Chemistry · Lesson
Reaction quotients and the direction of change
Compare a mixture’s current activity quotient with its equilibrium constant. Follow how changing the written equation changes both quantities without changing the same physical state.
Back to the Chemistry mapBy the end, you can
- Predict the net thermodynamic direction from Q and K°.
- Distinguish a composition change from a change in K°.
- Apply reversal and coefficient-scaling rules consistently.
Q describes the current mixture
Form Q from the current activities using the written reaction. Unlike K°, Q need not be an equilibrium value. For positive finite activities and the same convention, Q < K° favours the forward net reaction, Q > K° favours the reverse, and Q = K° identifies the equilibrium relation.
This direction criterion follows ΔᵣG = RT ln(Q/K°) under the stated temperature and reaction convention. It predicts a thermodynamic tendency, not its speed. Extremely small rate constants or nucleation barriers can delay a visible response.
More product need not make Q exceed K°
Our hypothetical A + B ⇌ AB mixture has Q = 1.25 and K° = 5.00 at fixed temperature. If a_AB doubles from 0.100 to 0.200 while the other activities are initially unchanged, Q becomes 2.50. It has moved closer to K° but remains below it.
The forward tendency persists in that newly stipulated state. Product addition did not automatically reverse the reaction. If starting at equilibrium instead, an isolated increase in product activity would make Q exceed K° initially; the starting state and the size of the perturbation matter.
| Current state | a_A, a_B, a_AB | Q = a_AB/(a_A a_B) | Net thermodynamic tendency |
|---|---|---|---|
| Before the stipulated addition | 0.200, 0.400, 0.100 | 1.25 | Forward: Q < K° |
| Immediately after addition | 0.200, 0.400, 0.200 | 2.50 | Still forward: Q < K° |
| Equilibrium relation | Activities must satisfy the relation | 5.00 | No net reaction tendency on this basis |
Original hypothetical A + B ⇌ AB activity states at fixed temperature and K° = 5.00. The second row stipulates only a_AB doubled before readjustment. All activity values and quotients are dimensionless.
Reverse and scale the expression with the equation
For A ⇌ B, Q = a_B/a_A. Writing B ⇌ A inverts the quotient and constant. If the original K° is 4.00, the reversed constant is 0.250. Multiplying every coefficient by 2 gives Q² and K°² = 16.0. Halving them gives square roots, with K° = 2.00.
These are different numerical descriptions of the same compatible equilibrium state. Reversing the written direction also reverses what the words forward and reverse refer to. Never compare a quotient for one equation with a constant for a differently scaled or reversed equation.
Direction alone does not supply the final composition
A quotient test tells which way a compatible reaction can reduce the thermodynamic mismatch. Calculating final composition also needs the conserved material inventory, other active equilibria and the stated constraints. Initial stock concentrations generally cannot be inserted as though they were final equilibrium concentrations.
At fixed temperature and unchanged standard-state definition, a concentration perturbation changes Q while K° stays the same. Changing temperature generally changes K°. Keep those two operations distinct when explaining a response.
Worked examples
Keep the constant with its written equation
The hypothetical A ⇌ B equilibrium has K° = 4.00 under unchanged conditions. Find constants for its reverse, double and half equations.
- For B ⇌ A, invert 4.00 to obtain 0.250.
- For 2 A ⇌ 2 B, square 4.00 to obtain 16.0.
- For (1/2) A ⇌ (1/2) B, take its square root to obtain 2.00.
The three constants are 0.250, 16.0 and 2.00. Their equations must accompany the values.
Try it yourself
For the written reaction, Q = 2.50 and K° = 5.00. Which net direction is favoured?
Show the answer and reasoning
Forward.
Q remains below K° in the same convention.
If A ⇌ B has K° = 4.00, what is K° for B ⇌ A?
Show the answer and reasoning
0.250.
Reversing the entire equation inverts its activity product.
What is K° for 2 A ⇌ 2 B when A ⇌ B has K° = 4.00?
Show the answer and reasoning
16.0.
Doubling all coefficients squares the original constant.
At fixed temperature and convention, does adding a product change K° itself?
Show the answer and reasoning
No.
It changes the current quotient; the equilibrium relation stays fixed under those conditions.
What to carry forward
- Compare Q with the constant for the exact same equation.
- A thermodynamic direction is different from a rate or final composition.
- Inversion and powers follow reversal and stoichiometric scaling.
- Distinguish changes in Q from changes in K°.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.