Question

The standard heat of formation listed for gaseous ${NH}_3$ is $11.02kcal/mol$ at $298K$. Given that $298K$ the constant pressure heat capacities of gaseous $N_2,H_2$ and ${NH}_3$ respectively $6.96,6.89,8.38$ $cal/mol$. Determine $\Delta H_{298K}^o$ and $\Delta H_{773K}$ for the reaction,
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)⟶{NH}_3\left(g\right)$

Answer 1

$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)⟶{NH}_3\left(g\right)$
$\Delta H_{298K}=\sum {\left(H_f\right)}_P-\sum {\left(H_f\right)}_R=(-11.02-0)=-11.02kcal{mol}^{-1}$
$\frac{\Delta H_2-\Delta H_1}{T_2-T_1}=\Delta C_P$
$\frac{\Delta H_2-(-11.02)}{773-298}=(8.38-\frac{1}{2}\times 6.96-\frac{3}{2}\times 6.89)\times {10}^{-3}$
$H_2=-13.6kcal{mol}^{-1}$