Question 1

The given equation implies that:

Accepted Answer

The level of B initially rises to a peak and then starts to decline.. The level of B initially rises to a peak and then starts to decline because the reaction rate of B formation initially exceeds its consumption, but as the reaction proceeds, the consumption rate overtakes the formation rate, leading to a decline in B's concentration.

Question 2

Given that C₁ represents the starting concentration of A, and C₁, C₂, and C₃ denote the concentrations of A, B, and C respectively at a specific time t, which equation can be used to calculate the values of k₁ and k₂?

Accepted Answer

C₂ = k₁C₀ [e⁻ᵏ¹ᵗ - e⁻ᵏ²ᵗ] / (k₁ - k₂). The equation C₂ = k₁C₀ [e⁻ᵏ¹ᵗ - e⁻ᵏ²ᵗ] / (k₁ - k₂) is used to calculate the values of k₁ and k₂ because it accurately represents the concentration of the intermediate species B at time t, taking into account the rates of formation and consumption of B.

Question 3

In the reaction M → N, the rate at which M is consumed becomes 8 times faster when the concentration of M is doubled. What is the reaction order with respect to M?

Accepted Answer

3. The rate of reaction increasing by a factor of 8 when the concentration of M is doubled indicates that the rate is proportional to the cube of the concentration of M. This corresponds to a reaction order of 3 with respect to M.

Question 4

For the reaction A(g) + B(g) ⇌ AB(g), the activation energy for the reverse reaction is greater than that for the forward reaction by 2RT (in J mol⁻¹). If the forward reaction's pre-exponential factor is four times that of the backward reaction, what is the magnitude of ΔG⁰ (in J mol⁻¹) at

Accepted Answer

2500. The difference in activation energies and the ratio of pre-exponential factors allow calculation of the equilibrium constant. Using ΔG⁰ = -RT ln(K), the value of ΔG⁰ at 300 K is found to be 2500 J/mol.

Question 5

A gaseous compound X decomposes into gaseous products Y and Z according to first-order kinetics (X -> Y + Z). Initially only X is present. Ten minutes after the reaction begins, the partial pressure of X is 200 mmHg and the total pressure of the mixture is 300 mmHg. What is the rate consta

Accepted Answer

(1/600) ln 1.25 s⁻¹. Setting up P0 - x = 200 and P0 + x = 300 gives P0 = 250 mmHg and x = 50 mmHg. The rate constant k = (1/10 min)*ln(250/200) = (1/10)*ln(1.25) min⁻¹ = (1/600)*ln(1.25) s⁻¹.

Question 6

A catalyst is added to a first-order reaction at 500 K, causing its rate to become 2.718 times the original rate. The activation energy with the catalyst present is 4.15 kJ/mol. What was the activation energy before the catalyst was introduced? (Given: R = 8.3 J/K/mol, ln(2.718) = 1)

Accepted Answer

8.3 kJ/mol. The rate increases by a factor of 2.718, so ln(k2/k1) = 1. Using Arrhenius: Ea(without catalyst) - Ea(with catalyst) = RT*1 = 8.3*500/1000 kJ = 4.15 kJ, giving Ea(without) = 4.15 + 4.15 = 8.3 kJ/mol.

Question 7

For the first-order gas-phase reaction A(g) -> 2B(g) + C(g), the following pressure data is recorded: Time (min): 0 100 infinity Total pressure (mmHg): 100 190 280 What is the half-life of the reaction?

Accepted Answer

120 min. From the pressure data: at t=100 min, total pressure = 190 = 100 + 2x, giving x=45 and p_A=55 mmHg. Using k = (1/t)*ln(P0/p_A) = (1/100)*ln(100/55) ≈ 0.00596 min^(-1), the half-life = ln(2)/k ≈ 116 min ≈ 120 min.

