NCERT Solutions for Class 12 Chemistry Chapter 3 – Chemical Kinetics (English Medium)
⏱️ Get NCERT Solutions for Class 12 Chemistry Chapter 3 – Chemical Kinetics with accurate answers and step-by-step explanations in a clean, exam-friendly style. This chapter includes rate of reaction, rate law & order, molecularity, integrated rate equations (zero/first/second order), half-life, Arrhenius equation & activation energy, plus collision theory and catalysis. Best for CBSE boards and NEET/JEE revision. ✅
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Q.1: From the rate expression for the following reactions, determine their order of reaction and the dimensions (units) of the rate constant (k).
(i) 3NO(g) → N2O(g) Rate = k[NO]2
(ii) H2O2(aq) + 3I−(aq) + 2H+ → 2H2O(l) + I3− Rate = k[H2O2][I−]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k[CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k[C2H5Cl]
(i) 3NO(g) → N2O(g) Rate = k[NO]2
(ii) H2O2(aq) + 3I−(aq) + 2H+ → 2H2O(l) + I3− Rate = k[H2O2][I−]
(iii) CH3CHO(g) → CH4(g) + CO(g) Rate = k[CH3CHO]3/2
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) Rate = k[C2H5Cl]
Answer
📌 Useful rules 👉
• Order of reaction = sum of powers of concentration terms in rate law.
• Rate unit = mol L−1 s−1
• So, unit of k = (mol L−1)1−n s−1 , where n = order .
(i) 3NO(g) → N2O(g) ; Rate = k[NO]2
Order 🤔 = 2
Unit of k 📌
👉 k = (mol L−1)1−2 s−1
= (mol L−1)−1 s−1
= L mol−1 s−1
✅ Order = 2, k unit = L mol−1 s−1
Order 🤔 = 2
Unit of k 📌
👉 k = (mol L−1)1−2 s−1
= (mol L−1)−1 s−1
= L mol−1 s−1
✅ Order = 2, k unit = L mol−1 s−1
(ii) H2O2(aq) + 3I−(aq) + 2H+ → 2H2O(l) + I3− ; Rate = k[H2O2][I−]
Order 🤔 = 1 + 1 = 2
Unit of k 📌
👉 k = (mol L−1)1−2 s−1
= L mol−1 s−1
✅ Order = 2, k unit = L mol−1 s−1
Order 🤔 = 1 + 1 = 2
Unit of k 📌
👉 k = (mol L−1)1−2 s−1
= L mol−1 s−1
✅ Order = 2, k unit = L mol−1 s−1
(iii) CH3CHO(g) → CH4(g) + CO(g) ; Rate = k[CH3CHO]3/2
Order 🤔 = 3/2
Unit of k 📌
👉 k = (mol L−1)1−3/2 s−1
= (mol L−1)−1/2 s−1
= L1/2 mol−1/2 s−1
✅ Order = 3/2, k unit = L1/2 mol−1/2 s−1
Order 🤔 = 3/2
Unit of k 📌
👉 k = (mol L−1)1−3/2 s−1
= (mol L−1)−1/2 s−1
= L1/2 mol−1/2 s−1
✅ Order = 3/2, k unit = L1/2 mol−1/2 s−1
(iv) C2H5Cl(g) → C2H4(g) + HCl(g) ; Rate = k[C2H5Cl]
Order 🤔 = 1
Unit of k 📌
👉 k = (mol L−1)1−1 s−1
= s−1
✅ Order = 1, k unit = s−1
Order 🤔 = 1
Unit of k 📌
👉 k = (mol L−1)1−1 s−1
= s−1
✅ Order = 1, k unit = s−1
Explanation (Quick logic)
🤔 In a rate law, the powers of concentration terms add up to give the overall order
(n)
.📌 The unit of k comes from dividing the rate unit by the concentration terms in the rate law. That’s why:
👉 k = (mol L−1)1−n s−1
Did You Know?
✅ First-order reactions are very common (for example, radioactive decay), so the unit of k is always
s−1
.✅ A fractional order (like 3/2 ) often suggests a complex reaction mechanism, not a single-step reaction.
Q.2: For the reaction:
2A + B → A2B
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1.
Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1.
Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
2A + B → A2B
the rate = k[A][B]2 with k = 2.0 × 10−6 mol−2 L2 s−1.
Calculate the initial rate of the reaction when [A] = 0.1 mol L−1, [B] = 0.2 mol L−1.
