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 …
Concept: Effect of Concentration — For a reaction aA+bB→ products, the rate law is Rate=k[A]m[B]n, where m and n are the orders with respect to A and B.
(i) Given: first order in A, second order in B.
The differential rate equation is:
−dtd[A]=k[A]1[B]2
(ii) If [B] is tripled ([B]→3[B]), while [A] stays constant:
New rate=k[A](3[B])2=9×k[A][B]2=9×original rate
So the rate increases 9 times.
(iii) If both [A] and [B] are doubled ([A]→2[A], [B]→2[B]): …
For a reaction first order in A and second order in B, the rate law is r=k[A]1[B]2. Tripling [B] multiplies the rate by 9; doubling both [A] and [B] multiplies the rate by 8.
The core idea: the rate law is a product of concentration terms, each raised to its order.
The rate of a reaction tells us how fast reactants are used up or products form. For a simple reaction, the rate depends on the concentrations of the reactants, each raised to a power (the order). The overall rate law is:
r=k[A]m[B]n
where m is the order in A, n is the order in B, and k is the rate constant (which depends only on temperature, not on concentration). Here, we are told the reaction is first order in A (m=1) and second order in B (n=2).
Step-by-step solution
1. Write the differential rate equation.
The differential rate equation directly expresses the instantaneous rate in terms of concentrations. Using the given orders:
r=k[A]1[B]2
That’s it. The exponent on [A] is 1 (often omitted), and on [B] it’s 2. This is the complete rate law.
r=k[A][B]2
2. Effect of tripling the concentration of B.
We want to see what happens to the rate when only [B] changes. Let the initial rate be r1=k[A][B]2.
Now, increase [B] to three times its original value: [B]new=3[B]. The concentration of A stays the same. The new rate r2 is:
r2=k[A](3[B])2=k[A]⋅9[B]2=9⋅k[A][B]2=9r1
So the rate becomes 9 times the original rate.
Why 9? Because the order in B is 2 — the rate is proportional to [B]2. Tripling B means squaring the factor: 32=9.
A common mistake is to multiply the factor by the order (e.g., 3×2=6). That’s wrong. The order is an exponent, not a multiplier. Always raise the concentration factor to the power of the order.
3. Effect of doubling both [A] and [B]. …
Method: Rate Law Analysis Using Order of Reaction
This method uses the differential rate law — the direct relationship between reactant concentrations and reaction rate, based on experimentally determined orders.
(i) Write the differential rate equation
Step 1: Recall the general form for a reaction with multiple reactants:
Rate=k[A]m[B]n
where m and n are the orders with respect to A and B.
Step 2: Substitute the given orders (m=1, n=2):
Rate=k[A]1[B]2
(ii) Effect when concentration of B is increased three times
Step 1: Let initial rate be r1=k[A][B]2.
Step 2: New concentration: [B]′=3[B], while [A] remains unchanged.
Step 3: New rate:
r2=k[A](3[B])2=k[A]⋅9[B]2=9⋅k[A][B]2
Step 4: Compare:
r1r2=k[A][B]29⋅k[A][B]2=9
Answer: The rate becomes 9 times the original rate.
(iii) Effect when concentrations of both A and B are doubled
Step 1: Initial rate: r1=k[A][B]2 …
Here are the common mistakes students make on this exact question, along with how to avoid each one.
(i) Writing the Differential Rate Equation
Mistake 1: Writing the wrong order for B
Students often write:
Rate = k[A]1[B]1
Why this happens: They confuse the stoichiometric coefficient (from a balanced equation) with the order of reaction. Here, no balanced equation is given — only the order is stated.
How to avoid:
- The order is an experimentally determined number.
- If the reaction is first order in A → exponent on [A] is 1.
- If second order in B → exponent on [B] is 2.
✓ Correct equation:
Rate=k[A]1[B]2
Mistake 2: Forgetting the proportionality constant k
Some students write:
Rate = [A][B]2
Why this happens: They omit the rate constant, thinking it’s optional.
How to avoid:
- The rate law always includes k.
- k is the proportionality constant that makes the equation an equality.
✓ Always write:
Rate=k[A][B]2
(ii) Effect of Tripling the Concentration of B
Mistake 3: Thinking the rate triples
Students say:
Rate becomes 3 times
Why this happens: They apply a linear (first-order) logic to a second-order reactant.
How to avoid:
- For a second-order reactant, if concentration is multiplied by x, the rate is multiplied by x2.