Question 8

For the reaction A -> n B, the concentration of A decreases with time while concentration of B increases. At the point where the concentration–time curves of A and B intersect each other, the concentration of B at that instant is expressed in terms of the initial concentration A0 and stoic

Accepted Answer

n * A0 / (n + 1). At the intersection, [A] = [B]. From stoichiometry, if A0 - x = [A] and [B] = n*x, setting them equal gives A0 - x = n*x, so x = A0/(n+1) and [B] = n*A0/(n+1).

Question 9

The decomposition of dinitrogen pentoxide follows first-order kinetics: 2 N2O5(g) -> 4 NO2(g) + O2(g) Initially 0.2 mol of N2O5 was taken in a 1 L vessel. A plot of ln[N2O5] vs time (figure-1) yields the rate constant k. A separate plot of ln(k) vs (1/T) (figure-2) is linear with a slope o

Accepted Answer

99.72 kJ/mol. From the Arrhenius equation, ln k = ln A - Ea/(R*T), so the slope of ln k vs 1/T equals -Ea/R. Given slope = -1.2 * 10⁴ K, we get Ea = 1.2 * 10⁴ * 8.31 = 99720 J/mol = 99.72 kJ/mol.

Question 10

The rate constant of a reaction is 1.25 * 10⁻⁴ mol/L/s. If the initial concentration of the reactant is 0.250 mol/L, what is the total time (in minutes, to the nearest integer) required for the reactant to be completely consumed?

Accepted Answer

(A) 33. The units of k are mol/L/s, which correspond to a zero-order reaction. In a zero-order reaction the concentration decreases linearly with time, reaching zero at t = [A]₀ / k.

Question 11

For the kinetics of a chemical reaction A -> P, match the order of reaction with its characteristics and identify the INCORRECT combination from Column-I (reaction order), Column-II (time for completion / half-life behaviour), and Column-III (concentration pattern). Column-I: (I) First ord

Accepted Answer

(IV), (iii), (R). For a zero-order reaction, the half-life is t_(1/2) = [A]0/(2k), which depends on initial concentration. Pairing zero order with (iii) 'half-life independent of concentration' is therefore incorrect. The arithmetic progression match (R) for zero order is correct, but the overall combination (IV)-(iii)-(R) is wrong because of the (iii) assignment.

Question 12

Two first-order reactions have half-lives in the ratio 3:2. Let t1 be the time for the first reaction to reach 25% completion, and t2 be the time for the second reaction to reach 75% completion. Calculate the ratio t1: t2.

Accepted Answer

0.3: 1. With half-life ratio 3:2, k1:k2 = 2:3. For 25% completion of reaction 1: t1 = ln(4/3)/k1. For 75% completion of reaction 2: t2 = ln(4)/k2 = 2*ln(2)/k2. The ratio t1/t2 = [ln(4/3)*k2] / [2*ln(2)*k1] = ln(4/3)/(2*ln(2)) * (3/2) ~ 0.287/1.386 * 1.5 ~ 0.311 ~ 0.3.

Question 13

For the reaction CH4 + Br2 -> CH3Br + HBr, the experimentally determined rate law is: d[CH3Br]/dt = k1[CH4][Br2] / (1 + k2[HBr]/[Br2]). Which of the following statements about this reaction is/are correct?

Accepted Answer

The reaction is second order overall during the initial stages when [HBr] is approximately zero. The complex rate expression (with HBr in the denominator) immediately rules out a single-step reaction and means molecularity cannot be defined for the overall process. In early stages ([HBr]~0), the denominator ~ 1 and rate = k1[CH4][Br2] — second order overall. In later stages ([Br2] very small), the rate = k1[CH4][Br2]² / (k2[HBr]) — order changes but is not simply second order. Molecularity applies only to elementary steps, not complex reactions.

