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Chemical Thermodynamics - Gibbs Free Energy and Spontaneity

Grade 11ICSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Gibbs Free Energy (GG) is a thermodynamic state function defined as G=H−TSG = H - TS. It represents the maximum useful work obtainable from a system at constant temperature and pressure.

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The change in Gibbs Free Energy (ΔG\Delta G) is the criterion for spontaneity: If ΔG<0\Delta G < 0, the process is spontaneous; if ΔG>0\Delta G > 0, the process is non-spontaneous; if ΔG=0\Delta G = 0, the system is at equilibrium.

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The Gibbs-Helmholtz equation, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, shows that spontaneity depends on the enthalpy change (ΔH\Delta H), entropy change (ΔS\Delta S), and absolute temperature (TT).

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Standard Gibbs Free Energy change (ΔG∘\Delta G^\circ) is the change in free energy when reactants in their standard states are converted to products in their standard states at 1 atm1 \text{ atm} and a specified temperature (usually 298 K298 \text{ K}).

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A reaction is always spontaneous if ΔH\Delta H is negative (exothermic) and ΔS\Delta S is positive (increased randomness), as this ensures ΔG\Delta G is always negative.

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For reactions where ΔH\Delta H and ΔS\Delta S have the same sign, spontaneity is temperature-dependent. If both are positive, the reaction is spontaneous only at high temperatures (TΔS>ΔHT\Delta S > \Delta H).

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The relationship between standard free energy and the equilibrium constant is given by ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K, linking thermodynamics to chemical equilibrium.

📐Formulae

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

ΔG∘=∑ΔGf∘(products)−∑ΔGf∘(reactants)\Delta G^\circ = \sum \Delta G_f^\circ(\text{products}) - \sum \Delta G_f^\circ(\text{reactants})

ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K

ΔG∘=−2.303RTlog⁡10K\Delta G^\circ = -2.303 RT \log_{10} K

💡Examples

Problem 1:

For a particular reaction, ΔH=−10.5 kJ mol−1\Delta H = -10.5 \text{ kJ mol}^{-1} and ΔS=−44.1 J K−1 mol−1\Delta S = -44.1 \text{ J K}^{-1} \text{ mol}^{-1} at 298 K298 \text{ K}. Calculate ΔG\Delta G and predict if the reaction is spontaneous.

Solution:

Given: ΔH=−10.5×103 J mol−1\Delta H = -10.5 \times 10^3 \text{ J mol}^{-1}, ΔS=−44.1 J K−1 mol−1\Delta S = -44.1 \text{ J K}^{-1} \text{ mol}^{-1}, T=298 KT = 298 \text{ K}. Using the formula: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S ΔG=(−10500)−(298×−44.1)\Delta G = (-10500) - (298 \times -44.1) ΔG=−10500+13141.8\Delta G = -10500 + 13141.8 ΔG=+2641.8 J mol−1 or +2.64 kJ mol−1\Delta G = +2641.8 \text{ J mol}^{-1} \text{ or } +2.64 \text{ kJ mol}^{-1}

Explanation:

Since the calculated value of ΔG\Delta G is positive (+2.64 kJ mol−1+2.64 \text{ kJ mol}^{-1}), the reaction is non-spontaneous at 298 K298 \text{ K}.

Problem 2:

At what temperature will a reaction become spontaneous if ΔH=170 kJ mol−1\Delta H = 170 \text{ kJ mol}^{-1} and ΔS=170 J K−1 mol−1\Delta S = 170 \text{ J K}^{-1} \text{ mol}^{-1}?

Solution:

A reaction becomes spontaneous when ΔG<0\Delta G < 0. First, find the temperature at which ΔG=0\Delta G = 0 (equilibrium). 0=ΔH−TΔS  ⟹  T=ΔHΔS0 = \Delta H - T\Delta S \implies T = \frac{\Delta H}{\Delta S} T=170×103 J mol−1170 J K−1 mol−1T = \frac{170 \times 10^3 \text{ J mol}^{-1}}{170 \text{ J K}^{-1} \text{ mol}^{-1}} T=1000 KT = 1000 \text{ K}

Explanation:

Since both ΔH\Delta H and ΔS\Delta S are positive, the TΔST\Delta S term must be larger than ΔH\Delta H for ΔG\Delta G to be negative. Therefore, the reaction will be spontaneous at temperatures above 1000 K1000 \text{ K}.

Gibbs Free Energy and Spontaneity Class 11 Notes & Examples