Chemical Kinetics and Nuclear Chemistry
A → B (first reaction)
C → D (second reaction)
Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 K, 50% of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is _______ × 10-1 hour-1 (nearest integer).
Explanation:
$ \text { For } \mathrm{A} \xrightarrow{\mathrm{~K}_1} \mathrm{~B} $
$ \begin{aligned} & \ln (2)=\frac{E_{a_1}}{R}\left[\frac{1}{300}-\frac{1}{500}\right] \\ & E_{a_1}=\frac{\ln 2 \times R \times 1500}{2} \\ & E_{a_2}=\frac{E_{a_1}}{2}=\frac{\ln 2 \times R \times 1500}{4} \\ & \left(K_1\right)_{\text {at } 500 \mathrm{~K}}=\frac{\ln 2}{2} \\ & \left(K_2\right)_{\text {at } 500 \mathrm{~K}}=\ln 2 \end{aligned} $
Now for $\mathrm{C} \xrightarrow{\mathrm{K}_2} \mathrm{D}$
$ \begin{aligned} & \ln \left[\frac{\left(\mathrm{K}_2\right)_{\text {at } 500 \mathrm{~K}}}{\left(\mathrm{~K}_2\right)_{\text {at } 300 \mathrm{~K}}}\right]=\left(\frac{\ln 2 \times \mathrm{R} \times 1500}{4}\right) \times \frac{1}{\mathrm{R}} \times\left[\frac{1}{300}-\frac{1}{500}\right] \\ & \left(\mathrm{K}_2\right)_{\text {at } 300 \mathrm{~K}}=\frac{\ln 2}{\sqrt{2}}=0.49 \\ & \left(\mathrm{~K}_2\right)_{\text {at } 300 \mathrm{~K}}=4.9 \times 10^{-1} \end{aligned} $
Ans is 5 .
The half-life of ${ }^{65} \mathrm{Zn}$ is 245 days. After $x$ days, $75 \%$ of original activity remained. The value of $x$ in days is $\_\_\_\_$ . (Nearest integer)
(Given: $\log 3=0.4771$ and $\log 2=0.3010$ )
Explanation:
$ \begin{aligned} & \mathrm{t}_{1 / 2}=\frac{\ln 2}{\mathrm{~K}} \end{aligned} $
Here, $\mathrm{t}_{1/2}=245$ days, so we find $K$ as:
$ \begin{aligned} & \mathrm{~K}=\frac{\ln 2}{245} \end{aligned} $
Now we use the radioactive decay relation for activity:
$ \begin{aligned} & \mathrm{t}=\frac{1}{\mathrm{~K}} \ln \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}} \end{aligned} $
Given that $75\%$ of the original activity remains, so $\mathrm{a}_t = 0.75\,\mathrm{a}_0 = \frac{3}{4}\mathrm{a}_0$.
So,
$ \begin{aligned} & \frac{\mathrm{a}_0}{\mathrm{a}_{\mathrm{t}}}=\frac{\mathrm{a}_0}{\frac{3}{4}\mathrm{a}_0}=\frac{4}{3} \end{aligned} $
Therefore,
$ \begin{aligned} & \mathrm{t}_{25 \%}=\frac{1}{\mathrm{~K}} \ln \frac{4}{3} \end{aligned} $
Substitute $\mathrm{K}=\frac{\ln 2}{245}$:
$ \begin{aligned} & \mathrm{t}_{25 \%}=\frac{1}{\frac{\ln 2}{245}} \ln \frac{4}{3} \end{aligned} $
Simplifying:
$ \begin{aligned} & \mathrm{t}_{25 \%}=245 \frac{\ln \frac{4}{3}}{\ell \mathrm{n} 2}=245\left[\frac{2 \log 2-\log 3}{\log 2}\right] \end{aligned} $
Now use the given values $\log 3=0.4771$ and $\log 2=0.3010$:
$ \begin{aligned} & =245\left[\frac{2 \times 0.3010-0.4771}{0.3010}\right]=101.66 \text { day. } \end{aligned} $
For the thermal decomposition of reactant $\mathrm{AB}(\mathrm{g})$, the following plot is constructed.
The half life of the reaction is ' $x^{\prime} \,\mathrm{min}$.
$x=$ $\_\_\_\_$ min. (Nearest integer)
Explanation:
The graph between concentration $[\mathrm{AB}]$ and time $t$ is a straight line that decreases with time. This type of graph shows that the reaction is zero order in $[\mathrm{AB}]$.
For a zero-order reaction, the integrated rate law is:
$\begin{aligned} & {[\mathrm{AB}]_0-[\mathrm{AB}]_{\mathrm{t}}=\mathrm{kt}} \\ & \mathbf{0 . 6 0 - 0 . 5 5 = k}(100) \\ & \mathrm{k}=5 \times 10^{-4} \\ & \text { Half life }\left(\mathrm{t}_{1 / 2}\right)=\frac{[\mathrm{AB}]_0}{2 \mathrm{k}} \\ & =\frac{0.60}{2 \times 5 \times 10^{-4}} \\ & =600 \mathrm{sec} \\ & =10 \mathrm{~min}\end{aligned}$
From the graph, the initial concentration is $[\mathrm{AB}]_0 = 0.60$ and after $100 \,\text{s}$ the concentration is $0.55$. Substituting these values in the zero-order equation gives the rate constant $k = 5 \times 10^{-4} \,\text{mol L}^{-1}\text{s}^{-1}$.
For a zero-order reaction, the half-life is given by $t_{1/2} = \dfrac{[\mathrm{AB}]_0}{2k}$. Putting the values, we get $t_{1/2} = 600 \,\text{s} = 10 \,\text{min}$. Therefore, $x = 10 \,\text{min}$ (nearest integer).
