Question

The heat of reaction $\frac{1}{2}{H}_2\left(g\right)+\frac{1}{2}{Cl}_2\left(g\right)⟶HCl\left(g\right)$ at ${27}^{o}C$ is $-22kcal$. Calculate the heat of reaction at ${77}^{o}C$. The molar heat capacities at constant pressure at ${27}^{o}C$ for hydrogen, chlorine and $HCl$ are $6.82,7.70$ and $6.80$ $cal.{mol}^{-1}$ respectively (only magnitude in nearest integer in kcal).

Answer 1

$\Delta H_2-\Delta H_1=\Delta C_p(T_2-T_1)$

$\Delta H_1=-22kcal;$ $T_2-T_1=77-27=50K$

$\Delta C_p=C_p$ of products $-C_p$ of reactants

$\Rightarrow C_p=C_{p\left(HCl\right)}-(\frac{1}{2}{C}_{p\left(H_2\right)}+\frac{1}{2}{C}_{p\left(C{l}_2\right)})=6.80-(0.5\times 6.82+0.5\times 7.70)=-0.46cal$

$\Delta H_2=\Delta H_1+\Delta C_p(T_2-T_1)=-22-0.46\times 50\times {10}^{-3}=-22.023kcal$