JEE Chemistry Chemical Kinetics MCQs Set 26

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Multiple Choice Questions

Question 1: A sample of moon rock contains equal numbers of atoms of uranium and lead (assume all lead came from uranium). The half-life of uranium is 4.5 × 109 years. The age of the rock is
(a) 9.0 × 109 years
(b) 4.5 × 109 years
(c) 2.25 × 109 years
(d) 13.5 × 109 years
Show Answer & Explanation

Answer: (b) 4.5 × 109 years

Explanation:
1. Equal numbers of U and Pb mean half of the uranium has decayed.
2. That takes exactly one half-life, 4.5 × 109 years.

Teacher's Note:
1. N(Pb) = N(U) means N(U) = N0/2.
2. One half-life.

Question 2: For a reaction xA → yB, log(-d[A]/dt) = log(d[B]/dt) + 0.3. The ratio x : y is
(a) 2 : 1
(b) 3 : 10
(c) 3 : 1
(d) 1 : 2
Show Answer & Explanation

Answer: (a) 2 : 1

Explanation:
1. Taking antilogarithms: (-d[A]/dt)/(d[B]/dt) = 100.3 = 2.
2. For xA → yB the ratio of these rates equals x/y, so x : y = 2 : 1.

Teacher's Note:
1. Rate of disappearance of a reactant divided by its coefficient equals the rate of the reaction.
2. 100.3 is about 2.

Question 3: The decomposition of NH3 on a heated tungsten surface gave half-lives of 290 s at 65 mm Hg, x at 105 mm Hg, 670 s at y mm Hg and 820 s at 185 mm Hg. The approximate values of x and y are
(a) x = 430 s, y = 105 mm Hg
(b) x = 490 s, y = 120 mm Hg
(c) x = 467 s, y = 150 mm Hg
(d) x = 410 s, y = 115 mm Hg
Show Answer & Explanation

Answer: (c) x = 467 s, y = 150 mm Hg

Explanation:
1. Half-life increases in proportion to the initial pressure (290/65 = 4.46 s per mm Hg, and 820/185 = 4.43), so the surface reaction is zero order.
2. x = 105 × 4.45 = 467 s and y = 670/4.45 = 150 mm Hg.

Teacher's Note:
1. For zero order t1/2 = a/2k.
2. Constant half-life divided by pressure confirms zero order.

Question 4: A first order reaction has a half-life of ten minutes. The fraction of the reactant left after one hour is
(a) 1/6
(b) 1/12
(c) 1/64
(d) 1/32
Show Answer & Explanation

Answer: (c) 1/64

Explanation:
1. One hour is 6 half-lives.
2. Fraction left = (1/2)6 = 1/64.

Teacher's Note:
1. Number of half-lives = time / t1/2.
2. Fraction = (1/2)^n.

Question 5: A drop of solution (volume 0.05 mL) contains 3.0 × 10-6 mol of H+. The rate constant for the disappearance of H+ is 1.0 × 107 mol L-1 s-1. The time for the H+ in the drop to disappear is
(a) 6 × 10-9 s
(b) 6 × 10-8 s
(c) 6 × 10-10 s
(d) 6 × 10-7 s
Show Answer & Explanation

Answer: (a) 6 × 10-9 s

Explanation:
1. The concentration of H+ is 3.0 × 10-6/(0.05 × 10-3 L) = 0.06 mol L-1.
2. The units of k show zero order, so t = [H+]/k = 0.06/(1.0 × 107) = 6 × 10-9 s.

Teacher's Note:
1. Convert mL to L.
2. Zero order: t = initial concentration / k.

Question 6: The reaction H2(g) + 2ICl(g) → 2HCl(g) + I2(g) is first order in H2 and first order in ICl. Mechanism I is a single elementary step H2 + 2ICl → 2HCl + I2. Mechanism II is H2 + ICl → HCl + HI (slow), HI + ICl → HCl + I2 (fast). Which mechanism is consistent with the rate law?
(a) I only
(b) neither I nor II
(c) II only
(d) both I and II
Show Answer & Explanation

Answer: (c) II only

Explanation:
1. Mechanism I would give rate = k[H2][ICl]2, third order.
2. In mechanism II the slow step gives rate = k[H2][ICl], first order in each, consistent with the data.

Teacher's Note:
1. An elementary step's rate law follows its stoichiometry.
2. The rate-determining step controls the rate law.

