JEE Chemistry Chemical Kinetics MCQs Set 37

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Kinetics

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Multiple Choice Questions

Question 1: Which of the following rate law expressions represents a zero order reaction?
(a) K(A)3/2(B)−1(C)1/2
(b) K(A)0(B)2
(c) K(A)1/2(B)1/2(C)−1
(d) K(A)2(B)1
Show Answer & Explanation

Answer: (c) K(A)1/2(B)1/2(C)−1

Explanation:
1. The overall order is the sum of the exponents.
2. Option 3: 1/2 + 1/2 − 1 = 0. The others give 1, 2 and 3.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 2: The ratio of t0.875 and t0.75 for a first order reaction is:
(a) 4 : 3
(b) 3 : 2
(c) 2 : 1
(d) 1 : 2
Show Answer & Explanation

Answer: (b) 3 : 2

Explanation:
1. 87.5% completion takes 3 half-lives and 75% takes 2 half-lives.
2. Ratio = 3 : 2.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 3: The correct expression of the Arrhenius equation is:
(a) ln k = ln A − E/RT
(b) k e−E/RT = A
(c) log k = log A − 2.303 E/RT
(d) k = A−1e−E/RT
Show Answer & Explanation

Answer: (a) ln k = ln A − E/RT

Explanation:
1. The Arrhenius equation is k = Ae−E/RT.
2. Taking natural logs gives ln k = ln A − E/RT; in log form the term would be E/(2.303RT).
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 4: For every 10°C increase in temperature, the rate becomes twice. If the temperature increases from 10°C to 100°C, the rate increases ... times:
(a) 400
(b) 512
(c) 112
(d) 614
Show Answer & Explanation

Answer: (b) 512

Explanation:
1. The 90°C rise contains 9 intervals of 10°C.
2. Rate increases by 29 = 512 times.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 5: The specific rate of reaction for a first order reaction depends upon:
(a) pressure
(b) temperature
(c) concentration of reactant
(d) concentration of the product
Show Answer & Explanation

Answer: (b) temperature

Explanation:
1. The rate constant depends on temperature (and catalyst) only.
2. It is independent of concentrations.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 6: The plot of ln K versus 1/T is linear with slope:
(a) −Ea/R
(b) +Ea/R
(c) −Ea/2.303R
(d) Ea/2.303R
Show Answer & Explanation

Answer: (a) −Ea/R

Explanation:
1. ln K = ln A − (Ea/R)(1/T).
2. The slope is −Ea/R.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 7: For a first order gaseous reaction A → 2B + C, the initial pressure is Pi and the total pressure after time t is Pt. The expression for the rate constant is:
(a) k = (2.303/t) log[2Pi/(3Pi − Pt)]
(b) k = (2.303/t) log[2Pi/(2Pt − Pi)]
(c) k = (2.303/t) log[Pi/(Pi − Pt)]
(d) none of these
Show Answer & Explanation

Answer: (a) k = (2.303/t) log[2Pi/(3Pi − Pt)]

Explanation:
1. If p of A reacts, Pt = Pi + 2p, so p = (Pt − Pi)/2.
2. PA = Pi − p = (3Pi − Pt)/2.
3. k = (2.303/t) log(Pi/PA) = (2.303/t) log[2Pi/(3Pi − Pt)].
4. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 8: The rate of a reaction increases with temperature:
(a) for any reaction
(b) for exothermic reactions only
(c) for endothermic reactions only
(d) for none
Show Answer & Explanation

Answer: (a) for any reaction

Explanation:
1. The fraction of molecules above the activation energy rises with temperature.
2. This applies to every reaction, exothermic or endothermic.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 9: If the concentration of a reactant is increased 2 times, then K becomes:
(a) ln(K/2)
(b) K/2
(c) 2K
(d) K
Show Answer & Explanation

Answer: (d) K

Explanation:
1. The rate constant does not depend on concentration.
2. K remains unchanged.
3. Answer: (d)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 10: The activation energy for a chemical reaction depends upon:
(a) temperature
(b) nature of reacting species
(c) concentration of reacting species
(d) collision frequency
Show Answer & Explanation

Answer: (b) nature of reacting species

Explanation:
1. Activation energy is characteristic of the reaction, i.e. of the bonds being broken and formed.
2. It is essentially independent of temperature and concentration.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 11: The activation energy of the reaction A → B (ΔH = −52 kcal) is 128 kcal. The activation energy of the reaction B → A will be:
(a) −76 kcal
(b) +76 kcal
(c) −180 kcal
(d) +180 kcal
Show Answer & Explanation

Answer: (d) +180 kcal

Explanation:
1. Ea(reverse) = Ea(forward) − ΔH.
2. = 128 − (−52) = +180 kcal.
3. Answer: (d)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 12: Which of the following represents the expression for the 3/4th life of a first order reaction?
(a) (K/2.303) log(4/3)
(b) (2.303/K) log(3/4)
(c) (2.303/K) log 4
(d) (2.303/K) log 3
Show Answer & Explanation

