Q.The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
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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 …
The key idea is that for a second-order reaction, the rate depends on the square of the concentration of the reactant.
Reasoning:
- For the conversion X→Y following second-order kinetics, the rate law is:
Rate=k[X]2
- Let the initial concentration be [X]0, so the initial rate is r0=k[X]02. …
For a second-order reaction, rate depends on the square of the concentration. Tripling [X] multiplies the rate by 32=9, so the rate of formation of Y becomes 9 times the original.
Why concentration matters this way
The rate of a chemical reaction tells us how fast reactants turn into products. For the conversion X→Y, the rate of formation of Y is exactly the rate at which X is consumed (assuming no side reactions). The problem states this follows second order kinetics. That means the rate law is:
Rate=k[X]2
where k is the rate constant (depends only on temperature, not on concentration). The exponent 2 is what makes it second order — and that exponent is the key to the entire question.
A common mistake is to think "second order" means the rate doubles when concentration doubles. That would be true only for first order. For second order, the effect is squared, not linear.
Step-by-step reasoning
- Write the original rate. Let the initial concentration of X be [X]0. Then the original rate r0 is:
r0=k[X]02
- Apply the change. The concentration is increased to three times its original value:
[X]new=3[X]0
- Write the new rate. …
Method: Rate Law Analysis for Second-Order Kinetics
Step 1: Write the general rate law for a second-order reaction
For a second-order reaction involving a single reactant X:
Rate=k[X]2
Here:
- k = rate constant (depends only on temperature, not on concentration)
- [X] = concentration of X
- Rate = rate of formation of Y (since X → Y)
Step 2: Identify the initial condition
Let the initial concentration be [X]0.
The initial rate is:
Rate1=k[X]02
Step 3: Apply the change in concentration
Concentration of X is increased to three times:
[X]new=3[X]0
Step 4: Calculate the new rate
Rate2=k(3[X]0)2=k⋅9[X]02
Step 5: Compare the rates …
Here are the common mistakes students make on this question, along with clear strategies to avoid them.
Mistake 1: Confusing Order of Reaction with Stoichiometry
- The Mistake: Students see "conversion of X to Y" and assume the rate law is r=k[X]1 (first order) because the balanced equation looks like a 1:1 conversion. They then incorrectly calculate the new rate as 3× the original.
- Why it's wrong: The order (second order) is given in the problem statement. It is an experimental fact, not derived from the balanced chemical equation. For a second-order reaction, the rate depends on [X]2, not [X].
- How to Avoid: Always underline the given order in the question. Before writing any formula, ask yourself: "What is the order? What is the exponent on concentration?" Here, second order means exponent = 2.
Mistake 2: Forgetting to Square the Concentration Change
- The Mistake: Students correctly identify the rate law as r=k[X]2, but then plug in the new concentration incorrectly. They write:
- Original rate: r1=k[X]2
- New concentration: [X]new=3[X]
- New rate: r2=k×3[X] (missing the square)
- Result: r2=3×r1 (wrong)
- Why it's wrong: The rate law says square the concentration, not multiply by the factor. You must substitute the entire new concentration into the squared term.
- How to Avoid: Write the substitution step explicitly:
r2=k(3[X])2=k×9[X]2=9×(k[X]2)=9r1
Bold the key step: (3[X])2=9[X]2.
Mistake 3: Misinterpreting "Rate of Formation of Y"
- The Mistake: Students think the rate of formation of Y is different from the rate of disappearance of X. They try to use stoichiometric ratios (e.g., −dtd[X]=dtd[Y]) and get confused.
- Why it's wrong: For the reaction X→Y, the rate of disappearance of X equals the rate of formation of Y (since 1 mole of X gives 1 mole of Y). The question asks for the effect on the rate of formation of Y, which is exactly the same as the rate of the reaction.
- How to Avoid: Remember: For a simple conversion A→B, the rate of reaction = rate of formation of product = rate of consumption of reactant. No extra steps needed. Just apply the rate law directly.
Mistake 4: Not Stating the Final Answer Clearly …
- CBSE 2026Set 56/3/11 markMCQQ.Assertion (A) : Order of reaction is applicable to elementary as well as complex reactions. Reason (R) : Order of a reaction is an experimental quantity. Options : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.
›Reveal solutionSolution
The assertion is false because order is defined only for simple rate laws (elementary steps or overall reactions with a simple rate expression), not for all complex reactions. The reason is true: order is experimental. So the correct option is (D).
Concept First: What "Order" Really Means
The order of a reaction is the sum of the exponents of concentration terms in the experimentally determined rate law. For an elementary reaction (a single step), the order equals the molecularity — that’s straightforward. But for a complex reaction (a sequence of steps), the overall rate law can be messy: it might involve fractional exponents, negative exponents, or even terms that don’t look like a simple power law at all. In such cases, the concept of "order" simply doesn’t apply in the usual sense.
The reason given is a fundamental truth: order is always found by experiment, never deduced from the balanced equation (except for elementary steps). That’s correct.
Now let’s examine the assertion carefully.
Step-by-Step Reasoning
-
What does "applicable" mean here?
The assertion says order is "applicable" to both elementary and complex reactions. If a reaction has a rate law of the form r=k[A]m[B]n, then we can define order =m+n. For an elementary reaction, this always works. For a complex reaction, it works only if the overall rate law happens to be a simple power law — which is not guaranteed.
-
Counterexample: a complex reaction where order is not defined
Consider the reaction 2NO+O2→2NO2. Its mechanism involves a pre-equilibrium, and the experimental rate law is r=k[NO]2[O2]. Here order = 3, so it is applicable. But take the decomposition of N2O5: the rate law is r=k[N2O5], so order = 1 — again applicable.
