Following reaction takes place in one step:

2A + B → 2C

How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

Q: Following reaction takes place in one step: 2A + B → 2C How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume? Will there be any change in the order of reaction with the reduced volume?

**Rate law (one-step reaction)** Because the reaction is elementary, the rate law follows the stoichiometry: $\text{Rate} = k[A]^2[B]$ **Effect of reducing volume to one-third** Reducing the volume to ⅓ makes each concentration **3 times** larger: - $\text{New rate} = k(3[A])^2(3[B]) = k \cdot 9[A]^2 \cdot 3[B] = $**$27 k[A]^2[B]$** So the new rate is **27 times the original rate**. **Order of reaction** $\text{Order} = 2 + 1 = $**3**. Order depends only on the rate law, not on concentration or volume, so it **remains unchanged (third order)**.

Chemical Kinetics

Effect of concentration/volume on rate and order of reaction

Official Answer

Rate law (one-step reaction)

Because the reaction is elementary, the rate law follows the stoichiometry:

$\text{Rate} = k[A]^2[B]$

Effect of reducing volume to one-third

Reducing the volume to ⅓ makes each concentration 3 times larger:

$\text{New rate} = k(3[A])^2(3[B]) = k \cdot 9[A]^2 \cdot 3[B] =$ $27 k[A]^2[B]$

So the new rate is 27 times the original rate.

Order of reaction

$\text{Order} = 2 + 1 =$ 3. Order depends only on the rate law, not on concentration or volume, so it remains unchanged (third order).

elementary reactionrate law k[A]²[B]concentration tripledrate becomes 27 timesthird order reactionorder unchangedmolecularity equals ordervolume reduced to one-third

Marking Scheme

11 mark: writing the correct rate law $\text{Rate} = k[A]^2[B]$ and recognising concentrations become 3× when volume is ⅓.
20.5 mark: new rate = 27 × original rate.
30.5 mark: stating the order (3) is unchanged.

Hint

Write $\text{Rate} = k[A]^2[B]$ for the one-step reaction; reducing volume to ⅓ triples each concentration, so rate becomes $3^2 \times 3 = 27$ times; order is unchanged.

Quick Oral Answer

For this one-step reaction the rate is $k[A]^2[B]$ ; shrinking the volume to one-third triples both concentrations, so the rate rises by $3^2 \times 3 = 27$ times, while the order stays third order because order does not depend on concentration.

Analysis & Explanation

Concept — elementary reactions

For a single-step (elementary) reaction the molecularity equals the order, so the exponents in the rate law are simply the stoichiometric coefficients: second order in A and first order in B.

Why 27×

Concentration = moles / volume. Halving or reducing volume raises every concentration in inverse proportion.
Volume → V/3 means [A] → 3[A] and [B] → 3[B].
The rate scales as $(3)^2$ for A times $(3)^1$ for B $= 9 \times 3 = 27$ .

Exam trap

Do not change the order — increasing concentration changes the rate, never the order, which is a fixed property of the rate law.
Remember the coefficient of A is 2, so its concentration factor is squared, not just tripled.

Real-world link

This is why compressing a gaseous fuel–air mixture (as in an engine cylinder) sharply speeds up combustion — squeezing the volume multiplies every reactant concentration.

Common Mistakes

1Increasing rate by only 9 or by 3 — forgetting that A is squared AND B is also tripled ( $9 \times 3 = 27$ ).
2Claiming the order changes to a new value — order is a constant fixed by the rate law, not by concentration.
3Writing the rate law with wrong exponents (e.g., $[A][B]$ ) because they forget it is a one-step reaction where coefficients become exponents.

Interesting Facts

Only for elementary reactions can you read the rate-law exponents straight off the balanced equation — for multi-step reactions they must be found experimentally.

A true termolecular step like 2A + B is rare in practice because a simultaneous three-body collision is statistically very unlikely.

The idea that compressing reactants speeds a reaction is Le Chatelier and kinetics working together — it underlies high-pressure industrial processes like ammonia synthesis.

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Frequently Asked Questions

Why does reducing the volume to one-third triple the concentration?

Concentration is moles divided by volume. The number of moles of each reactant stays the same, but the volume becomes one-third, so each concentration (moles/volume) becomes three times its original value.

Does changing volume ever change the order of a reaction?

No. The order is a fixed property of the rate law, determined by the reaction mechanism. Changing concentration or volume changes the reaction rate but never the order — here it stays third order regardless of the vessel volume.