Question

The equilibrium constant $K_p$ for the following reaction is $1$ at ${27}^{o}C$ and $4$ at ${47}^{o}C$. $A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$
For the reaction calculate enthalpy change for the :
$B\left(g\right)+C\left(g\right)\rightleftharpoons A\left(g\right)$ (Given:$R=2calmo{l}^{-1}{K}^{-1}$)

• A. $-13.31kcal/mol$
• B. $+13.31kcal/mol$
• C. $-19.2kcal/mol$
• D. $+55.63kcal/mol$

Answer 1

Answer: (A)

$-13.31kcal/mol$

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

$\ln \frac{4}{1}=\frac{\Delta H}{2}[\frac{1}{300}-\frac{1}{320}]$

$\therefore \Delta H=13.31Kcal/mol$ for reaction $A\rightleftharpoons B+C$

Hence, for reaction $B+C\rightleftharpoons A,\Delta H=-\Delta H$

Or enthalpy for this reaction $=-13.31kcal/mol$