Calculate the rate of reaction after [A] is reduced to 0.06 mol L−1.
Answer
(i) Initial rate
👉 Initial rate = 8.0 × 10−9 mol L−1 s−1
👉 Initial rate = 8.0 × 10−9 mol L−1 s−1
(ii) Rate when [A] becomes 0.06 mol L−1
👉 Rate = 3.89 × 10−9 mol L−1 s−1
👉 Rate = 3.89 × 10−9 mol L−1 s−1
Explanation (Step by Step)
📌 Given
• Rate law: r = k[A][B]2
• k = 2.0 × 10−6 mol−2 L2 s−1
• [A] = 0.1 mol L−1 , [B] = 0.2 mol L−1
• Rate law: r = k[A][B]2
• k = 2.0 × 10−6 mol−2 L2 s−1
• [A] = 0.1 mol L−1 , [B] = 0.2 mol L−1
(i) Calculate initial rate 🤔
👉 r = k[A][B]2
= (2.0 × 10−6) × (0.1) × (0.2)2
First, (0.2)2 = 0.04
So, r = (2.0 × 10−6) × 0.1 × 0.04
= (2.0 × 10−6) × 0.004
= 8.0 × 10−9 mol L−1 s−1 ✅
👉 r = k[A][B]2
= (2.0 × 10−6) × (0.1) × (0.2)2
First, (0.2)2 = 0.04
So, r = (2.0 × 10−6) × 0.1 × 0.04
= (2.0 × 10−6) × 0.004
= 8.0 × 10−9 mol L−1 s−1 ✅
(ii) Rate after [A] reduces to 0.06 mol L−1 📌
New [A] = 0.06 mol L−1
So, decrease in A = 0.10 − 0.06 = 0.04 mol L−1
Now use stoichiometry of: 2A + B → A2B
👉 For every 2 moles of A consumed, 1 mole of B is consumed.
So, B consumed = (1/2) × 0.04 = 0.02 mol L−1
New [B] = 0.20 − 0.02 = 0.18 mol L−1
Now apply rate law again:
👉 r = k[A][B]2
= (2.0 × 10−6) × (0.06) × (0.18)2
(0.18)2 = 0.0324
So, r = (2.0 × 10−6) × 0.06 × 0.0324
= (2.0 × 10−6) × 0.001944
= 3.888 × 10−9 mol L−1 s−1
≈ 3.89 × 10−9 mol L−1 s−1 ✅
New [A] = 0.06 mol L−1
So, decrease in A = 0.10 − 0.06 = 0.04 mol L−1
Now use stoichiometry of: 2A + B → A2B
👉 For every 2 moles of A consumed, 1 mole of B is consumed.
So, B consumed = (1/2) × 0.04 = 0.02 mol L−1
New [B] = 0.20 − 0.02 = 0.18 mol L−1
Now apply rate law again:
👉 r = k[A][B]2
= (2.0 × 10−6) × (0.06) × (0.18)2
(0.18)2 = 0.0324
So, r = (2.0 × 10−6) × 0.06 × 0.0324
= (2.0 × 10−6) × 0.001944
= 3.888 × 10−9 mol L−1 s−1
≈ 3.89 × 10−9 mol L−1 s−1 ✅
Did You Know?
✅ Even though the reaction consumes A and B, the rate drops mainly because
[B] is squared
in the rate law, so small changes in
[B]
can have a big effect on the rate.
Q.3: The decomposition of NH3 on a platinum surface is a zero-order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10−4 mol L−1 s−1?
Answer
👉 Rate of production of N2 =
2.5 × 10−4 mol L−1 s−1
👉 Rate of production of H2 = 7.5 × 10−4 mol L−1 s−1
Explanation (Step by Step)
📌 Reaction:
2NH3 → N2 + 3H2
2NH3 → N2 + 3H2
🤔 For a zero-order reaction:
👉 Rate (r) = k (constant)
👉 Rate (r) = k (constant)
Now relate the rate with stoichiometric coefficients:
👉 r = −(1/2) d[NH3]/dt = d[N2]/dt = (1/3) d[H2]/dt
👉 r = −(1/2) d[NH3]/dt = d[N2]/dt = (1/3) d[H2]/dt
Since
r = k = 2.5 × 10−4 mol L−1 s−1
, we get:
✅ Rate of formation of N2:
d[N2]/dt = r = 2.5 × 10−4 mol L−1 s−1
✅ Rate of formation of H2:
(1/3) d[H2]/dt = r
So, d[H2]/dt = 3r = 3 × 2.5 × 10−4
= 7.5 × 10−4 mol L−1 s−1
✅ Rate of formation of N2:
d[N2]/dt = r = 2.5 × 10−4 mol L−1 s−1
✅ Rate of formation of H2:
(1/3) d[H2]/dt = r
So, d[H2]/dt = 3r = 3 × 2.5 × 10−4
= 7.5 × 10−4 mol L−1 s−1
Did You Know?