Consider $\mathrm{A} \xrightarrow{\mathrm{k}_1} \mathrm{~B}$ and $\mathrm{C} \xrightarrow{\mathrm{k}_2} \mathrm{D}$ are two reactions. If the rate constant $\left(\mathrm{k}_1\right)$ of the $\mathrm{A} \longrightarrow \mathrm{B}$ reaction can be expressed by the following equation $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$ and activation energy of $C \longrightarrow D$ reaction $\left(E a_2\right)$ is $\frac{1}{5}$ th of the $A \longrightarrow B$ reaction $\left(E a_1\right)$, then the value of $\left(E a_2\right)$ is
$\_\_\_\_$ $\mathrm{kJ} \mathrm{mol}^{-1}$. (Nearest Integer)
Explanation:
Given: $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$
From the Arrhenius equation in base-10 form, we write:
$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$
On comparing this with $\log _{10} \mathrm{k}=14.34-\frac{1.5 \times 10^4}{\mathrm{~T} / \mathrm{K}}$, the coefficient of $\frac{1}{T}$ gives:
$\frac{\mathrm{E}_{\mathrm{a}_1}}{2.303 \mathrm{R}}=1.5 \times 10^4$
Now substitute $\mathrm{R}=8.314~\mathrm{J\,mol^{-1}\,K^{-1}}$:
$\mathrm{E}_{\mathrm{a}_1}=1.5 \times 10^4 \times 2.303 \times 8.314$
$\mathrm{E}_{\mathrm{a}_1}=28.7207 \times 10^4 \mathrm{~J}$
Convert joules to kilojoules ($1~\mathrm{kJ}=10^3~\mathrm{J}$):
$\mathrm{E}_{\mathrm{a}_1}=287.207 \mathrm{~kJ}$
Given $\mathrm{E}_{\mathrm{a}_2}=\frac{1}{5}\mathrm{E}_{\mathrm{a}_1}$, so:
$\mathrm{E}_{\mathrm{a}_2}=\frac{\mathrm{E}_{\mathrm{a}_1}}{5}=\frac{287.207}{5}=57.44 \mathrm{~kJ}$
The temperature at which the rate constants of the given below two gaseous reactions become equal is $\_\_\_\_$ K. (Nearest integer)
$ \begin{array}{ll} \mathrm{X} \longrightarrow \mathrm{Y}, & \mathrm{k}_1=10^6 e^{\frac{-30000}{\mathrm{~T}}} \\ \mathrm{P} \longrightarrow \mathrm{Q}, & \mathrm{k}_2=10^4 e^{\frac{-24000}{\mathrm{~T}}} \end{array} $
Given : $\ln 10=2.303$
Explanation:
Given,
$k_1=10^6 e^{-30000/T},\qquad k_2=10^4 e^{-24000/T}$
At the temperature $T$ when $k_1=k_2$:
$10^6 e^{-30000/T}=10^4 e^{-24000/T}$
Divide both sides by $10^4 e^{-30000/T}$:
$10^2=e^{(-24000/T)-(-30000/T)}=e^{6000/T}$
Take natural log on both sides:
$\ln(10^2)=\frac{6000}{T}$
$2\ln 10=\frac{6000}{T}$
Given $\ln 10=2.303$:
$2(2.303)=\frac{6000}{T}$
$4.606=\frac{6000}{T}$
$T=\frac{6000}{4.606}\approx 1302.65\ \text{K}$
Nearest integer:
$\boxed{1303\ \text{K}}$
Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by $20 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If $\mathrm{k}_1$ and $\mathrm{k}_2$ are the rate constants of first and second reaction respectively at 300 K , then $\ln \frac{\mathrm{k}_2}{\mathrm{k}_1}$ will be $\_\_\_\_$ . (nearest integer) $\left[\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Explanation:
Arrhenius equation:
$k=Ae^{-E_a/RT}$
Given pre-exponential factors are identical, so $A_1=A_2$.
Also, activation energy of first reaction exceeds that of second by $20\,\text{kJ mol}^{-1}$:
$E_{a1}=E_{a2}+20{,}000\ \text{J mol}^{-1}$
Now,
$ \frac{k_2}{k_1} =\frac{A e^{-E_{a2}/RT}}{A e^{-E_{a1}/RT}} =e^{-(E_{a2}-E_{a1})/RT} =e^{(E_{a1}-E_{a2})/RT} $
So,
$\ln\left(\frac{k_2}{k_1}\right)=\frac{E_{a1}-E_{a2}}{RT}=\frac{20{,}000}{8.3\times 300}$
Compute:
$8.3\times 300=2490$
$\ln\left(\frac{k_2}{k_1}\right)=\frac{20{,}000}{2490}\approx 8.03$
Nearest integer:
$\boxed{8}$
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) $
30
9
3
0.9
$\mathrm{A} \rightarrow \mathrm{D}$ 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 $A \rightarrow D$ ?
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)
1.741
0.1741
17.41
3.482
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 $\_\_\_\_$ .
2
5
4
3
Observe the following reactions at $\mathrm{T}(\mathrm{K})$.
I. $\mathrm{A} \rightarrow$ products.
II. $5 \mathrm{Br}^{-}(\mathrm{aq})+\mathrm{BrO}_3{ }^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 3 \mathrm{Br}_2(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of $-\frac{\Delta\left[\mathrm{Br}^{-}\right]}{\Delta \mathrm{t}}$ at 10.10 am is $2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$. The concentration of A at 10.10 am is $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the first order rate constant (in $\mathrm{min}^{-1}$ ) of reaction $I$ ?
