“Why the Rate-Determining Step Is Not Always Obvious” A-Level Chemistry

 

“Why the Rate-Determining Step Is Not Always Obvious”

A student sees a three-step mechanism and immediately chooses the slowest-looking step as the answer.

But in A Level Chemistry, the examiner is not asking which step looks slow.

They are asking whether the proposed mechanism matches the experimental rate equation.

That is where the trap begins.

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Hard non-calculation question

Question

The reaction between nitrogen monoxide and hydrogen is represented by the overall equation:

$$2NO(g) + 2H_2(g) \rightarrow N_2(g) + 2H_2O(g)$$

Experimental results show that the rate equation is:

$$rate=k[NO{]}^2[H_2]$$

A student suggests the following mechanism:

Step 1

$$NO+NO\rightleftharpoons N_2{O}_{\mathrm{2<span\; style="font-family:\; arial;"></span>}}$$

Step 2

$$N_2{O}_2+H_2\to N_{2}O+H_{2}O$$

Step 3

$$N_{2}O+H_2\to N_2+H_{2}O$$

The student says:

“Step 3 must be the rate-determining step because it produces the final product, nitrogen.”

Explain whether the student is correct.

Your answer should refer to:

• the meaning of the rate-determining step
• the experimental rate equation
• the species present in the proposed mechanism
• why Step 2 is more likely to be the rate-determining step

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Model answer

The student is not correct.

The rate-determining step is the slowest step in a reaction mechanism. It limits the overall rate of the reaction, rather like the slowest stage in a production line. However, it is not necessarily the final step, and it is not chosen simply because it produces one of the final products.

The experimental rate equation is:

$$rate=k[NO{]}^2[H_2]$$

This tells us that the rate depends on the concentration of nitrogen monoxide squared and the concentration of hydrogen to the power of one.

The first step in the mechanism involves two molecules of nitrogen monoxide combining:

$$NO+NO\rightleftharpoons N_2{O}_{\mathrm{2<span\; style="font-family:\; arial;"></span>}}$$

This produces the intermediate $N_2O_2$. If this step is fast and reversible, then the concentration of $N_2O_2$ depends on the concentration of $NO$ squared.

The second step is:

$$N_2{O}_2+H_2\to N_{2}O+H_{2}O$$

If this is the rate-determining step, then the rate depends on $N_2O_2$ and $H_2$. Since $N_2O_2$ is formed from two $NO$ molecules, this gives a rate equation consistent with:

$$rate = k[NO]^2[H_2]$$

This matches the experimental rate equation.

Step 3 involves:

$$N_2O + H_2 \rightarrow N_2 + H_2O$$

If Step 3 were the rate-determining step, the rate would be expected to depend on the concentration of $N_2O$ and $H_2$, not directly on $NO^2$ and $H_2$. Since $N_2O$ is an intermediate and does not appear in the experimental rate equation, Step 3 is not supported as the rate-determining step.

Therefore, Step 2 is more likely to be the rate-determining step because it explains the experimentally observed rate equation.

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Why this makes a strong blog

This question is difficult because it tests several layers of understanding at once.

Many students know that the rate-determining step is the slowest step, but they do not always realise that the proposed mechanism must agree with the experimentally determined rate equation.

The key teaching point is:

A mechanism is not proved just because it adds up to the overall equation.

It must also explain the rate equation.

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Suggested blog structure

1. Why rates questions catch out good students

Start by explaining that rates questions often look simple because students recognise familiar words: rate equation, mechanism, intermediate, slow step.

The problem is that the question is not really about memorising definitions. It is about using evidence.

You could write:

“In rates questions, the experimental rate equation is the evidence. The mechanism is the explanation. The job of the chemist is to decide whether the explanation fits the evidence.”

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2. What the rate-determining step really means

Use a simple analogy.

A reaction mechanism is like a queue at a ferry crossing. Cars may arrive quickly, tickets may be checked quickly, but if only one ferry can load slowly, that step controls the whole process.

The rate-determining step is the bottleneck.

But the bottleneck does not have to be the final step.

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3. Why the final step is not automatically the slow step

This is the misconception to attack directly.

Students often think:

• the final step must be important
• the final product appears there
• therefore it must control the rate

But the final step may actually be very fast. The reaction may be limited by how quickly an intermediate is produced or how quickly it reacts earlier in the mechanism.

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4. How the rate equation gives the clue

This is the most important section.

Explain that the rate equation tells us which reactants affect the rate.

For this example:

$$rate=k[NO{]}^2[H_2]$$

This means:

• two molecules of $NO$ are involved before or during the rate-determining step
• one molecule of $H_2$ is involved before or during the rate-determining step
• the mechanism must explain why the reaction is second order with respect to $NO$

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5. Intermediates are allowed in mechanisms, but not in final rate equations

This is a really useful exam point.

The rate equation should be written using species whose concentrations can be measured experimentally, normally reactants.

Intermediates such as $N_2O_2$ or $N_2\mathrm{O<span\; style="font-family:\; arial;"> </span><span\; style="font-family:\; arial;">appear\; during\; the\; mechanism\; but\; are\; not\; present\; in\; the\; overall\; equation.</span>}$

Students need to understand that intermediate concentrations may be linked back to reactant concentrations.

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6. The final explanation

Bring it together:

Step 1 produces $N_2O_2$ from two $NO$ molecules.

Step 2 uses $N_2O_2$ and $H_2$.

So if Step 2 is slow, the rate depends on:

$$[NO]^2[H_2]$$

This matches the experimental rate equation.

Therefore Step 2 is the best candidate for the rate-determining step.

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Extension challenge for stronger students

You could add this at the end of the blog:

Challenge question

A different reaction has the mechanism:

Step 1

$$A + B \rightleftharpoons C$$

Step 2

$$C + D \rightarrow E$$

Step 3

$$E + B \rightarrow F$$

The experimental rate equation is:

$$rate = k[A][B][D]$$

Which step is most likely to be the rate-determining step?

Answer

Step 2 is most likely to be the rate-determining step.

Step 1 forms $C$ from $A$ and $B$. Step 2 then uses $C$ and $D$.

So the rate depends on:

$$[A][B][D]$$

This matches the experimental rate equation.