\left(\frac{40}{20.1}\right)^{n} \\\\ &4=(2)^{n} \\\\ &\therefore n=2 \\\\ &\therefore \text { Order of reaction w.r.t. NO is } 2 . \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th July Morning Shift
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)
Correct Answer: 2
Explanation:
$t_{1 / 2} \propto \frac{1}{\left(a_{0}\right)^{n-1}}$
$ \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} $
$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$
$(2)=(2)^{n-1}$
$\mathrm{n}-1=1$
$n=2$
$ \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} $
$\frac{100}{50}=\left(\frac{1}{0 \cdot 5}\right)^{n-1}$
$(2)=(2)^{n-1}$
$\mathrm{n}-1=1$
$n=2$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Evening Shift
For the decomposition of azomethane.
CH3N2CH3(g) $\to$ CH3CH3(g) + N2(g), a first order reaction, the variation in partial pressure with time at 600 K is given as
The half life of the reaction is __________ $\times$ 10$-$5 s. [Nearest integer]
Correct Answer: 2
Explanation:
For first order reaction,
$ \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} $
$ \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} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th July Morning Shift
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)
Correct Answer: 1
Explanation:
$\left(\mathrm{t}_{1 / 2}\right)_{\mathrm{A}}=240 \mathrm{~s}$ when $\mathrm{P}=500$ torr
$\left(t_{1 / 2}\right)_{A}=4 \min =4 \times 60=240$ sec when $P=250$ torr
If means half-life is independent of concentration of reactant present.
$\therefore$ Order of reaction $=1$
$\left(t_{1 / 2}\right)_{A}=4 \min =4 \times 60=240$ sec when $P=250$ torr
If means half-life is independent of concentration of reactant present.
$\therefore$ Order of reaction $=1$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 30th June Morning Shift
For the reaction P $\to$ B, the values of frequency factor A and activation energy EA are 4 $\times$ 1013 s$-$1 and 8.3 kJ mol$-$1 respectively. If the reaction is of first order, the temperature at which the rate constant is 2 $\times$ 10$-$6 s$-$1 is _____________ $\times$ 10$-$1 K.
(Given : ln 10 = 2.3, R = 8.3 J K$-$1 mol$-$1, log2 = 0.30)
Correct Answer: 225
Explanation:
$
\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 \times 2.3} \\\\
&\Rightarrow-19.3=-\frac{10^{3}}{T \times 2.3} \Rightarrow T=225 \times 10^{-1} \mathrm{~K}
\end{aligned}
$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th June Evening Shift
The equation
k = (6.5 $\times$ 1012s$-$1)e$-$26000K/T
is followed for the decomposition of compound A. The activation energy for the reaction is ________ kJ mol$-$1. [nearest integer]
(Given : R = 8.314 J K$-$1 mol$-$1)
Correct Answer: 216
Explanation:
$k=A e^{\frac{-E_{a}}{R T}}$
$ \begin{aligned} \frac{E_{a}}{R T} &=\frac{26000}{T} \\\\ E_{a} &=26000 \times 8.314 \\\\ &=216164 \mathrm{~J} \\\\ &=216 \mathrm{~kJ} \end{aligned} $
$ \begin{aligned} \frac{E_{a}}{R T} &=\frac{26000}{T} \\\\ E_{a} &=26000 \times 8.314 \\\\ &=216164 \mathrm{~J} \\\\ &=216 \mathrm{~kJ} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 29th June Morning Shift
The activation energy of one of the reactions in a biochemical process is 532611 J mol$-$1. When the temperature falls from 310 K to 300 K, the change in rate constant observed is k300 = x $\times$ 10$-$3 k310. The value of x is ____________.
