NCERT Intext Questions 3.1 to 3.9 from Chapter 4 – Chemical Kinetics (Class 12

 

NCERT Intext Questions 3.1 to 3.9 from Chapter 4 – Chemical Kinetics (Class 12) along with their correct and detailed answers:

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Q3.1

From the rate expressions, determine the order of reaction and the dimensions of the rate constant.

Reaction	Rate Law	Order	Units of Rate Constant
(i) 3NO → N₂O	Rate = k[NO]²	2	mol⁻¹ L s⁻¹
(ii) H₂O₂ + 3I⁻ + 2H⁺ → products	Rate = k[H₂O₂][I⁻]	2	mol⁻¹ L s⁻¹
(iii) CH₃CHO → CH₄ + CO	Rate = k[CH₃CHO]³ᐟ²	1.5	mol⁻⁰.⁵ L⁰.⁵ s⁻¹
(iv) C₂H₅Cl → C₂H₄ + HCl	Rate = k[C₂H₅Cl]	1	s⁻¹

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Q3.2

For the reaction 2A + B → A₂B with rate = k[A][B]² and k = 2.0×10⁻⁶ mol⁻² L² s⁻¹, calculate:

• When [A] = 0.1 mol/L and [B] = 0.2 mol/L

$$\text{Rate} = 2.0 \times 10^{-6} \times (0.1) \times (0.2)^2 = 2.0 \times 10^{-6} \times 0.1 \times 0.04 = 8.0 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1}$$

• After [A] is reduced to 0.06 mol/L (assume [B] = 0.2 unchanged)

$$\text{Rate} = 2.0 \times 10^{-6} \times (0.06) \times (0.2)^2 = 4.8 \times 10^{-9}\ \text{mol L}^{-1}\text{s}^{-1}$$

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Q3.3

NH₃ decomposition (zero-order): N₂ + 3H₂
Given: k = 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹

$$\text{Rate of NH₃ decomposition} = 2.5 \times 10^{-4}$$

Using stoichiometry:

• Rate of N₂ = ½ × k = 1.25 × 10⁻⁴ mol L⁻¹ s⁻¹

• Rate of H₂ = (3/2) × k = 3.75 × 10⁻⁴ mol L⁻¹ s⁻¹

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Q3.4

CH₃OCH₃ decomposition follows:

$$\text{Rate} = k[P_{\text{CH₃OCH₃}}]^{3/2}$$

If pressure is in bar and time in minutes:

• Rate unit = bar/min

• k unit = bar⁻½ min⁻¹

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Q3.5

Factors affecting the rate of a chemical reaction:

1. Concentration of reactants

2. Temperature

3. Presence of catalyst

4. Surface area (for solids)

5. Nature of reactants

6. Light (for photochemical reactions)

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Q3.6

Second-order reaction: Rate ∝ [R]²

• (i) If [R] is doubled, rate becomes 4 times

• (ii) If [R] is halved, rate becomes 1/4th

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Q3.7

Effect of temperature on rate constant:

• As temperature increases, rate constant increases exponentially.

• Quantified by Arrhenius equation:

  $$k = A e^{-E_a/RT}$$

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Q3.8

Pseudo first-order reaction: Calculate average rate from t = 30 s to 60 s

$$[A]_{30} = 0.31\ \text{mol/L},\ [A]_{60} = 0.17\ \text{mol/L}$$ $$\text{Rate} = \frac{0.31 - 0.17}{60 - 30} = \frac{0.14}{30} = 4.67 \times 10^{-3}\ \text{mol L}^{-1} \text{s}^{-1}$$

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Q3.9

Given: Rate = k[A]¹[B]²

• (i) Rate law:

  $$\frac{d[R]}{dt} = k[A][B]^2$$

• (ii) If [B] is tripled:
  Rate ∝ [B]² ⇒ Rate becomes 9 times

• (iii) If [A] and [B] both are doubled:
  Rate ∝ 2 × (2)² = 2 × 4 = 8 times

NCERT Class 12 Chemistry – Chapter 4: Chemical Kinetics Intext Questions (Q.3.10 to Q.3.15) along with complete step-by-step solutions:

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✅ Q3.10

Data Table:

[A] (mol/L)	[B] (mol/L)	Initial Rate (mol L⁻¹ s⁻¹)
0.20	0.30	5.07 × 10⁻⁵
0.20	0.10	5.07 × 10⁻⁵
0.40	0.05	1.43 × 10⁻⁴

