JEE Chemistry Chemical Kinetics MCQs Set 38

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Multiple Choice Questions

Question 1: For the reaction 2A + B → C, the initial rates at different concentrations are: [A] = 0.05, [B] = 0.05, rate = 0.045; [A] = 0.10, [B] = 0.05, rate = 0.090; [A] = 0.20, [B] = 0.10, rate = 0.72 (concentrations in mol L⁻¹, rate in mol L⁻¹ s⁻¹). The rate law is:
(a) rate = k[A][B]²
(b) rate = k[A]²[B]
(c) rate = k[A][B]
(d) rate = k[A]²[B]²
Show Answer & Explanation

Answer: (a) rate = k[A][B]²

Explanation:
1. Experiments 1 and 2: doubling \( [A] \) doubles the rate, so the order in \( A \) is 1.
2. Experiments 2 and 3: \( [A] \) doubles (factor 2) and \( [B] \) doubles; the rate rises by 8, so \( 2 \times 2^y = 8 \) and \( y = 2 \).
3. Rate = \( k[A][B]^{2} \).

Teacher's Note:
1. Vary one concentration at a time, or divide two experiments to isolate one order. Orders are found only from experiments.

Question 2: The plots of the concentration of reactant R against time for two reactions are: for reaction (i) ln[R] versus time is a straight line with negative slope; for reaction (ii) [R] versus time is a straight line with negative slope. The orders of reactions (i) and (ii) are respectively:
(a) 1, 1
(b) 0, 2
(c) 0, 1
(d) 1, 0
Show Answer & Explanation

Answer: (d) 1, 0

Explanation:
1. A first order reaction obeys \( \ln[R] = \ln[R]_{0} \) − kt, so \( \ln[R] \) against \( t \) is linear.
2. A zero order reaction obeys \( [R] = [R]_{0} \) − kt, so \( [R] \) against \( t \) is linear.

Teacher's Note:
1. Learn the linear plot of each order: zero \( [R] \) vs \( t \), first \( \ln[R] \) vs \( t \), second \( 1/[R] \) vs \( t \).

Question 3: For the reaction H₂ + I₂ → 2HI, the rate constant is 2.5 × 10⁻⁴ dm³ mol⁻¹ s⁻¹ at 327°C and 1.0 dm³ mol⁻¹ s⁻¹ at 527°C. The activation energy in kJ mol⁻¹ is (R = 8.314 J K⁻¹ mol⁻¹):
(a) 59
(b) 72
(c) 150
(d) 166
Show Answer & Explanation

Answer: (d) 166

Explanation:
1. \( T_{1} = 600 K \) and \( T_{2} = 800 K \).
2. \( \log(k_{2}/k_{1}) = \log(4000) = 3.602 \) = Ea/(2.303 × \( 8.314) \times (200/(600 \times 800) \)).
3. Ea = \( 3.602 \times 2.303 \times 8.314 \times 2400 \approx 1.655 \times 10^{5} J \), about 166 kJ mol⁻¹.

Teacher's Note:
1. Use the two-temperature Arrhenius equation and keep temperatures in kelvin.

Question 4: NO₂ required for a reaction is produced by the decomposition of N₂O₅ in CCl₄: 2N₂O₅ → 4NO₂ + O₂. The initial concentration of N₂O₅ is 3.00 mol L⁻¹ and it is 2.75 mol L⁻¹ after 30 minutes. The rate of formation of NO₂ is:
(a) 4.167 × 10⁻³ mol L⁻¹ min⁻¹
(b) 1.667 × 10⁻² mol L⁻¹ min⁻¹
(c) 8.333 × 10⁻³ mol L⁻¹ min⁻¹
(d) 2.083 × 10⁻³ mol L⁻¹ min⁻¹
Show Answer & Explanation

Answer: (b) 1.667 × 10⁻² mol L⁻¹ min⁻¹

Explanation:
1. \( -\Delta[N_{2}O_{5}]/\Deltat = 0.25/30 \) mol \( L^{-1} \min^{-1} \).
2. (1/4) d[NO₂]/dt = \( -(1/2) d[N_{2}O_{5}]/dt \), so d[NO₂]/dt = \( 2 \times 0.25/30 = 1.667 \times 10^{-2} \) mol \( L^{-1} \min^{-1} \).

