Question

What is $\Delta G^o$ for this reaction?
$\frac{1}{2}{N}_{2\left(g\right)}+\frac{3}{2}{H}_{2\left(g\right)}⟶{NH}_{3\left(g\right)};K_p=4.42\times {10}^4$

• A. $-26.5kJ{mol}^{-1}$
• B. $-1.5kJ{mol}^{-1}$
• C. $-2.2kJ{mol}^{-1}$
• D. $-0.97kJ{mol}^{-1}$

Answer 1

The correct option is A $-26.5kJ{mol}^{-1}$

$\Delta G^o=-2.303RT\log K_p$

$=-2.303\times 8.314\times 298\log (4.42\times {10}^4)$

$=-26.5kJ$

Hence, option A is correct.