Question

Calculate the standard free energy change for the reaction,
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)at298K.$
Given $△H^o=-92.4kJmo{l}^{-1}$ and $△S^o=-198.3J{K}^{-1}mo{l}^{-1}$

• A. -85.50 kJ/mol
• B. 33.31 kJ/mol
• C. -33.31 kJ/mol
• D. 85.50 kJ/mol

Answer 1

The correct option is C -33.31 kJ/mol

$△G^o=△H^o-T△S^o$
$=-92.4-298\times (-198.3)\times {10}^{-3}(\because △S^o=-198.3\times {10}^{-3}kJ{K}^{-1}mo{l}^{-1})$
$=-33.306kJ/mol$
Since $△G^o$ is negative, it means that a mixture of $H_2$ and $N_2$ at ${25}^{o}C,$ each present at a pressure of 1 atm would react spontaneously to form ammonia.