Reaction Quotient and Le Châtelier's Principle用 Q 解释勒夏特列原理

Big Idea · Transformations (TRA)AP Science Practice 4 · Model AnalysisAP link · Unit 7~45 min

Learning goals

  • Use the comparison of Q and K to explain which way an equilibrium shifts after a change in concentration or volume
  • Explain that a temperature change alters K, not Q, and use the sign of ΔH to decide how K changes
  • Calculate the factor by which Q changes when the volume changes, and confirm from the new equilibrium that a shift partly counteracts a change but cannot cancel it

Before you start

  • Reaction quotient and equilibrium constant (U7.3)
  • Introduction to Le Châtelier's principle (U7.9)

Why it shifts: look at Q and K

Le Châtelier’s principle tells you which way an equilibrium shifts; Q and K tell you why. At equilibrium Q = K; after a change, as long as Q ≠ K, the reaction proceeds until Q equals K again.

  • Changing a concentration changes Q. Adding a reactant (larger denominator) or removing a product (smaller numerator) makes Q < K, so the reaction proceeds forward.
  • Changing the volume changes every gas concentration by the same factor. Halving the volume doubles each gas concentration and multiplies Q by 2Δn (Δn = gas moles of products − gas moles of reactants). When Δn < 0, Q decreases and the reaction proceeds forward; when Δn = 0, Q doesn’t change.
  • Changing the temperature leaves the concentrations, and so Q, unchanged; K changes instead. If the forward reaction is endothermic, heating increases K, so Q < K and the reaction proceeds forward.
  • A catalyst or an inert gas at constant volume changes neither Q nor K.

At the particle level: after a reactant is added, reactant molecules collide more often, so the forward rate rises first and the net reaction goes forward until the two rates are equal again (U7.2).

Apply a change and watch where Q and K sit on the number line:

Try it: use Q and K to predict shifts

Example 1: Adding a reactant

At 800 °C, CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g) has K = 0.640, and at equilibrium [CO] = [H₂O] = 0.111 M and [CO₂] = [H₂] = 0.0889 M. At constant volume, CO is added so that [CO] instantly becomes 0.211 M. Use Q to decide which way the equilibrium shifts, then find [CO] at the new equilibrium.

Step-by-step solution

  1. Calculate Q after the change. Only [CO] has changed; the other concentrations are unchanged for the moment:

    Q=[COX2][HX2][CO][HX2O]=(0.0889)2(0.211)(0.111)=7.90×10−32.34×10−2=0.337\begin{aligned} Q &= \frac{[\ce{CO2}][\ce{H2}]}{[\ce{CO}][\ce{H2O}]} \\ &= \frac{(0.0889)^2}{(0.211)(0.111)} \\ &= \frac{7.90 \times 10^{-3}}{2.34 \times 10^{-2}} = 0.337 \end{aligned}

  2. Compare Q with K. Q = 0.337 < K = 0.640: the denominator (reactants) is relatively too large, so the reaction proceeds forward.

  3. Set up an ICE table. Start from the concentrations right after CO is added, and let x more react (concentrations in M):

    COH₂OCO₂, H₂
    I0.2110.1110.0889
    C−x−x+x
    E0.211 − x0.111 − x0.0889 + x

  4. Substitute into K.

    (0.0889+x)2(0.211−x)(0.111−x)=0.640\frac{(0.0889 + x)^2}{(0.211 - x)(0.111 - x)} = 0.640

    The numerator and denominator are not squares of the same expression, so a square root won’t work; expand into a quadratic instead.

  5. Expand and rearrange. Multiply both sides by the denominator. The left side, (0.0889 + x)², expands to x² + 0.178x + 0.00790; the right side, 0.640(0.211 − x)(0.111 − x) = 0.640(x² − 0.322x + 0.0234), expands to 0.640x² − 0.206x + 0.0150:

    x2+0.178x+0.00790=0.640x2−0.206x+0.0150\begin{gathered} x^2 + 0.178x + 0.00790 \\ = 0.640x^2 - 0.206x + 0.0150 \end{gathered}

    Move everything to the left side:

    0.360x2+0.384x−0.00709=00.360x^2 + 0.384x - 0.00709 = 0

  6. Use the quadratic formula. a = 0.360, b = 0.384 and c = −0.00709:

    b2−4ac=0.1474+0.0102=0.1576\begin{aligned} b^2 - 4ac &= 0.1474 + 0.0102 \\ &= 0.1576 \end{aligned} x=−0.384+0.15762×0.360=−0.384+0.3970.720=0.018\begin{aligned} x &= \frac{-0.384 + \sqrt{0.1576}}{2 \times 0.360} \\ &= \frac{-0.384 + 0.397}{0.720} = 0.018 \end{aligned}

    The other root is large and negative and would make concentrations negative, so it is rejected.

  7. Find the new equilibrium concentrations (in M). [H₂] equals [CO₂]:

    [CO]=0.211−0.018=0.193[HX2O]=0.111−0.018=0.093[COX2]=0.0889+0.018=0.107\begin{aligned} [\ce{CO}] &= 0.211 - 0.018 = \boxed{0.193} \\ [\ce{H2O}] &= 0.111 - 0.018 = 0.093 \\ [\ce{CO2}] &= 0.0889 + 0.018 = 0.107 \end{aligned}

  8. Check by substituting back.

    (0.107)2(0.193)(0.093)=0.011450.01795=0.638≈0.640 ✓\begin{aligned} \frac{(0.107)^2}{(0.193)(0.093)} &= \frac{0.01145}{0.01795} \\ &= 0.638 \approx 0.640 \ \checkmark \end{aligned}

  9. Conclusion. At the new equilibrium [CO] = 0.193 M: lower than right after the addition (0.211 M) but still higher than at the original equilibrium (0.111 M). The shift only partly counteracts the change; it cannot undo it completely.

