Question

A first order reaction, $A\to B$; requires activation energy of 70 kJ $mo{l}^{-1}$. When a 20% solution of A was kept at ${25}^{\circ }C$ for 20 min, 25% decomposition took place. What will be the percentage decomposition in the same time a in a 30% solution maintained at ${40}^{\circ }C$ ? Assume that activation energy remains constant in this range of temperature.

• A. 33%
• B. 80%
• C. 20%
• D. 67%

Answer 1

The correct option is D 67%

Applying Arrhenius equation,
$k=A{e}^{-E_a}/RT$

$logk=logA-$ $\frac{E_a}{2.303RT}$

$\therefore log\frac{k_2}{k_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$=\frac{70\times 1000}{2.303\times 8.314}[\frac{1}{298}-\frac{1}{303}]$

$\Rightarrow \frac{k_2}{k_1}=3.872$

For a first order reaction,

$k_1=\frac{2.303}{t}log\frac{a}{a-x}$,

$\left[\begin{array}{cc}a=100\%& x=25\%\\ \therefore a-x=75\%& \\ t=20min& \end{array}\right]=\frac{2.303}{20}log\frac{100}{(100-25)}$

$=0.014386$ $mi{n}^{-1}$

Calculation of % decomposition at ${40}^{\circ }C$

From equation (i),

$\frac{k_2}{0.014386}=3.872$

$k_2=0.05571mi{n}^{-1}$

Again, for a first order reaction,

$k_2=\frac{2.303}{t}log\frac{a}{a-x}$

$0.05571=$$\frac{2.303}{20}log\frac{100}{100-x}$

$x=67.17$

At $313K,67.17$ % decomposition takes place