JEE Main Chemistry: Chemical Kinetics questions with solutions

Q1. For the reaction A → products, the half-life is 120 min when the initial concentration of A is 8.0 × 10⁻² M. If the starting concentration is changed to 4.0 × 10⁻² M, the half-life increases to 240 min. What is the reaction order?

1. zero
2. one
3. two
4. 0.5

Answer: two

t-half is proportional to [A]^(1-n). Here halving [A] doubled t-half, so t-half ~ 1/[A], meaning 1-n = -1, giving n = 2 (second order).

Q2. For a first-order process A → B, if the rate constant is k and the starting concentration of A is 0.5 M, what is the half-life?

1. log 2 / k
2. log 2 / k√0.5
3. ln 2 / k
4. 0.693 / 0.5k

Answer: ln 2 / k

The half-life of a first-order reaction is independent of the initial concentration and is given by the formula t1/2 = ln(2) / k, which reflects the constant rate at which the reactant A is converted to product B.

Q3. For the elementary gas-phase reaction A(g) + 2B(g) → C(g) + D(g), an experiment starts with initial partial pressures p_A = 0.60 atm and p_B = 0.80 atm. If the partial pressure of C becomes 0.20 atm, then the reaction rate compared with the initial rate is

1. 1/6
2. 1/12
3. 1/36
4. 1/18

Answer: 1/6

The reaction rate is determined by the change in concentration of the reactants over time. Since the formation of C at 0.20 atm indicates that 0.20 atm of A and 0.40 atm of B have reacted, the initial rate can be calculated based on the initial pressures, leading to a final rate that is 1/6 of the initial rate.

Q4. From the experimental data below for the reaction A + B → products, identify the rate law that fits the observations. Exp. 1: [A] = 0.012, [B] = 0.035, initial rate = 0.1 Exp. 2: [A] = 0.024, [B] = 0.070, initial rate = 0.8 Exp. 3: [A] = 0.024, [B] = 0.035, initial rate = 0.1 Exp. 4: [A] = 0.012, [B] = 0.070, initial rate = 0.8

1. rate = k [B]³
2. rate = k [B]⁴
3. rate = k [A][B]³
4. rate = k [A]²[B]²

Answer: rate = k [B]³

The correct option is supported by the data showing that when the concentration of B is increased while A remains constant, the rate increases significantly, indicating a higher order dependence on B. Specifically, the rate increases by a factor of 8 when [B] is doubled, suggesting a cubic relationship with respect to [B].

Q5. A radioactive substance is accidentally spread on the floor of a room. If its half-life is 30 days and the initial activity is 10 times the allowed safe level, after how many days will it become safe to enter the room?

1. 100 days
2. 1000 days
3. 300 days
4. 10 days

Answer: 100 days

Need activity to drop from 10x to 1x: 10 = 2^n, n = log10/log2 = 3.32 half-lives. Time = 3.32 x 30 = ~100 days.

Q6. For the reversible reaction A2 + B2 ⇌ 2AB, the activation energies are 180 kJ mol⁻¹ for the forward direction and 200 kJ mol⁻¹ for the reverse direction. If a catalyst reduces the activation energy of both directions by 100 kJ mol⁻¹, what will be the enthalpy change of the reaction A2 + B2 ⇌ 2AB in the catalyzed pathway (in kJ mol⁻¹)?

1. 20
2. 300
3. 120
4. 280

Answer: 20

deltaH = Ea(forward) - Ea(reverse) = 180 - 200 = -20 kJ/mol. A catalyst lowers both activation energies by the same amount (100), so the difference and hence deltaH is unchanged at -20 kJ/mol (magnitude 20).

Q7. For the reaction Cl₂(aq) + H₂S(aq) → S(s) + 2H^+(aq) + 2Cl^-(aq), the experimentally observed rate law is rate = k[Cl₂][H₂S]. Which mechanism below is consistent with this rate law?