[Given : $\ln 10 = 2.3$, R = 8.3 J K$-$1 mol$-$1]
Correct Answer: 1
Explanation:
$
\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}
$
where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$
$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $
So K = 1
where $\mathrm{K}_2$ is at $310 \mathrm{~K} \,\& \mathrm{~K}_1$ is at $300 \mathrm{~K}$
$ \begin{aligned} & \ln \left(\frac{\mathrm{K}_2}{\mathrm{~K}_1}\right)=6.9 \\\\ & =3 \times \ln 10 \\\\ & \ln \frac{\mathrm{K}_2}{\mathrm{~K}_1}=\ln 10^3 \\\\ & \mathrm{~K}_2=\mathrm{K}_1 \times 10^3 \\\\ & \mathrm{~K}_1=\mathrm{K}_2 \times 10^3 \end{aligned} $
So K = 1
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Evening Shift
A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is ___________. (Nearest integer)
(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)
Correct Answer: 75
Explanation:
$\lambda=\frac{2.303}{t} \log \frac{A_{0}}{A}$
$ \begin{aligned} &\frac{0.693}{200}=\frac{2.303}{83} \log \frac{A_{0}}{A} \\\\ &\frac{A}{A_{0}}=0.75 \end{aligned} $
Hence, percentage of original activity remaining after 83 days is $75 \%$
$ \begin{aligned} &\frac{0.693}{200}=\frac{2.303}{83} \log \frac{A_{0}}{A} \\\\ &\frac{A}{A_{0}}=0.75 \end{aligned} $
Hence, percentage of original activity remaining after 83 days is $75 \%$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
For a first order reaction A $\to$ B, the rate constant, k = 5.5 $\times$ 10$-$14 s$-$1. The time required for 67% completion of reaction is x $\times$ 10$-$1 times the half life of reaction. The value of x is _____________ (Nearest integer)
(Given : log 3 = 0.4771)
Correct Answer: 16
Explanation:
$
\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}_{1 / 2} \\\\
&\mathrm{x}=15.66
\end{aligned}
$
Nearest integer $=16$
Nearest integer $=16$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Evening Shift
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$-$1. [nearest integer]
(Given ln10 = 2.3, R = 8.3 J K$-$1 mol$-$1, log 2 = 0.30)
Correct Answer: 59
Explanation:
$T_{1}=300 \mathrm{~K}$
(Rate constant)
$\mathrm{K}_{2}=2 \mathrm{~K}_{1}$, on increase temperature by $9 \mathrm{~K}$
$\mathrm{T}_{2}=309 \mathrm{~K}$
$\mathrm{Ea}=?$
$\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \cdot \mathrm{T}_{1}}\right]$
$\log 2=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{9}{309 \times 300}\right]$
$\mathrm{Ea}=\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}$
$=58988.1 \mathrm{~J} / \mathrm{mole}$
$\simeq 59 \mathrm{~kJ} / \mathrm{mole}$
(Rate constant)
$\mathrm{K}_{2}=2 \mathrm{~K}_{1}$, on increase temperature by $9 \mathrm{~K}$
$\mathrm{T}_{2}=309 \mathrm{~K}$
$\mathrm{Ea}=?$
$\log \frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}=\frac{\mathrm{Ea}}{2.3 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \cdot \mathrm{T}_{1}}\right]$
$\log 2=\frac{\mathrm{Ea}}{2.3 \times 8.3}\left[\frac{9}{309 \times 300}\right]$
$\mathrm{Ea}=\frac{0.3 \times 309 \times 300 \times 2.3 \times 8.3}{9}$
$=58988.1 \mathrm{~J} / \mathrm{mole}$
$\simeq 59 \mathrm{~kJ} / \mathrm{mole}$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
The rate constant for a first order reaction is given by the following equation:
$\ln k = 33.24 - {{2.0 \times {{10}^4}\,K} \over T}$
The activation energy for the reaction is given by ____________ kJ mol$-$1. (In nearest integer) (Given : R = 8.3 J K$-$1 mol$-$1)
Correct Answer: 166
Explanation:
$\ln \mathrm{k}=\ln \mathrm{A}-\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{RT}}$
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$
$ \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 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned} $
Given: $\ln k=33.24-\frac{2.0 \times 10^4}{\mathrm{~T}}$
$ \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 10^4 \times \mathrm{R} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=2.0 \times 10^4 \times 8.3 \mathrm{~J} \\\\ &\Rightarrow \mathrm{E}_{\mathrm{A}}=16.6 \times 10^4 \mathrm{~J}=166 \mathrm{~kJ} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
Catalyst A reduces the activation energy for a reaction by 10 kJ mol$-$1 at 300 K. The ratio of rate constants, ${{{}^kT,\,Catalysed} \over {{}^kT,\,Uncatalysed}}$ is ex. The value of x is ___________. [nearest integer]
[Assume that the pre-exponential factor is same in both the cases. Given R = 8.31 J K$-$1 mol$-$1]
Correct Answer: 4
Explanation:
Activation Energy : It is the minimum amount of energy required to activate the molecules/ atoms so that they can undergo the chemical reaction.
