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Equilibrium - Applications of Equilibrium Constants

Grade 11CBSEChemistry

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

🔑Concepts

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Predicting the extent of a reaction: The numerical value of the equilibrium constant KcK_c or KpK_p indicates the extent of a reaction. If Kc>103K_c > 10^3, products predominate, and the reaction proceeds nearly to completion. If Kc<10−3K_c < 10^{-3}, reactants predominate, and the reaction proceeds to a very small extent. If 10−3<Kc<10310^{-3} < K_c < 10^3, appreciable concentrations of both reactants and products are present.

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Predicting the direction of the reaction: This is done using the Reaction Quotient (QcQ_c). For a general reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, Qc=[C]tc[D]td[A]ta[B]tbQ_c = \frac{[C]_t^c [D]_t^d}{[A]_t^a [B]_t^b} where subscripts 'tt' denote concentrations at any time tt not necessarily at equilibrium.

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Directional rules: If Qc<KcQ_c < K_c, the reaction proceeds in the forward direction (left to right). If Qc>KcQ_c > K_c, the reaction proceeds in the reverse direction (right to left). If Qc=KcQ_c = K_c, the system is at equilibrium.

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Calculating equilibrium concentrations: By knowing the initial concentrations and the value of KcK_c, the equilibrium concentrations of all species can be determined, often using an 'ICE' table (Initial, Change, Equilibrium).

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Relationship with Gibbs Free Energy: The standard Gibbs energy change ΔG∘\Delta G^{\circ} and the equilibrium constant KK are related, indicating the spontaneity of the reaction under standard conditions.

📐Formulae

Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}

Qc=[C]tc[D]td[A]ta[B]tbQ_c = \frac{[C]_t^c [D]_t^d}{[A]_t^a [B]_t^b}

Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}

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

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

💡Examples

Problem 1:

For the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g), the equilibrium constant KcK_c is 54.854.8 at 700K700 K. At a particular time, the concentrations are [H2]=0.10M[H_2] = 0.10 M, [I2]=0.20M[I_2] = 0.20 M, and [HI]=0.40M[HI] = 0.40 M. Predict the direction of the reaction.

Solution:

First, calculate the reaction quotient QcQ_c: Qc=[HI]2[H2][I2]Q_c = \frac{[HI]^2}{[H_2][I_2]} Qc=(0.40)2(0.10)(0.20)=0.160.02=8Q_c = \frac{(0.40)^2}{(0.10)(0.20)} = \frac{0.16}{0.02} = 8 Since Qc=8Q_c = 8 and Kc=54.8K_c = 54.8, we have Qc<KcQ_c < K_c.

Explanation:

Because the reaction quotient QcQ_c is less than the equilibrium constant KcK_c, the system is not at equilibrium and will proceed in the forward direction (towards products) to reach equilibrium.

Problem 2:

Calculate the value of ΔG∘\Delta G^{\circ} for the reaction A+B⇌C+DA + B \rightleftharpoons C + D at 300K300 K if the equilibrium constant KcK_c is 10210^2. (Use R=8.314J K−1 mol−1R = 8.314 J \, K^{-1} \, mol^{-1})

Solution:

Using the formula: ΔG∘=−2.303RTlog⁡Kc\Delta G^{\circ} = -2.303 RT \log K_c Substitute the values: ΔG∘=−2.303×8.314×300×log⁡(102)\Delta G^{\circ} = -2.303 \times 8.314 \times 300 \times \log(10^2) ΔG∘=−2.303×8.314×300×2\Delta G^{\circ} = -2.303 \times 8.314 \times 300 \times 2 ΔG∘≈−11488.2 J mol−1=−11.49 kJ mol−1\Delta G^{\circ} \approx -11488.2 \, J \, mol^{-1} = -11.49 \, kJ \, mol^{-1}

Explanation:

The negative value of ΔG∘\Delta G^{\circ} indicates that the reaction is feasible and favors the formation of products under standard conditions.