Question 7: The rate equation for 2A + B → C is rate = k[A][B]. The correct statement is that
(a) the value of k is independent of the initial concentrations of A and B
(b) the unit of k must be s-1
(c) t1/2 is a constant
(d) the rate of formation of C is twice the rate of disappearance of A
Show Answer & Explanation

Answer: (a) the value of k is independent of the initial concentrations of A and B

Explanation:
1. The reaction is second order overall, so k has units L mol-1 s-1 and t1/2 depends on concentration.
2. A disappears twice as fast as C forms.
3. The rate constant depends only on temperature (and catalyst), not on concentrations.

Teacher's Note:
1. k is constant at fixed temperature.
2. Check stoichiometry for rate relations.

Question 8: For the decay of a nucleus, the time of decay is given as t = 4 t1/2. The relation between the mean life T and t is
(a) t = 4T ln 2
(b) t = ln 2 / T2
(c) t = 2T4 ln 2
(d) t = 2T ln 2
Show Answer & Explanation

Answer: (a) t = 4T ln 2

Explanation:
1. Mean life T = t1/2/ln 2, so t1/2 = T ln 2.
2. t = 4 t1/2 = 4T ln 2.

Teacher's Note:
1. Mean life = 1/λ.
2. t1/2 = 0.693 T.

Question 9: For 2NO + O2 → 2NO2 the mechanism is NO + NO ⇌ N2O2 (fast, k1 forward and k-1 reverse), N2O2 + O2 → 2NO2 (slow, k2). The rate constant of the overall reaction is
(a) k2 k1 k-1
(b) k2
(c) k2 k1/k-1
(d) k2 k1
Show Answer & Explanation

Answer: (c) k2 k1/k-1

Explanation:
1. Rate = k2[N2O2][O2].
2. From the fast equilibrium, [N2O2] = (k1/k-1)[NO]2.
3. Rate = (k2 k1/k-1)[NO]2[O2], so the overall rate constant is k2 k1/k-1.

Teacher's Note:
1. Replace the intermediate using the equilibrium step.
2. Overall k is a combination of step constants.

Question 10: A zero order reaction has k = 0.025 M s-1 for the disappearance of A. If the initial concentration is 0.50 M, the concentration after 15 s is
(a) 0.060 M
(b) 0.375 M
(c) 0.50 M
(d) 0.125 M
Show Answer & Explanation

Answer: (d) 0.125 M

Explanation:
1. For zero order, [A] = [A]0 - kt = 0.50 - 0.025 × 15.
2. [A] = 0.50 - 0.375 = 0.125 M.

Teacher's Note:
1. Zero order concentration falls linearly with time.
2. Check that the reaction has not finished: it would take 20 s.

Question 11: At 373 K a gaseous reaction A → 2B + C is first order. Starting with pure A, the total pressure after 10 min was 176 mm Hg and after a very long time it was 270 mm Hg. The pressure of A at the end of 10 minutes was
(a) 94 mm Hg
(b) 90 mm Hg
(c) 43 mm Hg
(d) 47 mm Hg
Show Answer & Explanation

Answer: (d) 47 mm Hg

Explanation:
1. Complete decomposition triples the pressure, so P0 = 270/3 = 90 mm Hg.
2. If x of A has reacted, the total pressure is 90 + 2x = 176, so x = 43 mm Hg.
3. P(A) = 90 - 43 = 47 mm Hg.

Teacher's Note:
1. Each mole of A gives three moles of gas, a net gain of two.
2. Final pressure = 3 P0.

Question 12: In a first order reaction the concentration falls from 0.8 M to 0.4 M in 15 minutes. The time for the concentration to change from 0.1 M to 0.025 M is
(a) 7.5 minutes
(b) 15 minutes
(c) 60 minutes
(d) 30 minutes
Show Answer & Explanation

Answer: (d) 30 minutes

Explanation:
1. The half-life is 15 minutes.
2. 0.1 M to 0.025 M is a fall by a factor of 4 = two half-lives, so t = 30 minutes.

Teacher's Note:
1. Count half-lives.
2. Half-life is constant for first order.

Question 13: For H2(g) + Br2(g) → 2HBr(g), experiment shows rate = k[H2][Br2]1/2. The molecularity and order are respectively
(a) 3/2, 3/2
(b) 2, 3/2
(c) 1, 1/2
(d) not defined, 3/2
Show Answer & Explanation

Answer: (d) not defined, 3/2

Explanation:
1. The order is the sum of the exponents: 1 + 1/2 = 3/2.
2. A fractional order shows that the reaction is complex (several steps), and molecularity applies only to a single elementary step, so it is not defined.

Teacher's Note:
1. Molecularity is always a whole number and refers to elementary steps.
2. Order can be fractional.