Answer: (c) (2.303/K) log 4

Explanation:
1. Time for 75% completion: t = (2.303/K) log[a/(a − 0.75a)].
2. = (2.303/K) log 4.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 13: The rate constant for a first order reaction whose half life is 480 s is:
(a) 1.44 × 10−3 s−1
(b) 1.44 s−1
(c) 0.72 × 10−3 s−1
(d) 2.88 × 10−3 s−1
Show Answer & Explanation

Answer: (a) 1.44 × 10−3 s−1

Explanation:
1. k = 0.693/t1/2 = 0.693/480.
2. = 1.44 × 10−3 s−1.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 14: In a parallel first order reaction, A forms B with rate constant 2 × 10−5 and C with rate constant 3 × 10−5. Find the percentage of B formed.
(a) 40%
(b) 60%
(c) 50%
(d) 90%
Show Answer & Explanation

Answer: (a) 40%

Explanation:
1. In parallel reactions products are formed in the ratio of their rate constants.
2. % B = 2/(2 + 3) × 100 = 40%.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 15: For a reaction nA → B + C, three plots are drawn. Plot (I): t1/2 against [A] is a horizontal line. Plot (II): t1/2 against [A] is a straight line through the origin. Plot (III): t1/2 against 1/[A] is a straight line through the origin. These plots respectively correspond to the reaction orders:
(a) 0, 1, 2
(b) 1, 2, 3
(c) 1, 0, 2
(d) none of these
Show Answer & Explanation

Answer: (c) 1, 0, 2

Explanation:
1. (I) t1/2 independent of [A] means first order.
2. (II) t1/2 ∝ [A] means zero order. (III) t1/2 ∝ 1/[A] means second order.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 16: The half life of a first order reaction is 6 hours. How long will it take for the reactant to change from 0.8 M to 0.25 M?
(a) 1.07 hour
(b) 5.1 hour
(c) 2.7 hour
(d) 10.07 hour
Show Answer & Explanation

Answer: (d) 10.07 hour

Explanation:
1. k = 0.693/6 = 0.1155 h−1.
2. t = (2.303/k) log(0.8/0.25) = (2.303/0.1155) × 0.505 = 10.07 h.
3. Answer: (d)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 17: The concentration of a reactant in solution decreases from 0.5 M to 0.25 M in 5 hours and from 1.0 M to 0.25 M in 10 hours. The order of the reaction will be:
(a) 2
(b) 1
(c) 3
(d) 0
Show Answer & Explanation

Answer: (b) 1

Explanation:
1. 1.0 → 0.5 → 0.25 takes two half-lives.
2. Since 5 h + 5 h = 10 h, the half-life is constant, so the reaction is first order.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 18: Which of the following is the correct expression for the Arrhenius equation?
(a) ln(K2/K1) = (Ea/R)(1/T1 − 1/T2)
(b) log(K2/K1) = (Ea/2.303)(T1T2/(T2 − T1))
(c) ln(K2/K1) = (Ea/2.303R)(1/T1 − 1/T2)
(d) K2/K1 = (Ea/2.303R)((T2 − T1)/(T1T2))
Show Answer & Explanation

Answer: (a) ln(K2/K1) = (Ea/R)(1/T1 − 1/T2)

Explanation:
1. Subtracting ln k = ln A − Ea/RT at two temperatures gives ln(k2/k1) = (Ea/R)(1/T1 − 1/T2).
2. The 2.303 appears only with log, and R must be present.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 19: For a first order reaction A → B, the time taken to reduce to 1/4 of the initial concentration is 10 min. The time required to reduce to 1/16 of the initial concentration will be:
(a) 10 min
(b) 20 min
(c) 4.46 min
(d) 2.24 min
Show Answer & Explanation

Answer: (b) 20 min

Explanation:
1. 1/4 is two half-lives, so t1/2 = 5 min.
2. 1/16 is four half-lives = 20 min.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 20: For the reaction 2N2O5 → 4NO2 + O2, the rate and rate constant are 1.02 × 10−4 M s−1 and 3.4 × 10−5 s−1 respectively. The concentration of N2O5 at that time will be:
(a) 1.732 M
(b) 3 M
(c) 1.02 × 10−4 M
(d) 3.5 × 105 M
Show Answer & Explanation

Answer: (b) 3 M

Explanation:
1. The units of k (s−1) show the reaction is first order, so rate = k[N2O5].
2. [N2O5] = 1.02 × 10−4/3.4 × 10−5 = 3 M.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 21: For the reaction N2O5(g) → N2O4(g) + ½O2(g), the initial pressure is 114 mm and after 20 seconds the pressure of the reaction mixture becomes 133 mm. The average rate of reaction will be:
(a) 1.9 atm s−1
(b) 8.75 × 10−3 atm s−1
(c) 2.5 × 10−3 atm s−1
(d) 6.65 atm s−1
Show Answer & Explanation

