Direction of Reversible Reactions可逆反应的方向

Big Idea · Transformations (TRA)AP Science Practice 6 · ArgumentationAP link · Unit 7~40 min

Learning goals

  • Explain that the net direction of a reversible reaction depends on the relative rates of the forward and reverse reactions
  • Calculate forward and reverse rates from rate constants and concentrations, and predict the direction of reaction and how the concentrations change
  • Describe how the forward and reverse rates change as a system approaches equilibrium, and find equilibrium concentrations from the condition that the two rates are equal

Before you start

  • Introduction to equilibrium (U7.1)
  • Introduction to rate law (U5.2)
  • Elementary reactions (U5.4)

The faster direction wins

The forward and reverse reactions of a reversible process happen at the same time. What we observe is the difference between them, the net reaction净反应:

  • forward rate > reverse rate: reactants decrease and products increase overall, so the reaction proceeds forward;
  • reverse rate > forward rate: products decrease overall, so the reaction proceeds in reverse;
  • equal rates: the concentrations stop changing, and the system is at equilibrium (U7.1).

As long as the system is not at equilibrium, the net reaction slows the faster reaction and speeds up the slower one until they are equal. A reversible reaction therefore never “overshoots”: it always moves toward equilibrium and stays there (in the sense that concentrations stop changing).

Calculating the rates from rate laws

If the forward and reverse reactions are elementary, their rates come straight from rate laws (U5.4). For N₂O₄(g) ⇌ 2NO₂(g):

ratef=kf[NX2OX4]rater=kr[NOX2]2\begin{aligned} \text{rate}_\text{f} &= k_\text{f}[\ce{N2O4}] \\ \text{rate}_\text{r} &= k_\text{r}[\ce{NO2}]^2 \end{aligned}

At equilibrium ratef = rater, which rearranges to [NO₂]²/[N₂O₄] = kf/kr. This ratio is a constant: the equilibrium constant K of the next lesson. Before equilibrium, putting the current concentrations into the same expression gives Q, and you can show that ratef ÷ rater = K ÷ Q: when Q < K, the forward reaction is faster.

Drag the concentrations of the two gases to compare the rates, then “let it react” and watch the rates converge:

Try it: compare the forward and reverse rates
N₂O₄(g) ⇌ 2NO₂(g), K = 6.3 × 10−3 (25 °C)

Example 1: Which direction?

At 25 °C, N₂O₄(g) ⇌ 2NO₂(g) has kf = 0.080 s⁻¹ and kr = 12.8 M⁻¹ s⁻¹ (schematic values). At some moment [N₂O₄] = 0.050 M and [NO₂] = 0.010 M. Which way does the reaction proceed, and how do the two rates change next?

Step-by-step solution

  1. Write both rate laws. Treat the forward and reverse reactions as elementary steps: each rate is proportional to the reactant concentrations raised to their coefficients:

    ratef=kf [NX2OX4]rater=kr [NOX2]2\begin{aligned} \text{rate}_\text{f} &= k_\text{f}\,[\ce{N2O4}] \\ \text{rate}_\text{r} &= k_\text{r}\,[\ce{NO2}]^2 \end{aligned}

  2. Calculate the forward rate.

    ratef=0.080×0.050=4.0×10−3 M/s\begin{aligned} \text{rate}_\text{f} &= 0.080 \times 0.050 \\ &= 4.0 \times 10^{-3}\ \text{M/s} \end{aligned}

  3. Calculate the reverse rate. Two NO₂ molecules take part in the reverse reaction, so [NO₂] is squared:

    rater=12.8×(0.010)2=12.8×1.0×10−4=1.28×10−3 M/s\begin{aligned} \text{rate}_\text{r} &= 12.8 \times (0.010)^2 \\ &= 12.8 \times 1.0 \times 10^{-4} \\ &= 1.28 \times 10^{-3}\ \text{M/s} \end{aligned}

  4. Compare the rates to find the direction.

    ratefrater=4.0×10−31.28×10−3=3.1>1\frac{\text{rate}_\text{f}}{\text{rate}_\text{r}} = \frac{4.0 \times 10^{-3}}{1.28 \times 10^{-3}} = 3.1 > 1

    The forward reaction is faster, so the net reaction goes forward: N₂O₄ decreases and NO₂ increases.

