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Equilibrium - The state of dynamic equilibrium

Grade 12IBChemistry

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

🔑Concepts

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Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction (ratef=raterrate_{f} = rate_{r}).

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At the state of equilibrium, the macroscopic properties of the system (such as concentration, pressure, and color intensity) remain constant over time.

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The equilibrium is described as 'dynamic' because reactions continue to occur at the molecular level, even though no net change in concentration is observed.

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A closed system is essential for equilibrium to be established, meaning no matter can enter or leave the system, although energy can be exchanged with the surroundings.

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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 > 1, the equilibrium favors the products; if Kc<1K_c < 1, it favors the reactants.

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The reaction quotient, QQ, characterizes the relative amounts of products and reactants present in a reaction at any given time. If Q=KcQ = K_c, the system is at equilibrium.

📐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}

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

Kreverse=1KforwardK_{reverse} = \frac{1}{K_{forward}}

Knew=(Koriginal)nK_{new} = (K_{original})^n

💡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]=0.60 mol dm−3[N_2] = 0.60\ mol\ dm^{-3}, [H2]=0.40 mol dm−3[H_2] = 0.40\ mol\ dm^{-3}, and [NH3]=0.15 mol dm−3[NH_3] = 0.15\ mol\ dm^{-3}. Calculate the value of KcK_c.

Solution:

Kc=[NH3]2[N2][H2]3=(0.15)2(0.60)(0.40)3=0.02250.0384≈0.586K_c = \frac{[NH_3]^2}{[N_2][H_2]^3} = \frac{(0.15)^2}{(0.60)(0.40)^3} = \frac{0.0225}{0.0384} \approx 0.586

Explanation:

The equilibrium constant expression is derived from the balanced equation. We substitute the given equilibrium concentrations into the expression. Since KcK_c is less than 11 at this temperature, the reactants are favored over the products.

Problem 2:

Explain why the color of the mixture remains constant in the reaction N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g) (where N2O4N_2O_4 is colorless and NO2NO_2 is brown) once equilibrium is reached in a sealed tube.

Solution:

Once the rate of the forward reaction (decomposition of N2O4N_2O_4) equals the rate of the reverse reaction (dimerization of NO2NO_2), the concentration of NO2NO_2 molecules remains constant. Consequently, the intensity of the brown color remains unchanged.

Explanation:

This is a hallmark of dynamic equilibrium: macroscopic properties like color do not change because the net concentration of the species responsible for the property (NO2NO_2) is stable, despite the continuous interconversion of molecules.

The state of dynamic equilibrium - Revision Notes & Key Formulas | IB Grade 12 Chemistry