Question

The rate of disappearance of $A$ at two temperature is given by: $A\rightleftharpoons B$
I. $-\frac{d\left[A\right]}{dt}=2\times {10}^{-2}\left[A\right]-4\times {10}^{-3}\left[B\right]$ at $300K$
II. $-\frac{d\left[A\right]}{dt}=4\times {10}^{-2}\left[A\right]-16\times {10}^{-4}\left[B\right]$ at $400K$
The heat of reaction for the change is :

• A. $-3.863kcal$
• B. $6.93kcal$
• C. $1.68kcal$
• D. $1.68\times {10}^{-2}kcal$

Answer 1

The correct option is A $-3.863kcal$

$A\stackrel{K_1}{\underset{K_2}{\leftrightharpoons }}B$

$-\frac{dA}{dt}=K_1\left[A\right]-K_2\left[B\right]$

$K_{eq}$ at $300$ $K=\frac{K_1}{K_2}=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}=5$

$K_{eq}$ at $400$ $K=\frac{4\times {10}^{-2}}{16\times {10}^{-4}}=25$

$\ln \frac{K_1^{\prime }}{K_2^{\prime }}=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\therefore \ln \frac{5}{25}=\frac{\Delta H}{2}[\frac{1}{300}-\frac{1}{400}]$

$\therefore \Delta H=-3.863$ $kcal$