Teacher's Note:
1. Convert between rates of different species using the stoichiometric coefficients.

Question 5: For the elementary reaction A₂ ⇌ 2A (forward rate constant k₁, backward rate constant k₋₁), the expression for d[A]/dt is:
(a) 2k₁[A₂] − k₋₁[A]²
(b) k₁[A₂] − k₋₁[A]²
(c) 2k₁[A₂] − 2k₋₁[A]²
(d) k₁[A₂] + k₋₁[A]²
Show Answer & Explanation

Answer: (c) 2k₁[A₂] − 2k₋₁[A]²

Explanation:
1. Net rate of reaction \( r = k_{1}[A_{2}] \) − k₋₁[A]².
2. A is formed two at a time, so \( d[A]/dt = 2r = 2k_{1}[A_{2}] \) − 2k₋₁[A]².

Teacher's Note:
1. Multiply by the stoichiometric coefficient when expressing the rate for a particular species.

Question 6: If a reaction follows the Arrhenius equation, the plot of ln k against 1/(RT) gives a straight line with gradient −y unit. The energy required to activate the reactants is:
(a) y/R unit
(b) −y unit
(c) yR unit
(d) y unit
Show Answer & Explanation

Answer: (d) y unit

Explanation:
1. \( \ln k = \ln A \) − Ea/(RT).
2. The gradient of \( \ln k \) against 1/(RT) is −Ea, so Ea = \( y \).

Teacher's Note:
1. If the plot were against \( 1/T \), the slope would be −Ea/R instead.

Question 7: For a reaction, a plot of ln k against 1/T is a straight line of slope −4606 K. If the rate constant at 400 K is 10⁻⁵ s⁻¹, the rate constant at 500 K is:
(a) 4 × 10⁻⁴ s⁻¹
(b) 10⁻⁶ s⁻¹
(c) 10⁻⁴ s⁻¹
(d) 2 × 10⁻⁴ s⁻¹
Show Answer & Explanation

Answer: (c) 10⁻⁴ s⁻¹

Explanation:
1. \( \ln(k_{2}/k_{1}) = 4606 (1/400 - 1/500) = 4606 \times 0.0005 = 2.303 \).
2. \( k_{2}/k_{1} = 10 \), so \( k_{2} = 10^{-4} s^{-1} \).

Teacher's Note:
1. A slope of −Ea/R lets you use the Arrhenius equation directly.

Question 8: Two reactions R₁ and R₂ have identical pre-exponential factors. The activation energy of R₁ exceeds that of R₂ by 10 kJ mol⁻¹. If k₁ and k₂ are the rate constants at 300 K, then ln(k₂/k₁) is equal to (R = 8.314 J mol⁻¹ K⁻¹):
(a) 8
(b) 12
(c) 6
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. \( k_{2}/k_{1} \) = exp[(Ea₁ − Ea₂)/RT].
2. \( \ln(k_{2}/k_{1}) = 10000/(8.314 \times 300) \approx 4 \).

Teacher's Note:
1. With the same \( A \), the reaction of lower Ea is faster.

Question 9: Higher order (greater than 3) reactions are rare due to:
(a) low probability of simultaneous collision of all the reacting species
(b) increase in entropy and activation energy as more molecules are involved
(c) shifting of equilibrium towards reactants due to elastic collisions
(d) loss of active species on collision
Show Answer & Explanation

Answer: (a) low probability of simultaneous collision of all the reacting species

Explanation:
1. A reaction needs proper orientation and effective collision of all reactant molecules.
2. The chance of four or more molecules colliding together in the right way is very small.

Teacher's Note:
1. Most elementary steps are unimolecular or bimolecular.