$4 \times 10^{-3}$
$2 \times 10^{-3}$
$10^{-3}$
$10^{-2}$
Correct statements regarding Arrhenius equation among the following are :
A. Factor $e^{-\mathrm{Ea} / \mathrm{RT}}$ corresponds to fraction of molecules having kinetic energy less than Ea.
B. At a given temperature, lower the Ea, faster is the reaction.
C. Increase in temperature by about $10^{\circ} \mathrm{C}$ doubles the rate of reaction.
D. Plot of $\log \mathrm{k}$ vs $\frac{1}{\mathrm{~T}}$ gives a straight line with slope $=-\frac{\mathrm{Ea}}{\mathrm{R}}$.
Choose the correct answer from the options given below :
A and B Only
B and D Only
B and C Only
A and C Only
$\mathrm{A} \rightarrow$ product (First order reaction).
Three sets of experiment were performed for a reaction under similar experimental conditions:
Run $1 \Rightarrow 100 \mathrm{~mL}$ of 10 M solution of reactant A
Run $2 \Rightarrow 200 \mathrm{~mL}$ of 10 M solution of reactant A
Run $3 \Rightarrow 100 \mathrm{~mL}$ of 10 M solution of reactant $\mathrm{A}+100 \mathrm{~mL}$ of $\mathrm{H}_2 \mathrm{O}$ added.
The correct variation of rate of reaction is
Run $1=$ Run $2=$ Run 3
Run $3<\operatorname{Run} 1<\operatorname{Run} 2$
Run $1<$ Run $2<$ Run 3
Run $3<$ Run $1=$ Run 2
Decomposition of A is a first order reaction at T(K) and is given by A(g) → B(g) + C(g).
In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min−1) of the reaction? (log 2 = 0.3)
$6.9 × 10^{-4}$
$6.9 × 10^{-3}$
$6.9 × 10^{-1}$
$6.9 × 10^{-2}$
For the reaction $\mathrm{A} \rightarrow \mathrm{B}$ the following graph was obtained. The time required (in seconds) for the concentration of A to reduce to $2.5 \mathrm{~g} \mathrm{~L}^{-1}$ (if the initial concentration of A was $50 \mathrm{~g} \mathrm{~L}^{-1}$ ) is $\qquad$ . (Nearest integer)
Given : $\log 2=0.3010$
Explanation:
To determine the time required for the concentration of A to decrease from an initial value of $50 \, \mathrm{g} \, \mathrm{L}^{-1}$ to $2.5 \, \mathrm{g} \, \mathrm{L}^{-1}$ in the reaction $ \mathrm{A} \rightarrow \mathrm{B} $, we assume first-order kinetics. Although the graph does not provide a clear indication of the reaction order over the intervals $0-5$, $5-10$, and $10-15$ seconds, where the order appears to be zero, we'll proceed with the assumption of first-order kinetics, since the graph is not a straight line.
The rate constant $ \mathrm{K} $ for first-order reactions can be calculated using the formula:
$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{\mathrm{A}_0}{\mathrm{A}_{\mathrm{t}}} $
For the interval where $\mathrm{A}_0 = 40 \, \mathrm{g/L}$ and $\mathrm{A}_{\mathrm{t}} = 20 \, \mathrm{g/L}$ after 10 seconds:
$ \mathrm{K} = \frac{1}{10} \ln \frac{40}{20} $
Now, to find the time $ \mathrm{t} $ required for the concentration to reduce to $2.5 \, \mathrm{g/L}$:
$ \mathrm{K} = \frac{1}{\mathrm{t}} \ln \frac{50}{2.5} $
Equating the two expressions for $\mathrm{K}$:
$ \frac{1}{10} \ln 2 = \frac{1}{\mathrm{t}} \ln 20 $
Solving for $\mathrm{t}$:
$ \mathrm{t} = \frac{1.3010 \times 10}{0.3010} = 43.3 \, \mathrm{sec} $
Therefore, the time required for the concentration to decrease to $2.5 \, \mathrm{g/L}$ is approximately 43 seconds.
For the reaction A $\to$ products.
The concentration of A at 10 minutes is _________ $\times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$ (nearest integer). The reaction was started with $2.5 \mathrm{~mol} \mathrm{~L}^{-1}$ of A .
Explanation:
Order of Reaction:
Since $ t_{1/2} \propto [A]_0 $, the reaction follows a zero-order kinetics.
Half-life Equation for Zero-order Reaction:
The half-life ($ t_{1/2} $) is calculated as:
$ t_{1/2} = \frac{[A]_0}{2K} $
Given the slope from the graph is $76.92$, which equals $\frac{1}{2K}$. This implies:
$ K = \frac{1}{2 \times 76.92} $
Concentration of A at 10 Minutes:
Apply the zero-order kinetics equation:
$ [A]_{10} = -Kt + [A]_0 $
Substituting the values:
$ [A]_{10} = -\left(\frac{1}{2 \times 76.92}\right) \times 10 + 2.5 = 2.435 \ \text{mol L}^{-1} $
Final Concentration:
Convert to scientific notation:
$ [A]_{10} = 2435 \times 10^{-3} \ \text{mol L}^{-1} $
Thus, the concentration of A at 10 minutes is approximately $ 2435 \times 10^{-3} \ \text{mol L}^{-1} $.
