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Equilibrium - The equilibrium law (HL)

Grade 12IBChemistry

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

🔑Concepts

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The Equilibrium Constant (KcK_c) is a constant for a particular reaction at a specific temperature. For the general reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, the expression is Kc=[C]c[D]d[A]a[B]bK_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}.

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The Reaction Quotient (QQ) uses initial concentrations in the equilibrium expression. If Q<KcQ < K_c, the reaction proceeds to the right (products); if Q>KcQ > K_c, it proceeds to the left (reactants); if Q=KcQ = K_c, the system is at equilibrium.

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If a reaction is reversed, the new equilibrium constant is the reciprocal of the original: Knew=1KoldK_{new} = \frac{1}{K_{old}}.

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If the coefficients of a balanced equation are multiplied by a factor nn, the new equilibrium constant is Knew=(Kold)nK_{new} = (K_{old})^n.

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The relationship between the standard Gibbs free energy change and the equilibrium constant is given by ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K.

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When ΔG⊖\Delta G^\ominus is negative, ln⁡K\ln K is positive and K>1K > 1, meaning products are favored at equilibrium. When ΔG⊖\Delta G^\ominus is positive, K<1K < 1, meaning reactants are favored.

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The value of KcK_c is only affected by temperature changes. For exothermic reactions (ΔH<0\Delta H < 0), KcK_c decreases as temperature increases. For endothermic reactions (ΔH>0\Delta H > 0), KcK_c increases as temperature increases.

📐Formulae

Kc=∏[Products]coefficients∏[Reactants]coefficientsK_c = \frac{\prod [Products]^{coefficients}}{\prod [Reactants]^{coefficients}}

ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K

Q=[C]tc[D]td[A]ta[B]tbQ = \frac{[C]^c_{t} [D]^d_{t}}{[A]^a_{t} [B]^b_{t}}

R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}

💡Examples

Problem 1:

Calculate the standard Gibbs free energy change, ΔG⊖\Delta G^\ominus, for a reaction at 298 K298 \text{ K} if the equilibrium constant KcK_c is 1.5×1051.5 \times 10^5.

Solution:

Using the formula ΔG⊖=−RTln⁡K\Delta G^\ominus = -RT \ln K:

  1. T=298 KT = 298 \text{ K}
  2. R=8.31 J K−1 mol−1R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}
  3. K=1.5×105K = 1.5 \times 10^5 ΔG⊖=−(8.31)×(298)×ln⁡(1.5×105)\Delta G^\ominus = -(8.31) \times (298) \times \ln(1.5 \times 10^5) ΔG⊖=−8.31×298×11.918\Delta G^\ominus = -8.31 \times 298 \times 11.918 ΔG⊖≈−29514 J mol−1\Delta G^\ominus \approx -29514 \text{ J mol}^{-1} or −29.5 kJ mol−1-29.5 \text{ kJ mol}^{-1}.

Explanation:

Since Kc>1K_c > 1, the reaction is spontaneous under standard conditions, resulting in a negative ΔG⊖\Delta G^\ominus value.

Problem 2:

For the reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), Kc=0.060K_c = 0.060 at a certain temperature. Calculate KcK_c for the reaction NH3(g)⇌12N2(g)+32H2(g)NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) at the same temperature.

Solution:

  1. The original reaction is reversed: K′=1Kc=10.060=16.67K' = \frac{1}{K_c} = \frac{1}{0.060} = 16.67.
  2. The coefficients are then multiplied by 12\frac{1}{2}: K′′=(K′)1/2=16.67≈4.08K'' = (K')^{1/2} = \sqrt{16.67} \approx 4.08.

Explanation:

To find the new KcK_c, we first take the reciprocal because the reaction was reversed, then take the square root because the coefficients were halved.