Question 10: The rate of a reaction doubles when its temperature changes from 300 K to 310 K. The activation energy of such a reaction will be (R = 8.314 J K⁻¹ mol⁻¹ and log 2 = 0.301):
(a) 53.6 kJ mol⁻¹
(b) 48.6 kJ mol⁻¹
(c) 58.5 kJ mol⁻¹
(d) 60.5 kJ mol⁻¹
Show Answer & Explanation

Answer: (a) 53.6 kJ mol⁻¹

Explanation:
1. \( \log 2 \) = Ea/(2.303 × \( 8.314) \times (310 - 300)/(300 \times 310 \)).
2. Ea = \( 0.301 \times 2.303 \times 8.314 \times 9300 \approx 5.36 \times 10^{4} J = 53.6 \) kJ mol⁻¹.

Teacher's Note:
1. Rate doubling for a ten degree rise corresponds to Ea of about 50 kJ near room temperature.

Question 11: For a first order reaction A → products, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is:
(a) 1.73 × 10⁻⁵ M min⁻¹
(b) 3.47 × 10⁻⁴ M min⁻¹
(c) 3.47 × 10⁻⁵ M min⁻¹
(d) 1.73 × 10⁻⁴ M min⁻¹
Show Answer & Explanation

Answer: (b) 3.47 × 10⁻⁴ M min⁻¹

Explanation:
1. 0.1 to 0.025 is two half-lives, so \( t\frac{1}{2} = 20 \min \) and \( k = 0.0347 \min^{-1} \).
2. Rate = \( k[A] = 0.0347 \times 0.01 = 3.47 \times 10^{-4} M \min^{-1} \).

Teacher's Note:
1. Concentration falling to one quarter means two half-lives for a first order reaction.

Question 12: A reactant A forms two products: A → B (rate constant k₁, activation energy Ea₁) and A → C (k₂, Ea₂). If Ea₂ = 2Ea₁, then k₁ and k₂ are related as:
(a) \( k_{1} = 2k_{2} e^{Ea_{2}/RT} \)
(b) \( k_{1} = k_{2} e^{Ea_{1}/RT} \)
(c) \( k_{2} = k_{1} e^{Ea_{2}/RT} \)
(d) \( k_{1} = A k_{2} e^{Ea_{1}/RT} \)
Show Answer & Explanation

Answer: (b) \( k_{1} = k_{2} e^{Ea_{1}/RT} \)

Explanation:
1. With the same \( A, k_{1} = A e^{-Ea_{1}/RT} \) and \( k_{2} = A e^{-2Ea_{1}/RT} \).
2. Dividing, \( k_{1}/k_{2} = e^{Ea_{1}/RT} \), so \( k_{1} = k_{2} e^{Ea_{1}/RT} \).

Teacher's Note:
1. Competing reactions with the same \( A \) differ in rate only by their activation energies.

Question 13: For the reaction Cl₂(aq) + H₂S(aq) → S(s) + 2H⁺(aq) + 2Cl⁻(aq), the rate equation is rate = k[Cl₂][H₂S]. Mechanism I: Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁺ + HS⁻ (slow); Cl⁺ + HS⁻ → H⁺ + Cl⁻ + S (fast). Mechanism II: H₂S ⇌ H⁺ + HS⁻ (fast equilibrium); Cl₂ + HS⁻ → 2Cl⁻ + H⁺ + S (slow). Which mechanism is consistent with the rate equation?
(a) II only
(b) Both I and II
(c) Neither I nor II
(d) I only
Show Answer & Explanation

Answer: (d) I only

Explanation:
1. The rate follows the slow step.
2. In I, rate = k[Cl₂][H₂S], which matches.
3. In II, rate = k[Cl₂][HS⁻] with [HS⁻] = \( K[H_{2}S]/[H^{+}] \), giving k[Cl₂][H₂S][H⁺]⁻¹, which does not match.

Teacher's Note:
1. Substitute any intermediates using the fast equilibrium before comparing rate laws.