  5. How the rates change next. As [N₂O₄] falls, ratef decreases; as [NO₂] rises, rater increases (as the square). They move toward each other until ratef = rater and equilibrium is reached.

  6. Another check. At equilibrium ratef = rater, so kf[N₂O₄] = kr[NO₂]², which means that at equilibrium

    [NOX2]2[NX2OX4]=kfkr=0.08012.8=6.25×10−3\begin{aligned} \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} &= \frac{k_\text{f}}{k_\text{r}} = \frac{0.080}{12.8} \\ &= 6.25 \times 10^{-3} \end{aligned}

    Right now this ratio is 0.010² ÷ 0.050 = 2.0 × 10⁻³, smaller than at equilibrium. There is not yet enough NO₂, so the reaction must move toward NO₂, which agrees with step 4.

Example 2: Equilibrium concentrations from rate constants

For the elementary reaction A ⇌ B, kf = 0.30 s⁻¹ and kr = 0.10 s⁻¹. At some moment [A] = 0.20 M and [B] = 0.80 M. Which way does the reaction proceed, and what are [A] and [B] at equilibrium?

Step-by-step solution

  1. Compare the two rates.

    ratef=0.30×0.20=0.060 M/srater=0.10×0.80=0.080 M/s\begin{aligned} \text{rate}_\text{f} &= 0.30 \times 0.20 = 0.060\ \text{M/s} \\ \text{rate}_\text{r} &= 0.10 \times 0.80 = 0.080\ \text{M/s} \end{aligned}

    rater > ratef, so the net reaction goes in reverse: B decreases and A increases.

  2. Write the equilibrium condition. At equilibrium the two rates are equal:

    kf [A]=kr [B][B][A]=kfkr=0.300.10=3.0\begin{aligned} k_\text{f}\,[\ce{A}] &= k_\text{r}\,[\ce{B}] \\ \frac{[\ce{B}]}{[\ce{A}]} &= \frac{k_\text{f}}{k_\text{r}} = \frac{0.30}{0.10} = 3.0 \end{aligned}

  3. Set up a table. The reaction runs in reverse, so let B decrease by x; A then increases by x (both coefficients are 1; concentrations in M):

    AB
    Now0.200.80
    Change+x−x
    Equilibrium0.20 + x0.80 − x

  4. Solve the equation. Substitute the equilibrium concentrations into the ratio from step 2:

    0.80−x0.20+x=3.00.80−x=0.60+3.0x0.20=4.0xx=0.050\begin{aligned} \frac{0.80 - x}{0.20 + x} &= 3.0 \\ 0.80 - x &= 0.60 + 3.0x \\ 0.20 &= 4.0x \\ x &= 0.050 \end{aligned}

  5. Find the equilibrium concentrations.

    [A]=0.20+0.050=0.25 M[B]=0.80−0.050=0.75 M\begin{aligned} [\ce{A}] &= 0.20 + 0.050 = \boxed{0.25\ \text{M}} \\ [\ce{B}] &= 0.80 - 0.050 = \boxed{0.75\ \text{M}} \end{aligned}

    B has indeed decreased, as predicted in step 1.