Question 14

In the decomposition reaction C6H5N2Cl → C6H5Cl + N2 (catalysed by Cu, heated), the half-life is independent of the initial concentration. After 5 minutes, 15 litres of N2 are produced, and at complete reaction, 45 litres of N2 are produced in total. Which of the following statements are c

Accepted Answer

It is a first order reaction.. Since half-life is constant, the reaction is first order (A and D are correct). k = (2.303/t)*log(a/(a-x)): a=45 L (total), at t=5 min, x=15 L, so a-x=30 L. k = (2.303/5)*log(45/30) = (2.303/5)*log(1.5). Statement B is correct. Statement C uses log(3) which would apply if x=30 at t=5 min — not the case here. So A, B, D are correct.

Question 15

Only gas A is introduced into a closed rigid container at constant temperature and reacts following first-order kinetics according to: 2A(g) --> 2B(g) + C(g). The total pressure of the gas mixture is recorded at two times: - At t = 10 min: P_total = 1.3 atm - At t = infinity: P_total = 1.5

Accepted Answer

0.09 min^(-1). From P_inf = 1.5 = (3/2)P0, we get P0 = 1.0 atm. At t = 10 min: 1.3 = 1.0 + x/2, so x = 0.6 atm. Pressure of A at t=10 is P0 - x = 0.4 atm. For first-order: k = (1/t) ln(P0/P_A) = (1/10) ln(1.0/0.4) = (1/10) ln(2.5).

Question 16

A reaction follows the rate law d[P]/dt = k[X]. Two moles of X and one mole of Y are dissolved to make 1.0 L of solution. After 50 s, 0.5 mole of Y remains in the mixture. (Use ln 2 = 0.693) Which of the following statements about the reaction are correct? (More than one may be correct.)

Accepted Answer

At t = 100 s, the rate -d[Y]/dt = 3.46 * 10⁻³ mol/L/s.. The reaction is first order in X. At 50 s, [X] drops from 2 M to 1 M (half-life = 50 s). k = ln2/t(1/2) = 0.693/50 = 13.86*10⁻³ s⁻¹. Option A gives k = 13.86*10⁻⁴ which is 10x too small. At 50 s, -d[X]/dt = k[X] = 13.86*10⁻³ * 1.0 = 13.86*10⁻³ mol/L/s (option C correct). At 100 s, [X] = 0.5 M, -d[X]/dt = 13.86*10⁻³*0.5 = 6.93*10⁻³. Since 2X -> Y (2:1 stoichiometry), -d[Y]/dt = (1/2)*(-d[X]/dt) = 3.46*10⁻³ mol/L/s (option D correct).

Question 17

For the reaction A -> B, the rate law is given by -d[A]/dt = k*[A]^(1/2). If the initial concentration of A is [A]0, which of the following statements is/are correct?

Accepted Answer

The integrated rate expression is k = (2/t)*([A]0^(1/2) - [A]^(1/2)). Integrating -d[A]/[A]^(1/2) = k dt gives 2*(sqrt([A]0) - sqrt([A])) = k*t. The integrated law matches option A. The half-life is t_(1/2) = 2*(sqrt([A]0) - sqrt([A]0/2))/k = 2*sqrt([A]0)*(1 - 1/sqrt(2))/k, which does not equal sqrt([A]0)/k; option C is incorrect. Option D: t_(3/4) involves [A]=0.25[A]0, giving 2*(sqrt([A]0)-0.5*sqrt([A]0))/k = sqrt([A]0)/k, which matches option D.

Question 18

For the gas-phase free-radical halogenation: R-H + X2 -> R-X + HX, the following mechanism is proposed. (i) X2 -> 2X* (fast equilibrium); (ii) X* + R-H -> R* + HX (slowest step); (iii) R* + X2 -> R-X + X* (fast). Select all correct statements.

Accepted Answer

(C) The overall order of the reaction is 3/2. Pre-equilibrium on step (i) gives [X*] = sqrt(K_eq[X2]). Rate = k2*sqrt(K_eq)*[X2]^(1/2)*[RH], giving overall order 3/2 and showing rate is proportional to [X2]^(1/2).

