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Equilibrium - Dynamic equilibrium

Grade 11IBChemistry

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

🔑Concepts

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A reaction is in dynamic equilibrium when the rate of the forward reaction is equal to the rate of the reverse reaction (rateforward=ratereverserate_{forward} = rate_{reverse}).

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Dynamic equilibrium can only be achieved in a closed system, where neither matter nor energy (typically in the form of gas) can enter or leave.

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At equilibrium, the concentrations of reactants and products remain constant over time, though they are rarely equal to each other.

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Macroscopic properties, such as color intensity, pressure, and density, remain constant when a system reaches equilibrium.

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The equilibrium constant KcK_c is a temperature-dependent value that indicates the extent of a reaction. If Kc≫1K_c \gg 1, the equilibrium lies to the right (products favored). If Kc≪1K_c \ll 1, it lies to the left (reactants favored).

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Le Chatelier's Principle states that if a system at equilibrium is subjected to a change in conditions (concentrationconcentration, temperaturetemperature, or pressurepressure), the position of equilibrium will shift to oppose the change.

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A catalyst increases the rate of both forward and reverse reactions equally; therefore, it helps a system reach equilibrium faster but does not change the position of equilibrium or the value of KcK_c.

📐Formulae

aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD

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

Qc=[C]initialc[D]initiald[A]initiala[B]initialbQ_c = \frac{[C]_{initial}^c [D]_{initial}^d}{[A]_{initial}^a [B]_{initial}^b}

ΔG=−RTln⁡K\Delta G = -RT \ln K

💡Examples

Problem 1:

For the Haber process reaction N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g), the equilibrium concentrations at a specific temperature are [N2]=0.45 mol dm−3[N_2] = 0.45\text{ mol dm}^{-3}, [H2]=0.63 mol dm−3[H_2] = 0.63\text{ mol dm}^{-3}, and [NH3]=0.27 mol dm−3[NH_3] = 0.27\text{ mol dm}^{-3}. Calculate the value of KcK_c.

Solution:

Kc=[NH3]2[N2][H2]3K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} Kc=(0.27)2(0.45)(0.63)3K_c = \frac{(0.27)^2}{(0.45)(0.63)^3} Kc=0.07290.45×0.250047K_c = \frac{0.0729}{0.45 \times 0.250047} Kc≈0.648K_c \approx 0.648

Explanation:

The equilibrium constant expression is derived using the coefficients of the balanced chemical equation as exponents. The units for KcK_c in IB Chemistry are generally omitted unless specifically requested.

Problem 2:

Consider the endothermic reaction: N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) where ΔH=+57 kJ mol−1\Delta H = +57\text{ kJ mol}^{-1}. Predict the effect of increasing the temperature on the position of equilibrium and the value of KcK_c.

Solution:

Position: Shifts to the right (towards NO2NO_2). Value of KcK_c: Increases.

Explanation:

According to Le Chatelier's Principle, increasing the temperature favors the endothermic direction to absorb the excess heat. Since the forward reaction is endothermic (ΔH>0\Delta H > 0), the equilibrium shifts toward the products, increasing the numerator of the KcK_c expression and thus increasing the value of KcK_c.

Dynamic equilibrium - Revision Notes & Key Formulas | IB Grade 11 Chemistry