Q52 / 200 · NEET 2004 · Chemistry
Standard enthalpy and standard entropy changes for the oxidation of ammonia at $298 \mathrm{~K}$ are $-382.64 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and -145.6 $\mathrm{JK}^{-1} \mathrm{~mol}^{-1}$, respectively. Standard Gibbs energy change for the same reaction at $298 \mathrm{~K}$ is:
- A$-221.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- Correct answerB$-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- C$-439.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
- D$-523.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Solution
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=-382.64-298$
$$
\left(\frac{-145.6}{1000}\right)
$$
$$
\begin{aligned}
& =-382.64+43.38 \\
& =-339.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
Related Theory
Free energy is an extensive quantity, but the standard free energy is an intensive quantity.
Hence, the correct answer is option: 'B'.