Question

The equilibrium constant $K_p$ for the reaction $A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$ is $1$ at ${27}^{o}C$ and $4$ at ${47}^{o}C$. Calculate enthalpy change for: $B\left(g\right)+C\left(g\right)\rightleftharpoons A\left(g\right)$ (Given $R=2cal/mol.K$).

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

Answer 1

Answer: (A)

$-13.31kcal/mol$

For $A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$

$log\left(4/1\right)=log4=\left(\frac{\Delta H}{2.303R}\right)\frac{(320-300)}{(320\times 300)};$

$\Rightarrow \Delta H=13.31kcal/mol$

So for reverse reaction:

$B\left(g\right)+C\left(g\right)\rightleftharpoons A\left(g\right)$

$\Delta H^{\prime }=-13.31kcal/mol$