- Here, x=3, so rate multiplies by 32=9.
✓ Correct reasoning:
New rate=k[A][3B]2=k[A]⋅9[B]2=9×(original rate)
Key result: Rate becomes 9 times the original.
(iii) Effect of Doubling Both A and B
Mistake 4: Adding the effects instead of multiplying
Students say:
Rate becomes 2+4=6 times
Why this happens: They treat the effect as additive rather than multiplicative.
How to avoid:
- The rate law is multiplicative.
- Doubling [A] (order 1) multiplies rate by 21=2. …
- AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQQ.At T (K), the following data was obtained for the following reaction, S2O82−+3I−→2SO42−+I3−From the data, the rate constant of the reaction (in M−1s−1) is (A) 8.0×10−2 (B) 4.0×10−2 (C) 8.0×10−3 (D) 6.0×10−3
Ex. No. [S2O82−] [I−] Initial rate (M/s) 1. 0.080 0.034 2.2×10−4 2. 0.080 0.017 1.1×10−4 3. 0.160 0.017 2.2×10−4 ›Reveal solutionSolution
Tests deriving a rate law and rate constant from tabulated initial-rate data; the reaction is
first order in each reactant (second order overall), and k≈8.0×10−2 M−1s−1.
Concept and Intuition
The method of initial rates finds the order in each reactant by comparing pairs of experiments in
which only one concentration changes while the other is held fixed — the ratio of rates then
directly reveals the order with respect to the varied species. Once both orders are known, the
rate constant is obtained by substituting any one experiment's data into the full rate law.
Step-by-Step Solution
- Compare Exp 1 and Exp 2: [S2O82−] is unchanged (0.080), while [I−] halves from 0.034 to 0.017. The rate also halves, from 2.2×10−4 to 1.1×10−4. So rate ∝[I−]1.
- Compare Exp 2 and Exp 3: [I−] is unchanged (0.017), while [S2O82−] doubles from 0.080 to 0.160. The rate also doubles, from 1.1×10−4 to 2.2×10−4. So rate ∝[S2O82−]1.
- Hence rate =k[S2O82−][I−] (overall second order).
- Substitute Exp 1 data: 2.2×10−4=k×0.080×0.034=k×2.72×10−3. …
- AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQQ.For the reaction A+B→C, the following data were obtained Expt. [A] M [B] M Initial rate (Mmin−1)
-
0.1 0.1 $1.0\times10^{-4}$ -
0.1 0.3 $9.0\times10^{-4}$ -
0.3 0.3 $2.7\times10^{-3}$
›Reveal solutionSolution
Use the method of initial rates, comparing pairs of experiments where only one concentration changes, to find the order in each reactant. Answer: order in A = 1, order in B = 2.
Concept and Intuition
For a rate law rate=k[A]m[B]n, comparing two experiments where only one concentration changes isolates that reactant's order: the ratio of rates equals the ratio of concentrations raised to that reactant's order.
Step-by-Step Solution
- Compare Expt 1 and Expt 2: [A] stays at 0.1, [B] goes 0.1→0.3 (×3); rate goes 1.0×10−4→9.0×10−4 (×9).
- So 3n=9⇒n=2 (order in B). …
-
- AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQQ.Observe the following equilibrium at T(K): H2(g)+I2(g)⇌2HI(g). Which one of the following does not disturb the above equilibrium? (A) Addition of H2(g) (B) Removal of HI(g) (C) Addition of I2(g) (D) Addition of He(g)
›Reveal solutionSolution
Adding an inert gas at constant volume doesn't change the concentrations of the reacting species, so it leaves the equilibrium (with equal moles of gas on each side) undisturbed.
Concept and Intuition
Le Chatelier's principle: a system at equilibrium responds to changes that alter concentration/partial pressure of the species actually involved in the equilibrium. Adding an inert gas at constant volume doesn't change any reactant/product concentration (each species still occupies the same volume), so at constant volume it has no effect on the position of equilibrium — especially clear here since Δng=0 for this reaction anyway.
Step-by-Step Solution
- Adding H2 increases [H2], shifting equilibrium forward (toward HI) — this does disturb equilibrium.