Question 14: The half-life of a first order reaction is 6.93 min. The time required for 99% completion of the reaction will be (log 2 = 0.301):
(a) 230.3 min
(b) 23.03 min
(c) 46.06 min
(d) 460.6 min
Show Answer & Explanation

Answer: (c) 46.06 min

Explanation:
1. \( k = 0.693/6.93 = 0.1 \min^{-1} \).
2. \( t = (2.303/k) \log(100/1) = 23.03 \times 2 = 46.06 \min \).

Teacher's Note:
1. Ninety-nine percent completion takes about 6.64 half-lives.

Question 15: The activation energies of the forward and reverse reactions for A₂ + B₂ ⇌ 2AB are 180 and 200 kJ mol⁻¹ respectively. A catalyst lowers the activation energy of both by 100 kJ mol⁻¹. The magnitude of the enthalpy change of the reaction in the presence of the catalyst is (in kJ mol⁻¹):
(a) 300
(b) 120
(c) 280
(d) 20
Show Answer & Explanation

Answer: (d) 20

Explanation:
1. With the catalyst, Ea(forward) = 80 and Ea(reverse) = 100 kJ mol⁻¹.
2. \( \DeltaH = 80 - 100 = -20 \) kJ mol⁻¹, magnitude 20.

Teacher's Note:
1. A catalyst does not change \( \DeltaH \). It lowers both activation energies equally.

Question 16: A radioactive element is spilled over the floor of a room. Its half-life is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?
(a) 1000 days
(b) 300 days
(c) 10 days
(d) 100 days
Show Answer & Explanation

Answer: (d) 100 days

Explanation:
1. \( N/N_{0} = 1/10 = (1/2)^n \), so \( n = \log 10/\log 2 = 3.32 \).
2. \( t = 3.32 \times 30 \approx 100 \) days.

Teacher's Note:
1. Radioactive decay is a first order process.

Question 17: In the Arrhenius equation \( k = A e^{-E/RT}, E \) represents:
(a) the energy above which all the colliding molecules will react
(b) the energy below which colliding molecules will not react
(c) the total energy of the reacting molecules at a temperature T
(d) the fraction of molecules with energy greater than the activation energy
Show Answer & Explanation

Answer: (b) the energy below which colliding molecules will not react

Explanation:
1. \( E \) is the activation energy, the threshold energy barrier.
2. Colliding molecules with energy below it do not react.

Teacher's Note:
1. Not all molecules with energy above \( E \) need react, since orientation matters; the exponential term is the fraction with energy greater than \( E \).

Question 18: A reaction involving two different reactants can never be:
(a) a bimolecular reaction
(b) a second order reaction
(c) a first order reaction
(d) a unimolecular reaction
Show Answer & Explanation

Answer: (d) a unimolecular reaction

Explanation:
1. Molecularity is the number of species colliding in an elementary step.
2. Two different reactants must collide, so molecularity is at least 2 and it cannot be unimolecular.

Teacher's Note:
1. Order can be any value, but molecularity is a whole number of 1, 2 or 3.

Question 19: For an endothermic reaction X → Y with activation energies Eb (backward) and Ef (forward), in general:
(a) there is no definite relation between Eb and Ef
(b) Eb = Ef
(c) Eb > Ef
(d) Eb < Ef
Show Answer & Explanation

Answer: (d) Eb < Ef

Explanation:
1. \( \DeltaH \) = Ef − Eb.
2. For an endothermic reaction \( \DeltaH \) is positive, so Ef > Eb.

Teacher's Note:
1. In an exothermic reaction the backward reaction has the higher activation energy.

Question 20: The rate equation for the reaction 2A + B → C is found to be rate = k[A][B]. The correct statement is that:
(a) the unit of k must be s⁻¹
(b) t½ is a constant
(c) the rate of formation of C is twice the rate of disappearance of A
(d) the value of k is independent of the initial concentrations of A and B
Show Answer & Explanation

Answer: (d) 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 unit \( L \) mol⁻¹ \( s^{-1} \) and \( t\frac{1}{2} \) depends on concentration.
2. Rate of formation of \( C \) is half the rate of disappearance of \( A \).
3. \( k \) depends on temperature only, not on concentration.

