Question

$K_c$ for $\frac{3}{2}{H}_2+\frac{1}{2}{N}_2\rightleftharpoons N{H}_3$ are 0.0266 and 0.0129 $at{m}^{-1}$ respectively at ${350}^{o}C$ and ${400}^{o}C$. Calculate heat of formation of $N{H}_3$.

• A. 12.140 Kcal
• B. 1.214 Kcal
• C. $-$12.140 Kcal
• D. $-$1.214 Kcal

Answer 1

Answer: (C)

$-$12.140 Kcal

The van't Hoff equation is $log\frac{K_2}{K_1}=+\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]=\frac{\Delta H}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$.
Here, $K_1$ and $K_2$ are the values of the equilibrium constants at temperatures $T_1$ and $T_2$ respectively.
$\Delta H$ is the enthalpy of the reaction and R is the ideal gas constant.
Substitute values in the above expression.
$2.303log\left(\frac{0.0129}{0.0266}\right)=\frac{\Delta H}{2}\left[\frac{673-623}{673\times 623}\right]\Delta H=-12140cal=-12.140kcal$.