At a certain temperature, T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

Related Formula

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