Reversible Reactions and Dynamic Equilibrium可逆反应与动态平衡

Big Idea · Transformations (TRA)AP Science Practice 6 · ArgumentationAP link · Unit 7~35 min

Learning goals

  • Describe reversible reactions and write them with a double arrow
  • Describe dynamic equilibrium, where forward and reverse rates are equal and concentrations stay constant
  • Decide whether a reversible reaction has reached equilibrium

Before you start

  • Theoretical and percent yield (2.6.7)
  • Factors that affect reaction rate (2.11.2)

Why is there always nitrogen and hydrogen left over?

As you saw in 2.6.7, making ammonia never gives a 100% yield. Seal 1 mol of N₂ and 3 mol of H₂ in a container and wait as long as you like: you never get 2 mol of NH₃. The container always holds a mixture of N₂, H₂, and NH₃, because as NH₃ forms, some of it breaks down again.

Reversible reactions

A reversible reaction可逆反应 runs in both directions under the same conditions. It is written with a double arrow:

NX2(g)+3 HX2(g)⇌2 NHX3(g)\ce{N2(g) + 3H2(g) <=> 2NH3(g)}

Left to right is the forward reaction; right to left is the reverse reaction. The ionization of a weak acid in 2.7.5 (CH₃COOH ⇌ CH₃COO⁻ + H⁺) is another example. In a reversible reaction the reactants are never completely used up.

Dynamic equilibrium

When a reversible reaction starts in a closed container, the reactant concentrations are highest, so the forward rate is at its maximum and the reverse rate is zero. As products build up, the forward rate falls and the reverse rate rises. When the two rates become equal, the concentrations of reactants and products stop changing: the system has reached chemical equilibrium化学平衡.

Equilibrium is a dynamic equilibrium动态平衡. Both reactions keep going at the same rate, so there is no overall change, but molecules are still reacting in both directions. The concentrations at equilibrium are constant but usually not equal.

Under the same conditions, the same equilibrium is reached whether you start from reactants or from products. Here is N₂O₄ ⇌ 2NO₂ at 25 °C (NO₂ is red-brown, N₂O₄ is colorless), started three different ways:

Try it: how N₂O₄ ⇌ 2NO₂ reaches equilibrium
Start with
Color of the gas (NO₂ is red-brown)

How to tell that a system is at equilibrium

  • The forward and reverse rates are equal. When you compare rates for two different substances, one must be a forward rate and the other a reverse rate, in the ratio of their coefficients.
  • Concentrations (and amounts) stop changing. They don’t have to be equal, or in the ratio of the coefficients.
  • In a rigid container, if the moles of gas differ on the two sides, a constant total pressure means equilibrium. If the moles of gas are the same on both sides, the pressure never changes, so it tells you nothing.

How Chinese textbooks differ

  • Chinese textbooks sum up the features of equilibrium in five characters, 逆、等、动、定、变: a reversible reaction, equal rates, dynamic, constant concentrations, and a shift when conditions change.
  • Chinese problem sets include many multiple-choice questions asking which observation shows that a system is at equilibrium, like Example 1. US and AP courses more often test the idea with concentration–time graphs and particle diagrams.

Example 1: is it at equilibrium?

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) runs in a rigid, sealed container. Which of these show that it has reached equilibrium? (a) N₂ is used up at 1 mol/min while NH₃ forms at 2 mol/min. (b) H₂ is used up at 3 mol/min while NH₃ is used up at 2 mol/min. (c) [N₂] : [H₂] : [NH₃] = 1 : 3 : 2. (d) The total pressure stays constant.

Step-by-step solution
  1. Two tests. Either the forward and reverse rates are equal (one forward process and one reverse process, matched through the coefficients), or something that changes as the reaction runs has stopped changing.
  2. (a) No. Using up N₂ and forming NH₃ are both the forward reaction; they are always in a 1 : 2 ratio.
  3. (b) Yes. Using up H₂ is the forward reaction and using up NH₃ is the reverse reaction. Using up 3 mol/min of H₂ forms 2 mol/min of NH₃, exactly as fast as NH₃ is used up, so the forward and reverse rates are equal.
  4. (c) No. Equilibrium concentrations don’t have to be in the ratio of the coefficients; they depend on the starting amounts.
  5. (d) Yes. The moles of gas change (4 → 2), so in a rigid container the pressure changes as the reaction runs. Once it stops changing, every amount has stopped changing.
  6. A common mistake. Using constant pressure as the test for H₂(g) + I₂(g) ⇌ 2HI(g). The moles of gas are the same on both sides, so the pressure never changes and tells you nothing.