  6. Check. At equilibrium the two rates should be equal:

    ratef=0.30×0.25=0.075rater=0.10×0.75=0.075 ✓\begin{aligned} \text{rate}_\text{f} &= 0.30 \times 0.25 = 0.075 \\ \text{rate}_\text{r} &= 0.10 \times 0.75 = 0.075 \ \checkmark \end{aligned}
Practice: at each moment, which way does the net reaction go?
  1. ratef = 3.0 × 10⁻³ M/s, rater = 1.0 × 10⁻³ M/s
  2. ratef = rater
  3. The container holds only products, no reactants
  4. The reactant concentration is decreasing
  5. The product concentration is decreasing
  6. rater is twice ratef
  7. No concentration is changing
  8. Two reactants have just been mixed; there is no product yet

Common mistakes

Mistake 1: “A reversible reaction always proceeds forward.” The direction depends on which rate is larger. When there is relatively a lot of product, the reverse reaction is faster and the reaction proceeds in reverse.

Mistake 2: “While a reaction proceeds forward, the reverse reaction isn’t happening.” Whenever product is present, the reverse reaction occurs; it is just slower than the forward one. What we observe is the difference.

Mistake 3: “The reaction continues until a reactant runs out.” A reversible reaction stops changing overall when the two rates become equal, with reactants and products both present.

Quick check

In a closed container, the concentration of a product of a reversible reaction is decreasing. At this moment:

At some moment in A ⇌ B, the forward rate is twice the reverse rate. Over the next while:

Pure B is placed in a container, where A ⇌ B. At the very start:

For the elementary reaction A ⇌ B, kf = 2kr. At equilibrium, [B] ÷ [A] equals:

On a concentration–time graph, the curve for P starts at zero, rises, and levels off, while the curve for R falls and levels off. Which statement is correct?

A reversible reaction is at equilibrium. At constant temperature, more reactant is suddenly added. At that instant:

AP-style free response

AP-style free response

For the elementary reaction A(g) ⇌ B(g), kf = 0.20 s⁻¹ and kr = 0.050 s⁻¹ (schematic values). At a constant temperature, A is added to a container so that [A] = 1.00 M and [B] = 0.

(a) Calculate the initial forward and reverse rates.

(b) At some moment [A] = 0.60 M and [B] = 0.40 M. Calculate both rates and determine the direction of reaction.

(c) Calculate [A] and [B] at equilibrium.

(d) In another experiment only B is added ([B] = 1.00 M). Which way does the reaction proceed at first? How do the equilibrium concentrations compare with (c)?

(e) Describe how the forward and reverse rates change from the start until equilibrium, and give their values at equilibrium.

Model answer and scoring

(a) ratef = 0.20 × 1.00 = 0.20 M/s; rater = 0.050 × 0 = 0.

(b) ratef = 0.20 × 0.60 = 0.12 M/s and rater = 0.050 × 0.40 = 0.020 M/s; the forward reaction is faster, so the reaction proceeds forward.

(c) At equilibrium 0.20[A] = 0.050[B], so [B] = 4.0[A]; with [A] + [B] = 1.00, [A] = 0.20 M and [B] = 0.80 M.

(d) In reverse (at first only the reverse reaction occurs). The equilibrium concentrations are the same: [A] = 0.20 M and [B] = 0.80 M, because the temperature and total amount are the same.

(e) The forward rate decreases from 0.20 M/s and the reverse rate increases from 0; at equilibrium both are 0.20 × 0.20 = 0.050 × 0.80 = 0.040 M/s.

Scoring: (a)(b) rates from rate laws and the direction; (c) equal rates at equilibrium; (d) same equilibrium from either side; (e) trends in the rates and their equilibrium value.

Summary

  • The net reaction goes in the direction of the faster reaction; equilibrium is reached when the two rates are equal.
  • For elementary reactions, rates come from rate laws: ratef = kf × (reactant concentrations to their powers), rater = kr × (product concentrations to their powers).
  • As a system approaches equilibrium, the faster reaction slows and the slower one speeds up; at equilibrium kf/kr equals a constant ratio of concentrations (the equilibrium constant).

AP tip: When predicting direction, say what is being compared: “the forward rate is greater than the reverse rate, so reactants decrease and products increase overall.” Don’t just say “it shifts right.”

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.2 · Direction of Reversible Reactions · 5 questions

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