Question

For a particular reaction at 135.4 °C, Δ=−775.41 kJ/mol, and Δ=817.91 J/(mol⋅K).

Calculate ΔG for this reaction at 12.7 °C.

Δ=

Answer 1

Answer:

$-675.053\ \text{kJ/mol}$

Explanation:

$\Delta G$ = Gibbs free energy = $-775.41\ \text{kJ/mol}$

$\Delta S$ = Change in entropy = $817.91\ \text{J/mol K}=0.81791\ \text{kJ/mol K}$

$T$ = Temperature = $135.4^{\circ}\text{C}=408.55\ \text{K}$

$\Delta H$ = Change in enthalpy

Gibbs free energy is given by

$\Delta G=\Delta H-T\Delta S\\\Rightarrow \Delta H=\Delta G+T\Delta S\\\Rightarrow \Delta H=-775.41+408.55\times 0.81791\\\Rightarrow \Delta H=-441.253\ \text{kJ/mol}$

$T=12.7^{\circ}\text{C}=285.85\ \text{K}$

$\Delta G=\Delta H-T\Delta S\\\Rightarrow \Delta G=-441.253-285.85\times 0.81791\\\Rightarrow \Delta G=-675.053\ \text{kJ/mol}$

The required Gibbs free energy is $-675.053\ \text{kJ/mol}$.