Consider a complex reaction taking place in three steps with rate constants $\mathrm{k}_1, \mathrm{k}_2$ and $\mathrm{k}_3$ respectively. The overall rate constant $k$ is given by the expression $k=\sqrt{\frac{k_1 k_3}{k_2}}$. If the activation energies of the three steps are 60, 30 and $10 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively, then the overall energy of activation in $\mathrm{kJ} \mathrm{mol}^{-1}$ is _________ . (Nearest integer)
Explanation:
To determine the overall energy of activation for the given complex reaction with rate constants $k_1$, $k_2$, and $k_3$, we start with the expression for the overall rate constant $k$:
$ k = \sqrt{\frac{k_1 k_3}{k_2}} $
The rate constant can also be expressed in terms of the Arrhenius equation:
$ A \cdot e^{-E_a / RT} = \sqrt{\frac{A_1 e^{-E_{a_1} / RT} \cdot A_3 e^{-E_{a_3} / RT}}{A_2 e^{-E_{a_2} / RT}}} $
By comparing the exponential terms from both sides, we have:
$ \frac{E_a}{RT} = \frac{1}{2} \left( \frac{E_{a_1}}{RT} + \frac{E_{a_3}}{RT} - \frac{E_{a_2}}{RT} \right) $
This simplifies to:
$ E_a = \frac{E_{a_1} + E_{a_3} - E_{a_2}}{2} $
Substituting the given activation energies—$E_{a_1} = 60 \, \text{kJ mol}^{-1}$, $E_{a_2} = 30 \, \text{kJ mol}^{-1}$, and $E_{a_3} = 10 \, \text{kJ mol}^{-1}$:
$ E_a = \frac{60 + 10 - 30}{2} = \frac{40}{2} = 20 \, \text{kJ mol}^{-1} $
Therefore, the overall energy of activation is 20 kJ/mol.
For the thermal decomposition of $\mathrm{N}_2 \mathrm{O}_5(\mathrm{~g})$ at constant volume, the following table can be formed, for the reaction mentioned below.
$2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \rightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
| Sr. No. | Time/s | Total pressure/(atm) |
|---|---|---|
| 1 | 0 | 0.6 |
| 2 | 100 | '$\mathrm{x}$' |
$\mathrm{x}=$ __________ $\times 10^{-3} \mathrm{~atm}$ [nearest integer]
Given : Rate constant for the reaction is $4.606 \times 10^{-2} \mathrm{~s}^{-1}$.
Explanation:
$\begin{aligned} & \mathrm{K}_{\mathrm{N}_2 \mathrm{O}_5}=2 \times 4.606 \times 10^{-2} \mathrm{~S}^{-1} \\ & 2 \mathrm{~N}_2 \mathrm{O}_5(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2 \mathrm{O}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \end{aligned}$
$\matrix{ {{P_i}} & {0.6} & 0 & 0 \cr {{P_f}} & {0.6 - P} & P & {{P \over 2}} \cr } $
$\begin{aligned} & 2 \times 4.606 \times 10^{-2}=\frac{2.303}{100} \log \frac{0.6}{0.6-\mathrm{P}} \\ & \quad 4 \log _{10} \frac{0.6}{0.6-\mathrm{P}} \\ & \quad 10^4=\frac{0.6}{0.6-\mathrm{P}} \\ & \Rightarrow 0.6 \times 10^4-10^4 \mathrm{P}=0.6 \end{aligned}$
$\begin{aligned} \begin{aligned} & \Rightarrow 10^4 \mathrm{P}=0.6\left(10^4-1\right) \\ & \mathrm{P}=(6000-0.6) \times 10^{-4} \\ &=5999 . \times 10^{-4} \\ &=0.59994 \\ & \mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{P}}{2} \\ &= 0.6+0.29997 \\ &= 0.89997 \\ &=899.97 \times 10^{-3} \\ & \text { Ans. } 900 \end{aligned} \end{aligned}$
$\mathrm{P}_{\text {Total }}=0.6+\frac{\mathrm{x}}{2}$
As given in equation
$\mathrm{K}_{\mathrm{r}}=4.606 \times 10^{-2} \mathrm{sec}^{-1}$
(Here language conflict in question)
($\mathrm{K}_{\mathrm{r}}=\frac{\mathrm{KA}}{2}$ not considered)
$\begin{aligned} \mathrm{K}_{\mathrm{r}} \mathrm{t} & =\ln \frac{0.6}{0.6-\mathrm{x}} \\ 4.606 & \times 10^{-2} \times 100=2.303 \log \frac{0.6}{0.6-\mathrm{x}} \\ \mathrm{P}_{\text {Total }} & =0.6+\frac{0.594}{2}=0.897 \mathrm{~atm} \\ \quad & =897 \times 10^{-3} \mathrm{~atm} \end{aligned}$
$\mathrm{A \rightarrow B}$
The molecule A changes into its isomeric form B by following a first order kinetics at a temperature of 1000 K . If the energy barrier with respect to reactant energy for such isomeric transformation is $191.48 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and the frequency factor is $10^{20}$, the time required for $50 \%$ molecules of A to become B is __________ picoseconds (nearest integer). $\left[\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right]$
Explanation:
To determine the time required for 50% of molecule A to change into its isomeric form B, follow these steps:
Half-life Formula for First Order Kinetics:
The half-life ($ t_{1/2} $) for a first-order reaction is given by:
$ t_{1/2} = \frac{0.693}{K} $
Calculate the Rate Constant (K):
The rate constant $ K $ can be calculated using the Arrhenius equation:
$ K = A \cdot e^{-\frac{E_a}{RT}} $
Given:
$ A $ (frequency factor) = $ 10^{20} $
$ E_a $ (activation energy) = $ 191.48 \, \text{kJ/mol} = 191.48 \times 10^3 \, \text{J/mol} $
$ R $ (universal gas constant) = $ 8.314 \, \text{J/mol} \cdot \text{K} $
$ T = 1000 \, \text{K} $
Substitute the values into the Arrhenius equation:
$ K = 10^{20} \times e^{-\frac{191.48 \times 10^3}{8.314 \times 1000}} $
Simplify this calculation:
$ K = 10^{20} \times e^{-23.031} $
Simplifying further by recognizing that $ e^{-23.031} $ is a very small number, gives:
$ K \approx \frac{10^{20}}{10^{10}} = 10^{10} \, \text{sec}^{-1} $
Calculate the Half-life:
Using the calculated value of $ K $:
$ t_{1/2} = \frac{0.693}{10^{10}} = 6.93 \times 10^{-11} \, \text{seconds} $
Convert to Picoseconds:
Since $ 1 \, \text{second} = 10^{12} \, \text{picoseconds} $:
$ t_{1/2} = 6.93 \times 10^{-11} \times 10^{12} \, \text{picoseconds} = 69.3 \, \text{picoseconds} $
Therefore, the time required for 50% of the molecules of A to become B is approximately 69 picoseconds (nearest integer).
