Exercise: Questions 4.1–4.5 (Page 117–118) Class 12 Chemical Kinetics

 

🔹 Exercise: Questions 4.1–4.5 (Page 117–118)

1. Q4.1
   Find the order and dimensions of rate constants for these rate laws:
   a) Rate = k [NO]² → Order = 2, k has units M⁻¹ s⁻¹
   b) Rate = k [H₂O₂][I⁻] → Order = 2, k has units M⁻¹ s⁻¹
   c) Rate = k [CH₃CHO]³⁄² → Order = 1.5, units of k: M⁻½ s⁻¹
   d) Rate = k [C₂H₅Cl] → Order = 1, units M⁰ s⁻¹ = s⁻¹ 

2. Q4.2
   For 2A + B → A₂B, Rate = k[A][B]² (k = 2×10⁻⁶ M⁻² s⁻¹). Calculate initial rate with [A]=0.1, [B]=0.2 and then with [A]=0.06, assuming B drops to 0.18.

   • Initial: Rate = 2×10⁻⁶ ×0.1×(0.2)² = 8.0×10⁻⁹ M s⁻¹

   • After change: Rate ≈ 3.89×10⁻⁹ M s⁻¹ 

3. Q4.3
   Decomposition of NH₃ on Pt (zero-order, k=2.5×10⁻⁴ M s⁻¹). Find rates of N₂ and H₂ formation.
   For each mole NH₃ decomposed: produces ½ N₂ and 3/2 H₂ →

   • Rate(N₂) = ½ k = 1.25×10⁻⁴ M s⁻¹

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

4. Q4.4
   Dimethyl ether decomposition, Rate = k [CH₃OCH₃]³⁄². Units if pressure in bar and time in min?

   • Unit of rate: bar min⁻¹

   • k has units = bar^(-½) min⁻¹ 

5. Q4.5
   List factors that affect chemical reaction rate.

• Concentration (or pressure for gases)

• Temperature

• Presence of catalyst 

🔹 Exercise Questions and Answers (Q.4.6 to Q.4.15)

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Q4.6

A reaction is first-order in A. After 15 minutes, 75% of A has reacted. What is the half-life?

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Q4.7

What will be the effect of temperature on rate constant?

Q4.8

Why is the rate constant k independent of initial concentrations?

Answer:
Because k depends only on temperature and activation energy, not on concentrations. The rate, however, does depend on concentrations.

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Q4.9

A reaction is first-order in A. Time for 60% completion is 10 minutes. What is the time for 90% completion?

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Q4.10

Plot log₁₀[R] vs time for a first-order reaction. What does the slope represent?

Answer:
A straight-line graph is obtained.
Slope = –k/2.303, where k is the rate constant.

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Q4.11

The rate constant for a first-order reaction is 60 s⁻¹. How much time will it take to reduce the reactant to 1/16th of its initial value?

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Q4.12

Write the Arrhenius equation. What does each symbol mean?

Answer:
k = A e^(-Eₐ/RT)
Where:

• k = rate constant

• A = frequency factor

• Eₐ = activation energy

• R = gas constant

• T = temperature (K)

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Q4.13

Arrhenius plot of log k vs 1/T gives a straight line. What does the slope represent?

Answer:
Slope = –Ea / 2.303R
From log k = log A – Ea / (2.303RT)

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Q4.14

The rate of a reaction doubles for every 10°C rise in temperature. Explain.

Answer:
According to the Arrhenius equation, rate constant increases exponentially with temperature, approximately doubles for every 10°C rise due to more molecules having energy ≥ Ea.

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Q4.15

The activation energy of a reaction is 75.2 kJ/mol. Calculate rate constant at 298 K if A = 5 × 10⁹ s⁻¹.