Question

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

• A. 67.21%
• B. 68.43%
• C. 70.78%
• D. 72.00%

Answer 1

The correct option is A 67.21%

For 25% decomposition,

$k=\frac{2.303}{20}log\frac{100}{100-25}=0.0144$

The arrehenius equation at two different temperatures is

$log\frac{k^{\prime }}{k}=\frac{E_a}{2.303R}\frac{T^{\prime }-T}{T{T}^{\prime }}$

Substitute values in the above expression.

$log\frac{k^{\prime }}{0.0144}=\frac{70000}{2.303\times 8.314}\frac{313-298}{298\ast 313}$

$k^{\prime }=0.0557$

The expression for the rate constant is

$0.0557=\frac{2.303}{20}log\frac{100}{100-x}$

$x=67.21$. This is the percentage dissociation.