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What does it mean for a reaction to reach dynamic equilibrium, and how does Kc describe it?

Describe dynamic equilibrium in closed systems and use the equilibrium constant expression Kc to relate equilibrium concentrations and reaction extent.

What dynamic equilibrium means in a closed system, how to write the equilibrium constant expression Kc, and how the size of Kc indicates the extent of a reversible reaction.

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  1. What this dot point is asking
  2. What dynamic equilibrium is
  3. The equilibrium constant expression
  4. What the value of Kc tells you

What this dot point is asking

You must describe dynamic equilibrium, write a correct KcK_c expression for a given equation, and interpret what the value of KcK_c tells you about the extent of reaction.

What dynamic equilibrium is

In a closed system (no matter enters or leaves), a reversible reaction starts with reactants. As products build up, the reverse reaction speeds up while the forward reaction slows. Eventually the two rates become equal.

At equilibrium nothing appears to change, but molecules are constantly reacting in both directions.

The equilibrium constant expression

For the general reaction:

aA+bB⇌cC+dDa\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}

the equilibrium constant is:

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

Products go on top, reactants on the bottom, each concentration raised to the power of its coefficient. Pure solids and pure liquids are omitted (their concentration is effectively constant).

What the value of Kc tells you

KcK_c measures the extent of reaction at equilibrium:

  • Kc≫1K_c \gg 1: products dominate - the reaction goes nearly to completion.
  • Kc≪1K_c \ll 1: reactants dominate - little product forms.
  • Kc≈1K_c \approx 1: significant amounts of both are present.

KcK_c is constant for a given reaction at a fixed temperature. Changing concentrations or pressure shifts the position of equilibrium but does not change KcK_c; only a change in temperature changes its value.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 SACE Stage 24 marksFor the decomposition COBr2(g) <=> CO(g) + Br2(g), 1.00 mol of COBr2 was placed into an empty 5.00 L flask at 77 degrees C and sealed. After equilibrium, 0.60 mol of bromine was present. Calculate the value of Kc for the reaction at 77 degrees C.
Show worked answer →

Build an ICE table in moles, then convert to concentrations.

  1. At equilibrium n(Br2) = 0.60 mol, so n(CO) = 0.60 mol (1:1) and n(COBr2) = 1.00 - 0.60 = 0.40 mol.

  2. Divide by 5.00 L: [Br2] = [CO] = 0.60 / 5.00 = 0.120 mol L-1; [COBr2] = 0.40 / 5.00 = 0.080 mol L-1.

  3. Kc = ([CO][Br2]) / [COBr2] = (0.120 x 0.120) / 0.080.

  4. Kc = 0.0144 / 0.080 = 0.18 (mol L-1). One mark each for the equilibrium moles, the concentrations, the correct expression, and the value.

2023 SACE Stage 22 marksFor CO2(g) + 3H2(g) <=> CH3OH(g) + H2O(g), write the Kc expression for this reaction.
Show worked answer →

The equilibrium constant expression is products over reactants, each concentration raised to the power of its coefficient:

Kc = ([CH3OH][H2O]) / ([CO2][H2]^3)

One mark for the correct products-over-reactants form, one mark for the correct power of 3 on [H2].

2024 SACE Stage 25 marksFor H2O(g) + CO(g) <=> H2(g) + CO2(g), steam and CO each at initial concentration 0.50 mol L-1 were placed in a vessel at 730 degrees C. After some time [CO2] was 0.24 mol L-1. Determine whether the system has reached equilibrium, given Kc = 1.44 at this temperature.
Show worked answer →

Find the current concentrations, calculate the reaction quotient Q, then compare with Kc.

  1. CO2 formed = 0.24 mol L-1, so by the 1:1:1:1 stoichiometry: [H2] = 0.24, [CO] = 0.50 - 0.24 = 0.26, [H2O] = 0.50 - 0.24 = 0.26 mol L-1.

  2. Q = ([H2][CO2]) / ([H2O][CO]) = (0.24 x 0.24) / (0.26 x 0.26).

  3. Q = 0.0576 / 0.0676 = 0.85.

  4. Q (0.85) is less than Kc (1.44).

  5. Since Q does not equal Kc, the system has NOT yet reached equilibrium; because Q is less than Kc, the reaction will continue in the forward direction. Marks for the concentrations, Q expression, Q value, the comparison, and the conclusion.

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