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Energetics / Thermochemistry - Entropy and spontaneity (HL only)

Grade 12IBChemistry

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

🔑Concepts

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Entropy (SS) is a thermodynamic function that measures the distribution of available energy among particles or the degree of disorder in a system. The units are J K−1 mol−1J\ K^{-1}\ mol^{-1}.

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A system becomes more disordered (positive entropy change, ΔS>0\Delta S > 0) when there is an increase in the number of moles of gas, a change of state from solid→liquid→gassolid \rightarrow liquid \rightarrow gas, or an increase in temperature.

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The Second Law of Thermodynamics states that for a process to be spontaneous, the total entropy of the universe must increase.

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Gibbs Free Energy (ΔG\Delta G) relates enthalpy and entropy to determine reaction spontaneity. For a reaction to be spontaneous at a given temperature, ΔG\Delta G must be negative (ΔG<0\Delta G < 0).

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The standard Gibbs free energy of formation (ΔGf⊖\Delta G_f^{\ominus}) is the free energy change when one mole of a compound is formed from its elements in their standard states. For elements in their standard states, ΔGf⊖=0\Delta G_f^{\ominus} = 0.

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Spontaneity depends on the balance between ΔH\Delta H and TΔST\Delta S. If ΔH\Delta H and ΔS\Delta S have the same sign, the spontaneity of the reaction will change as the temperature (TT) changes.

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Thermodynamic spontaneity does not guarantee a fast reaction. A reaction with ΔG<0\Delta G < 0 may be kinetically stable if it has a very high activation energy (EaE_a).

📐Formulae

ΔSreaction⊖=∑S⊖(products)−∑S⊖(reactants)\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}(products) - \sum S^{\ominus}(reactants)

ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}

ΔGreaction⊖=∑ΔGf⊖(products)−∑ΔGf⊖(reactants)\Delta G^{\ominus}_{reaction} = \sum \Delta G_f^{\ominus}(products) - \sum \Delta G_f^{\ominus}(reactants)

Ttransition=ΔH⊖ΔS⊖T_{transition} = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}

💡Examples

Problem 1:

For the reaction N2(g)+3H2(g)→2NH3(g)N_2(g) + 3H_2(g) \rightarrow 2NH_3(g), the enthalpy change ΔH⊖=−92.2 kJ mol−1\Delta H^{\ominus} = -92.2\ kJ\ mol^{-1} and the entropy change ΔS⊖=−198.7 J K−1 mol−1\Delta S^{\ominus} = -198.7\ J\ K^{-1}\ mol^{-1}. Calculate the Gibbs free energy change (ΔG⊖\Delta G^{\ominus}) at 298 K298\ K and determine if the reaction is spontaneous.

Solution:

  1. Convert units so they are consistent: ΔS⊖=−198.7÷1000=−0.1987 kJ K−1 mol−1\Delta S^{\ominus} = -198.7 \div 1000 = -0.1987\ kJ\ K^{-1}\ mol^{-1}.
  2. Use the formula ΔG⊖=ΔH⊖−TΔS⊖\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}.
  3. ΔG⊖=−92.2 kJ mol−1−(298 K×−0.1987 kJ K−1 mol−1)\Delta G^{\ominus} = -92.2\ kJ\ mol^{-1} - (298\ K \times -0.1987\ kJ\ K^{-1}\ mol^{-1}).
  4. ΔG⊖=−92.2+59.2=−33.0 kJ mol−1\Delta G^{\ominus} = -92.2 + 59.2 = -33.0\ kJ\ mol^{-1}.

Explanation:

Since ΔG⊖\Delta G^{\ominus} is negative (−33.0 kJ mol−1-33.0\ kJ\ mol^{-1}), the reaction is spontaneous at 298 K298\ K. Note that because ΔS\Delta S is negative, the reaction will become non-spontaneous at higher temperatures.

Problem 2:

A reaction has ΔH⊖=+178 kJ mol−1\Delta H^{\ominus} = +178\ kJ\ mol^{-1} and ΔS⊖=+160 J K−1 mol−1\Delta S^{\ominus} = +160\ J\ K^{-1}\ mol^{-1}. At what temperature (in KK) does the reaction become spontaneous?

Solution:

  1. The reaction becomes spontaneous when ΔG<0\Delta G < 0. Set ΔG=0\Delta G = 0 to find the threshold temperature: 0=ΔH−TΔS0 = \Delta H - T\Delta S.
  2. Rearrange for TT: T=ΔH⊖ΔS⊖T = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}.
  3. Convert ΔS\Delta S to kJkJ: +0.160 kJ K−1 mol−1+0.160\ kJ\ K^{-1}\ mol^{-1}.
  4. T=1780.160=1112.5 KT = \frac{178}{0.160} = 1112.5\ K.

Explanation:

At temperatures above 1112.5 K1112.5\ K, the TΔST\Delta S term (which is positive) will outweigh the positive ΔH\Delta H term, making ΔG\Delta G negative and the reaction spontaneous.