- Removing HI decreases [HI], shifting equilibrium forward to replace it — disturbs equilibrium. …
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.Consider the reaction carried out at T(K) A(g)+B(g)→C(g) The rate law for this reaction is r=k[A]1[B]2. The concentration of A in experiment 2 and rate in experiment 3 shown as x and z in the table. x and z are respectively
Experiment [A] / mol L−1 [B] / mol L−1 Initial rate / mol L−1 s−1 1 0.05 0.05 R 2 x 0.05 2R 3 0.20 0.10 z (A) x = 0.10, z = 8R (B) x = 0.05, z = 4R (C) x = 0.10, z = 16R (D) x = 0.20, z = 16R ›Reveal solutionSolution
This tests using a given rate law r=k[A][B]2 to back-calculate an unknown concentration and an unknown rate from a data table. Answer: x = 0.10, z = 16R.
Concept and Intuition
The rate law tells us the reaction is first order in A and second order in B. Since [B] is unchanged between experiments 1 and 2, any change in rate there is due to [A] alone (first-order, so rate scales linearly with [A]). Between experiments 1 and 3, both concentrations change, so we substitute directly using k eliminated as a common factor.
Step-by-Step Solution
- Write the rate law: r=k[A]1[B]2.
- Experiment 1: R=k(0.05)(0.05)2=k(0.05)(0.0025)=1.25×10−4k.
- Experiment 2: [B] is still 0.05, only [A]=x changes. 2R=kx(0.0025).
- Divide: R2R=k(0.05)(0.0025)kx(0.0025)=0.05x, so 2=0.05x⇒x=0.10. …
- AP EAPCET 2021Set ap-2021-09-03-FN1 markMCQQ.For the reaction 2A+B2⟶2C, the following data is provided. Find the overall order of this reaction. Experiment 1: [A]=0.5, [B]=0.5, Rate =1×10−4 mol.L−1.s−1; Experiment 2: [A]=0.5, [B]=1.0, Rate =2×10−4 mol.L−1.s−1; Experiment 3: [A]=1.0, [B]=1.0, Rate =2×10−4 mol.L−1.s−1 (A) 2 (B) 0 (C) 3 (D) 1
›Reveal solutionSolution
A classic method-of-initial-rates problem: isolate each concentration's effect by comparing pairs of experiments where only one concentration changes. Overall order = 1 (answer D).
Concept and Intuition
For a rate law Rate=k[A]x[B]y, we can find x and y separately by picking pairs of experiments where only one reactant's concentration changes while the other stays fixed — any change in rate must then be due to that one reactant alone.
Step-by-Step Solution
- Find order in B: Compare Experiment 1 ([A]=0.5,[B]=0.5, rate =1×10−4) and Experiment 2 ([A]=0.5,[B]=1.0, rate =2×10−4). Here [A] is unchanged, [B] doubles, and rate doubles. Since 2y=2⇒y=1. …
- AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQQ.For a reaction, the initial rate is given as Ro=k[Ao]2[Bo]. By what factor, the initial rate of reaction will increase if the concentration of A taken is increased to 1.5 times the original, and that of B is tripled? (A) 4.50 (B) 2.25 (C) 6.75 (D) 3.45
›Reveal solutionSolution
Plugging the new concentrations into the rate law R=k[A]2[B]: the rate
increases by a factor of (1.5)2×3=6.75.
Concept and Intuition
When a rate law has the form R=k[A]m[B]n, scaling each concentration by a
factor scales the rate by that factor raised to its respective order — and
since these effects multiply independently, the overall factor is the product
of each individual scaling factor raised to its order.
Step-by-Step Solution
- Given rate law: Ro=k[Ao]2[Bo]1.
- New concentrations: [A]=1.5[Ao], [B]=3[Bo].
- New rate: R=k(1.5[Ao])2(3[Bo])=k×(1.5)2[Ao]2×3[Bo]. …
- AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQQ.For a certain reaction between X and Y, the order with respect to X is 3 and that with respect to Y is 4. If the concentrations of both X and Y are tripled, the rate of reaction will increase by a factor of ________ (A) 47 (B) 37 (C) 73 (D) 127
›Reveal solutionSolution
Tripling a reactant raised to the power n multiplies the rate by 3n;
tripling both X (order 3) and Y (order 4) multiplies the rate by
33×34=37.
Concept and Intuition
The rate law expresses rate as k[X]a[Y]b where a,b are the individual
orders. If a concentration is scaled by a factor f, the rate contribution from
that species scales by forder; when several concentrations change
simultaneously, their scaling factors multiply.
Step-by-Step Solution
- Rate law: Rate=k[X]3[Y]4 (orders given as 3 and 4 respectively).
