JEE Chemistry Chemical Kinetics MCQs Set 11

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Chapter-wise Objective Questions: Chemical Kinetics

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Multiple Choice Questions

Question 1: A graph between t1/2 and concentration for an nth order reaction is a straight line parallel to the concentration axis. The reaction is 50% complete in 10 minutes when the concentration is 2 mol L−1. It is 50% decomposed in t minutes at 4 mol L−1. n and t respectively are
(a) 0, 20 min
(b) 1, 10 min
(c) 1, 20 min
(d) 0, 5 min
Show Answer & Explanation

Answer: (b) 1, 10 min

Explanation:
1. A constant t1/2 independent of concentration means first order, n = 1.
2. Hence t1/2 at 4 mol L−1 is also 10 min.

Teacher's Note:
1. Straight line parallel to the x-axis: t1/2 independent of a.

Question 2: In a first order reaction, 75% of the reactant disappeared in 1.388 s. The rate constant of the reaction is
(a) 1 s−1
(b) 2.8 × 10−4 s−1
(c) 17.2 × 10−3 s−1
(d) 1.8 × 10−3 s−1
Show Answer & Explanation

Answer: (a) 1 s−1

Explanation:
1. \( k = \dfrac{2.303}{1.388}\log 4 = \dfrac{1.386}{1.388} \approx 1 \) s−1.

Teacher's Note:
1. 75% completion means 25% remains, a/(a−x) = 4.

Question 3: For a first order reaction, the ratio of the time required for 99.9% completion to that for 50% completion is
(a) 2
(b) 5
(c) 10
(d) none of these
Show Answer & Explanation

Answer: (c) 10

Explanation:
1. t99.9 = (2.303/k) log 1000 = 6.909/k.
2. t50 = 0.693/k.
3. Ratio = 9.97, about 10.

Teacher's Note:
1. Ratio is independent of k.

Question 4: From different sets of data of t1/2 at different initial concentrations a for a given reaction, t1/2 × a is found to be constant. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t1/2 ∝ 1/a means second order.

Teacher's Note:
1. Second order: t1/2 = 1/(ka).

Question 5: The reaction 2N2O5(g) → 4NO2(g) + O2(g) is first order with respect to N2O5. Which plot would give a straight line?
(a) log P(N2O5) versus time with negative slope
(b) 1/P(N2O5) versus time
(c) P(N2O5) versus time
(d) log P(N2O5) versus time with positive slope
Show Answer & Explanation

Answer: (a) log P(N2O5) versus time with negative slope

Explanation:
1. First order: ln P = ln P0 − kt, a straight line of negative slope.

Teacher's Note:
1. Pressure is proportional to concentration at constant temperature.

Question 6: Which curves represent a first order reaction? (i) log(a − x) versus t, a decreasing straight line; (ii) log(a − x) versus 1/t, an increasing straight line; (iii) t1/2 versus a, a horizontal line.
(a) only (i)
(b) only (ii)
(c) only (iii)
(d) (i) and (iii) both
Show Answer & Explanation

Answer: (d) (i) and (iii) both

Explanation:
1. First order gives a linear decreasing log(a − x) against t, and t1/2 independent of a.
2. log(a − x) versus 1/t is not linear for first order.

Teacher's Note:
1. Check both the integrated law and the half-life property.

Question 7: The following data were obtained at a certain temperature for the decomposition of ammonia: p (mm) = 50, 100, 200 and t1/2 = 3.64, 1.82, 0.91. The order of the reaction is
(a) 0
(b) 1
(c) 2
(d) 3
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. t1/2 halves when p doubles, so t1/2 ∝ 1/p.
2. The reaction is second order.

Teacher's Note:
1. Compare t1/2p: 182, 182, 182, constant.

Question 8: The rate constant for a first order reaction is x s−1. By what factor is the rate increased if the initial concentration of A is tripled?
(a) 3
(b) 9
(c) x
(d) remains the same
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. Rate = k[A], so tripling [A] triples the rate.

Teacher's Note:
1. k itself does not change with concentration.

