Question

The enthalpy change (ΔH) and entropy change(ΔS) for the reaction of 30.54 kj mol-¹ and 0.06 kj mol-¹ respectively. Calculate the temperature at equilibrium. Also predict the spontaneity below the temperature at which Gibb's free energy is zero. Justify your answer with a valid reason.

Answer 1

Answer on Question #73551, Chemistry / General Chemistry :

The enthalpy change $(\Delta H)$ and entropy change $(\Delta S)$ for the reaction of $30.54~\mathrm{kJ~mol^{-1}}$ and $0.06~\mathrm{kJ~mol^{-1}}$ respectively. Calculate the temperature at equilibrium. Also predict the spontaneity below the temperature at which Gibb's free energy is zero. Justify your answer with a valid reason.

Solution.

$$\begin{array}{l} \Delta H = 30.54~\mathrm{kJ/mol} \\ \Delta S = 0.06~\mathrm{kJ/mol} \cdot K \\ \Delta G = 0 \\ T - ? \end{array}$$

Gibb's free energy is:

$$\Delta G = \Delta H - T \cdot \Delta S$$

When $\Delta G = 0$, and:

$$\begin{array}{l} 0 = \Delta H - T \cdot \Delta S \\ T \cdot \Delta S = \Delta H \\ T = \frac{\Delta H}{\Delta S} = \frac{30.54~\mathrm{kJ/mol}}{0.06~\mathrm{kJ/mol} \cdot K} \\ T = 509~\mathrm{K} \\ \end{array}$$

If $T > 509~\mathrm{K}$ :

$$\begin{array}{l} \Delta H - T \cdot \Delta S < 0 \\ \Delta G < 0 \\ \end{array}$$

Reaction with a positive Gibbs free energy will not proceed spontaneously.

When $\Delta G < 0$, the process is exergonic and will proceed spontaneously in the forward direction to form more products.

Answer: $T = 509~\mathrm{K}$

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