Question

Standard enthalpy and standard entropy for the oxidation of $N{H}_3$ at $298$ K are $-382.64$ $\text{kJ}{\text{mol}}^{-1}$ and $-145.6$ $\text{J}{\text{mol}}^{-1}{\text{K}}^{-1}$ respectively. Standard free energy change (in ${\text{kJ mol}}^{-1}$) for the same reaction at $298$ K is:

• A. $-221.1$
• B. $-339.3$
• C. $-439.3$
• D. $-523.2$

Answer 1

The correct option is B $-339.3$

We know the relation:

$\Delta G=\Delta H-T\Delta S$
$\Delta H=-382.64$ kJ mol${}^{-1}$ (given)
$\Delta S=-145.6$ J mol${}^{-1}$ K${}^{-1}$ (given)
$\Delta G=-382.64+298\times 145.6\times {10}^{-3}$ kJ mol${}^{-1}$ $=-339.25$ kJ mol${}^{-1}$
$\Delta G\simeq -339.3$ kJ mol${}^{-1}$

Hence, the free energy change for the reaction at $298$ K is $-339.3$ kJ mol${}^{-1}$.

Option B is correct.