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 energy barrier.
In the Question,
Given : $\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}$
$ \mathrm{T}=300 \mathrm{~K} $
According to Arrhenius Equation,
$ \begin{aligned} & \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\ & \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}} \end{aligned} $
$ \begin{aligned} & =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\ & =e^4 \\\\ & =4 \end{aligned} $
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 energy barrier.
In the Question,
Given : $\mathrm{E}_{\text {cat }}-\mathrm{E}_{\text {uncat }}=10 \mathrm{~kJ} / \mathrm{mol}$
$ \mathrm{T}=300 \mathrm{~K} $
According to Arrhenius Equation,
$ \begin{aligned} & \mathrm{K}=A \mathrm{e}^{-\mathrm{Ea} / \mathrm{RT}} \\\\ & \frac{\mathrm{E}_{\text {cat }}}{\mathrm{E}_{\text {uncat }}}=e^{\frac{E a-E_a^1}{R T}} \end{aligned} $
$ \begin{aligned} & =e^{\frac{10 \times 10^3}{8.21 \times 300}} \\\\ & =e^4 \\\\ & =4 \end{aligned} $
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 $=-\frac{E_{a}}{R \times 10^{3}}=-18.5$
$\therefore E_{a}=18.5 \times 8.31 \times 1000 \simeq 154 \mathrm{~kJ} \mathrm{~mol}^{-1}$
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
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
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 28th January Morning Shift
An organic compound undergoes first order decomposition. The time taken for decomposition to $\left(\frac{1}{8}\right)^{\text {th }}$ and $\left(\frac{1}{10}\right)^{\text {th }}$ of its initial concentration are $\mathrm{t}_{1 / 8}$ and $\mathrm{t}_{1 / 10}$ respectively.
What is the value of $\frac{\mathrm{t}_{1 / 8}}{\mathrm{t}_{1 / 10}} \times 10$ ?
$ (\log 2=0.3) $
A.
30
B.
9
C.
3
D.
0.9
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 24th January Morning Shift
$\mathrm{A} \rightarrow \mathrm{D}$ is an endothermic reaction occurring in three steps (elementary).
(i) $\mathrm{A} \rightarrow \mathrm{B} \Delta \mathrm{H}_i=+\mathrm{ve}$
(ii) $\mathrm{B} \rightarrow \mathrm{C} \Delta \mathrm{H}_{i i}=-\mathrm{ve}$
(iii) $\mathrm{C} \rightarrow \mathrm{D} \Delta \mathrm{H}_{i i i}=-\mathrm{ve}$
Which of the following graphs between potential energy ( $y$-axis) vs reaction coordinate ( $x$-axis) correctly represents the reaction profile of $A \rightarrow D$ ?
A.
B.
C.
D.
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 24th January Morning Shift
At $27^{\circ} \mathrm{C}$ in presence of a catalyst, activation energy of a reaction is lowered by $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The logarithm of ratio of $\frac{\mathrm{k} \text { (catalysed) }}{\mathrm{k} \text { (uncatalysed) }}$ is….
(Consider that the frequency factor for both the reactions is same)
A.
1.741
B.
0.1741
C.
17.41
D.
3.482
2026
JEE Mains
MCQ
JEE Main 2026 (Online) 23rd January Evening Shift
Given above is the concentration vs time plot for a dissociation reaction : $\mathrm{A} \rightarrow \mathrm{nB}$.
Based on the data of the initial phase of the reaction (initial 10 min ), the value of n is $\_\_\_\_$ .
A.
2
B.
5
C.
4
D.
3