NCERT Solutions for Class 12 Chemistry Chapter 3 – Chemical Kinetics - MyNoteswala

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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NCERT Solutions for Class 12 Chemistry Chapter 3 – Chemical Kinetics

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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]

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

(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

(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

= L^1/2 mol^−1/2 s⁻¹

✅ Order = 3/2, k unit = L^1/2 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

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.

Answer

(i) Initial rate

👉 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

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

(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 ✅

(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 ✅

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

🤔 For a zero-order reaction:

👉 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

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

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 p_CH₃OCH₃^3/2.

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) .

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 ✅

Step 2: Unit of k

Rearrange:

👉 k = Rate / (p)^3/2

So units:

• Rate unit = bar min⁻¹

• p^3/2 unit = (bar)^3/2

Therefore,

👉 unit of k = (bar min⁻¹) / (bar)^3/2

= bar^{1 − 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.

2) Temperature

🤔 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.

4) Catalyst

✅ A catalyst provides an alternative pathway with lower activation energy (E_a).
👉 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 E_a ).

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!