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Equilibrium - Law of Mass Action and Equilibrium Constant

Grade 11ICSEChemistry

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

🔑Concepts

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The Law of Mass Action states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations (active masses) of the reactants, with each concentration term raised to a power equal to its stoichiometric coefficient in the balanced chemical equation.

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For a general reversible reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the equilibrium constant KcK_c is defined as the ratio of the product of the molar concentrations of the products to the product of the molar concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

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The term 'Active Mass' refers to the molar concentration of a substance, expressed as mol L−1mol \, L^{-1} or MM. For pure solids and pure liquids, the active mass is taken as unity (11).

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Equilibrium Constant in terms of partial pressures (KpK_p) is used for reactions involving gases. It is expressed using the partial pressures of the reactants and products at equilibrium.

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The relationship between KpK_p and KcK_c is given by the equation Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}, where Δng\Delta n_g is the difference between the number of moles of gaseous products and gaseous reactants.

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The value of the equilibrium constant is independent of the initial concentrations of reactants, the volume of the container, or the presence of a catalyst; it depends only on the temperature of the system.

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If K>103K > 10^3, the reaction proceeds nearly to completion. If K<10−3K < 10^{-3}, the reaction hardly proceeds. If KK is between 10−310^{-3} and 10310^3, appreciable concentrations of both reactants and products are present at equilibrium.

📐Formulae

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

Kp=(PC)c(PD)d(PA)a(PB)bK_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}

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

Δng=(nproducts)g−(nreactants)g\Delta n_g = (n_{products})_g - (n_{reactants})_g

Active Mass=Number of moles (n)Volume in Litres (V)=wM×V\text{Active Mass} = \frac{\text{Number of moles (n)}}{\text{Volume in Litres (V)}} = \frac{w}{M \times V}

💡Examples

Problem 1:

For the synthesis of ammonia: N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), calculate Δng\Delta n_g and state the relationship between KpK_p and KcK_c.

Solution:

  1. Identify gaseous moles of products: np=2n_p = 2 (from 2NH32NH_3).
  2. Identify gaseous moles of reactants: nr=1+3=4n_r = 1 + 3 = 4 (from 1N21N_2 and 3H23H_2).
  3. Calculate Δng=2−4=−2\Delta n_g = 2 - 4 = -2.
  4. Relationship: Kp=Kc(RT)−2K_p = K_c(RT)^{-2} or Kp=Kc(RT)2K_p = \frac{K_c}{(RT)^2}.

Explanation:

In this reaction, the number of moles decreases. Therefore, Δng\Delta n_g is negative, making Kc>KpK_c > K_p.

Problem 2:

In the reaction H2(g)+I2(g)⇌2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g), if the equilibrium concentrations are [H2]=0.2 mol L−1[H_2] = 0.2 \, mol \, L^{-1}, [I2]=0.2 mol L−1[I_2] = 0.2 \, mol \, L^{-1}, and [HI]=0.8 mol L−1[HI] = 0.8 \, mol \, L^{-1}, find KcK_c.

Solution:

Kc=[HI]2[H2][I2]K_c = \frac{[HI]^2}{[H_2][I_2]} Kc=(0.8)2(0.2)(0.2)K_c = \frac{(0.8)^2}{(0.2)(0.2)} Kc=0.640.04=16K_c = \frac{0.64}{0.04} = 16

Explanation:

The equilibrium constant is calculated by plugging the equilibrium concentrations into the KcK_c expression derived from the balanced equation. Since Δng=0\Delta n_g = 0 in this case, KcK_c is dimensionless and Kp=KcK_p = K_c.

Law of Mass Action and Equilibrium Constant Revision - Class 11 Chemistry ICSE