Question 14: In the inversion of cane sugar (first order) the total angle of rotation is +40° at t = 0, +15° at 10 min, and -10° after a long time. The rate constant in s-1 is (ln 2 = 0.693)
(a) 1.15 × 10-3
(b) 0.693
(c) 0.0693
(d) 1.15 × 10-2
Show Answer & Explanation

Answer: (a) 1.15 × 10-3

Explanation:
1. k = (1/t) ln[(r0 - r∞)/(rt - r∞)] = (1/600 s) ln[50/25].
2. k = 0.693/600 = 1.155 × 10-3 s-1.

Teacher's Note:
1. Convert 10 min to 600 s.
2. Differences from the final rotation measure the reactant left.

Question 15: Two radionuclides A and B have disintegration constants in the ratio 1 : 2. For the probability of detecting particles from A and from B at t = 0 to be the same, the ratio of the numbers of atoms NA : NB at t = 0 must be
(a) 4 : 1
(b) 2 : 1
(c) 1 : 4
(d) 1 : 2
Show Answer & Explanation

Answer: (b) 2 : 1

Explanation:
1. Initial activity = λ N, so equal activities need λA NA = λB NB.
2. NA/NB = λB/λA = 2/1.

Teacher's Note:
1. Activity is proportional to λ x N.
2. The less stable nuclide needs fewer atoms.

Question 16: For A2 + B2 → 2AB with rate = k[A2]^a[B2]^b, the data are: [A2] = 0.2, [B2] = 0.2, rate 0.04; [A2] = 0.1, [B2] = 0.4, rate 0.04; [A2] = 0.2, [B2] = 0.4, rate 0.08. The value of a + b is
(a) 3
(b) 1
(c) 2
(d) 4
Show Answer & Explanation

Answer: (c) 2

Explanation:
1. Runs 1 and 3: [B2] doubles and the rate doubles, so b = 1.
2. Runs 1 and 2: [A2] halves and [B2] doubles with unchanged rate: (1/2)^a x 2 = 1, so a = 1.
3. a + b = 2.

Teacher's Note:
1. Compare runs where only one concentration changes.
2. Then use the third run to find the other order.

Question 17: Decomposition of H2O2 is first order: in fifty minutes the concentration falls from 0.5 M to 0.125 M. When the concentration reaches 0.05 M, the rate of formation of O2 is (2H2O2 → 2H2O + O2)
(a) 2.66 L min-1 at STP
(b) 1.34 × 10-2 mol min-1
(c) 6.93 × 10-2 mol min-1
(d) 6.93 × 10-4 mol L-1 min-1
Show Answer & Explanation

Answer: (d) 6.93 × 10-4 mol L-1 min-1

Explanation:
1. A fall to one-quarter in 50 min gives t1/2 = 25 min and k = 0.693/25 = 0.0277 min-1.
2. Rate of decomposition of H2O2 = 0.0277 × 0.05 = 1.386 × 10-3 M min-1.
3. O2 forms at half this rate (2H2O2 → O2): 6.93 × 10-4 mol L-1 min-1.

Teacher's Note:
1. Rate of O2 formation = (1/2)(-d[H2O2]/dt).
2. Find k from the half-life.

Question 18: Which of the following statements is NOT true for a second order reaction?
(a) Its half-life is inversely proportional to its initial concentration
(b) It can have a rate constant of 1 × 10-2 L mol-1 s-1
(c) t50 = 1/(k x initial concentration)
(d) The time to complete 75% of the reaction is twice the half-life
Show Answer & Explanation

Answer: (d) The time to complete 75% of the reaction is twice the half-life

Explanation:
1. For second order t1/2 = 1/(k a), so the second and fourth statements are true.
2. t75 = (1/k)(1/(0.25a) - 1/a) = 3/(ka) = 3 t1/2, not 2 t1/2.
3. (t75 = 2 t1/2 is true for a first order reaction.)

Teacher's Note:
1. For first order t75 = 2 t1/2; for second order t75 = 3 t1/2.
2. Derive t75 from the integrated rate law.

Question 19: The activation energy of a reaction is 8.3 kJ mol-1 without a catalyst and 4.15 kJ mol-1 with one. The slope of a plot of ln k against 1/T in the absence of the catalyst is (R = 8.3 J mol-1 K-1)
(a) -1000
(b) +1
(c) -1
(d) +1000
Show Answer & Explanation

Answer: (a) -1000

Explanation:
1. ln k = ln A - Ea/(RT), so the slope is -Ea/R.
2. Slope = -8300/8.3 = -1000 K.

