Question

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

Calculate :
(i) equilibrium constants at $300K$ and $400K$ and
(ii) heat of reaction.

• A. $0.2,0.04,-3.86kcal$
• B. $5,25,-3.86kcal$
• C. $0.2,25,11.56kcal$
• D. $5,25,3.86kcal$

Answer 1

The correct option is D $5,25,3.86kcal$

At $300K$ : $K_f=2\times {10}^{-2};K_b=4\times {10}^{-3}$

$\therefore K_{c_1}=\frac{K_f}{K_b}=\frac{2\times {10}^{-2}}{4\times {10}^{-3}}=5$

At $400K$ : $K_f=4\times {10}^{-2};K_b=16\times {10}^{-4}$

$\therefore K_{c_2}=\frac{K_f}{K_b}=\frac{4\times {10}^{-2}}{16\times {10}^{-4}}=25$

Now, $2.303\log \frac{K_{c_2}}{K_{c_1}}=\frac{\Delta H}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

$\therefore 2.303\log \frac{25}{5}=\frac{\Delta H}{2}\left[\frac{400-300}{400\times 300}\right]$

$\therefore \Delta H=3863.3cal=3.86kcal$