Question

$K_C$ for $3/2H_2+1/2N_2\rightleftharpoons {NH}_3$ are 0.0266 and $0.0129{atm}^{-1}$ respectively, at 350${}^o$C and 400${}^o$C. Calculate the heat of formation of ${NH}_3$.

• A. $\therefore △H=-50462cal$
• B. $\therefore △H=-8133cal$
• C. $\therefore △H=12140cal$
• D. $\therefore △H=-12140cal$

Answer 1

The correct option is D $\therefore △H=-12140cal$

As we know,
$2.303log[\frac{K_{P_2}}{K_{P_1}}]=\frac{△H}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$2.303log[\frac{0.0129}{0.0266}]=\frac{△H}{2}\left[\frac{673-623}{673\times 623}\right]$
$\therefore △H=12140cal=-12140cal$