Teacher's Note:
1. Use the uncatalysed Ea.
2. Slope of ln k vs 1/T is negative.

Question 20: The activity per mL of a radioactive solution is x. How much water must be added to 200 mL of the solution so that the activity per mL falls to x/20 after 4 half-lives?
(a) 150 mL
(b) 100 mL
(c) 50 mL
(d) 80 mL
Show Answer & Explanation

Answer: (c) 50 mL

Explanation:
1. After 4 half-lives the total activity is 1/16 of the original: 200x/16.
2. This is spread over a volume V so that activity per mL is x/20: V = (200x/16)/(x/20) = 250 mL.
3. Water to be added = 250 - 200 = 50 mL.

Teacher's Note:
1. Total activity = activity per mL x volume.
2. Dilution does not change the total activity.

Question 21: Po-218 (half-life 183 s) decays by α emission to Pb-214 (half-life 161 s), a β emitter. Starting with pure Po-218, the time at which the number of Pb-214 nuclei is a maximum is closest to
(a) 147.5 s
(b) 182 s
(c) 247.5 s
(d) 304 s
Show Answer & Explanation

Answer: (c) 247.5 s

Explanation:
1. Number of daughter nuclei is maximum when its rate of formation equals its rate of decay: λ1 N1 = λ2 N2.
2. This occurs at t = ln(λ1/λ2)/(λ1 - λ2).
3. With λ1 = 0.693/183 and λ2 = 0.693/161, t = ln(161/183)/[0.693(1/183 - 1/161)] = 247.5 s.

Teacher's Note:
1. Set d N2/dt = 0.
2. Use the half-lives to find λ values.

Question 22: The rate constant of the reaction A → 2B is 1.0 × 10-3 mol L-1 min-1. If the initial concentration of A is 1.0 mol L-1, the concentration of B after 100 minutes is
(a) 0.1 mol L-1
(b) 0.9 mol L-1
(c) 0.2 mol L-1
(d) 1.8 mol L-1
Show Answer & Explanation

Answer: (c) 0.2 mol L-1

Explanation:
1. The units of k show a zero-order reaction.
2. A consumed = kt = 1.0 × 10-3 x 100 = 0.1 mol L-1.
3. Each A gives two B, so [B] = 0.2 mol L-1.

Teacher's Note:
1. Units mol L-1 min-1 indicate zero order.
2. Do not forget the stoichiometric factor 2.

Question 23: The volume of O2 from the first order decomposition H2O2(aq) → H2O + (1/2)O2(g) is 5 L after 20 minutes and 50 L after completion (same conditions). The rate constant is
(a) (1/20) log 10 min-1
(b) (2.303/20) log(45/50) min-1
(c) (2.303/20) log 10 min-1
(d) (2.303/20) log(50/45) min-1
Show Answer & Explanation

Answer: (d) (2.303/20) log(50/45) min-1

Explanation:
1. The reactant left is proportional to V∞ - Vt = 50 - 5 = 45.
2. k = (2.303/20) log[V∞/(V∞ - Vt)] = (2.303/20) log(50/45) min-1.

Teacher's Note:
1. Gas volume measures how much has reacted.
2. Reactant left is proportional to the unevolved volume.

Question 24: In the presence of acid, the concentration of cane sugar fell from 0.2 M to 0.1 M in 5 h and to 0.05 M in 10 h. The order of the reaction is
(a) two
(b) three
(c) one
(d) zero
Show Answer & Explanation

Answer: (c) one

Explanation:
1. The concentration halves in each 5 h interval, so the half-life is constant (5 h) regardless of the concentration.
2. A constant half-life is characteristic of a first order reaction.

Teacher's Note:
1. Check two successive half-lives.
2. Inversion of cane sugar is pseudo-first order.

Question 25: The rate constants of two parallel reactions are 1.0 × 10-2 and 3.0 × 10-2 dm3 mol-1 s-1 with activation energies 60.0 and 70.0 kJ mol-1. The apparent overall activation energy is
(a) 100.0 kJ mol-1
(b) 65.0 kJ mol-1
(c) 130.0 kJ mol-1
(d) 67.5 kJ mol-1
Show Answer & Explanation

Answer: (d) 67.5 kJ mol-1

Explanation:
1. For parallel reactions E = (k1 E1 + k2 E2)/(k1 + k2).
2. E = (1 × 60 + 3 × 70)/4 = 67.5 kJ mol-1.

Teacher's Note:
1. The overall activation energy is the weighted mean.
2. It lies closer to the faster reaction's value.

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