Question

What is $\Delta G^o$ for the following reaction?
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\rightleftharpoons {NH}_3\left(g\right);K_p=4.42\times {10}^4$ at ${25}^{o}C$.

• A. $-26.5kJ{mol}^{-1}$
• B. $-11.5kJ{mol}^{-1}$
• C. $-2.2kJ{mol}^{-1}$
• D. $-0.97kJ{mol}^{-1}$

Answer 1

Answer: (A)

$-26.5kJ{mol}^{-1}$

$\Delta G^o$ for the following reaction is $-26.5kJ{mol}^{-1}$
$\frac{1}{2}{N}_2\left(g\right)+\frac{3}{2}{H}_2\left(g\right)\rightleftharpoons {NH}_3\left(g\right);K_p=4.42\times {10}^4$ at ${25}^{o}C$
$\Delta G^o=-2.303RT\log K_p$
$\Delta G^o=-2.303\times 8.314J/mol/K\times (25+273)K\times \log 4.42\times {10}^4$
$\Delta G^o=-26506J{mol}^{-1}$
$\Delta G^o=-26506J{mol}^{-1}\times \frac{1kJ}{1000J}$
$\Delta G^o=-26.5kJ{mol}^{-1}$