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Equilibrium - Law of Chemical Equilibrium and Equilibrium Constant

Grade 11CBSEChemistry

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 given temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants, each raised to a power equal to its stoichiometric coefficient.

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For a general reversible reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the Law of Chemical Equilibrium defines the equilibrium constant KcK_c as the ratio of the product of molar concentrations of the products to that of the reactants at equilibrium.

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The equilibrium constant can be expressed in terms of molar concentration (KcK_c) or partial pressure (KpK_p). KpK_p is used specifically for reactions involving gases.

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In heterogeneous equilibrium, the concentrations (or activities) of pure solids and pure liquids are taken as unity (11) and are omitted from the KcK_c and KpK_p expressions.

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The value of the equilibrium constant is independent of the initial concentrations of reactants/products, pressure, or the presence of a catalyst; it depends only on the temperature for a specific reaction.

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If K>103K > 10^3, the equilibrium favors the products (reaction proceeds nearly to completion). If K<10−3K < 10^{-3}, the equilibrium favors the reactants (reaction proceeds only to a very small extent).

📐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=(moles of gaseous products)−(moles of gaseous reactants)\Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}

Qc=[C]tc[D]td[A]ta[B]tb (where Qc is the reaction quotient at any time t)Q_c = \frac{[C]_t^c [D]_t^d}{[A]_t^a [B]_t^b} \text{ (where } Q_c \text{ is the reaction quotient at any time } t)

💡Examples

Problem 1:

For the equilibrium N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the value of KcK_c is 0.0610.061 at 500K500 K. Calculate the value of KpK_p for the same reaction. (R=0.0821 L⋅atm⋅K−1⋅mol−1R = 0.0821\ L \cdot atm \cdot K^{-1} \cdot mol^{-1})

Solution:

  1. Identify Δng\Delta n_g: Δng=2−(1+3)=−2\Delta n_g = 2 - (1 + 3) = -2.
  2. Use the relation: Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}.
  3. Substitute values: Kp=0.061×(0.0821×500)−2K_p = 0.061 \times (0.0821 \times 500)^{-2}.
  4. Kp=0.061×(41.05)−2≈3.6×10−5K_p = 0.061 \times (41.05)^{-2} \approx 3.6 \times 10^{-5}.

Explanation:

The relationship between KpK_p and KcK_c involves the change in the number of moles of gaseous components (Δng\Delta n_g). Since Δng\Delta n_g is negative here, KpK_p is significantly smaller than KcK_c.

Problem 2:

Write the equilibrium constant expression (KcK_c) for the reaction: CaCO3(s)⇌CaO(s)+CO2(g)CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g).

Solution:

Kc=[CO2]K_c = [CO_2]

Explanation:

In this heterogeneous equilibrium, CaCO3CaCO_3 and CaOCaO are pure solids. Their molar concentrations are constant and are incorporated into the equilibrium constant, so they do not appear in the final expression. Only the concentration of the gas CO2CO_2 is included.

Law of Chemical Equilibrium and Equilibrium Constant Revision - Class 11 Chemistry CBSE