Question 9: Which is incorrect?
(a) The half life of a first order reaction is independent of initial concentration
(b) The rate of reaction is constant for a first order reaction
(c) The unit of k for a second order reaction is mol−1 L s−1
(d) The half life of a zero order reaction is proportional to the initial concentration
Show Answer & Explanation

Answer: (b) The rate of reaction is constant for a first order reaction

Explanation:
1. Rate = k[A] decreases as [A] falls; the rate constant is what stays constant.
2. The other three statements are correct.

Teacher's Note:
1. Do not confuse rate with rate constant.

Question 10: Hydrolysis of an ester in alkaline medium is
(a) first order with molecularity one
(b) second order with molecularity two
(c) first order with molecularity two
(d) second order with molecularity one
Show Answer & Explanation

Answer: (b) second order with molecularity two

Explanation:
1. Rate = k[ester][OH−] is second order.
2. Two species collide, so molecularity is two.

Teacher's Note:
1. In alkaline hydrolysis, OH− is not in large excess relative to the ester.

Question 11: The expression that gives the 1/4th life of a first order reaction is
(a) \( \dfrac{k}{2.303}\log\dfrac43 \)
(b) \( \dfrac{2.303}{k}\log 3 \)
(c) \( \dfrac{2.303}{k}\log\dfrac34 \)
(d) \( \dfrac{2.303}{k}\log\dfrac43 \)
Show Answer & Explanation

Answer: (d) \( \dfrac{2.303}{k}\log\dfrac43 \)

Explanation:
1. 1/4th life means 1/4 of the reactant has reacted: a/(a − x) = a/(3a/4) = 4/3.
2. t = (2.303/k) log (4/3).

Teacher's Note:
1. Option (c) has the log of a ratio less than 1 and gives a negative time.

Question 12: The rate constant of a zero order reaction is 0.2 mol dm−3 h−1. If the concentration of the reactant after 30 minutes is 0.05 mol dm−3, its initial concentration would be
(a) 6.05 mol dm−3
(b) 0.15 mol dm−3
(c) 0.25 mol dm−3
(d) 4.00 mol dm−3
Show Answer & Explanation

Answer: (b) 0.15 mol dm−3

Explanation:
1. 30 min = 0.5 h.
2. [A]0 = 0.05 + 0.2 × 0.5 = 0.15 mol dm−3.

Teacher's Note:
1. Convert the time to the unit used in k.

Question 13: A reaction is of first order. After 100 minutes 75 g of reactant A are decomposed when 100 g are taken initially. The time required when 150 g of the reactant A are decomposed, if the initial weight taken is 200 g, is
(a) 100 minutes
(b) 200 minutes
(c) 150 minutes
(d) 175 minutes
Show Answer & Explanation

Answer: (a) 100 minutes

Explanation:
1. 75 g from 100 g is 75% decomposition in 100 min.
2. 150 g from 200 g is also 75%, so the time is again 100 min.

Teacher's Note:
1. First order: the time for a given percentage is independent of the initial amount.

Question 14: For which reaction are the units of rate and rate constant identical?
(a) first order
(b) zero order
(c) second order
(d) fractional order
Show Answer & Explanation

Answer: (b) zero order

Explanation:
1. Zero order: rate = k, so both have units mol L−1 s−1.

Teacher's Note:
1. Unit of k = (mol L−1)1−n s−1.

Question 15: A reaction 2A → Product follows first order kinetics. If the half life at [A]initial = 0.2 mol L−1 is 20 min, the value of the rate constant is
(a) 4 s−1
(b) 20 s−1
(c) 4 L mol−1 min−1
(d) 0.034 min−1
Show Answer & Explanation

Answer: (d) 0.034 min−1

Explanation:
1. k = 0.693/t1/2 = 0.693/20 = 0.0347 min−1.

Teacher's Note:
1. The unit of k for first order is time−1.

Question 16: The correct statement about a first order reaction is
(a) tcompletion is finite
(b) t1/2 ∝ 1/a
(c) the unit of K is mol L−1 s−1
(d) t1/2 × k is constant at a given temperature
Show Answer & Explanation

Answer: (d) t1/2 × k is constant at a given temperature

Explanation:
1. t1/2 × k = 0.693, a constant.
2. Completion time is infinite, t1/2 is independent of a, and k has unit s−1.

Teacher's Note:
1. 0.693 = k t1/2 holds for every first order reaction.