1. A. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow) Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)
2. B. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (fast) Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (slow)
3. C. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow) Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (slow)
4. D. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (fast) Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)

Answer: A. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁻ + HS⁻ (slow) Cl⁻ + HS⁻ → H⁺ + Cl⁻ + S (fast)

Observed rate = k[Cl2][H2S] is first order in each reactant, so the slow step must be the bimolecular reaction Cl2 + H2S. That is the mechanism whose FIRST step (Cl2 + H2S -> H+ + 2Cl- + HS-) is slow, giving rate = k[Cl2][H2S].

Q8. For the reaction 3A + 2B + C → products, the following initial-rate data were obtained at different starting concentrations: Initial rate (M s⁻¹) | [A]0 (M) | [B]0 (M) | [C]0 (M) 5.0 × 10⁻³ | 0.010 | 0.005 | 0.010 5.0 × 10⁻³ | 0.010 | 0.005 | 0.015 1.0 × 10⁻² | 0.010 | 0.010 | 0.010 1.25 × 10⁻³ | 0.005 | 0.005 | 0.010 The reaction orders with respect to A, B and C, in that order, are:

1. 3, 2, 0
2. 3, 2, 1
3. 2, 2, 0
4. 2, 1, 0

Answer: 2, 1, 0

Rows 1->2: C changes, rate same -> order C=0. Rows 1->3: B doubled, rate doubled -> order B=1. Rows 1->4: A halved, rate quartered -> order A=2. So orders are 2, 1, 0.

Q9. A first-order reaction has a half-life of 1386 s. What is the rate constant for this reaction?

1. 0.5 × 10⁻² s⁻¹
2. 0.5 × 10⁻³ s⁻¹
3. 5.0 × 10⁻² s⁻¹
4. 5.0 × 10⁻³ s⁻¹

Answer: 0.5 × 10⁻³ s⁻¹

The rate constant for a first-order reaction can be calculated using the formula k = 0.693/t₁/₂, where t₁/₂ is the half-life. Substituting 1386 s into the formula gives a rate constant of approximately 0.5 × 10⁻³ s⁻¹.

Q10. For a first-order reaction, the term 1/4 may be defined as the time required for the concentration of a reactant to decrease to three-fourths of its initial value. If the rate constant is k, this 1/4 is expressed as:

1. 0.75/k
2. 0.69/k
3. 0.29/k
4. 0.10/k

Answer: 0.29/k

For a first-order reaction, t = (2.303/k) log([R0]/[R]). For [R] = (3/4)[R0]: t = (2.303/k) log(4/3) = (2.303/k)(0.1249) = 0.29/k.

Q11. For the reaction aG + bH → products, the rate becomes eight times larger when the concentrations of both G and H are doubled. If only the concentration of G is doubled while H is kept unchanged, the rate becomes twice as large. What is the overall order of the reaction?

1. 0
2. 1
3. 2
4. 3

Answer: 3

Doubling both -> 8x means 2^(a+b)=8, so a+b=3. Doubling G only -> 2x means a=1, hence b=2. Overall order = a+b = 3.

Q12. For a reaction involving A and B, the rate expression is Rate = k[A]ⁿ[B]^m If the concentration of A is made twice and the concentration of B is reduced to one-half, then the new rate compared with the original rate is given by:

1. m + n
2. n - m
3. 2^(n - m)
4. 1/2^(m + n)

Answer: 2^(n - m)

New rate / old rate = [2A]^n [B/2]^m / ([A]^n [B]^m) = 2^n * (1/2)^m = 2^(n-m). The factor by which the rate changes is 2^(n-m).

Q13. For two reactions, the rate constants are given by k₁ = 10¹⁶ e⁻²⁰⁰⁰/T and k₂ = 10¹⁵ e⁻¹⁰⁰⁰/T. At what temperature do the two rate constants become equal?

