Related Formula
Δ H = Δ U + Δ ng RT$$\Delta H^{\ominus} = \Delta U^{\ominus} + \Delta n_g RT$$
Δ G = Δ H - T Δ S$$\Delta G^{\ominus} = \Delta H^{\ominus} - T \Delta S^{\ominus}$$
Core Logic
First, calculate Δ ng$\Delta n_g$ (change in gaseous moles):
Δ ng = Σ ng(products) - Σ ng(reactants) = 2 - (2 + 1) = -1$\Delta n_g = \sum n_{g\text{(products)}} - \sum n_{g\text{(reactants)}} = 2 - (2 + 1) = -1$
Calculate Δ H$\Delta H^{\ominus}$:
Δ U = -10 kJ/mol = -10000 J/mol$\Delta U^{\ominus} = -10 \text{ kJ/mol} = -10000 \text{ J/mol}$
R = 8.31 J/(mol K)$R = 8.31 \text{ J/(mol K)}$
T = 298 K$T = 298 \text{ K}$
Δ H = -10000 + (-1)(8.31)(298)$\Delta H^{\ominus} = -10000 + (-1)(8.31)(298)$
Δ H = -10000 - 2476.38 = -12476.38 J/mol$\Delta H^{\ominus} = -10000 - 2476.38 = -12476.38 \text{ J/mol}$
Step 1: Calculate Delta G
Δ S = -44 J/(K mol)$\Delta S^{\ominus} = -44 \text{ J/(K mol)}$
Δ G = -12476.38 - (298)(-44)$\Delta G^{\ominus} = -12476.38 - (298)(-44)$
Δ G = -12476.38 + 13112$\Delta G^{\ominus} = -12476.38 + 13112$
Δ G = +635.62 J/mol = +0.63562 kJ/mol$\Delta G^{\ominus} = +635.62 \text{ J/mol} = +0.63562 \text{ kJ/mol}$
Step 2: Determine Spontaneity
Since Δ G$\Delta G^{\ominus}$ is positive (>0$>0$), the reaction is non-spontaneous at 298 K.
Pattern Recognition
Always convert Δ U$\Delta U$ from kJ to J before adding the RT$RT$ term (which is in Joules), or convert R$R$ to kJ. A positive Δ G$\Delta G$ invariably signifies a non-spontaneous process.
Chapter Mix
Class 11 Chemistry: Thermodynamics