Answer: (c) 2.5 × 10−3 atm s−1

Explanation:
1. Each mole of N2O5 consumed raises the total pressure by 1/2, so x/2 = 133 − 114 = 19 mm and x = 38 mm.
2. Rate = 38/20 = 1.9 mm s−1 = 1.9/760 = 2.5 × 10−3 atm s−1.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 22: The half lives of a first order and a zero order reaction are the same. Then the ratio of the initial rates of the first order reaction to that of the zero order reaction is (initial concentration of reactant = 1 molar):
(a) 1/0.693
(b) 2 × 0.693
(c) 2/0.693
(d) 6.93
Show Answer & Explanation

Answer: (b) 2 × 0.693

Explanation:
1. First order: t1/2 = 0.693/k1. Zero order: t1/2 = [A]0/2k0 = 1/2k0.
2. Equal t1/2 gives k1/k0 = 2 × 0.693.
3. Initial rates are k1[1] and k0, so the ratio is 2 × 0.693.
4. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 23: The rate constant for a first order reaction is 5.78 × 10−5 s−1. What percentage of the initial reactant will react in 10 hours?
(a) 12.5%
(b) 25%
(c) 87.5%
(d) 75%
Show Answer & Explanation

Answer: (c) 87.5%

Explanation:
1. kt = 5.78 × 10−5 × 36000 = 2.081.
2. Remaining fraction = e−2.081 = 0.125, so 87.5% has reacted.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 24: The half life period and initial concentration for a reaction are: initial concentration 350, 540, 158 and t1/2 425, 275, 941 respectively. The order of the reaction is:
(a) 2
(b) 1
(c) 0
(d) 4
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. The product (initial concentration × t1/2) is about constant: 350 × 425 ≈ 540 × 275 ≈ 158 × 941 ≈ 1.49 × 105.
2. t1/2 ∝ 1/a means second order.
3. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 25: The activation energy for the reaction H2O2 → H2O + ½O2 is 18 kcal/mol at 300 K. Calculate the fraction of molecules having energy equal to or greater than the activation energy (R = 2 cal K−1 mol−1; antilog(−13.02) = 9.36 × 10−14).
(a) 9.36 × 10−14
(b) 1.2 × 10−12
(c) 4.2 × 10−16
(d) 5.2 × 10−15
Show Answer & Explanation

Answer: (a) 9.36 × 10−14

Explanation:
1. Fraction = e−Ea/RT, and log of it = −Ea/(2.303RT).
2. = −18000/(2.303 × 2 × 300) = −13.02.
3. Fraction = antilog(−13.02) = 9.36 × 10−14.
4. Answer: (a)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 26: For the reaction N2(g) + 3H2(g) → 2NH3(g) under certain conditions of temperature and partial pressure of the reactants, the rate of formation of NH3 is 0.001 kg lit−1 h−1. The rate of conversion of H2 under the same condition is ... kg lit−1 h−1. (N = 14, H = 1)
(a) 0.001
(b) 0.0015
(c) 0.00017
(d) 0.0017
Show Answer & Explanation

Answer: (c) 0.00017

Explanation:
1. 0.001 kg NH3 = 1/17 mol (in kmol units: 5.88 × 10−5 kmol).
2. H2 consumed = (3/2) × 5.88 × 10−5 = 8.82 × 10−5 kmol = 1.76 × 10−4 kg ≈ 0.00017 kg lit−1 h−1.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 27: For a first order reaction:
(a) the degree of dissociation is equal to e−kt
(b) a plot of reciprocal concentration of the reactant vs time gives a straight line
(c) the time taken for the completion of 75% reaction is thrice the t1/2 of the reaction
(d) the pre-exponential factor in the Arrhenius equation has the dimension of time−1
Show Answer & Explanation

Answer: (d) the pre-exponential factor in the Arrhenius equation has the dimension of time−1

Explanation:
1. For first order A = Ae−E/RT has the same units as k, i.e. s−1.
2. The fraction left is e−kt (not the degree of dissociation), ln[A] vs t is linear, and 75% completion takes two half-lives.
3. Answer: (d)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 28: In the reaction xA → yB, log[−d[A]/dt] = log[d[B]/dt] + 0.3010, where the negative sign indicates the rate of disappearance of the reactant. Thus x : y is:
(a) 1 : 2
(b) 2 : 1
(c) 3 : 1
(d) 3 : 10
Show Answer & Explanation

Answer: (b) 2 : 1

Explanation:
1. log of the ratio = 0.3010, so (−d[A]/dt)/(d[B]/dt) = 2.
2. This ratio equals x/y, so x : y = 2 : 1.
3. Answer: (b)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

Question 29: The Arrhenius equation may be written as:
(a) d(ln K)/dT = −Ea/RT
(b) d(ln K)/dT = −Ea/RT2
(c) d(ln K)/dT = +Ea/RT2
(d) d(ln K)/dT = +Ea/RT
Show Answer & Explanation

Answer: (c) d(ln K)/dT = +Ea/RT2

Explanation:
1. Differentiating ln K = ln A − Ea/RT with respect to T.
2. d(ln K)/dT = +Ea/RT2.
3. Answer: (c)

Teacher's Note:
1. Identify the order first and use the matching integrated rate law. Check that the units of k agree with the order.

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Kinetics

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