1. 1000 K
2. 2000/2.303 K
3. 2000 K
4. 1000/2.303 K

Answer: 1000/2.303 K

10^16 e^(-2000/T) = 10^15 e^(-1000/T) -> 10 = e^(1000/T) -> ln10 = 1000/T -> T = 1000/2.303 K.

Q14. In an Arrhenius plot, the slope is given by:

1. -Ea / 2.303R
2. Ea / R
3. R / 2.303Ea
4. None of these

Answer: -Ea / 2.303R

From log k = log A - Ea/(2.303 R T), a plot of log k versus 1/T is linear with slope = -Ea/(2.303 R).

Q15. For the reaction A + B → products, the following observations are made: (1) If only the initial concentration of A is doubled, the reaction rate becomes twice its original value. (2) If the initial concentrations of both A and B are doubled together, the reaction rate increases by a factor of 8. The rate law for this reaction is:

1. rate = k [A][B]²
2. rate = k [A]² [B]²
3. rate = k [A][B]
4. rate = k [A]² [B]

Answer: rate = k [A][B]²

Doubling A -> 2x means order in A = 1. Doubling both -> 8x means 2^(1+b)=8, so b=2. Rate = k[A][B]^2.

Q16. For the decomposition reaction 2N2O5 → 4NO2 + O2, the rate and the rate constant are 1.02 × 10⁻⁴ mol L⁻¹ s⁻¹ and 3.4 × 10⁻⁵ s⁻¹, respectively. What is the concentration of N2O5 at that instant?

1. 1.732 M
2. 3 M
3. 3.4 × 10⁵ M
4. 1.02 × 10⁻⁴ M

Answer: 3 M

For first-order decomposition, rate = k[N2O5], so [N2O5] = rate/k = (1.02 x 10^-4)/(3.4 x 10^-5) = 3 M.

Q17. Identify the statement that is not correct.

1. The activation energy of the forward process is equal to that of the reverse process.
2. In a reversible reaction, raising the temperature increases the rates of both the forward and backward reactions.
3. For a second-order reaction, a higher starting concentration of reactant leads to a smaller half-life.
4. If the change in A is infinitesimally small, the average rate is equal to the instantaneous rate.

Answer: The activation energy of the forward process is equal to that of the reverse process.

Ea(forward) - Ea(reverse) = delta H of reaction, so they are equal only if delta H = 0; in general they differ, making this statement the incorrect one. The other three statements are true.

Q18. For the reaction 3A → B + C, under which condition will it behave as a zero-order reaction?

1. The reaction rate is proportional to the square of the concentration of A
2. The reaction rate stays constant at all concentrations
3. The rate does not change with any concentration of B and C
4. The reaction rate becomes double when the concentration of B is doubled

Answer: The reaction rate stays constant at all concentrations

A zero-order reaction has a rate that does not depend on reactant concentration, i.e. the rate stays constant at all concentrations of A.

Q19. A substance undergoes decomposition according to first-order kinetics. If its concentration falls to one-eighth of the initial concentration in 24 minutes, what is the rate constant for the process?

1. 1/24 min⁻¹
2. 0.692 / 24 min⁻¹
3. 2.303 / 24 log(1/8) min⁻¹
4. 2.303 / 24 log(8/1) min⁻¹

Answer: 2.303 / 24 log(8/1) min⁻¹

The correct option uses the first-order kinetics formula, which relates the rate constant to the logarithm of the ratio of initial and final concentrations. Since the concentration falls to one-eighth, the logarithm of 8/1 is used, reflecting the change in concentration over the given time period.

Q20. While investigating the rate of reaction between potassium iodate (KIO3) and sodium sulphite (Na2SO3) with starch solution as the indicator at different concentrations and temperatures, which precaution should be followed?

1. The sodium thiosulphate solution must always be kept at a lower concentration than the potassium iodate solution.
2. Use starch solution that has been prepared freshly.
3. Carry out the experiments using fresh solutions of H2O2 and KI.
4. All of the above.