In a first order decomposition reaction, the time taken for the decomposition of reactant to one fourth and one eighth of its initial concentration are $t_1$ and $t_2$ (s), respectively. The ratio $t_1/t_2$ will be:
$\frac{4}{3}$
$\frac{3}{2}$
$\frac{3}{4}$
$\frac{2}{3}$
A(g) → B(g) + C(g) is a first order reaction.
| Time | t | ∞ |
|---|---|---|
| Psystem | Pt | P∞ |
The reaction was started with reactant A only. Which of the following expressions is correct for rate constant k?
A person's wound was exposed to some bacteria and then bacterial growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay(r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine?
[Given: $N=$ No. of bacteria, $t=$ time, bacterial growth follows $1^{\text {st }}$ order kinetics.]
Reaction $\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$ is a first order reaction. It was started with pure A
| t/min | Pressure of system at time t/mm Hg |
|---|---|
| 10 | 160 |
| $\infty$ | 240 |
Which of the following option is incorrect?
Consider the following plots of $\log$ of rate constant $\mathrm{k}(\log \mathrm{k})$ vs $\frac{1}{\mathrm{~T}}$ for three different reactions. The correct order of activation energies of these reactions is :
Half life of zero order reaction $\mathrm{A} \rightarrow$ product is 1 hour, when initial concentration of reactant is $2.0 \mathrm{~mol} \mathrm{~L}{ }^{-1}$. The time required to decrease concentration of A from 0.50 to $0.25 \mathrm{~mol} \mathrm{~L}^{-1}$ is :
For $\mathrm{A}_2+\mathrm{B}_2 \rightleftharpoons 2 \mathrm{AB}$
$\mathrm{E}_{\mathrm{a}}$ for forward and backward reaction are 180 and $200 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively
If catalyst lowers $\mathrm{E}_{\mathrm{a}}$ for both reaction by $100 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Which of the following statement is correct?
Rate law for a reaction between $A$ and $B$ is given by
$\mathrm{r}=\mathrm{k}[\mathrm{~A}]^{\mathrm{n}}[\mathrm{~B}]^{\mathrm{m}}$
If concentration of $A$ is doubled and concentration of $B$ is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{r_2}{r_1}\right)$ is
Consider the following statements related to temperature dependence of rate constants.
Identify the correct statements.
A. The Arrhenius equation holds true only for an elementary homogenous reaction.
B. The unit of $A$ is same as that of $k$ in Arrhenius equation.
C. At a given temperature, a low activation energy means a fast reaction.
D. A and Ea as used in Arrhenius equation depend on temperature.
E. When $\mathrm{Ea} \gg \mathrm{RT}, \mathrm{A}$ and Ea become interdependent.
Choose the correct answer from the options given below:
In a reaction $A+B \rightarrow C$, initial concentrations of $A$ and $B$ are related as $[A]_0=8[B]_0$. The half lives of $A$ and $B$ are 10 min and 40 min , respectively. If they start to disappear at the same time, both following first order kinetics, after how much time will the concentration of both the reactants be same?
Reactant A converts to product D through the given mechanism (with the net evolution of heat):
A → B slow; ΔH = +ve
B → C fast; ΔH = -ve
C → D fast; ΔH = -ve
Which of the following represents the above reaction mechanism?
Drug $X$ becomes ineffective after $50 \%$ decomposition. The original concentration of drug in a bottle was $16 \mathrm{mg} / \mathrm{mL}$ which becomes $4 \mathrm{mg} / \mathrm{mL}$ in 12 months. The expiry time of the drug in months is _________.
Assume that the decomposition of the drug follows first order kinetics.
12
3
6
2
The reaction $A_2 + B_2 \rightarrow 2AB$ follows the mechanism:
$A_2 \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} A + A$ (fast)
$A + B_2 \xrightarrow{k_2} AB + B$ (slow)
$A + B \rightarrow AB$ (fast)
The overall order of the reaction is:
3
2.5
1.5
2
Consider an elementary reaction
$ \mathrm{A}(\mathrm{~g})+\mathrm{B}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{~g})+\mathrm{D}(\mathrm{~g}) $
If the volume of reaction mixture is suddenly reduced to $\frac{1}{3}$ of its initial volume, the reaction rate will become ' $x^{\prime}$ times of the original reaction rate. The value of $x$ is :
3
9
$\frac{1}{3}$
$\frac{1}{9}$
For bacterial growth in a cell culture, growth law is very similar to the law of radioactive decay. Which of the following graphs is most suitable to represent bacterial colony growth ?
Where N - Number of Bacteria at any time, $\mathrm{N}_0$ - Initial number of Bacteria.
For a given reaction $\mathrm{R} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2}$ is related to $[\mathrm{A}]_0$ as given in table.
Given: $\log 2=0.30$
Which of the following is true?
