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Difficulty: MediumEntropy, Free Energy and Reaction Spontaneity

An industrial synthesis reaction carried out at 27∘C27^\circ\text{C} has a standard enthalpy change (ΔH∘\Delta H^\circ) of −45.0 kJ mol−1-45.0\text{ kJ mol}^{-1} and a standard Gibbs free energy change (ΔG∘\Delta G^\circ) of −15.0 kJ mol−1-15.0\text{ kJ mol}^{-1}. What is the standard entropy change (ΔS∘\Delta S^\circ) for this reaction?

  1. A
    +100.0 J K−1 mol−1+100.0\text{ J K}^{-1}\text{ mol}^{-1}
  2. −100.0 J K−1 mol−1-100.0\text{ J K}^{-1}\text{ mol}^{-1}Answer
  3. C
    −1111.1 J K−1 mol−1-1111.1\text{ J K}^{-1}\text{ mol}^{-1}
  4. D
    −0.10 J K−1 mol−1-0.10\text{ J K}^{-1}\text{ mol}^{-1}

Answer

−100.0 J K−1 mol−1-100.0\text{ J K}^{-1}\text{ mol}^{-1}
Using the standard thermodynamic relation ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ, convert 27∘C27^\circ\text{C} to 300 K300\text{ K}. Solving for ΔS∘\Delta S^\circ yields −45.0−(−15.0)300=−0.100 kJ K−1 mol−1\frac{-45.0 - (-15.0)}{300} = -0.100\text{ kJ K}^{-1}\text{ mol}^{-1}, which equals −100.0 J K−1 mol−1-100.0\text{ J K}^{-1}\text{ mol}^{-1}.

Step-by-Step Solution

1
Convert temperature from Celsius to Kelvin
T=27+273.15=300 KT = 27 + 273.15 = 300\text{ K}
Thermodynamic calculations involving temperature require absolute temperature in Kelvin.
2
Rearrange the Gibbs free energy equation ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ to solve for ΔS∘\Delta S^\circ
ΔS∘=ΔH∘−ΔG∘T\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}
To isolate the standard entropy change variable.
3
Substitute given values and calculate ΔS∘\Delta S^\circ in kJ K−1 mol−1\text{kJ K}^{-1}\text{ mol}^{-1}
ΔS∘=−45.0 kJ mol−1−(−15.0 kJ mol−1)300 K=−30.0300=−0.100 kJ K−1 mol−1\Delta S^\circ = \frac{-45.0\text{ kJ mol}^{-1} - (-15.0\text{ kJ mol}^{-1})}{300\text{ K}} = \frac{-30.0}{300} = -0.100\text{ kJ K}^{-1}\text{ mol}^{-1}
Evaluate standard free energy and enthalpy difference divided by temperature.
4
Convert ΔS∘\Delta S^\circ to standard units of J K−1 mol−1\text{J K}^{-1}\text{ mol}^{-1}
ΔS∘=−0.100×1000=−100.0 J K−1 mol−1\Delta S^\circ = -0.100 \times 1000 = -100.0\text{ J K}^{-1}\text{ mol}^{-1}
Entropy changes are conventionally reported in joules per kelvin per mole.

Key Concept

The relationship between Gibbs free energy, enthalpy, absolute temperature, and entropy is governed by ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ.
Estimated Time:1m 30s
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Topics

Entropy, Free Energy and Reaction Spontaneity

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