Question

For the reaction at 300 K, A(g)+B(g) → C(g). $\Delta$U = -3 kcal; $\Delta$S = 10 cal/K

The value of $\Delta$G is: [Take: R = 2 cal mol$^{-1}$ K$^{-1}$]

Answer 1

-6.6 kcal

Answer 2

The Gibbs Free Energy change ($\Delta$G) is related to enthalpy change ($\Delta$H), temperature (T), and entropy change ($\Delta$S) by the Gibbs-Helmholtz equation: $\Delta G = \Delta H - T\Delta S$ The enthalpy change ($\Delta$H) can be calculated from the internal energy change ($\Delta$U) using the relation: $\Delta H = \Delta U + \Delta n_g RT$ where $\Delta$n$_g$ is the change in the number of moles of gas in the reaction.

For the given reaction A(g) + B(g) → C(g): The change in the number of moles of gas is: $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$ $\Delta n_g = 1 - (1 + 1) = 1 - 2 = -1$

Given values: Temperature, $T = 300$ K Internal energy change, $\Delta U = -3$ kcal Entropy change, $\Delta S = 10$ cal/K Gas constant, $R = 2$ cal mol$^{-1}$ K$^{-1}$

To maintain consistency in units, we convert all values to kilocalories (kcal): $\Delta U = -3$ kcal $\Delta S = 10 \text{ cal/K} = \frac{10}{1000} \text{ kcal/K} = 0.01 \text{ kcal/K}$ $R = 2 \text{ cal mol}^{-1} \text{ K}^{-1} = \frac{2}{1000} \text{ kcal mol}^{-1} \text{ K}^{-1} = 0.002 \text{ kcal mol}^{-1} \text{ K}^{-1}$

Now, calculate the enthalpy change ($\Delta$H): $\Delta H = \Delta U + \Delta n_g RT$ $\Delta H = -3 \text{ kcal} + (-1) \times (0.002 \text{ kcal mol}^{-1} \text{ K}^{-1}) \times (300 \text{ K})$ $\Delta H = -3 \text{ kcal} + (-0.6 \text{ kcal})$ $\Delta H = -3.6 \text{ kcal}$

Next, calculate the $T\Delta S$ term: $T\Delta S = (300 \text{ K}) \times (0.01 \text{ kcal/K})$ $T\Delta S = 3 \text{ kcal}$

Finally, calculate the Gibbs Free Energy change ($\Delta$G): $\Delta G = \Delta H - T\Delta S$ $\Delta G = -3.6 \text{ kcal} - 3 \text{ kcal}$ $\Delta G = -6.6 \text{ kcal}$