- Tripling [X]: contributes a factor of 33.
- Tripling [Y]: contributes a factor of 34. …
- AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQQ.For a A+B→ products the rate of the reaction is given by Rate=K[A][B]2. The units of rate constant (K) will be ________ (A) mol L−1 S−1 (B) L mol−1 S−1 (C) mol2 L−2 S−1 (D) mol−2 L2 S−1
›Reveal solutionSolution
The reaction is third order overall (1+2), so rearranging the rate law for K gives units mol−2 L2 s−1.
Concept and Intuition
The rate constant's units always depend on the overall order of the reaction, since Rate=K×(concentration)n must be dimensionally consistent, with Rate always in mol L−1 s−1 (or mol L−1 time−1). Solving for K's units directly from the given rate law is the safest approach — no need to memorize a general formula, just do the algebra.
Step-by-Step Solution
- Given: Rate=K[A][B]2, with Rate in mol L−1 s−1, and [A],[B] in mol L−1.
- Rearranging: K=[A][B]2Rate.
- Units: K=(mol L−1)×(mol L−1)2mol L−1 s−1=mol3 L−3mol L−1 s−1. …
- AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQQ.The reaction N2(g)+3H2(g)⇌2NH3(g) is exothermic and reversible. A mixture of N2(g), H2(g) and NH3(g) is at equilibrium in a closed container. When a certain quantity of extra H2(g) is introduced into the container, while keeping the volume constant, then which statement among the following is true? (A) The pressure inside the container will not change (B) Equilibrium condition will not change (C) The temperature will increase (D) The temperature will decrease
›Reveal solutionSolution
Adding extra H2 to the exothermic ammonia equilibrium (constant volume, closed container) shifts the reaction forward by Le Chatelier's principle, and since the forward direction releases heat, the system's temperature rises.
Concept and Intuition
Le Chatelier's principle says a system at equilibrium responds to a disturbance (here, adding more H2) by shifting in the direction that partially counteracts the change — i.e. consuming the added H2 by shifting the N2+3H2⇌2NH3 equilibrium to the RIGHT (forward). Because the forward reaction is stated to be exothermic, this net forward conversion of reactants to products liberates heat. In a closed container (not necessarily in contact with a large external heat sink that instantly removes this heat), that liberated heat raises the temperature of the system.
Step-by-Step Solution
- Adding extra H2 increases [H2], disturbing the existing equilibrium.
- By Le Chatelier's principle, the equilibrium shifts in the direction that consumes some of the added H2 — i.e. the FORWARD direction (toward NH3).
- Volume is held constant while more gas (H2) has been added, so total pressure necessarily increases initially — ruling out option (A).
- The equilibrium position (concentrations) DOES change as the system re-equilibrates — ruling out option (B). …
- AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQQ.Compounds A and B react according to the equation 2A(g)+B(g)→2C(g)+D(g). The initial rate of formation was determined at different initial concentrations of A and B. The following results were obtained. The rate law for this reaction may be _________ [All concentrations are in mol/lit](A) Rate=K[A]2[B] (B) Rate=K[A][B]2 (C) Rate=K[A][B] (D) Rate=K[A]2[B]0
Exp No Initial [A] Initial [B] Initial rate of formation of C 1 0.1 0.1 6×10−3 2 0.3 0.2 7.2×10−2 3 0.3 0.4 2.88×10−1 4 0.4 0.1 2.4×10−2 ›Reveal solutionSolution
Tests determining reaction order from experimental rate data; the answer is (B), Rate=K[A][B]2.
Concept and Intuition
To find the order with respect to each reactant, compare pairs of experiments where only ONE concentration changes while the other is held constant, then see how the rate scales.
Step-by-Step Solution
- Order in A: Compare Exp 1 ([A]=0.1, [B]=0.1, rate=6×10−3) and Exp 4 ([A]=0.4, [B]=0.1, rate=2.4×10−2) — [B] is constant, [A] increases by a factor of 4. Rate ratio =6×10−32.4×10−2=4. If order in A is x: 4x=4⇒x=1.
- Order in B: Compare Exp 2 ([A]=0.3, [B]=0.2, rate=7.2×10−2) and Exp 3 ([A]=0.3, [B]=0.4, rate=2.88×10−1) — [A] is constant, [B] doubles. Rate ratio =7.2×10−22.88×10−1=4. If order in B is y: 2y=4⇒y=2. …
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