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Zorluk: OrtaEntropy, Free Energy and Reaction Spontaneity

A chemical reaction has an enthalpy change (ΔH\Delta H) of +40.0 kJ mol−1+40.0\text{ kJ mol}^{-1} and an entropy change (ΔS\Delta S) of +100 J K−1 mol−1+100\text{ J K}^{-1}\text{ mol}^{-1}. Above what minimum temperature, in degrees Celsius (∘C^\circ\text{C}), will the reaction become spontaneous?

  1. 127∘C127^\circ\text{C}Cevap
  2. B
    400∘C400^\circ\text{C}
  3. C
    673∘C673^\circ\text{C}
  4. D
    −272.6∘C-272.6^\circ\text{C}

Cevap

The minimum temperature above which the reaction becomes spontaneous is 127∘C127^\circ\text{C}.
According to the Gibbs free energy relationship ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, a reaction is spontaneous when ΔG<0\Delta G < 0. At the transition temperature between spontaneous and non-spontaneous states, ΔG=0\Delta G = 0. Substituting ΔH=40,000 J mol−1\Delta H = 40,000\text{ J mol}^{-1} and ΔS=100 J K−1 mol−1\Delta S = 100\text{ J K}^{-1}\text{ mol}^{-1} gives T=40000100=400 KT = \frac{40000}{100} = 400\text{ K}. Converting to degrees Celsius by subtracting 273273 yields 127∘C127^\circ\text{C}. Therefore, above 127∘C127^\circ\text{C}, the reaction becomes spontaneous.

Adım Adım Çözüm

1
Convert the enthalpy change from kilojoules to joules to ensure consistent units with entropy change.
ΔH=+40.0 kJ mol−1=+40,000 J mol−1\Delta H = +40.0\text{ kJ mol}^{-1} = +40,000\text{ J mol}^{-1}
ΔS\Delta S is given in J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}, so ΔH\Delta H must be expressed in Joules.
2
Determine the threshold temperature (TT) at equilibrium where ΔG=0\Delta G = 0 using the Gibbs free energy equation ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S.
0=ΔH−TΔS  ⟹  T=ΔHΔS=40000 J mol−1100 J K−1 mol−1=400 K0 = \Delta H - T\Delta S \implies T = \frac{\Delta H}{\Delta S} = \frac{40000\text{ J mol}^{-1}}{100\text{ J K}^{-1}\text{ mol}^{-1}} = 400\text{ K}
A reaction is spontaneous when ΔG<0\Delta G < 0, which occurs when temperature exceeds the threshold temperature T=ΔHΔST = \frac{\Delta H}{\Delta S} for endothermic reactions with positive entropy change.
3
Convert the temperature from Kelvin (K\text{K}) to degrees Celsius (∘C^\circ\text{C}).
T(∘C)=400 K−273=127∘CT(^\circ\text{C}) = 400\text{ K} - 273 = 127^\circ\text{C}
The question specifically requests the temperature in degrees Celsius.

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