A. The order of the reaction is $1 / 2$.
B. If $[\mathrm{A}]_0$ is 1 M , then $\mathrm{t}_{1 / 2}$ is $200 \sqrt{10} \mathrm{~min}$
C. The order of the reaction changes to 1 if the concentration of reactant changes from 0.100 M to 0.500 M.
D. $\mathrm{t}_{1 / 2}$ is 800 min for $[\mathrm{A}]_0=1.6 \mathrm{M}$
Choose the correct answer from the options given below:
Given below are two statements :
Statement (I) : 
is valid for first order reaction.
Statement (II) : 
is valid for first order reaction.
In the light of the above statements, choose the correct answer from the options given below :
For a reaction, $\mathrm{N}_2 \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}$ in a constant volume container, no products were present initially. The final pressure of the system when $50 \%$ of reaction gets completed is
Which of the following graphs most appropriately represents a zero order reaction?
Consider the given figure and choose the correct option :
Which of the following statement is not true for radioactive decay?
Consider the following first order gas phase reaction at constant temperature $ \mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
If the total pressure of the gases is found to be 200 torr after 23 $\mathrm{sec}$. and 300 torr upon the complete decomposition of A after a very long time, then the rate constant of the given reaction is ________ $\times 10^{-2} \mathrm{~s}^{-1}$ (nearest integer)
[Given : $\log _{10}(2)=0.301$]
Explanation:
To find the rate constant of the first-order reaction indicated, let us first understand the reaction dynamics based on the total pressure change over time. The reaction is:
$\mathrm{A(g)} \rightarrow 2\mathrm{B(g)} + \mathrm{C(g)}$
Initially, only A is present. As the reaction progresses, A decreases, and B and C are formed. For each mole of A that reacts, a total of 3 moles of gas are produced (2 moles of B and 1 mole of C), leading to an increase in the total pressure of the system if the volume remains constant.
At the start: Total pressure due to A only.
Finally, when A has completely decomposed: Total pressure is due to 3 times the initial amount of A, as no A is left and for every mole of A decomposed, 3 moles of gas (2B + 1C) are produced.
The total pressure change thus directly relates to the extent of reaction, i.e., how much A has decomposed into B and C.
The total pressure at 23 sec is 200 torr, and the final total pressure is 300 torr after a very long time, indicating that the pressure increases by 100 torr due to the complete decomposition of A.
To use this information, we need to understand the relationship between pressure change and concentration in a closed system, especially for a first-order reaction. For a first-order reaction:
$\ln\left(\frac{[A]_0}{[A]}\right) = kt$
Where: $[A]_0$ is the initial concentration (or in this context, partial pressure), $[A]$ is the concentration (or partial pressure) at time t, $k$ is the rate constant, and $t$ is the time.
Let's denote the initial total pressure due to A as $P_0$. At complete decomposition, the pressure increase is from the conversion of A to 2B + C which increases the total pressure to 300 torr, indicating that initially, $P_0$ must have been 100 torr since the pressure increase is due to the tripling effect of the reaction, completing to 300 torr.
At 23 sec, the total pressure is at 200 torr. This means 100 torr is due to the unreacted A, and the additional 100 torr is from the formation of 2B + C. Given the nature of the reaction (1:3 stoichiometry of A to the total gas), we can deduce that at 23 seconds, half of A has reacted because the pressure due to the products matches the pressure due to the unreacted A.
Hence, at 23 sec, $[A] = \frac{1}{2}[A]_0$. Plugging this into the first-order reaction formula:
$\ln\left(\frac{[A]_0}{\frac{1}{2}[A]_0}\right) = kt$
$\ln(2) = k \times 23$
Given that $\log_{10}(2) = 0.301$, and knowing that $\ln(2) = \log_{e}(2)$ (where $\log_{e}(2) \approx 0.693$, for conversion from base 10 to e we can use the natural logarithm of 2 directly), we have:
$0.693 = k \times 23$
$k = \frac{0.693}{23} = 0.0301 \, \mathrm{s}^{-1}$
When expressed in the format requested, $k = 3.01 \times 10^{-2} \, \mathrm{s}^{-1}$. Rounding to the nearest integer, $k = 3 \times 10^{-2} \, \mathrm{s}^{-1}$.
Given below are two statements :
Statement I : The rate law for the reaction $A+B \rightarrow C$ is rate $(r)=k[A]^2[B]$. When the concentration of both $\mathrm{A}$ and $\mathrm{B}$ is doubled, the reaction rate is increased "$x$" times.
Statement II : 
The figure is showing "the variation in concentration against time plot" for a "$y$" order reaction.
The Value of $x+y$ is __________.Explanation:
$\begin{aligned} & x=8, y=0 \\ & \frac{r_2}{r_1}=\left(\frac{C_{\mathrm{A}_2}}{C_{\mathrm{A}_1}}\right)^2\left(\frac{\mathrm{C}_{\mathrm{B}_2}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\left(\frac{2 \mathrm{C}_{\mathrm{A}_1}}{\mathrm{C}_{\mathrm{A}_1}}\right)^2\left(\frac{2 \mathrm{C}_{\mathrm{B}_1}}{\mathrm{C}_{\mathrm{B}_1}}\right) \\ & \mathrm{r}_2=8 \mathrm{r}_1 \\ & x=8 \end{aligned}$
$\begin{aligned} & \Rightarrow \text { Zero order } \\ & y=0 \end{aligned}$
Consider the following reaction
$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$
The time taken for A to become $1 / 4^{\text {th }}$ of its initial concentration is twice the time taken to become $1 / 2$ of the same. Also, when the change of concentration of B is plotted against time, the resulting graph gives a straight line with a negative slope and a positive intercept on the concentration axis.
The overall order of the reaction is ________.
