For a chemical reaction carried out at 27^\circ\text{C}, the standard Gibbs free energy ch… — JAMB UTME

For a chemical reaction carried out at $27^\circ\text{C}$ , the standard Gibbs free energy change ( $\Delta G^\circ$ ) is $-54.0\text{ kJ mol}^{-1}$ . Given that the standard entropy change ( $\Delta S^\circ$ ) for the reaction is $-120.0\text{ J K}^{-1}\text{ mol}^{-1}$ , calculate the standard enthalpy change ( $\Delta H^\circ$ ) in $\text{kJ mol}^{-1}$ .

Answer: -90 kJ mol^-1

Answer

The standard enthalpy change ($\Delta H^\circ$) for the reaction is $-90.0\text{ kJ mol}^{-1}$.

The standard enthalpy change is calculated using the rearranged Gibbs free energy equation

\Delta H^\circ = \Delta G^\circ + T\Delta S^\circ

. Converting

27^\circ\text{C}

300\text{ K}

and

-120.0\text{ J K}^{-1}\text{ mol}^{-1}

-0.120\text{ kJ K}^{-1}\text{ mol}^{-1}

yields

\Delta H^\circ = -54.0 + (300 \times -0.120) = -90.0\text{ kJ mol}^{-1}

Step-by-Step Solution

Convert the temperature from degrees Celsius to Kelvin

T = 300 K

Thermodynamic calculations require absolute temperature in Kelvin: T = 27 + 273 = 300 K.

Convert the standard entropy change units from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹

ΔS° = -0.120 kJ K⁻¹ mol⁻¹

Since ΔG° is given in kJ mol⁻¹, ΔS° must be converted to kJ K⁻¹ mol⁻¹ by dividing by 1000.

Rearrange the Gibbs free energy equation to express ΔH° and substitute the given values

ΔH° = -90.0 kJ mol⁻¹

From ΔG° = ΔH° - TΔS°, rearranging gives ΔH° = ΔG° + TΔS° = -54.0 + (300 × -0.120) = -90.0 kJ mol⁻¹.

Key Concept

Gibbs Free Energy Equation and Unit Consistency

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