Answer: Use starch solution that has been prepared freshly.

In the iodate-sulphite (iodine clock) experiment the starch indicator must be freshly prepared, since old starch loses sensitivity and gives a weak or delayed blue end-point. The other options mention sodium thiosulphate and H2O2/KI, which belong to different experiments, so 'All of the above' is not correct.

Q21. The time t1/4 is defined as the duration required for a reactant’s concentration to fall to three-fourths of its starting value. For a first-order reaction with rate constant K, t1/4 is given by

1. 0.75/K
2. 0.69/K
3. 0.29/K
4. 0.10/K

Answer: 0.29/K

For a first-order reaction, the time required for the concentration of a reactant to decrease to three-fourths of its initial value is derived from the first-order kinetics equation, leading to the specific relationship of t1/4 = 0.29/K.

Q22. A reaction is second order in carbon monoxide. If the concentration of carbon monoxide is made twice as large while all other conditions are unchanged, what happens to the reaction rate?

1. It becomes four times as large
2. It becomes twice as large
3. It stays the same
4. It becomes three times as large

Answer: It becomes four times as large

In a second-order reaction, the rate is proportional to the square of the concentration of the reactant. Therefore, if the concentration of carbon monoxide is doubled, the rate increases by a factor of two squared, resulting in a fourfold increase in the reaction rate.

Q23. According to the Arrhenius relation for reaction rate, k = A e^(−Ea/RT), what does Ea denote?

1. the total energy possessed by the reacting molecules at temperature T
2. the proportion of molecules whose energy exceeds the activation energy of the reaction
3. the minimum energy threshold above which every colliding molecule will undergo reaction
4. the energy level below which all colliding molecules will react

Answer: the minimum energy threshold above which every colliding molecule will undergo reaction

Ea represents the activation energy, which is the minimum energy required for reactants to successfully collide and form products. This threshold ensures that only those molecules with sufficient energy can overcome the energy barrier for the reaction to occur.

Q24. For the reaction between nitric oxide and bromine forming nitrosyl bromide, the following sequence is suggested: NO(g) + Br2(g) → NOBr2(g) NOBr2(g) + NO(g) → 2NOBr(g) If the second step controls the rate, what is the reaction order with respect to NO(g)?

1. 3
2. 2
3. 1
4. 0

Answer: 2

The reaction order with respect to NO(g) is 2 because the rate-determining step involves two molecules of NO(g) reacting with NOBr2(g), indicating that the concentration of NO(g) is squared in the rate law.

Q25. Consider the reaction, 2A + B → products. When concentration of B alone was doubled, the half-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

1. s⁻¹
2. L mol⁻¹ s⁻¹
3. no unit
4. mol L⁻¹ s⁻¹

Answer: L mol⁻¹ s⁻¹

Half-life independent of [B] means first order in B; doubling [A] doubling rate means first order in A. Overall second order, so k has units L mol^-1 s^-1 (index 1).

Q26. A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial velocity is ten times the permissible value, after how many days will it be safe to enter the room?

1. 100 days
2. 1000 days
3. 300 days
4. 10 days

Answer: 100 days

Activity must fall to 1/10 of its current value. (1/2)^(t/T) = 1/10 with T = 30 days gives t = 30 * log10/log2 = 30 * 3.32 ~ 99.7 = 100 days. Correct answer: 100 days.

Q27. For a reaction 1/2 A → 2B, rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression

1. −d[A]/dt = 1/2 d[B]/dt
2. −d[A]/dt = 1/4 d[B]/dt
3. −d[A]/dt = d[B]/dt
4. −d[A]/dt = 4 d[B]/dt

Answer: −d[A]/dt = 1/4 d[B]/dt

Rate = -(1/(1/2)) d[A]/dt = (1/2) d[B]/dt, so -2 d[A]/dt = (1/2) d[B]/dt, giving -d[A]/dt = (1/4) d[B]/dt.

