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

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

Step-by-step 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* 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* 0.81791\\\Rightarrow \Delta G=-675.053\ \text{kJ/mol}`

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