Example 2: starting from either side

At 25 °C, flask A (1.00 L) starts with 0.040 mol of N₂O₄ and flask B (1.00 L) starts with 0.080 mol of NO₂. How do the two equilibrium states compare?

Step-by-step solution
  1. Same atoms. 0.080 mol of NO₂ would make exactly 0.040 mol of N₂O₄, so both flasks contain the same amounts of N and O.

  2. Different directions. Flask A starts with the forward reaction, so N₂O₄ breaks down and the color deepens. Flask B starts with the reverse reaction, so NO₂ combines and the color fades.

  3. Same result. At the same temperature and volume both flasks reach the same equilibrium, with the same color.

  4. The numbers, with an ICE table (come back after lesson 2.11.4). At 25 °C, K = 6.25 × 10⁻³ for N₂O₄ ⇌ 2NO₂. Let x M of N₂O₄ break down in flask A:

    (M)N₂O₄NO₂
    I0.0400
    C−x+2x
    E0.040 − x2x

    Substitute into K:

    (2x)20.040−x=6.25×10−3\frac{(2x)^2}{0.040 - x} = 6.25 \times 10^{-3}
  5. Solve. Clearing the fraction gives a quadratic:

    4x2+6.25×10−3 x−2.50×10−4=0\begin{gathered} 4x^2 + 6.25 \times 10^{-3}\,x \\ - 2.50 \times 10^{-4} = 0 \end{gathered}

    The positive root is x = 7.16 × 10⁻³ M:

    [NX2OX4]=0.040−0.00716≈0.033 M[NOX2]=2×0.00716≈0.014 M\begin{aligned} [\ce{N2O4}] &= 0.040 - 0.00716 \\ &\approx \boxed{0.033\ \text{M}} \\ [\ce{NO2}] &= 2 \times 0.00716 \\ &\approx \boxed{0.014\ \text{M}} \end{aligned}
  6. Flask B too. Let y M of N₂O₄ form in flask B, so [N₂O₄] = y and [NO₂] = 0.080 − 2y. The same K gives y = 0.0328, exactly matching flask A.

Practice: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) in a rigid container at constant temperature. Does this show that equilibrium has been reached?
  1. [N₂] stops changing.
  2. [N₂] : [H₂] : [NH₃] = 1 : 3 : 2
  3. N₂ forms at 1 mol/min while NH₃ forms at 2 mol/min.
  4. N₂ is used up at 1 mol/min while NH₃ forms at 2 mol/min.
  5. The total pressure stops changing.
  6. The density of the gas mixture stops changing.
  7. The forward rate for H₂ equals 3 times the reverse rate for N₂.
  8. All of the N₂ and H₂ has turned into NH₃.

Common mistakes

Mistake 1: “At equilibrium the reaction stops.” Equilibrium is dynamic. The forward and reverse reactions continue at equal rates, so concentrations don’t change.

Mistake 2: “At equilibrium, reactant and product concentrations are equal.” The concentrations are constant, not equal. Equilibrium can lie toward reactants or toward products.

Mistake 3: “Burning hydrogen makes water and electrolysis turns water back into hydrogen, so this is a reversible reaction.” The two reactions need different conditions (a spark and an electric current) and don’t happen at the same time, so this isn’t a reversible reaction.

Quick check

Which of these is a reversible reaction?

What is true at chemical equilibrium?

Summary

  • A reversible reaction runs in both directions under the same conditions, written with ⇌. The reactants are never completely used up.
  • At equilibrium, forward rate = reverse rate ≠ 0 and the concentrations stay constant. The equilibrium is dynamic and can be reached from either direction.
  • To show equilibrium, use one forward and one reverse rate in the ratio of the coefficients, or a quantity that would otherwise change but has stopped changing.
On to AP (for later)

AP topic 7.1 asks you to read concentration–time and rate–time graphs and identify when equilibrium is reached, explaining that the forward and reverse rates are equal while the concentrations usually aren’t. Topic 7.2 covers direction: equilibrium can be reached from either side. A standard phrase for answers is “the rates of the forward and reverse reactions are equal.”

Practice this chapter: 2.11 · Reaction Rates and Equilibrium · 10 questions

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