Chemical Kinetics: 35 Marks Test Paper (HSC Pattern)
Targeting HSC Board Exam 2025? "Chemical Kinetics" is a high-weightage chapter. This 35 Marks Unit Test Paper is designed exactly as per the Maharashtra State Board pattern (Half Paper). It is perfect for college terminal exams or practice tests. It includes MCQs, VSA, Numericals, and Derivations.
SECTION A (7 Marks)
Q.1. Select and write the correct answer: [4 × 1 = 4 Marks]
i. The molecularity of an elementary reaction:
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Correct Answer: (a) Is always a whole number
ii. The slope of the graph of ln[A]t versus t for a first-order reaction is: March 2022
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Correct Answer: (a) -k
[attachment_0](attachment)t vs time]
Equation: ln[A]t = -kt + ln[A]0
[attachment_0](attachment)t vs time]
Equation: ln[A]t = -kt + ln[A]0
iii. For a zero-order reaction, the unit of rate constant (k) is:
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Correct Answer: (b) mol L⁻¹ s⁻¹
iv. In a first order reaction, the concentration of reactant decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is: July 2023
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Correct Answer: (a) 30 min
(0.8→0.4 is 1st half-life = 15m. 0.1→0.05→0.025 is 2 half-lives = 30m).
(0.8→0.4 is 1st half-life = 15m. 0.1→0.05→0.025 is 2 half-lives = 30m).
Q.2. Answer the following: [3 × 1 = 3 Marks]
i. Define 'Average rate of reaction'.
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The change in concentration of reactant or product divided by the time interval over which the change occurs.
ii. Write the integrated rate law for a zero-order reaction.
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[A]₀ - [A]â‚œ = kt
iii. What is the effect of a catalyst on Activation Energy?
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A catalyst lowers the activation energy.
SECTION B (8 Marks)
Attempt any FOUR of the following: [4 × 2 = 8 Marks]
Q.3. Distinguish between Order and Molecularity of a reaction. March 2018
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1. Values: Order can be zero/fractional; Molecularity is always integer.
2. Origin: Order is experimental; Molecularity is theoretical.
2. Origin: Order is experimental; Molecularity is theoretical.
Q.4. Define 'Pseudo-first order reaction' with an example. July 2022
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A higher molecularity reaction that follows first-order kinetics because one reactant is in excess.
Ex: Acid hydrolysis of methyl acetate.
Ex: Acid hydrolysis of methyl acetate.
Q.5. A first-order reaction takes 40 minutes for 30% decomposition. Calculate its half-life period.
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k = (2.303/40) log(100/70) = 0.00891 min⁻¹.
t1/2 = 0.693 / 0.00891 = 77.78 min.
t1/2 = 0.693 / 0.00891 = 77.78 min.
Q.6. Derive the relation between half-life (t1/2) and rate constant (k) for a zero-order reaction.
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[A]₀ - [A]â‚œ = kt. At t1/2, [A]â‚œ = [A]₀/2.
[A]₀/2 = k × t1/2 ⇒ t1/2 = [A]₀ / 2k.
[A]₀/2 = k × t1/2 ⇒ t1/2 = [A]₀ / 2k.
Q.7. Rate = k[A]²[B]. How is the rate affected if (i) [A] is doubled, (ii) [B] is halved?
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(i) Rate becomes 4 times (2²).
(ii) Rate becomes Half (1/2).
(ii) Rate becomes Half (1/2).
Q.8. Define Activation Energy. Represent it graphically.
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Minimum extra energy required by reactants to form activated complex.
[attachment_1](attachment)
(Draw Potential Energy vs Reaction Coordinate graph).
[attachment_1](attachment)
(Draw Potential Energy vs Reaction Coordinate graph).
SECTION C (12 Marks)
Attempt any FOUR of the following: [4 × 3 = 12 Marks]
Q.9. Derive the integrated rate equation for a first-order reaction. March 2023
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Rate = -d[A]/dt = k[A] ⇒ ∫d[A]/[A] = -k∫dt.
ln[A] = -kt + I. At t=0, I = ln[A]₀.
ln([A]₀/[A]) = kt.
k = (2.303/t) log([A]₀/[A]â‚œ).
ln[A] = -kt + I. At t=0, I = ln[A]₀.
ln([A]₀/[A]) = kt.
k = (2.303/t) log([A]₀/[A]â‚œ).
Q.10. The rate constant for a first-order reaction is 1.54 × 10⁻³ s⁻¹. Calculate (i) Half-life, (ii) Time for 90% completion.
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(i) t1/2 = 0.693/k = 450 s.
(ii) t = (2.303/k) log(100/10) = (2.303/0.00154) × 1 = 1495 s.
(ii) t = (2.303/k) log(100/10) = (2.303/0.00154) × 1 = 1495 s.
Q.11. Explain graphical representation for first-order reaction: (i) ln[A]â‚œ vs t, (ii) log([A]₀/[A]â‚œ) vs t.
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(i) ln[A]ₜ vs t: Straight line, negative slope (-k).
(ii) log([A]₀/[A]â‚œ) vs t: Straight line through origin, slope = k/2.303.
(ii) log([A]₀/[A]â‚œ) vs t: Straight line through origin, slope = k/2.303.
Q.12. Activation energy is 50 kJ/mol. Rate constant at 300 K is 2 × 10⁻³ s⁻¹. Calculate Frequency Factor (A).
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log A = log k + Ea/(2.303RT).
log A = log(2×10⁻³) + 50000/(5744).
log A = -2.699 + 8.70 = 6.001.
A = 10⁶ s⁻¹.
log A = log(2×10⁻³) + 50000/(5744).
log A = -2.699 + 8.70 = 6.001.
A = 10⁶ s⁻¹.
Q.13. Show that for a first-order reaction, time for 99.9% completion is 10 times the half-life. July 2023
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t99.9% = (2.303/k) log(1000) = 3 × 2.303/k = 6.909/k.
t1/2 = 0.693/k.
Ratio = 6.909 / 0.693 ≈ 10.
t1/2 = 0.693/k.
Ratio = 6.909 / 0.693 ≈ 10.
Q.14. For 2N₂O₅ → 4NO₂ + O₂, relate rates of reactant consumption and product formation.
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Rate = -½ d[N₂O₅]/dt = +¼ d[NO₂]/dt = + d[O₂]/dt.
SECTION D (8 Marks)
Attempt any TWO of the following: [2 × 4 = 8 Marks]
Q.15. (a) Derive Integrated Rate Law for Zero Order reaction. (3M)
(b) Give one example of zero order reaction. (1M)
(b) Give one example of zero order reaction. (1M)
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(a) Rate = k. d[A] = -k dt. [A] = -kt + C. At t=0, C=[A]₀. So, k = ([A]₀ - [A]â‚œ)/t.
(b) Decomposition of Ammonia on Platinum at high pressure.
(b) Decomposition of Ammonia on Platinum at high pressure.
Q.16. (a) Reaction is 1st order in A, 2nd in B. (i) Write rate law. (ii) Effect if [B]
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