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Chemical Thermodynamics - Entropy and Second Law of Thermodynamics

Grade 11CBSEChemistry

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

🔑Concepts

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Entropy (SS) is a measure of the degree of randomness or disorder of a system. It is a state function and an extensive property.

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The Second Law of Thermodynamics states that the entropy of the universe (DeltaStotal\\Delta S_{total}) always increases during a spontaneous process: DeltaStotal=DeltaSsystem+DeltaSsurrounding>0\\Delta S_{total} = \\Delta S_{system} + \\Delta S_{surrounding} > 0.

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Gibbs Free Energy (GG) is a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. It is defined as G=H−TSG = H - TS.

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The spontaneity of a reaction at constant temperature and pressure is determined by the Gibbs Free Energy change (DeltaG\\Delta G): If DeltaG<0\\Delta G < 0, the process is spontaneous; if DeltaG>0\\Delta G > 0, the process is non-spontaneous; if DeltaG=0\\Delta G = 0, the process is at equilibrium.

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Standard Gibbs Energy Change (DeltaG∘\\Delta G^\circ) is related to the equilibrium constant (KK) by the equation DeltaG∘=−RTlnK\\Delta G^\circ = -RT \\ln K.

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The Third Law of Thermodynamics states that the entropy of any perfectly crystalline substance approaches zero as the absolute temperature approaches zero (0,K0\\,K).

📐Formulae

DeltaS=fracqrevT\\Delta S = \\frac{q_{rev}}{T}

DeltaStotal=DeltaSsys+DeltaSsurr\\Delta S_{total} = \\Delta S_{sys} + \\Delta S_{surr}

DeltaSsurr=−fracDeltaHsysT\\Delta S_{surr} = -\\frac{\\Delta H_{sys}}{T}

DeltaG=DeltaH−TDeltaS\\Delta G = \\Delta H - T\\Delta S

DeltaG∘=sumDeltaGf∘(products)−sumDeltaGf∘(reactants)\\Delta G^\circ = \\sum \\Delta G_f^\circ(products) - \\sum \\Delta G_f^\circ(reactants)

DeltaG∘=−2.303RTlogK\\Delta G^\circ = -2.303 RT \\log K

DeltaSvap=fracDeltaHvapTb\\Delta S_{vap} = \\frac{\\Delta H_{vap}}{T_b}

💡Examples

Problem 1:

For the reaction 2A(g)+B(g)rightarrow2D(g)2A(g) + B(g) \\rightarrow 2D(g), DeltaH∘=−10.5,kJ\\Delta H^\circ = -10.5\\,kJ and DeltaS∘=−44.1,J,K−1\\Delta S^\circ = -44.1\\,J\\,K^{-1}. Predict whether the reaction is spontaneous at 298,K298\\,K.

Solution:

Given: DeltaH∘=−10.5,kJ=−10500,J\\Delta H^\circ = -10.5\\,kJ = -10500\\,J, DeltaS∘=−44.1,J,K−1\\Delta S^\circ = -44.1\\,J\\,K^{-1}, and T=298,KT = 298\\,K. Using the Gibbs equation: DeltaG∘=DeltaH∘−TDeltaS∘\\Delta G^\circ = \\Delta H^\circ - T\\Delta S^\circ. DeltaG∘=−10500−(298times−44.1)=−10500+13141.8=+2641.8,J\\Delta G^\circ = -10500 - (298 \\times -44.1) = -10500 + 13141.8 = +2641.8\\,J. Since DeltaG∘\\Delta G^\circ is positive, the reaction is non-spontaneous at 298,K298\\,K.

Explanation:

Spontaneity depends on the sign of DeltaG\\Delta G. Even though the reaction is exothermic (DeltaH<0\\Delta H < 0), the decrease in entropy (DeltaS<0\\Delta S < 0) is significant enough at this temperature to make the overall free energy change positive.

Problem 2:

The equilibrium constant for a reaction is 1010. What will be the value of DeltaG∘\\Delta G^\circ at 300,K300\\,K? (Given R=8.314,J,K−1mol−1R = 8.314\\,J\\,K^{-1}mol^{-1})

Solution:

Using the relation DeltaG∘=−2.303RTlogK\\Delta G^\circ = -2.303 RT \\log K. Substitute the values: DeltaG∘=−2.303times8.314times300timeslog(10)\\Delta G^\circ = -2.303 \\times 8.314 \\times 300 \\times \\log(10). Since log(10)=1\\log(10) = 1, DeltaG∘=−2.303times8.314times300=−5744.1,J,mol−1\\Delta G^\circ = -2.303 \\times 8.314 \\times 300 = -5744.1\\,J\\,mol^{-1} or −5.744,kJ,mol−1-5.744\\,kJ\\,mol^{-1}.

Explanation:

A positive equilibrium constant K>1K > 1 results in a negative DeltaG∘\\Delta G^\circ, indicating that the products are favored at standard state conditions.