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IB DP Chemistry -R2.3.5 Reaction quotient (Q) – Study Notes – New Syllabus

IB DP Chemistry -R2.3.5 Reaction quotient (Q) - Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – R2.3.5 Reaction quotient (Q) – Study Notes – New Syllabus

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Reactivity 2.3.5 –Reaction Quotient \( Q \) and Predicting Direction of Reaction

Reactivity 2.3.5 –Reaction Quotient \( Q \) and Predicting Direction of Reaction

The reaction quotient \( Q \) is calculated the same way as the equilibrium constant \( K \), but using the concentrations (or partial pressures) of the species at a particular moment, not necessarily at equilibrium.

 

Expression for \( Q \):

For the general reaction:

\( aA + bB \rightleftharpoons cC + dD \)

The reaction quotient is given by:

\( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

Important:

  • Use current (measured) concentrations or pressures.
  • Include only species that appear in the equilibrium expression — exclude solids and pure liquids.
  • Units are often omitted, as \( Q \) is usually used for qualitative comparisons.

Comparing \( Q \) with \( K \) to Predict Reaction Direction

The value of \( Q \) tells us whether the system is at equilibrium and, if not, which direction the reaction must shift to reach equilibrium.

  • \( Q < K \): The forward reaction is favored. More products will form.
  • \( Q = K \): The system is at equilibrium. No net change occurs.
  • \( Q > K \): The reverse reaction is favored. More reactants will form.

This comparison helps us determine how the system must shift to reach equilibrium.

Effect on Equilibrium:

  • Direction of shift depends on whether the current ratio \( Q \) matches the value of \( K \).
  • As the reaction proceeds, the concentrations change and \( Q \) gradually adjusts to equal \( K \).

Example

For the reaction:
\( \text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \),
the equilibrium constant at 400°C is \( K_c = 0.5 \).
Suppose at a given moment: [N2] = 0.4 mol/L, [H2] = 0.6 mol/L, [NH3] = 0.1 mol/L.
Calculate \( Q_c \) and determine the direction of the shift.

▶️Answer/Explanation

Use:
\( Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.1)^2}{(0.4)(0.6)^3} = \frac{0.01}{0.4 \times 0.216} = \frac{0.01}{0.0864} \approx 0.116 \)

Since \( Q_c < K_c \), the reaction is not yet at equilibrium. It will shift to the right to produce more NH3 and reach equilibrium.

Example

For the reaction:
\( \text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g) \),
the equilibrium constant at 450°C is \( K_c = 50.0 \).
At a given point: [H2] = 0.2 mol/L, [I2] = 0.2 mol/L, [HI] = 3.0 mol/L.
Determine \( Q_c \) and predict the direction of shift.

▶️Answer/Explanation

Calculate:
\( Q_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(3.0)^2}{(0.2)(0.2)} = \frac{9.0}{0.04} = 225 \)

Since \( Q_c > K_c \), the reaction has too much product and not enough reactant. It will shift to the left to produce more H2 and I2 and reduce [HI].

Example

At 600 K, the equilibrium constant for the decomposition of dinitrogen tetroxide is:
\( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \quad K_c = 0.36 \)
A reaction vessel is filled such that:
[N2O4] = 0.40 mol/L and [NO2] = 0.30 mol/L
Calculate the reaction quotient \( Q_c \), compare it to \( K_c \), and predict the direction the reaction will shift.

▶️Answer/Explanation

Step 1 – Write the expression for \( Q_c \):
\( Q_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \)

Step 2 – Substitute the values:
\( Q_c = \frac{(0.30)^2}{0.40} = \frac{0.09}{0.40} = 0.225 \)

Step 3 – Compare \( Q_c \) and \( K_c \):
\( Q_c = 0.225 \), \( K_c = 0.36 \)
Since \( Q_c < K_c \), the reaction is not at equilibrium. It will shift to the right to produce more NO2.

Conclusion: The forward reaction is favored — more N2O4 will decompose to restore equilibrium.
Value of \( K \): Remains constant, as temperature is unchanged.

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