Q28. The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301)

1. 23.03 minutes
2. 46.06 minutes
3. 406.6 minutes
4. 230.03 minutes

Answer: 46.06 minutes

k = 0.693/6.93 = 0.1 per min. For 99% completion, t = (2.303/k)*log(100/1) = (2.303/0.1)*2 = 46.06 minutes.

Q29. The time for half life period of a certain reaction Products is 1 hour. When the initial concentration of the reactant ‘A’ is 2.0 mol L⁻¹, how much time does it take for this concentration to come from 0.50 to 0.25 mol L⁻¹ if it is a zero order reaction ?

1. 4 h
2. 0.5 h
3. 0.25 h
4. 1 h

Answer: 0.25 h

From t1/2 = [A]0/(2k): 1 = 2.0/(2k) -> k = 1.0 mol L^-1 h^-1. Going from 0.50 to 0.25 needs Δ[A] = 0.25, so t = 0.25/1.0 = 0.25 h.

Q30. Consider the reaction: Cl₂(aq) + H₂S(aq) → S(s) + 2H⁺(aq) + 2Cl⁻(aq) The rate equation for this reaction is rate = k[Cl₂][H₂S] Which of these mechanisms is/are consistent with this rate equation? A. Cl₂ + H₂S → H⁺ + Cl⁻ + Cl⁺ + HS⁻ (slow) Cl⁺ + HS⁻ → H⁺ + Cl⁻ + S (fast) B. H₂S ⇌ H⁺ + HS⁻ (fast equilibrium) Cl₂ + HS⁻ → 2Cl⁻ + H⁺ + S (slow)

1. A only
2. Both A and B
3. Neither A nor B
4. B only

Answer: A only

In A the slow step is Cl2 + H2S, giving rate = k[Cl2][H2S], which matches. In B the pre-equilibrium gives [HS-] = K[H2S]/[H+], so the slow step gives rate = k'[Cl2][H2S]/[H+] (an extra 1/[H+] dependence), which does not match. Hence only A is consistent.

Q31. 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:

1. 1.73 × 10⁻⁵ M/min
2. 3.47 × 10⁻⁴ M/min
3. 3.47 × 10⁻⁵ M/min
4. 1.73 × 10⁻⁴ M/min

Answer: 3.47 × 10⁻⁴ M/min

0.1 -> 0.025 M is two halvings, so 2*t_half = 40 min -> t_half = 20 min and k = 0.693/20 = 0.0347 /min. Rate at [A] = 0.01 M is k[A] = 0.0347*0.01 = 3.47x10^-4 M/min.

Q32. The rate of reaction doubles when its temperature changes from 300 K to 310 K. The activation energy of the reaction will be: (R = 8.314 JK⁻¹ mol⁻¹ and log 2 = 0.301)

1. 53.6 kJ mol⁻¹
2. 48.6 kJ mol⁻¹
3. 58.5 kJ mol⁻¹
4. 60.5 kJ mol⁻¹

Answer: 53.6 kJ mol⁻¹

log(k2/k1) = Ea/(2.303 R) x (1/T1 - 1/T2). 0.301 = Ea/(2.303 x 8.314) x (10/(300x310)); Ea = 0.301 x 19.15 / 1.075e-4 ~ 53.6 kJ/mol (index 0).

Q33. For the non-stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K. Initial Concentration (A) | Initial Concentration (B) | Initial rate of formation of C (mol L⁻¹ s⁻¹) 0.1 M | 0.1 M | 1.2 × 10⁻³ 0.1 M | 0.2 M | 1.2 × 10⁻³ 0.2 M | 0.1 M | 2.4 × 10⁻³ The rate law for the formation of C is:

1. dc/dt = k[A][B]
2. dc/dt = k[A]²[B]
3. dc/dt = k[A][B]²
4. dc/dt = k[A]

Answer: dc/dt = k[A]

The rate of formation of C is independent of the concentration of B, as evidenced by the constant rate observed when the concentration of A is held constant while varying B. This indicates that the reaction rate depends solely on the concentration of A, confirming that the correct rate law is dc/dt = k[A].

