Question

Standard enthalpy and standard entropy changes for the oxidation of ammonia at $298 \,K$ are $-382.64 \,kJ\, mol ^{-1}$ and $-145.6\, J K ^{-1} mol ^{-1}$, respectively. Standard Gibbs energy change for the same reaction at $298 \,K$ is

• A. $ -2221.1kJ\,mo{{l}^{-1}} $
• B. $ -339.3\,kJ\,mo{{l}^{-1}} $
• C. $ -439.3\,kJ\,mo{{l}^{-1}} $
• D. $ -523.2\,kJ\,mo{{l}^{-1}} $

Answer 1

Answer: (B)

For the oxidation of ammonia at $298 \,K$ the standard enthalpy change $(\Delta H)$ and standard entropy change $(\Delta S)$ are $-382.64 \,kJ\, mol ^{-1}$ and $-145.6\, JK ^{-1} mol ^{-1}$ respectively. The relationship of Gibbs free energy $(\Delta G)$ with $\Delta H$ and $\Delta S$ is represented in form of following equation. $\Delta G=\Delta H-T \cdot \Delta S$
Given that
$\Delta H =-382.64\, kJ \,mol ^{-1} $
$\Delta S =-145.6 \,J K ^{-1} mol ^{-1} $
$=-145.6 \times 10^{-3} kJ K ^{-1}$
or $ \Delta G =-382.64-\left(298 \times-145.6 \times 10^{-3}\right) $
$=-339.3 \,kJ \,mol ^{-1}$