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

Hint: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.