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Chemical Thermodynamics - Spontaneity

Grade 11CBSEChemistry

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

🔑Concepts

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A spontaneous process is an irreversible process that has a natural tendency to occur on its own or after proper initiation under a given set of conditions.

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Entropy (SS) is a thermodynamic state function that measures the degree of randomness or disorder in a system. For a spontaneous process in an isolated system, the change in entropy ΔS\Delta S is always positive.

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The Second Law of Thermodynamics states that the entropy of the universe (Stotal=Ssys+SsurrS_{total} = S_{sys} + S_{surr}) increases in all spontaneous processes.

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Gibbs Free Energy (GG) is defined as the maximum amount of energy available from a system that can be converted into useful work. It is given by G=H−TSG = H - TS.

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The Gibbs-Helmholtz equation, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S, is used to predict the spontaneity of a process at constant temperature and pressure.

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Criteria for Spontaneity: If ΔG<0\Delta G < 0 (negative), the process is spontaneous; if ΔG>0\Delta G > 0 (positive), the process is non-spontaneous; if ΔG=0\Delta G = 0, the system is at equilibrium.

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The Third Law of Thermodynamics states that the entropy of a perfectly crystalline substance is zero at absolute zero temperature (0 K0 \text{ K}).

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The standard Gibbs energy change of a reaction is related to the equilibrium constant (KK) by the equation ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K.

📐Formulae

ΔS=qrevT\Delta S = \frac{q_{rev}}{T}

ΔStotal=ΔSsystem+ΔSsurrounding>0 (for spontaneity)\Delta S_{total} = \Delta S_{system} + \Delta S_{surrounding} > 0 \text{ (for spontaneity)}

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

💡Examples

Problem 1:

For a certain reaction, ΔH=−110 kJ mol−1\Delta H = -110 \text{ kJ mol}^{-1} and ΔS=40 J K−1 mol−1\Delta S = 40 \text{ J K}^{-1} \text{ mol}^{-1} at 298 K298 \text{ K}. Predict whether the reaction is spontaneous at this temperature.

Solution:

Given: ΔH=−110 kJ mol−1\Delta H = -110 \text{ kJ mol}^{-1}, ΔS=40 J K−1 mol−1=0.040 kJ K−1 mol−1\Delta S = 40 \text{ J K}^{-1} \text{ mol}^{-1} = 0.040 \text{ kJ K}^{-1} \text{ mol}^{-1}, and T=298 KT = 298 \text{ K}. Using the Gibbs-Helmholtz equation: ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S ΔG=−110−(298×0.040)\Delta G = -110 - (298 \times 0.040) ΔG=−110−11.92\Delta G = -110 - 11.92 ΔG=−121.92 kJ mol−1\Delta G = -121.92 \text{ kJ mol}^{-1}

Explanation:

Since the calculated value of ΔG\Delta G is negative (−121.92 kJ mol−1-121.92 \text{ kJ mol}^{-1}), the reaction is spontaneous at 298 K298 \text{ K}.

Problem 2:

Calculate the temperature at which the reaction Ag2O(s)→2Ag(s)+12O2(g)Ag_2O(s) \rightarrow 2Ag(s) + \frac{1}{2}O_2(g) becomes spontaneous, given ΔH=30.56 kJ mol−1\Delta H = 30.56 \text{ kJ mol}^{-1} and ΔS=66 J K−1 mol−1\Delta S = 66 \text{ J K}^{-1} \text{ mol}^{-1}.

Solution:

At the point of transition between spontaneity and non-spontaneity, ΔG=0\Delta G = 0. Set ΔG=ΔH−TΔS=0\Delta G = \Delta H - T\Delta S = 0, which gives T=ΔHΔST = \frac{\Delta H}{\Delta S}. T=30.56×103 J mol−166 J K−1 mol−1T = \frac{30.56 \times 10^3 \text{ J mol}^{-1}}{66 \text{ J K}^{-1} \text{ mol}^{-1}} T≈463 KT \approx 463 \text{ K}

Explanation:

Since both ΔH\Delta H and ΔS\Delta S are positive, the reaction becomes spontaneous at temperatures higher than 463 K463 \text{ K} because the TΔST\Delta S term must outweigh the ΔH\Delta H term to make ΔG\Delta G negative.

Spontaneity - Revision Notes & Key Formulas | CBSE Class 11 Chemistry