2004 AIPMT ThermodynamicsGibbs Energy Change Easy Question 66 of 82 in Thermodynamics

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 –339.3 kJ mol$^{-1}$
B –439.3 kJ mol$^{-1}$
C –523.2 kJ mol$^{-1}$
D –221.1 kJ mol$^{-1}$

The answer is A.

Solution

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
Convert $\Delta S^\circ$ to kJ: $-145.6$ J K$^{-1}$ mol$^{-1}$ $= -0.1456$ kJ K$^{-1}$ mol$^{-1}$
$T\Delta S^\circ = 298 \times (-0.1456) = -43.39$ kJ mol$^{-1}$
$\Delta G^\circ = -382.64 - (-43.39) = -382.64 + 43.39$
$\Delta G^\circ = -339.25 \approx -339.3$ kJ mol$^{-1}$

Concept Gibbs-Helmholtz equation

Thermodynamics in past papers

82 questions from this chapter have appeared across 26 exam years.