Q34. Higher order (>3) reactions are rare due to:

1. shifting of equilibrium towards reactants due to elastic collisions
2. loss of active species on collision
3. low probability of simultaneous collision of all the reacting species
4. increase in entropy and activation energy as more molecules are involved

Answer: low probability of simultaneous collision of all the reacting species

Reactions of order greater than 3 are rare because the simultaneous collision of four or more correctly oriented molecules has a very low probability.

Q35. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:

1. 2.66 L min−1 at STP
2. 1.34 × 10−2 mol min−1
3. 6.96 × 10−2 mol min−1
4. 6.93 × 10−4 mol min−1

Answer: 6.93 × 10−4 mol min−1

The correct option is derived from the first-order kinetics of the decomposition reaction, where the rate of reaction is proportional to the concentration of H2O2. At 0.05 M concentration, the calculated rate of formation of O2, based on the stoichiometry of the reaction and the rate constant, results in 6.93 × 10−4 mol min−1.

Q36. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by 10 kJ mol−1. If k1 and k2 are respectively rate constants for reactions R1 and R2 at 300 K, then ln(k1/k2) is equal to (R = 8.314 J mol−1 K−1)

1. 8
2. 12
3. 6
4. 4

Answer: 4

The difference in activation energies leads to a specific relationship between the rate constants, which can be derived from the Arrhenius equation. Given that the activation energy of R1 is 10 kJ mol−1 higher than that of R2, the natural logarithm of the ratio of their rate constants can be calculated, resulting in ln(k1/k2) = -10,000 J/mol / (8.314 J/mol K * 300 K) = -4, which simplifies to 4 when considering the absolute value.

Q37. At 518°C, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s−1 when 5% had reacted and 0.5 Torr s−1 when 33% had reacted. The order of the reaction is:

1. 1
2. 2
3. 3
4. 0

Answer: 2

Remaining pressures: at 5% reacted P = 0.95(363) = 344.9 Torr (rate 1.00); at 33% reacted P = 0.67(363) = 243.2 Torr (rate 0.50). For rate = k P^n, 1.00/0.50 = (344.9/243.2)^n gives 2 = 1.418^n, so n ~ 2. The reaction is second order.

Q38. The following results were obtained during kinetic studies of the reaction; 2A + B → Products Experiment I: [A] in mol L−1 = 0.1, [B] in mol L−1 = 0.2, Initial Rate of reaction (in mol L−1 min−1) = 6.93 × 10−3 Experiment II: [A] in mol L−1 = 0.1, [B] in mol L−1 = 0.25, Initial Rate of reaction (in mol L−1 min−1) = 6.93 × 10−3 Experiment III: [A] in mol L−1 = 0.2, [B] in mol L−1 = 0.3, Initial Rate of reaction (in mol L−1 min−1) = 1.386 × 10−2 The time (in minutes) required to consume half of A is

1. 5
2. 10
3. 1
4. 100

Answer: 10

Exp I vs II ([A] fixed, [B] changes, rate unchanged) -> order in B = 0. Exp I vs III ([A] doubled, rate doubled) -> order in A = 1. So rate = k[A]; k = 6.93e-3/0.1 = 0.0693 min^-1. t1/2 = 0.693/0.0693 = 10 min.

Q39. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such reaction will be: (R = 8.314 J K−1 mol−1 and log 2 = 0.301)

1. 58.5 kJ mol−1
2. 60.5 kJ mol−1
3. 53.6 kJ mol−1
4. 48.5 kJ mol−1

Answer: 53.6 kJ mol−1

The correct option is derived using the Arrhenius equation, which relates the rate of reaction to temperature and activation energy. Given that the rate doubles with a temperature increase of 10 K, we can apply the formula to calculate the activation energy, leading to the value of 53.6 kJ mol−1.

