Question

The following reaction has the thermodynamic values at 298 K: ΔH° = -136.9 kJ/mol and ΔS° = -120.6 J/mol K. Calculate ΔG° at 298 K for this reaction in kJ/mol (Enter your answer to four significant figures.):

Answer 1

`\Delta G^(0)` at 298 K is -101.0 kJ/mol

Step-by-step explanation:

According to thermodynamic of state,
`\Delta G^(0)=\Delta H^(0)-T\Delta S^(0)`

where,
`\Delta G^(0)`,
`\Delta H^(0)`, T and
`\Delta S^(0)` represent change in free energy in standard state, change in enthalpy in standard state, temperature in kelvin scale and change in entropy in standard state.

Here,
`\Delta H^(0)=-136.9kJ/mol`, T = 298 K and
`\Delta S^(0)=-0.1206kJ/(mol.K)`

So,
`\Delta G^(0)=(-136.9kJ/mol)-(298K* -0.1206(kJ)/(mol.K))`

= -101.0 kJ/mol

So,
`\Delta G^(0)` at 298 K is -101.0 kJ/mol