✅ Zero-order reactions are common on catalyst surfaces (like Pt) because once the surface is fully covered, increasing reactant concentration doesn’t increase the rate.
Q.4: The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by
Rate = k [CH3OCH3]3/2.
The rate is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether:
Rate = k pCH3OCH33/2.
If pressure is measured in bar and time in minutes, what are the units of rate and rate constant (k)?
Rate = k [CH3OCH3]3/2.
The rate is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether:
Rate = k pCH3OCH33/2.
If pressure is measured in bar and time in minutes, what are the units of rate and rate constant (k)?
Answer
👉 Unit of rate =
bar min−1
👉 Unit of rate constant (k) = bar−1/2 min−1
Explanation (Step by Step)
📌 Here, the rate is expressed using partial pressure:
👉 Rate = k (p)3/2
🤔 Since pressure is measured in bar , the “concentration term” is replaced by p (in bar) .
👉 Rate = k (p)3/2
🤔 Since pressure is measured in bar , the “concentration term” is replaced by p (in bar) .
Step 1: Unit of rate
Rate is the change in pressure with time:
👉 Rate ∝ dp/dt
So, unit of rate = bar / min = bar min−1 ✅
Rate is the change in pressure with time:
👉 Rate ∝ dp/dt
So, unit of rate = bar / min = bar min−1 ✅
Step 2: Unit of k
Rearrange:
👉 k = Rate / (p)3/2
So units:
• Rate unit = bar min−1
• p3/2 unit = (bar)3/2
Therefore,
👉 unit of k = (bar min−1) / (bar)3/2
= bar1 − 3/2 min−1
= bar−1/2 min−1 ✅
Rearrange:
👉 k = Rate / (p)3/2
So units:
• Rate unit = bar min−1
• p3/2 unit = (bar)3/2
Therefore,
👉 unit of k = (bar min−1) / (bar)3/2
= bar1 − 3/2 min−1
= bar−1/2 min−1 ✅
Did You Know?
✅ When you write a rate law in terms of partial pressure, the unit of k changes compared to concentration-based rate laws.👉 That’s why it’s always important to check the units used (bar, atm, Pa, seconds, minutes, etc.) before comparing rate constants.
Q.5: Mention the factors that affect the rate of a chemical reaction.
Answer
📌 The main factors that affect the rate of a chemical reaction are:
1) Concentration of reactants
👉 When the concentration of reactants increases, the number of effective collisions per second increases, so rate increases.
👉 When the concentration of reactants increases, the number of effective collisions per second increases, so rate increases.
2) Temperature
🤔 Increasing temperature increases the kinetic energy of molecules, so collisions become more frequent and more energetic.
👉 इसलिए rate increases (often very noticeably).
🤔 Increasing temperature increases the kinetic energy of molecules, so collisions become more frequent and more energetic.
👉 इसलिए rate increases (often very noticeably).
3) Surface area (especially for solids)
📌 Greater surface area means more particles are exposed for collision.
👉 इसलिए powdered solids react faster than lumps/granules.
📌 Greater surface area means more particles are exposed for collision.
👉 इसलिए powdered solids react faster than lumps/granules.
4) Catalyst
✅ A catalyst provides an alternative pathway with lower activation energy (Ea).
👉 So the reaction becomes faster without the catalyst being consumed.
✅ A catalyst provides an alternative pathway with lower activation energy (Ea).
👉 So the reaction becomes faster without the catalyst being consumed.
Explanation
🤔 Reaction rate mainly depends on
how often particles collide
and how many collisions are
effective
.📌 Concentration, temperature, and surface area increase collisions, while a catalyst increases the fraction of collisions that form products (by lowering Ea ).
Did You Know?
✅ Many everyday processes use catalysts:
car catalytic converters,
enzyme reactions in our body,
and industrial manufacture of ammonia
(Haber process).👉 Without catalysts, some reactions would take years instead of seconds or minutes!