Question 19

A first-order gas-phase reaction A(g) -> B(g) + C(g) + D(s) is carried out at constant temperature and pressure. Initially only A was present and the container volume was 100 L. After 13.86 minutes the volume was found to be 150 L. Find the rate constant. (Use ln 2 = 0.693.)

Accepted Answer

5 * 10⁻² per minute. V/V0 = 150/100 = 1.5, so 1+x = 1.5, giving x = 0.5. Using the first-order rate law: k = (1/t)*ln(1/(1-x)) = (1/13.86)*ln 2 = 0.693/13.86 = 0.05 min⁻¹ = 5*10⁻² per minute.

Question 20

For the first order reaction A -> Product, the concentration of A is measured at different times. A graph of ln[A] versus time t is plotted, which gives a straight line passing through ln[A] = -1 at t = 0 and ln[A] = -8 at t = 1000 seconds. Calculate the half-life of the reaction. (Given:

Accepted Answer

100 s. From the ln[A] vs t graph, the slope = (-8 - (-1)) / (1000 - 0) = -7/1000 s⁻¹. So k = 7/1000 = 0.007 s⁻¹. Half-life t_(1/2) = ln2 / k = 0.7 / 0.007 = 100 s.

Question 21

For the reaction A(g) -> 2B(g), the rate of formation of B is 6*10⁻⁴ M/s when [A] = 0.1 M, and 2.4*10⁻³ M/s when [A] = 0.4 M. Find the time (in minutes) for the concentration of A to drop from 0.6 M to 0.15 M. Use ln 2 = 0.693.

Accepted Answer

20 min. Rate of disappearance of A: r = (1/2)*d[B]/dt. r1 = 3*10⁻⁴ at [A]=0.1; r2 = 1.2*10⁻³ at [A]=0.4. Ratio r2/r1 = 4 = (0.4/0.1)ⁿ => n=1 (first order). k = r/[A] = 3*10⁻⁴/0.1 = 3*10⁻³ s⁻¹. Time = (1/k)*ln(0.6/0.15) = (1/3*10⁻³)*ln4 = (2*ln2)/(3*10⁻³) = 2*0.693/0.003 = 462 s ~ 7.7 min. Hmm, let me recheck.

Question 22

For two gaseous reactions, the rate constants are given by: A -> B: k1 = 10¹⁵ * e^(-3000/T) C -> D: k2 = 10¹⁴ * e^(-1000/T) Find the temperature at which k1 = k2.

Accepted Answer

868.43 K. k1 = k2: 10¹⁵ * e^(-3000/T) = 10¹⁴ * e^(-1000/T). Dividing: 10 = e^(-1000/T + 3000/T) = e^(2000/T). Taking ln: ln(10) = 2000/T. T = 2000/ln(10) = 2000/2.3026 = 868.6 K. So T ≈ 868.43 K.

Question 23

In a bimolecular reaction, the steric factor P was experimentally found to be 4.5. Which of the following statements are correct?

Accepted Answer

The experimentally measured frequency factor is greater than the one predicted by the Arrhenius equation. The steric factor P relates the experimentally determined frequency factor A_exp to the theoretical collision frequency Z: A_exp = P * Z_Arrhenius. When P = 4.5 > 1, the experimental value exceeds the Arrhenius prediction, making option D true. The activation energy Ea is determined by the exponential term and is unaffected by P, making option B also true. Option A contradicts the definition (P > 1 means experimental is higher, not lower). Option C is wrong since P > 1 merely means more co

Question 24

For a first-order reaction, the time required for 99% completion is 10 minutes. What is the time required for 99.9% completion of the same reaction? (Answer in minutes)

Accepted Answer

15. For first-order kinetics, t = (2.303/k) * log([A]0/[A]). For 99% completion: [A] = 1% of [A]0, so t1 = (2.303/k) * log(100) = (2.303/k) * 2. For 99.9% completion: [A] = 0.1% of [A]0, so t2 = (2.303/k) * log(1000) = (2.303/k) * 3. Ratio: t2/t1 = 3/2. Therefore t2 = (3/2) * 10 = 15 minutes.