Explanation:
To determine the overall order of the reaction given by $\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$, we can derive the information based on the given details about the kinetics of reactant A and the graphical behavior of reactant B.
The first piece of information tells us that the time for the concentration of A to reduce to $1/4$ of its initial concentration ($[A]_0/4$) is twice the time it takes to reduce to half of its initial concentration ($[A]_0/2$). This characteristic is a hallmark of first-order reactions. In first-order reactions, the time it takes for the concentration of the reactant to reduce to half of its initial value (known as the half-life, $t_{1/2}$) is constant and does not depend on the initial concentration. Specifically, the fact that the concentration decreases by a factor of 4 (to $1/4$ its initial value) in twice the time it takes to decrease by a factor of 2 (to $1/2$ its initial value) indicates a constant half-life, consistent with first-order kinetics for reactant A.
As for reactant B, the information provided is that a plot of its concentration change over time is a straight line with a negative slope and a positive intercept on the concentration axis. This description matches the behavior of a reactant in a reaction of zero-order kinetics with respect to B, where the rate of reaction is constant and independent of the concentration of B. In zero-order reactions, the concentration of the reactant decreases linearly over time, since the rate of decrease is constant.
Therefore, considering the kinetics of both A and B:
- Reactant A shows first-order kinetics.
- Reactant B shows zero-order kinetics.
The overall order of the reaction is the sum of the individual orders with respect to each reactant, which gives us:
$\text{Overall order} = \text{Order with respect to A} + \text{Order with respect to B} = 1 (A) + 0 (B) = 1.$
Hence, the overall order of the reaction is 1.
Consider the two different first order reactions given below
$\begin{aligned} & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \text { (Reaction 1) } \\ & \mathrm{P} \rightarrow \mathrm{Q} \text { (Reaction 2) } \end{aligned}$
The ratio of the half life of Reaction 1 : Reaction 2 is $5: 2$ If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {rd }}$ and $4 / 5^{\text {th }}$ of Reaction 1 and Reaction 2 , respectively, then the value of the ratio $t_1: t_2$ is _________ $\times 10^{-1}$ (nearest integer). [Given : $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$]
Explanation:
$\mathrm{A+B \rightarrow C}$ Reaction .... (1)
$\mathrm{P} \rightarrow \mathrm{Q} \quad$ Reaction .... (2)
$\frac{\left(t_{\frac{1}{2}}\right)_1}{\left(t_{\frac{1}{2}}\right)_2}=\frac{k}{k_1}= \frac{5}{2}$
$\begin{aligned} & \frac{\frac{t_2}{3}}{t_{\frac{4}{5}}^5}=\frac{k_2}{k_1} \frac{\log 3}{\log 5} \\ & =\left(\frac{5}{2}\right) \frac{0.477}{0.699}=1.7 \\ & =17 \times 10^{-1} \\ & =x=17 \end{aligned}$
Time required for $99.9 \%$ completion of a first order reaction is _________ times the time required for completion of $90 \%$ reaction.(nearest integer)
Explanation:
To determine the time required for a certain level of completion ($x\%$) of a first-order reaction, we can use the formula that relates the time $ t $, the rate constant $ k $, and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ and the concentration at time $t$, $[A]_t$, is given by the integrated rate law for first-order reactions:
$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$
For a given percentage of completion, we use the fact that $[A]_t = [A]_0(1-\frac{x}{100})$, where $x$ is the percentage completion. Thus, the equation becomes:
$\ln\left(\frac{[A]_0}{[A]_0(1-\frac{x}{100})}\right) = kt$
Simplifying, we get:
$\ln\left(\frac{1}{1-\frac{x}{100}}\right) = kt$
Let's calculate the time required for $99.9\%$ and $90\%$:
- For $99.9\%$ completion ($x = 99.9\%$):
$\ln\left(\frac{1}{1-\frac{99.9}{100}}\right) = kt_{99.9}$
$\ln\left(\frac{1}{0.001}\right) = kt_{99.9}$
$\ln(1000) = kt_{99.9}$
- For $90\%$ completion ($x = 90\%$):
$\ln\left(\frac{1}{1-\frac{90}{100}}\right) = kt_{90}$
$\ln\left(\frac{1}{0.1}\right) = kt_{90}$
$\ln(10) = kt_{90}$
Since $k$ is a constant for a given reaction at a constant temperature, we can compare the times $t_{99.9}$ and $t_{90}$ directly by comparing the logarithmic values:
$\frac{t_{99.9}}{t_{90}} = \frac{\ln(1000)}{\ln(10)}$
Using the property of logarithms, we know that $\ln(1000) = 3\ln(10)$ because $1000 = 10^3$, so:
$\frac{t_{99.9}}{t_{90}} = \frac{3\ln(10)}{\ln(10)} = 3$
Therefore, the time required for $99.9\%$ completion of a first-order reaction is $3$ times the time required for completion of $90\%$ of the reaction. The nearest integer to this value is $3$.
Consider the following single step reaction in gas phase at constant temperature.