Step 1: Order with respect to B

Compare Experiments 1 and 2:

• [A] is constant (0.20), [B] changes from 0.30 to 0.10 (i.e., 1/3)

• Rate remains same → Order w.r.t. B = 0

Step 2: Order with respect to A

Compare Experiments 2 and 3:

• [A] changes from 0.20 to 0.40 (i.e., ×2), [B] changes (but irrelevant, since order in B = 0)

• Rate increases from 5.07×10⁻⁵ to 1.43×10⁻⁴

$$\frac{1.43\times10^{-4}}{5.07\times10^{-5}} \approx 2.82 \approx 2.8 ≈ 2^1.5 \Rightarrow \text{Order w.r.t. A} ≈ 1.5$$

✅ Answer:

• Order w.r.t. A = 1.5

• Order w.r.t. B = 0

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✅ Q3.11

Given:

Exp.	[A] (mol/L)	[B] (mol/L)	Rate (mol L⁻¹ min⁻¹)
I	0.1	0.1	6.0 × 10⁻³
II	0.3	0.2	7.2 × 10⁻²
III	0.3	0.4	2.88 × 10⁻¹
IV	0.4	0.1	2.40 × 10⁻²

Step 1: Order w.r.t. A

Compare I and IV:

• [B] constant (0.1), [A] changes from 0.1 to 0.4 (4×)

• Rate changes from 6.0×10⁻³ to 2.4×10⁻² (4×)
  → Order w.r.t. A = 1

Step 2: Order w.r.t. B

Compare II and III:

• [A] constant (0.3), [B] changes from 0.2 to 0.4 (2×)

• Rate increases from 7.2×10⁻² to 2.88×10⁻¹ (4×)
  → Order w.r.t. B = 2

Rate Law:

$$\text{Rate} = k[A]^1[B]^2$$

Find k using any experiment (say, I):

$$6.0 \times 10^{-3} = k(0.1)(0.1)^2 \Rightarrow k = \frac{6.0 \times 10^{-3}}{0.1 \times 0.01} = 6.0$$

✅ Rate Law: Rate = 6.0 [A][B]²

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✅ Q3.12

Given:
Reaction is 1st order in A, 0 order in B
So, Rate = k[A]
Use this to fill the missing data:

Exp	[A]	[B]	Rate
I	0.1	0.1	2.0 × 10⁻²
II	0.2	0.2	4.0 × 10⁻²
III	0.4	0.4	8.0 × 10⁻²
IV	0.1	0.2	2.0 × 10⁻²

✅ Answers filled in bold.

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✅ Q3.13

Use:

$$t_{1/2} = \frac{0.693}{k}$$

(i) For k = 200 s⁻¹

$$t_{1/2} = \frac{0.693}{200} = 3.465 \times 10^{-3}\ \text{s}$$

(ii) For k = 2 min⁻¹

$$t_{1/2} = \frac{0.693}{2} = 0.3465\ \text{min}$$

(iii) For k = 4 yr⁻¹

$$t_{1/2} = \frac{0.693}{4} = 0.1733\ \text{years}$$

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✅ Q3.14

Use formula for 1st-order decay:

$$t = \frac{2.303}{k} \log\left(\frac{N_0}{N}\right),\quad \text{where } k = \frac{0.693}{5730} ≈ 1.209 \times 10^{-4}\ \text{yr}^{-1}$$ $$N/N_0 = 80\% = 0.8 \Rightarrow t = \frac{2.303}{1.209 \times 10^{-4}} \log\left(\frac{1}{0.8}\right) ≈ 1894\ \text{years}$$

✅ Age of sample = 1894 years

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✅ Q3.15

Given:
[N₂O₅] data over time.

(i) Plot [N₂O₅] vs time

You can plot the values as given. The plot is non-linear (shows exponential decay) ⇒ Not zero-order.

(ii) Half-life:

Find time where [N₂O₅] becomes half of its initial value:

Initial: 1.63 × 10⁻² mol/L
Half = 0.815 × 10⁻² mol/L

From data:

• At 1200 s → 0.93

• At 1600 s → 0.78
  So halfway between → t₁/₂ ≈ 1400 s

(iii) Plot log[N₂O₅] vs time

This plot will be linear → Confirms first-order reaction

(iv) Rate law:

Since log[N₂O₅] vs t is linear ⇒
✅ Rate = k[N₂O₅], i.e., first-order reaction