Q40. For the non-stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments at 298 K. Initial Concentration (A) | Initial Concentration (B) | Initial rate of formation of C (mol L S) 0.1 M | 0.1 M | 1.2 × 10⁻³ 0.1 M | 0.2 M | 1.2 × 10⁻³ 0.2 M | 0.1 M | 2.4 × 10⁻³ The rate law for the formation of C is -

1. dc/dt = k[A]²[B]
2. dc/dt = k[A][B]²
3. dc/dt = k[A]
4. dc/dt = k[A][B]

Answer: dc/dt = k[A]

The correct option indicates that the rate of formation of C is directly proportional to the concentration of A only, as evidenced by the data showing that varying the concentration of B does not affect the rate when A is held constant. This suggests that the reaction is first-order with respect to A and zero-order with respect to B.

Q41. Higher order (> 3) reactions are rare due to: (1) Low probability of simultaneous collision of all the reacting species (2) Increase in entropy and activation energy as more molecules are involved (3) Shifting of equilibrium towards reactants due to elastic collisions (4) Loss of active species on collision

1. Low probability of simultaneous collision of all the reacting species
2. Increase in entropy and activation energy as more molecules are involved
3. Shifting of equilibrium towards reactants due to elastic collisions
4. Loss of active species on collision

Answer: Low probability of simultaneous collision of all the reacting species

Higher order reactions are uncommon because the likelihood of multiple molecules colliding at the exact same time decreases significantly as the number of reactants increases, making such reactions less probable.

Q42. Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be:

1. 6.93 × 10⁻² mol min⁻¹
2. 6.93 × 10⁻⁴ mol min⁻¹
3. 2.66 L min⁻¹ at STP
4. 1.34 × 10⁻² mol min⁻¹

Answer: 6.93 × 10⁻⁴ mol min⁻¹

The correct option is right because, in a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. At 0.05 M concentration of H2O2, the rate constant can be calculated from the initial data, leading to the formation rate of O2 being 6.93 × 10⁻⁴ mol min⁻¹.

Q43. For a first order reaction, A → P, t1/2 (half-life) is 10 days. The time required for 1/4th conversion of A(in days) is – [ln2 = 0.693, ln3 = 1.1] [JEE-Main On line-2018]

1. 3.2
2. 2.5
3. 4.1
4. 5

Answer: 4.1

For first order, k = ln2/t_half = 0.693/10. For 1/4 (25%) conversion, 75% remains: t = (1/k) ln(1/0.75) = (1/k) ln(4/3) = (10/0.693)(2x0.693 - 1.1) ~ 4.1 days.

Q44. N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mm Hg to 87.5 mm Hg. The pressure of the gaseous mixture after 100 minutes at constant temperature will be ________. [JEE-Main On line-2018]

1. (A) 136.25 mm Hg
2. (B) 106.25 mm Hg
3. (C) 175.0 mm Hg
4. (D) 116.25 mm Hg

Answer: (B) 106.25 mm Hg

For N2O5 -> 2NO2 + 1/2 O2, total pressure = P0 + (3/2)x where x is N2O5 decomposed. At 50 min total=87.5 gives x=25 (half of 50), so t(1/2)=50 min. After 100 min (two half-lives) x=37.5, total = 50 + (3/2)(37.5) = 106.25 mm Hg.

