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Zorluk: ZorEntropy, Free Energy and Reaction Spontaneity

For a particular gasification process, a chemical reaction has a standard enthalpy change (ΔH∘\Delta H^\circ) of +136.5 kJ mol−1+136.5\text{ kJ mol}^{-1} and a standard entropy change (ΔS∘\Delta S^\circ) of +325.0 J K−1 mol−1+325.0\text{ J K}^{-1}\text{ mol}^{-1}. What is the minimum temperature, in degrees Celsius (∘C^\circ\text{C}), above which the reaction becomes thermodynamically spontaneous?

Cevap: 147 °C

Cevap

The minimum temperature above which the reaction becomes spontaneous is 147 °C.
At the boundary of spontaneity, ΔG∘=0\Delta G^\circ = 0. Substituting this into ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ yields T=ΔH∘ΔS∘T = \frac{\Delta H^\circ}{\Delta S^\circ}. Converting ΔH∘\Delta H^\circ to Joules yields +136,500 J mol−1+136,500\text{ J mol}^{-1}. Dividing by +325.0 J K−1 mol−1+325.0\text{ J K}^{-1}\text{ mol}^{-1} gives T=420 KT = 420\text{ K}. Converting to Celsius gives 420−273=147∘C420 - 273 = 147^\circ\text{C}.

Adım Adım Çözüm

1
Convert enthalpy change from kilojoules per mole to joules per mole
ΔH∘=+136.5 kJ mol−1=+136,500 J mol−1\Delta H^\circ = +136.5\text{ kJ mol}^{-1} = +136,500\text{ J mol}^{-1}
ΔH∘\Delta H^\circ must be expressed in Joules to match the unit of ΔS∘\Delta S^\circ (325.0 J K−1 mol−1325.0\text{ J K}^{-1}\text{ mol}^{-1}).
2
Determine the threshold condition for reaction spontaneity using Gibbs free energy equation
ΔG∘=0  ⟹  T=ΔH∘ΔS∘\Delta G^\circ = 0 \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}
A reaction is spontaneous when ΔG∘<0\Delta G^\circ < 0. The minimum temperature where spontaneity begins occurs when ΔG∘=0\Delta G^\circ = 0.
3
Calculate the absolute temperature in Kelvin
T=136,500 J mol−1325.0 J K−1 mol−1=420 KT = \frac{136,500\text{ J mol}^{-1}}{325.0\text{ J K}^{-1}\text{ mol}^{-1}} = 420\text{ K}
Dividing the enthalpy change in Joules by the entropy change in Joules per Kelvin yields the temperature in Kelvin.
4
Convert temperature from Kelvin to degrees Celsius
T(∘C)=420−273=147∘CT(^\circ\text{C}) = 420 - 273 = 147^\circ\text{C}
The question explicitly requests the answer in degrees Celsius (T(∘C)=T(K)−273T(^\circ\text{C}) = T(\text{K}) - 273).

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Entropy, Free Energy and Reaction Spontaneity