AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/7/7-2)
Unit 7 · Topic 7.2
7.2 Direction of Reversible Reactions
A reversible reaction moves in whichever direction is currently faster. If the forward rate is larger, reactants turn into products overall; if the reverse rate is larger, products turn back into reactants; when the two rates are equal, the system is at equilibrium.
Key terms
- forward rate
- reverse rate
- net change
- equilibrium
Net change comes from the difference in rates
Both the forward and reverse reactions run at the same time. What you observe is the net change, which is the difference between them.
When the forward reaction outpaces the reverse, reactants turn into products faster than products turn back, so the amount of product grows. When the reverse reaction is faster, the amount of reactant grows instead. When the two rates match, nothing changes overall: that's equilibrium.
Why the rates change on their own
Reaction rates depend on concentration (topic 5.2). The forward rate depends on the reactant concentrations; the reverse rate depends on the product concentrations.
Suppose the forward rate is bigger. Reactants get used up, so the forward rate slows. Products build up, so the reverse rate speeds up. The gap between the rates shrinks until they match. This self-correcting behavior is why every reversible reaction in a closed system heads toward equilibrium.
Equilibrium can be approached from either side
You can start with only reactants, only products, or a mixture. Starting with only H₂ and I₂, the forward reaction runs first to make HI. Starting with only HI, the reverse reaction runs first to make H₂ and I₂. Either way, the system ends at a mixture where the rates are equal.
Which direction runs depends on the current mixture, not on how the equation happens to be written on paper.
Picturing it on a rate graph
Starting with only reactants: the forward-rate curve starts high and drops; the reverse-rate curve starts at zero and climbs. They meet, then run together as one flat line.
Starting with only products: the curves swap roles. The reverse rate starts high and falls; the forward rate starts at zero and rises.
In Topic 7.7 you'll learn a number-based way to predict direction, by comparing Q with K. It describes the same thing as comparing rates.
Rates and the equilibrium constant
For a simple one-step reversible reaction A ⇌ B, the forward rate is k_f[A] and the reverse rate is k_r[B], where k_f and k_r are rate constants (topic 5.4). At equilibrium the rates are equal, so k_f[A] = k_r[B], which rearranges to [B]/[A] = k_f/k_r. That ratio is the equilibrium constant, K.
This shows why a catalyst doesn't change the equilibrium: it speeds up the forward and reverse reactions by the same factor, so the ratio k_f/k_r stays the same. The system just reaches equilibrium sooner.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Predicting the net direction from rates
At one moment during the reaction 2NO(g) + O₂(g) ⇌ 2NO₂(g), the forward reaction uses up NO at 3.0 × 10⁻⁴ M/s and the reverse reaction makes NO at 1.0 × 10⁻⁴ M/s. Describe the net change in [NO], and how the rates will change over time.
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- Step 1: NO is consumed faster (3.0 × 10⁻⁴ M/s) than it is made (1.0 × 10⁻⁴ M/s), so the forward rate is greater.
- Step 2: Net change in [NO]: −3.0 × 10⁻⁴ + 1.0 × 10⁻⁴ = −2.0 × 10⁻⁴ M/s. Overall, reactants are being turned into products.
- Step 3: As NO and O₂ are used up, the forward rate decreases. As NO₂ builds up, the reverse rate increases.
- Step 4: This continues until the two rates are equal and [NO] stops changing.
Answer: [NO] is falling at a net 2.0 × 10⁻⁴ M/s; the forward rate will decrease and the reverse rate will increase until they are equal at equilibrium.
- Example 2
Trap: starting from the product side
A sealed flask is filled with only NO₂ gas. A student says that because the reaction is written N₂O₄(g) ⇌ 2NO₂(g), nothing will happen in a flask that has no N₂O₄. Is the student right?
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- Step 1: The reverse reaction, 2NO₂ → N₂O₄, can happen whenever NO₂ is present.
- Step 2: At the start, there's no N₂O₄, so the forward rate is zero and the reverse rate is greater.
- Step 3: There is a net conversion of NO₂ into N₂O₄, and the brown color fades somewhat, until the forward and reverse rates become equal.
Answer: No. The reverse reaction runs faster at first, converting NO₂ to N₂O₄, until the rates become equal at equilibrium.
Common mistakes
- Assuming a reaction can only go in the direction the equation is written.
- Saying the forward reaction 'stops' when products build up. It slows down; the reverse speeds up.
- Confusing a large rate with a large amount. Direction depends on which rate is larger right now, not on which side has more stuff.
On the exam
- When asked which way a reaction proceeds, compare the forward and reverse rates (or Q and K) and then say plainly which way the overall change goes, for example 'reactants are being converted to products overall'.
- Rate–time graphs may ask you to identify the moment equilibrium is reached: it's where the forward and reverse rate curves meet and stay together.
Connected topics
Videos
Check yourself
5 questions on 7.2 Direction of Reversible Reactions. Pick an answer to see if you got it, and why.
| Time (min) | Forward rate, A → B (M/min) | Reverse rate, B → A (M/min) |
|---|---|---|
| 0 | 0.120 | 0.000 |
| 5 | 0.080 | 0.020 |
| 10 | 0.058 | 0.035 |
| 15 | 0.046 | 0.046 |
| 20 | 0.046 | 0.046 |
Hypothetical data for the reaction A(g) ⇌ B(g), which starts with only A in a rigid container at constant temperature
Which of the following is true at 5 min?
Which of the following best explains why the reverse rate starts at zero and then increases?
The experiment is repeated at the same temperature, but the container starts with only B, at the same concentration that A had before. Which of the following would be observed?
| Time (s) | [N₂O₄] (M) | [NO₂] (M) |
|---|---|---|
| 0 | 0.100 | 0.000 |
| 10 | 0.070 | 0.060 |
| 20 | 0.055 | 0.090 |
| 30 | 0.050 | 0.100 |
| 40 | 0.050 | 0.100 |
| 50 | 0.050 | 0.100 |
Hypothetical data. Concentrations in a sealed flask for the reaction N₂O₄(g) ⇌ 2NO₂(g) at constant temperature
Which of the following correctly compares the forward and reverse reaction rates at 10 s?
Which of the following best describes the system at 40 s?
0 of 5 answered