Q.A reaction is first order in A and second order in B
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Effect of Concentration
Effect of Concentration: The Intuition
Imagine you're in a large, empty hall with just one other person. The two of you are trying to bump into each other accidentally. It'll take a while, right? Now imagine the same hall packed with a thousand people. Bumping into someone becomes almost certain within seconds.
That's the core idea behind the effect of concentration on reaction rates. Concentration simply means how much of a substance is packed into a given space. Higher concentration = more particles in the same volume.
When particles are more crowded, they collide more frequently. And since chemical reactions happen only when particles collide with enough energy and the right orientation, more collisions mean more reactions per second. The reaction speeds up.
The Precise Statement
For most chemical reactions, the rate of a reaction is directly proportional to the molar concentration of the reactants (raised to some power, which we'll get to).
Rate∝[Reactant]n
Here, [ ] means "concentration in moles per litre" (mol/L or M), and n is the order of reaction with respect to that reactant.
What does "order" mean?
For a simple reaction like A→Products:
- First order (n=1): Double the concentration of A → double the rate.
- Second order (n=2): Double the concentration of A → quadruple the rate (22=4).
- Zero order (n=0): Changing concentration has no effect on the rate. This happens when the reaction is limited by something else (like a catalyst surface that's already fully covered).
The order n is not the same as the stoichiometric coefficient from the balanced equation. It must be determined experimentally. For example, the reaction 2A→B could be first order in A, not second order.
Why does this happen? The collision theory
The rate depends on two things:
- Collision frequency — how often particles meet.
- Fraction of effective collisions — how many of those collisions have enough energy (activation energy) and the right orientation.
Doubling the concentration doubles the number of particles per unit volume. This roughly doubles the collision frequency. For a first-order reaction, that directly doubles the rate. For higher orders, the effect compounds because multiple reactant particles must meet simultaneously.
A concrete example
Consider the reaction between hydrochloric acid and sodium thiosulphate:
Na2S2O3(aq)+2HCl(aq)→2NaCl(aq)+S(s)+SO2(g)+H2O(l) …
Why this formula?
Effect of Concentration on Reaction Rate — The Reasoning
The Effect of Concentration is rooted in collision theory. The key idea is simple:
More particles in the same volume → more frequent collisions → higher reaction rate.
Let's break down why the mathematical relationships hold.
1. The Rate Law — Why r=k[A]m[B]n?
This is not derived from theory alone — it is empirical (found experimentally). But the reasoning behind its form comes from collision probability.
For an elementary reaction (one step):
Consider:
A+B→products
- The rate depends on how often A and B molecules meet.
- In a given volume, the number of A molecules is proportional to [A], and the number of B molecules is proportional to [B].
- The number of A–B collisions per second is proportional to the product:
Collision frequency∝[A]×[B]
- Therefore:
r∝[A][B]
or
r=k[A][B]
For a reaction with coefficient aA+bB:
If the reaction is elementary, the stoichiometric coefficients become the exponents:
r=k[A]a[B]b
Why? Because for 2A to react, two A molecules must collide simultaneously — the probability of that happening is proportional to [A]×[A]=[A]2.
2. The Integrated Rate Laws — Why These Forms?
These come from solving the differential equation r=−dtd[A]=k[A]n.
Zero-order (n=0):
−dtd[A]=k
- Reasoning: Rate is independent of concentration. This happens when the reaction is limited by something else (e.g., a saturated catalyst surface).
- Integrate:
∫[A]0[A]d[A]=−k∫0tdt
⇒[A]=[A]0−kt
First-order (n=1):
−dtd[A]=k[A]
- Reasoning: Rate is directly proportional to [A]. Each molecule has a constant probability of reacting per unit time (like radioactive decay).
- Integrate:
∫[A]0[A][A]d[A]=−k∫0tdt
⇒ln[A]=ln[A]0−kt
or
[A]=[A]0e−kt
Second-order (n=2):
−dtd[A]=k[A]2
- Reasoning: Rate depends on two molecules of A colliding. Doubling [A] quadruples the collision frequency.
- Integrate:
∫[A]0[A][A]2d[A]=−k∫0tdt
⇒[A]1=[A]01+kt
3. The Half-Life — Why It Depends on Order …
With order 1 in A and 2 in B, the rate law is rate=k[A][B]2; changing concentrations scales the rate by the corresponding powers. …
Rate law =k[A][B]2; [B]×3⇒ rate ×9; both ×2⇒ rate ×8.
Concept. Order and the effect of concentration changes on rate — CBSE Class-12 chemical-kinetics.
(i) Differential rate equation. First order in A, second order in B:
rate=−dtd[A]=k[A]1[B]2.
(ii) Effect of tripling [B]. New rate =k[A](3[B])2=9k[A][B]2, i.e. the rate becomes 9 times the original.
…
- GUJCET 2024Set 131 markMCQQ.For any reaction the rate constant K=2.3×10−5 mol−3/2 L3/2S−1; then the order of reaction will be ________. (A) 0.0 (zero) (B) 1.5 (C) 0.5 (D) 2.5
›Reveal solutionSolution
The units of the rate constant are mol1−nLn−1s−1 for an n-th order reaction; match the exponents.
Concept: For an n-order reaction, k has units mol1−nLn−1s−1.
Given units mol−3/2 L3/2 s−1: …
- GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQQ.A reaction is first order in A and second order in B. How many times the rate constant affected on increasing the concentration of B three times.(a) 9 times decreases(b) 9 times increases(c) 6 times increases(d) 6 times decreases
›Reveal solutionSolution
Rate = k[A]^1[B]^2, so tripling [B] multiplies the rate by 3^2 = 9; note the rate constant k is a temperature-dependent constant and does not itself change with concentration - it is the reaction RATE that changes.
Rate1 = k[A][B]^2
If [B] becomes 3[B]: Rate2 = kA^2 = 9 x k[A][B]^2 = 9 x Rate1
…
- GUJCET 2023Set 091 markMCQQ.A reaction is first order in terms of A and second order in terms of B. What will be the rate of reaction, if concentration of B is increased two times? (A) 4-Times (B) 2-Times (C) 8-Times (D) 16-Times
›Reveal solutionSolution
[!TLDR]
With rate =k[A][B]2, doubling [B] increases the rate 4 times.
Concept
The order of a reactant is the exponent on its concentration in the rate law. If concentration is changed by a factor x, the rate changes by xorder.
Solution
Given first order in A and second order in B:
rate=k[A]1[B]2 …
- GUJCET 2022Set 171 markMCQQ.A reaction is first order with respect to a reactant A and second order with respect to reactant B. What is the effect of rate when concentration of both A and B increased by doubled? (A) Eight times (B) Quadrupled (C) Doubled (D) Sixteen times
›Reveal solutionSolution
21×22=8 ⇒ rate becomes eight times.
Concept. First order in A, second order in B:
rate=k[A][B]2 …
- GUJCET 2019Set 131 markMCQQ.In a reaction A→B, if the concentration of reactant is increased by 9 times then rate of reaction increases 3 times. What is the order of reaction? (A) 31 (B) 21 (C) 3 (D) 2
›Reveal solutionSolution
If multiplying concentration by 9 multiplies rate by 3, then 9n=3⇒n=21.
Concept — order of reaction. Rate =k[A]n. When [A] scales by a factor, the rate scales by (factor)n.
Steps. …
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