Question 17: The half life of a first order reaction is 1386 s. The specific rate constant of the reaction is
(a) 5.0 × 10−2 s−1
(b) 5.0 × 10−3 s−1
(c) 0.5 × 10−2 s−1
(d) 0.5 × 10−3 s−1
Show Answer & Explanation

Answer: (d) 0.5 × 10−3 s−1

Explanation:
1. k = 0.693/1386 = 5 × 10−4 = 0.5 × 10−3 s−1.

Teacher's Note:
1. Divide 0.693 by the half life.

Question 18: For the first order reaction A(g) → 2B(g) + C(s), if P0 is the initial pressure of A and Pt the total pressure at time t, the rate constant is
(a) \( \dfrac1t\ln\dfrac{P_0}{P_0-P_t} \)
(b) \( \dfrac1t\ln\dfrac{P_0}{2P_0-P_t} \)
(c) \( \dfrac1t\ln\dfrac{2P_0}{3P_0-P_t} \)
(d) \( \dfrac1t\ln\dfrac{P_0}{3P_0-P_t} \)
Show Answer & Explanation

Answer: (b) \( \dfrac1t\ln\dfrac{P_0}{2P_0-P_t} \)

Explanation:
1. If x is the pressure of A that reacts, gas B has pressure 2x and C is solid.
2. Pt = (P0 − x) + 2x = P0 + x, so x = Pt − P0.
3. PA = P0 − x = 2P0 − Pt, and k = (1/t) ln [P0/(2P0 − Pt)].

Teacher's Note:
1. Track the change in total moles of gas.
2. Solids do not contribute to pressure.

Question 19: For a first order gas phase reaction A(g) → 2B(g) + C(g), with initial pressure P0 and total pressure Pt at time t, the integrated rate equation is
(a) \( \dfrac{2.303}{t}\log\dfrac{P_0}{P_0-P_t} \)
(b) \( \dfrac{2.303}{t}\log\dfrac{2P_0}{3P_0-P_t} \)
(c) \( \dfrac{2.303}{t}\log\dfrac{P_0}{2P_0-P_t} \)
(d) \( \dfrac{2.303}{t}\log\dfrac{2P_0}{2P_0-P_t} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{2.303}{t}\log\dfrac{2P_0}{3P_0-P_t} \)

Explanation:
1. Let pressure x of A react: Pt = (P0 − x) + 2x + x = P0 + 2x.
2. x = (Pt − P0)/2, PA = P0 − x = (3P0 − Pt)/2.
3. k = (2.303/t) log [2P0/(3P0 − Pt)].

Teacher's Note:
1. The total number of moles rises from 1 to 3, giving a net +2 per mole of A reacted.

Question 20: In a zero order reaction, for every 10° rise of temperature the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become
(a) 64 times
(b) 128 times
(c) 256 times
(d) 512 times
Show Answer & Explanation

Answer: (d) 512 times

Explanation:
1. Rise of 90° gives 9 steps of 10°.
2. Rate becomes 29 = 512 times.

Teacher's Note:
1. Rate factor = 2ΔT/10.

Question 21: For a reaction rate = k[CH3COCH3]3/2, the unit of rate and rate constant respectively are
(a) mol L−1 s−1, mol−1/2 L1/2 s−1
(b) mol−1 L−1 s−1, mol−1/2 L1/2 s−1
(c) mol L−1 s−1, mol1/2 L1/2 s−1
(d) mol L s, mol1/2 L1/2 s
Show Answer & Explanation

Answer: (a) mol L−1 s−1, mol−1/2 L1/2 s−1

Explanation:
1. Rate is always in mol L−1 s−1.
2. k = rate/conc3/2 has units mol L−1 s−1 / (mol L−1)3/2 = mol−1/2 L1/2 s−1.

Teacher's Note:
1. Unit of k = (mol L−1)1−n s−1.

Question 22: For a reaction, a plot of t1/2 versus 1/a2 is a straight line. The order of the reaction is
(a) zero order
(b) first order
(c) second order
(d) third order
Show Answer & Explanation

Answer: (d) third order

Explanation:
1. t1/2 ∝ a1−n = a−2, so 1 − n = −2 and n = 3.

Teacher's Note:
1. The exponent of 1/a is n − 1.

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