Q45. The following results were obtained during kinetic studies of the reaction: 2A + B → Products Experiment I: [A] = 0.10 mol L⁻¹, [B] = 0.20 mol L⁻¹, Initial Rate of reaction = 6.93 × 10⁻³ mol L⁻¹ min⁻¹ Experiment II: [A] = 0.10 mol L⁻¹, [B] = 0.25 mol L⁻¹, Initial Rate of reaction = 6.93 × 10⁻³ mol L⁻¹ min⁻¹ Experiment III: [A] = 0.20 mol L⁻¹, [B] = 0.30 mol L⁻¹, Initial Rate of reaction = 1.386 × 10⁻² mol L⁻¹ min⁻¹ The time (in minutes) required to consume half of A is -

1. 5
2. 1
3. 100
4. 10

Answer: 10

Exp I vs II: changing [B] leaves rate unchanged -> order in B = 0. Exp I vs III: doubling [A] doubles rate -> order in A = 1. So rate = k[A], k = 6.93e-3/0.10 = 0.0693 min^-1. t(1/2) = 0.693/k = 10 min.

Q46. For a reaction scheme A —k1→ B —k2→ C, if the rate of formation of B is set to be zero then the concentration of B is given by -

1. (k1 − k2)[A]
2. k1k2[A]
3. (k1 + k2)[A]
4. (k1/k2)[A]

Answer: (k1/k2)[A]

Setting the rate of formation of B to zero implies that the rate of its production and consumption are balanced, leading to the relationship where the concentration of B is proportional to the ratio of the rate constants k1 and k2 multiplied by the concentration of A.

Q47. Decomposition of X exhibits a rate constant of 0.05 μg/year. How many year are required for the decomposition of 5 μg of X into 2.5 μg?

1. (1) 50
2. (2) 20
3. (3) 25
4. (4) 40

Answer: (1) 50

The rate constant has units ug/year, so the decomposition is zero order. Amount to decompose = 5 - 2.5 = 2.5 ug. Time = 2.5/0.05 = 50 years.

Q48. For the reaction 2A + B → C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is: [A] (mol L⁻¹) | [B] (mol L⁻¹) | Initial Rate (mol L⁻¹ s⁻¹) 0.05 | 0.05 | 0.045 0.10 | 0.05 | 0.090 0.20 | 0.10 | 0.72

1. Rate = k [A][B]
2. Rate = k [A][B]²
3. Rate = k [A]² [B]²
4. Rate = k [A]² [B]

Answer: Rate = k [A][B]²

Doubling [A] (B fixed) doubles the rate -> first order in A. From row 2 to 3, [A] doubles (x2) and [B] doubles while rate increases 8-fold; A accounts for x2, so B accounts for x4 -> second order in B. Rate = k[A][B]^2.

Q49. If a reaction follows the Arrhenius equation, the plot ln k vs 1/(RT) gives straight line with a gradient (-y) unit. The energy required to activate the reactant is:

1. y unit
2. yR unit
3. y/R unit
4. -y unit

Answer: y unit

The gradient of the plot ln k vs 1/(RT) is equal to -Ea/R, where Ea is the activation energy. Therefore, the activation energy can be expressed as y unit when considering the negative sign in the gradient.

Q50. The results given in the below table were obtained during kinetic studies of the following reaction: 2A + B → C + D Experiment | [A]/mol L⁻¹ | [B]/mol L⁻¹ | Initial rate/mol L⁻¹ min⁻¹ I | 0.1 | 0.1 | 6.00 × 10⁻³ II | 0.1 | 0.2 | 2.40 × 10⁻² III | 0.2 | 0.1 | 1.20 × 10⁻² IV | X | 0.2 | 7.20 × 10⁻² V | 0.3 | Y | 2.88 × 10⁻¹ X and Y in the given table are respectively:

1. 0.3, 0.4
2. 0.4, 0.3
3. 0.4, 0.4
4. 0.3, 0.3

Answer: 0.3, 0.4

Comparing I and III gives order 1 in A; comparing I and II gives order 2 in B, so rate = k[A][B]^2 with k = 6e-3/((0.1)(0.1)^2) = 6. For IV: 7.2e-2 = 6*X*(0.2)^2 -> X = 0.3. For V: 2.88e-1 = 6*(0.3)*Y^2 -> Y^2 = 0.16 -> Y = 0.4. So X,Y = 0.3, 0.4.