However, consider a reaction like H2+Br2→2HBr. The experimental rate law is:
r=1+k′[HBr]/[Br2]k[H2][Br2]1/2
This is not of the form k[A]m[B]n — it has a denominator with a concentration term. You cannot assign a single "order" to this reaction. The concept of order is simply not applicable here. …
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- CBSE 2025Set ANNUAL1 markQ.The conversion of molecule X to Y follows third order kinetics. If concentration of X is increased to three times, how will it affect the rate of formation of Y?
›Reveal solutionSolution
For a third-order reaction, Rate ∝ [X]^3, so tripling [X] multiplies the rate by 3^3 = 27.
Rate law: Rate = k[X]^3 (third order in X).
Original rate: Rate1 = k[X]^3 …
- CBSE 2024Set D1 markMCQQ.The rate of reaction of a substance depends upon(a) Atomic mass(b) Equivalent mass(c) Molecular mass(d) Active mass
›Reveal solutionSolution
Rate depends on active mass (concentration), per the law of mass action.
The law of mass action states that the rate of a chemical reaction is proportional to the product of the ACTIVE MASSES (molar concentrations) of the reacting substances, each raised to a power. Atomic, equivalent and molecular masses are fixed properties of a substance and do not govern how fast a reaction …
- CBSE 2024Set ANNUAL1 markMCQQ.A reaction in which reactant (R) are converted into products (P) follows second order kinetics. If concentration of reactant (R) is increased by 4 times, what will be the increase in rate of formation of P?(a) 9 times(b) 4 times(c) 16 times(d) 8 times
›Reveal solutionSolution
Second order kinetics means Rate = k[R]^2; scaling the concentration by a factor scales the rate by that factor squared.
Rate = k[R]^2 (second order in R)
…
- CBSE 2024Set ANNUAL1 markMCQQ.Rate law for the reaction A + 2B → C is found to be Rate = K[A][B]. Concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will be :(a) the same(b) doubled(c) quadroupled(d) halved
›Reveal solutionSolution
The rate constant k is an intrinsic property of a reaction at a given temperature — it does NOT depend on the concentrations of reactants; only the rate itself changes when concentration changes.
For Rate =k[A][B], doubling [B] while keeping [A] constant doubles the RATE of the reaction (since rate ∝[B]), but k itself is unaffected — k is a constant determined only by the nature of the reaction, the temperature, and the …
- CBSE 2024Set ANNUAL1 markQ.In general, what happens to the rate of reaction as the reaction progresses?
›Reveal solutionSolution
As reactants are used up over time, their concentration drops, and since rate depends on concentration, the rate falls too.
For most reactions, the rate law has the form Rate=k[reactant]n (n > 0), i.e. the rate is directly related to the concentration of the reactant(s).
As a reaction proceeds, reactant molecules are continuously converted to products, so the concentration of the reactants keeps decreasing with time. Since the rate depends on this concentration, the rate of reaction generally decreases as the reaction progresses, being fastest at the very start (when reactant concentration is highest) and slowing down as the reaction approaches completion.
…
- CBSE 2023Set F1 markMCQQ.The rate law equation of a chemical reaction is represented as Rate = K[A][B]^2. If the concentration of B is trebled keeping that of A constant, then rate becomes(a) double(b) trebled(c) quadrupled(d) nine times
›Reveal solutionSolution
Rate depends on [B]^2, so tripling [B] increases the rate by a factor of 9.
Given Rate = K[A][B]^2 and [A] kept constant, only the [B]^2 term changes. If [B] is trebled (multiplied by 3):
…
- CBSE 2023Set ANNUAL1 markMCQQ.A reaction is first order in A and second order in B. How is the rate affected when concentrations of both A and B are doubled?(a) It increases 4 times(b) It increases 6 times(c) It increases 8 times(d) It reduces 8 times
›Reveal solutionSolution
Substituting the doubled concentrations into the rate law Rate = k[A]^1[B]^2 shows the rate scales by a factor of 8.
Given: order in A = 1, order in B = 2, so Rate = k[A][B]^2.
If [A] -> 2[A] and [B] -> 2[B]: …
- CBSE 2022Set HE2181 markQ.Fill in the blank: The rate of reaction is ______ of concentration of reactant.
›Reveal solutionSolution
The rate law states that the rate of a reaction is proportional to the concentration(s) of the reactant(s) raised to some power (the order), not necessarily equal to the stoichiometric coefficient.
For a general reaction A -> products, the experimentally determined rate law is written as:
Rate = k[A]^n
where k is the rate constant and n is the order of the reaction with respect to A (found experimentally, not simply read off the balanced equation). This shows the rate of reaction is a function of (depends on / is proportional to) the concentration of the reactant — as concentration increases, the frequency of effective molecular collisions increases, so the rat …
- CBSE 2020Set ANNUAL1 markQ.State the rate law for chemical reaction.
›Reveal solutionSolution
The rate law is the experimentally determined equation expressing reaction rate as proportional to reactant concentrations raised to their (experimental) orders.
For a general reaction, aA+bB→Products, the rate law (or rate equation) expresses how the rate of reaction depends on the concentration of each reactant:
Rate=k[A]x[B]y …
- CBSE 2018Set ANNUAL1 markQ.What is rate law ?
›Reveal solutionSolution
The rate law connects reaction rate to reactant concentrations via experimentally-found exponents (orders).
For a reaction, the rate law (or rate equation) is the mathematical expression relating the instantaneous rate of the reaction to the molar concentrations of the reactants, each raised to a power (the order with respect to that reactant), determined experimentally (not necessarily from the stoichiometric coefficients):
Rate=k[A]x[B]y …
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