For the thermal decomposition of a compound, the standard enthalpy change (\Delta H^\circ)… — JAMB UTME

For the thermal decomposition of a compound, the standard enthalpy change ( $\Delta H^\circ$ ) is $+117.0\text{ kJ mol}^{-1}$ and the standard entropy change ( $\Delta S^\circ$ ) is $+180.0\text{ J K}^{-1}\text{ mol}^{-1}$ . What is the minimum temperature, in kelvin ( $\text{K}$ ), above which the reaction becomes spontaneous?

Answer: 650 K

Answer

The minimum temperature above which the reaction becomes spontaneous is 650 K.

For a reaction with positive

\Delta H^\circ

and positive

\Delta S^\circ

, spontaneity depends on temperature. Spontaneity occurs when

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ < 0

, which simplifies to

T > \frac{\Delta H^\circ}{\Delta S^\circ}

. Converting

\Delta H^\circ

to joules gives

117000\text{ J mol}^{-1}

, so

T = \frac{117000}{180.0} = 650\text{ K}

Step-by-Step Solution

Convert enthalpy change units from kilojoules per mole to joules per mole

\Delta H^\circ = 117.0\text{ kJ mol}^{-1} \times 1000\text{ J/kJ} = 117000\text{ J mol}^{-1}

Enthalpy and entropy must be in consistent energy units (joules) before performing thermodynamic calculations.

Apply the condition for reaction spontaneity threshold

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0

A reaction is spontaneous when

\Delta G^\circ < 0

. The threshold temperature occurs exactly when

\Delta G^\circ = 0

Calculate the threshold temperature T

T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{117000\text{ J mol}^{-1}}{180.0\text{ J K}^{-1}\text{ mol}^{-1}} = 650\text{ K}

Solving the threshold condition for

T

yields the absolute temperature in kelvin above which

T\Delta S^\circ > \Delta H^\circ

, making

\Delta G^\circ

negative.

Key Concept

Relationship between Gibbs free energy, enthalpy, entropy, and reaction spontaneity threshold

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