$2 \mathrm{~A}_{(\mathrm{g})}+\mathrm{B}_{(\mathrm{g})} \rightarrow \mathrm{C}_{(\mathrm{g})}$
The initial rate of the reaction is recorded as $\mathrm{r}_1$ when the reaction starts with $1.5 \mathrm{~atm}$ pressure of $\mathrm{A}$ and $0.7 \mathrm{~atm}$ pressure of B. After some time, the rate $r_2$ is recorded when the pressure of C becomes $0.5 \mathrm{~atm}$. The ratio $\mathrm{r}_1: \mathrm{r}_2$ is _________ $\times 10^{-1}$. (Nearest integer)
Explanation:
$2 \mathrm{~A}(\mathrm{~g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})$
As this is single step reaction,
$\mathrm{r}=\mathrm{k}_{\mathrm{f}}[\mathrm{A}]^2[\mathrm{~B}]^2$
$\begin{array}{lllll} & 2 \mathrm{~A}(\mathrm{~g})+ & \mathrm{B}(\mathrm{g}) \rightarrow & \mathrm{C}(\mathrm{g}) \\ \mathrm{t}=0 & 1.5 & 0.7 & 0 \\ \mathrm{t}=\mathrm{t}^{\prime} & 1.5-2 \mathrm{x} \quad & 0.7-\mathrm{x} & \mathrm{x} \end{array}$
$\begin{aligned} & \Rightarrow x=0.5 \mathrm{~atm} \\ & \mathrm{r}_1=\mathrm{k}_{\mathrm{f}}(1.5)^2(0.7) \\ & \mathrm{r}_2=\mathrm{k}_{\mathrm{f}}(0.5)^2(0.2) \\ & \frac{r_1}{r_2}=\frac{9 \times 7}{2}=31.5 \\ & =315 \times 10^{-1} \end{aligned}$
During Kinetic study of reaction $\mathrm{2 A+B \rightarrow C+D}$, the following results were obtained :
| $\mathrm{A [M]}$ | $\mathrm{B [M]}$ | initial rate of formation of $\mathrm{D}$ | |
|---|---|---|---|
| I | 0.1 | 0.1 | $6.0\times10^{-3}$ |
| II | 0.3 | 0.2 | $7.2\times10^{-2}$ |
| III | 0.3 | 0.4 | $2.88\times10^{-1}$ |
| IV | 0.4 | 0.1 | $2.40\times10^{-2}$ |
Based on above data, overall order of the reaction is _________.
Explanation:
$\begin{aligned} & \text { Rate }=k[A]^x[B]^y \\ & \frac{6 \times 10^{-}}{2.4 \times 10^{-}}=\left(\frac{0.1}{0.4}\right)^x \Rightarrow x=1 \\ & \frac{7.2 \times 10^{-}}{2.88 \times 10^{-}}=\left(\frac{0.2}{0.4}\right)^y \Rightarrow y=2 \end{aligned}$
Consider the following reaction, the rate expression of which is given below
$\begin{aligned} & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{C} \\ & \text { rate }=\mathrm{k}[\mathrm{A}]^{1 / 2}[\mathrm{~B}]^{1 / 2} \end{aligned}$
The reaction is initiated by taking $1 \mathrm{~M}$ concentration of $\mathrm{A}$ and $\mathrm{B}$ each. If the rate constant $(\mathrm{k})$ is $4.6 \times 10^{-2} \mathrm{~s}^{-1}$, then the time taken for $\mathrm{A}$ to become $0.1 \mathrm{~M}$ is _________ sec. (nearest integer)
Explanation:
$\begin{aligned} & A+B \rightarrow C \\ & \frac{-d[A]}{d t}=k[A]^{1 / 2}[B]^{1 / 2} \end{aligned}$
Since, $[A]=[B]$
$\begin{aligned} & \Rightarrow \quad \frac{-d[A]}{d t}=k[A] \\ & \Rightarrow \quad k t=\ln \frac{[A]_0}{[A]} \\ & \Rightarrow \quad t=\frac{1}{4.6 \times 10^{-2}} \times \ln \left(\frac{1}{0.1}\right) \\ & =\frac{2.303}{4.6} \times 100 \approx 50 \end{aligned}$
Consider the following transformation involving first order elementary reaction in each step at constant temperature as shown below.
Some details of the above reactions are listed below.
| Step | Rate constant (sec$^{-1}$) | Activation energy (kJ mol$^{-1}$) |
|---|---|---|
| 1 | $\mathrm{k_1}$ | 300 |
| 2 | $\mathrm{k_2}$ | 200 |
| 3 | $\mathrm{k_3}$ | $\mathrm{Ea_3}$ |
If the overall rate constant of the above transformation (k) is given as $\mathrm{k=\frac{k_1 k_2}{k_3}}$ and the overall activation energy $(\mathrm{E}_{\mathrm{a}})$ is $400 \mathrm{~kJ} \mathrm{~mol} \mathrm{~m}^{-1}$, then the value of $\mathrm{Ea}_3$ is ________ integer)
Explanation:
$\begin{aligned} & k=\frac{k_1 k_2}{k_3} \\ & E_{a_{e f f}}=E_{a_1}+E_{a_2}-E_{a_3} \end{aligned}$
$\begin{aligned} & 400=300+200-E_{\mathrm{a}_3} \\ & 400=500-E_{\mathrm{a}_3} \\ & E_{\mathrm{a}_3}=100 \mathrm{~kJ} \mathrm{~mole}^{-1} \end{aligned}$
$\mathrm{A}(\mathrm{g}) \rightarrow 2 \mathrm{~B}(\mathrm{~g})+\mathrm{C}(\mathrm{g})$
| S.No. | Time /s | Total pressure /(atm) |
|---|---|---|
| 1. | 0 | 0.1 |
| 2. | 115 | 0.28 |
The rate constant of the reaction is ________ $\times 10^{-2} \mathrm{~s}^{-1}$ (nearest integer)
Explanation:
$\begin{aligned} & 0.1+2 \mathrm{x}=0.28 \\\\ & 2 \mathrm{x}=0.18 \\\\ & \mathrm{x}=0.09 \\\\ & \mathrm{~K}=\frac{1}{115} \ln \frac{0.1}{0.1-0.09} \\\\ & =0.0200 \mathrm{sec}^{-1} \\\\ & =2 \times 10^{-2} \mathrm{sec}^{-1}\end{aligned}$