Question

The reaction for $N_2+3H_2\to 2{NH}_3$ at ${27}^{o}C$ is $-91.94kJ$. What will be its value at ${50}^{o}C$? (The molar heat capacities at constant $P$ and ${27}^{o}C$ for $N_2,H_2$ and ${NH}_3$ are $28.45,28.32$ and $37.07kJ$ respectively)

• A. $-92.843kJ$
• B. $-47.7723kJ$
• C. $-132.5kJ$
• D. $-176,11kJ$

Answer 1

The correct option is A $-92.843kJ$

$\Delta H_{323K}-\Delta H_{300K}=C_p{\int }_{300}^{323}dT$

$C_{p_{reaction}}=2\times 37.07-3\times 28.32-28.45=-39.37$ $J/K$

$\Delta H_{323}+91.94=\frac{-39.37\times (323-300)}{1000}$

$⟹\Delta H=-91.94-0.9055$

$⟹\Delta H=-92.843$ $KJ$