Question

First order reaction, ($A \to B$) requires activation of $70KJmo{l^{ - 1}}$ . When a $20\%$ solution of A was kept at $25^\circ C$ for $20$ minutes, $25\%$ decomposition took place. Assume that activation energy remains constant in this range of temperature. The percentage decomposition at the same time in a $30\%$ solution maintained at $40^\circ C$ will be
A. $67.2\%$
B. $71.4\%$
C. $69.3\%$
D. None of these

Answer 1

The first order reaction is the reaction which depends on the concentration of only one reactant. Activation energy is related to the rate of a reaction by Arrhenius equation.

Complete step by step answer:
The first order reaction of the conversion of $A$ to $B$ requires activation energy of $70KJmo{l^{ - 1}}$. The rate of the first order reaction is written as,
$rate = - \dfrac{{d[A]}}{{dt}} = k[A]$, where $\dfrac{{d[A]}}{{dt}}$, is the rate of disappearance of reactant $A$ and $k$ is the reaction rate coefficient. The activation energy for this conversion is $Ea = 70KJmo{l^{ - 1}}$ .
The integrated form of first order reaction is written as
$k = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$ where $a$ is the initial concentration of reactant $A$ and $a - x$ is the remaining amount of reactant $A$ after time $t$.
Let the initial concentration of reactant $A$ be $100$, i.e.$a = 100$. After $20$ minutes the amount of $A$ decomposed is $25\%$, i.e.$x = 25$ . Therefore $a - x = 100 - 25 = 75$ . Thus the rate of the reaction at this time is
${k_1} = \dfrac{{2.303}}{{20}}\log \dfrac{{100}}{{75}}$
${k_1} = 0.0144{\min ^{ - 1}}$.
The rate of reaction is related to the activation energy of a reaction by Arrhenius equation as,
$k = A{e^{\dfrac{{ - Ea}}{{RT}}}}$
The logarithmic form of the Arrhenius equation is
$\log k = \log A - \dfrac{{Ea}}{{2.303RT}}$
Thus for the two solutions the ratio of the rate of reactions,
$\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{Ea}}{{2.303R}}\left[ {\dfrac{1}{{{T_1}}} - \dfrac{1}{{{T_2}}}} \right]$
Given, $Ea = 70KJmo{l^{ - 1}} = 70000Jmo{l^{ - 1}}$ , $R = 8.314Jmo{l^{ - 1}}{K^{ - 1}}$ , ${T_1} = 25^\circ C = 298K$ ,${T_2} = 40^\circ C = 313K$.
Inserting the values,
$\log \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{{70000}}{{2.303 \times 8.314}}\left[ {\dfrac{1}{{298}} - \dfrac{1}{{313}}} \right]$
$\dfrac{{{k_2}}}{{{k_1}}} = 3.874$
But ${k_1} = 0.0144{\min ^{ - 1}}$,
Thus $\dfrac{{{k_2}}}{{0.0144}} = 3.874$
${k_2} = 0.0558{\min ^{ - 1}}$
Using the value of rate constant for the reaction at $40^\circ C$ , the percentage of decomposition is calculated as
${k_2} = \dfrac{{2.303}}{t}\log \dfrac{a}{{a - x}}$
Here $t = 20min$, $a = 100$ and $x = ?$
Thus, $0.0558 = \dfrac{{2.303}}{{20}}\log \dfrac{{100}}{{100 - x}}$
$\log \dfrac{{100}}{{100 - x}} = 0.485$
$\dfrac{{100}}{{100 - x}} = 3.052$
$x = 67.23.$
Hence option A is the correct answer.

Note:
The rate of reaction depends on the temperature and the activation energy of a reaction. As the activation energy is the same in this case so it solely depends on the temperature and the percentage of decomposition.