Teacher's Note:
1. A rate constant changes with temperature and catalysts but not with concentration.

Question 21: In a first order reaction, the concentration of the reactant decreases from 0.8 M to 0.4 M in 15 min. The time taken for the concentration to change from 0.1 M to 0.025 M is:
(a) 30 min
(b) 15 min
(c) 7.5 min
(d) 60 min
Show Answer & Explanation

Answer: (a) 30 min

Explanation:
1. \( t\frac{1}{2} = 15 \min \).
2. 0.1 to \( 0.025 M \) is two half-lives, so \( t = 30 \min \).

Teacher's Note:
1. Each half-life of a first order reaction is the same, whatever the starting concentration.

Question 22: The rate law for a reaction between A and B is rate = k[A]ⁿ[B]ᵐ. On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate is:
(a) \( 1/2^{m+n} \)
(b) (m + n)
(c) (n − m)
(d) \( 2^{n-m} \)
Show Answer & Explanation

Answer: (d) \( 2^{n-m} \)

Explanation:
1. New rate = k(2[A])ⁿ([B]/2)ᵐ.
2. Ratio = 2ⁿ × 2⁻ᵐ = \( 2^{n-m} \).

Teacher's Note:
1. Substitute the new concentrations in the rate law and divide.

Question 23: For the reaction H₂ + I₂ → 2HI, the differential rate law is:
(a) −d[H₂]/dt = −d[I₂]/dt = 2 d[HI]/dt
(b) −2 d[H₂]/dt = −2 d[I₂]/dt = d[HI]/dt
(c) −d[H₂]/dt = −d[I₂]/dt = d[HI]/dt
(d) −d[H₂]/(2dt) = −d[I₂]/(2dt) = d[HI]/dt
Show Answer & Explanation

Answer: (b) −2 d[H₂]/dt = −2 d[I₂]/dt = d[HI]/dt

Explanation:
1. Rate = \( -d[H_{2}]/dt = -d[I_{2}]/dt = \frac{1}{2} \) d[HI]/dt.
2. Multiplying by 2 gives \( -2d[H_{2}]/dt = -2d[I_{2}]/dt \) = d[HI]/dt.

Teacher's Note:
1. HI forms twice as fast as \( H_{2} \) is used up.

Question 24: For a first order reaction with initial concentration C₀ and concentration Cₜ at time t, kt = log C₀ − log Cₜ. A straight line graph is obtained by plotting:
(a) t versus log C₀
(b) t versus log Cₜ
(c) t⁻¹ versus log Cₜ
(d) log C₀ versus log Cₜ
Show Answer & Explanation

Answer: (b) t versus log Cₜ

Explanation:
1. Rearranging, \( \log \) Cₜ = \( \log C_{0} \) − kt.
2. So \( \log \) Cₜ is linear in \( t \); plotting \( t \) against \( \log \) Cₜ gives a straight line with negative slope.

Teacher's Note:
1. Any equation of the form \( y \) = mx \( + c \) gives a straight line.

Question 25: Consider two statements for a reaction obeying the Arrhenius equation (0°C < T < 300°C). I: the rate constant k decreases exponentially as the activation energy Ea increases (at fixed temperature). II: when k is plotted against temperature in °C, it increases exponentially. Choose the correct option.
(a) Both I and II are wrong
(b) Both I and II are correct
(c) I is wrong but II is correct
(d) I is right but II is wrong
Show Answer & Explanation

Answer: (b) Both I and II are correct

Explanation:
1. \( k = A e^{-Ea/RT} \), so \( k \) falls exponentially as Ea rises, and statement I is correct.
2. As \( T \) (in kelvin, and hence in \( ^{\circ}C \) over this range) increases, \( k \) rises exponentially, so statement II is also correct.

Teacher's Note:
1. A higher barrier gives a smaller \( k \), and a higher temperature gives a larger \( k \).

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