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Equilibrium - Relationship between Equilibrium Constant, Reaction Quotient and Gibbs Energy

Grade 11CBSEChemistry

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

🔑Concepts

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The Gibbs energy change ΔG\Delta G at any point in a reaction is related to the standard Gibbs energy ΔG∘\Delta G^\circ and the reaction quotient QQ by the equation: ΔG=ΔG∘+RTln⁡Q\Delta G = \Delta G^\circ + RT \ln Q.

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At equilibrium, the system is at its lowest energy state, meaning ΔG=0\Delta G = 0 and the reaction quotient QQ is equal to the equilibrium constant KK.

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Substituting equilibrium conditions into the thermodynamic equation gives: 0=ΔG∘+RTln⁡K0 = \Delta G^\circ + RT \ln K, which simplifies to ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K.

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To use base-10 logarithms, the relationship is expressed as: ΔG∘=−2.303RTlog⁡K\Delta G^\circ = -2.303 RT \log K.

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If ΔG∘<0\Delta G^\circ < 0 (negative), then log⁡K\log K must be positive, implying K>1K > 1. This indicates a spontaneous reaction that favors product formation at equilibrium.

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If ΔG∘>0\Delta G^\circ > 0 (positive), then log⁡K\log K is negative, implying K<1K < 1. This indicates a non-spontaneous reaction that favors reactants at equilibrium.

📐Formulae

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

ΔG∘=−RTln⁡K\Delta G^\circ = -RT \ln K

ΔG∘=−2.303RTlog⁡K\Delta G^\circ = -2.303 RT \log K

K=e−ΔG∘RTK = e^{-\frac{\Delta G^\circ}{RT}}

💡Examples

Problem 1:

The equilibrium constant for a reaction is 1010 at 300 K300\text{ K}. Calculate the standard Gibbs energy change ΔG∘\Delta G^\circ for the reaction. (Given R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} and log⁡10=1\log 10 = 1)

Solution:

  1. Identify the given values: K=10K = 10, T=300 KT = 300\text{ K}, R=8.314 J K−1 mol−1R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}.
  2. Use the formula: ΔG∘=−2.303RTlog⁡K\Delta G^\circ = -2.303 RT \log K.
  3. Substitute the values: ΔG∘=−2.303×8.314×300×log⁡(10)\Delta G^\circ = -2.303 \times 8.314 \times 300 \times \log(10)
  4. Since log⁡(10)=1\log(10) = 1, the calculation becomes: ΔG∘=−2.303×2494.2×1\Delta G^\circ = -2.303 \times 2494.2 \times 1 ΔG∘=−5744.14 J mol−1\Delta G^\circ = -5744.14\text{ J mol}^{-1} To find the difference between a reference energy of 8000 J8000\text{ J} and the magnitude of this result (rounded to 57445744): 8000−57442256\begin{array}{r} 8000 \\ - 5744 \\ \hline 2256 \end{array}

Explanation:

The negative value of ΔG∘\Delta G^\circ (−5.744 kJ/mol-5.744\text{ kJ/mol}) indicates that the reaction is spontaneous under standard conditions. Since K>1K > 1, the equilibrium mixture will have a higher concentration of products than reactants.