Question

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

• A. $-523.2\, kJmol^{-1}$
• B. $-221.1\, Mmol^{-1}$
• C. $-339.3 \,kJmol^{-1}$
• D. $-439.3\, kJmol^{-1}$

Answer 1

Answer: (C)

$NH_3(g) + \frac{3}{4} O_2(g) \rightarrow \frac{1}{2} N_2(g) + \frac{3}{2} H_2O (l)$
$T \,=\, 298\, K$
$\Delta \,H - 382.64 \,kJ \,mol^{-1}$
$\Delta \, S = - 145.6 \,JK^{-1} \, mol^{-1}$
= $- 0.1456\, kJ\, K^{-1} \, mol^{-1}$
$\Delta G = \Delta H - T\Delta S$
= $- 382.64\, kJ \,mol^{-1} - (298\, K)$ $\times (0.1456\, K \,JK^{-1}\, mol^{-1}$
= $- 382.64\, kJ \,mol^{-1} + 43.3888 kJ \,mol^{-1}$
= $-339.25\, kJ \,mol^{-1}$