Example 2: Halving the volume

At 25 °C, N₂O₄(g) ⇌ 2NO₂(g) is at equilibrium with K = 6.3 × 10⁻³. The volume is suddenly compressed to half. What does Q become, and which way does the equilibrium shift?

Step-by-step solution

  1. How the concentrations change. At the instant of compression the amounts haven’t changed but the volume is halved, so every gas concentration doubles:

    c′=nV/2=2×nV=2cc' = \frac{n}{V/2} = 2 \times \frac{n}{V} = 2c

  2. Q after compression. Substitute the doubled concentrations, where [ ] means the original equilibrium values:

    Q=(2[NOX2])22[NX2OX4]=4[NOX2]22[NX2OX4]=2×[NOX2]2[NX2OX4]=2K\begin{aligned} Q &= \frac{(2[\ce{NO2}])^2}{2[\ce{N2O4}]} \\ &= \frac{4[\ce{NO2}]^2}{2[\ce{N2O4}]} \\ &= 2 \times \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = 2K \end{aligned}

  3. Substitute the value of K.

    Q=2×6.3×10−3=1.26×10−2\begin{aligned} Q &= 2 \times 6.3 \times 10^{-3} \\ &= \boxed{1.26 \times 10^{-2}} \end{aligned}

  4. Compare Q with K. Q = 2K > K: there is relatively too much product, so the reaction proceeds in reverse: NO₂ decreases and N₂O₄ increases until Q is back to K.

  5. The general rule. When the volume is halved, Q is multiplied by 2Δn, where Δn is the change in gas coefficients (products − reactants). Here Δn = 2 − 1 = +1, so Q doubles. This agrees with “compressing shifts toward fewer gas moles,” but now with a quantitative reason.

Practice: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH < 0, is at equilibrium. How do Q and K change?
  1. Add O₂
  2. Add SO₃
  3. Halve the volume
  4. Double the volume
  5. Raise the temperature
  6. Lower the temperature
  7. Add a catalyst
  8. Add argon at constant volume

Common mistakes

Mistake 1: “Whenever the total pressure rises, the equilibrium shifts toward fewer gas moles.” Only changes in the concentrations (partial pressures) of the reacting species affect Q. Adding an inert gas at constant volume raises the total pressure, but Q is unchanged and there is no shift.

Mistake 2: “Heating changes Q.” At the instant of heating the concentrations haven’t changed, so Q is the same; K is what changes.

Mistake 3: “Changing a concentration changes K.” Concentrations change only Q. K changes only with temperature.

Quick check

A(g) ⇌ 2B(g) is at equilibrium. The volume is compressed to 1/3. At that instant, Q is how many times K?

A reaction has an endothermic forward reaction. After the temperature is raised and equilibrium is re-established, compared with the original equilibrium:

H₂(g) + I₂(g) ⇌ 2HI(g) is at equilibrium. After the volume is doubled, which statement is correct?

Some product is removed from a system at equilibrium. At that instant, the forward and reverse rates:

Which change alters K?

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is at equilibrium. The volume is halved. At that instant, Q becomes:

AP-style free response

AP-style free response

At a certain temperature, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) is at equilibrium in a container with [SO₂] = 0.20 M, [O₂] = 0.10 M, and [SO₃] = 0.40 M. For the forward reaction, ΔH = −197.8 kJ/mol (from NBS enthalpies of formation).

(a) Calculate K at this temperature.

(b) At constant temperature, the volume is suddenly compressed to half. Calculate Q at that instant and determine the direction of reaction.

(c) Explain the direction in (b) in terms of the forward and reverse rates.

(d) If the temperature were raised instead, would Q or K change? Which way would the equilibrium shift?

(e) More O₂ is added to the original equilibrium. How does the fraction of SO₂ converted change? Justify your answer.

Model answer and scoring

(a) K = 0.40² ÷ (0.20² × 0.10) = 0.16 ÷ 0.0040 = 40.

(b) Every concentration doubles: Q = 0.80² ÷ (0.40² × 0.20) = 0.64 ÷ 0.032 = 20 < 40, so the reaction proceeds forward.

(c) After compression all concentrations double. The forward rate depends on [SO₂]²[O₂] (if treated as elementary, it grows 8 times), while the reverse rate depends on [SO₃]² (it grows 4 times). The forward rate increases more, so the net reaction goes forward.

(d) K changes; Q doesn’t. The forward reaction is exothermic, so heating lowers K, making Q > K, and the equilibrium shifts in reverse.

(e) It increases. More O₂ makes Q < K, so the equilibrium shifts forward and more SO₂ is converted to SO₃.

Scoring: (a) K; (b) comparing Q with K; (c) comparing the rates; (d) temperature changes K; (e) adding another reactant raises the conversion.

Summary

  • Changes in concentration or volume change only Q; when Q ≠ K, the reaction proceeds until Q = K.
  • Halving the volume multiplies Q by 2Δn; when Δn = 0, changing the volume doesn’t shift the equilibrium.
  • Temperature changes only K: heating raises K for an endothermic reaction and lowers it for an exothermic one.
  • Catalysts and inert gases at constant volume change neither Q nor K.

AP tip: To explain which way an equilibrium shifts and why, use three steps: which quantity changed (Q or K) → how Q now compares with K → therefore the direction of reaction. This earns more credit than “by Le Châtelier’s principle” alone.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.10 · Reaction Quotient and Le Châtelier's Principle · 6 questions

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