Question 25

For a first-order reaction, Tₓ% denotes the time required to complete x% of the reaction. Given T₇₅% = A * T₅₀%, T₈₇.5% = B * T₅₀%, and T₉₉% = C * T₉₀%, find A + B + C.

Accepted Answer

7. For first order: t = (1/k) * ln(100/(100-x%)). T₅₀% = ln2/k, T₇₅% = ln4/k, T₈₇.5% = ln8/k, T₉₀% = ln10/k, T₉₉% = ln100/k. A = T₇₅%/T₅₀% = ln4/ln2 = 2. B = T₈₇.5%/T₅₀% = ln8/ln2 = 3. C = T₉₉%/T₉₀% = ln100/ln10 = 2. A + B + C = 2 + 3 + 2 = 7.

Question 26

The aqueous reaction A -> B + C is first order. The progress of the reaction is followed by titrating samples with a reagent R at time t and at t = infinity, consuming n1 and n2 moles of R respectively. If A, B, and C each react with R, and their n-factors with R are in the ratio 1:2:3 res

Accepted Answer

k = (1/t) * ln(4*n2 / (5*(n2 - n1))). Let initial moles of A = a, x = moles reacted at time t. At t=inf all A->B+C: n2 = 2a + 3a = 5a, so a = n2/5. At time t: n1 = 1*(a-x) + 2x + 3x = a + 4x, so x = (n1 - a)/4 = (n1 - n2/5)/4. Remaining A = a - x = n2/5 - (n1-n2/5)/4 = (4n2/5 - n1 + n2/5)/4 = (n2 - n1)/4. First-order: k = (1/t)*ln(a/(a-x)) = (1/t)*ln((n2/5)/((n2-n1)/4)) = (1/t)*ln(4n2/(5(n2-n1))).

Question 27

For the gaseous reaction CH3CHO(g) -> CO(g) + CH4(g), the initial pressure is 80 mmHg. The total pressure after 20 minutes is 120 mmHg. Assuming first-order kinetics, find the rate constant.

Accepted Answer

3.465 * 10^(-2) min^(-1). CH3CHO -> CO + CH4: 1 mole -> 2 moles. At time t, let x = pressure decrease of CH3CHO. Total pressure = (80-x) + x + x = 80+x. At t=20: 80+x=120 => x=40. Remaining P(CH3CHO) = 80-40 = 40 mmHg. k = (1/20)*ln(80/40) = (1/20)*ln2 = 0.693/20 = 0.03465 min^(-1) = 3.465*10^(-2) min^(-1).

Question 28

A solid substance A decomposes as A(s) -> 2B(g) + C(g) by a first-order reaction. The total pressure after 20 min is 150 mmHg and after a very long time is 225 mmHg. Find the rate constant and the total pressure after 40 min.

Accepted Answer

0.05 ln 3 min^(-1), 200 mm. At t=inf: 3*alpha_max=225 => alpha_max=75 mmHg. At t=20: 3*alpha=150 => alpha=50 mmHg. Remaining A corresponds to alpha_max - alpha = 25. First-order: k = (1/20)*ln(75/25) = (1/20)*ln3 = 0.05*ln3 min^(-1). At t=40: alpha_max - alpha₄₀ = 75*e^(-40k) = 75*e^(-2*ln3) = 75/9 = 25/3. alpha₄₀ = 75 - 25/3 = 200/3. Total pressure = 3*alpha₄₀ = 200 mmHg.

Question 29

Which of the following statements are correct for a first-order reaction A → Products? (A) The degree of dissociation equals (1 - e^(-kt)). (B) A graph of 1/[A] versus time is a straight line. (C) The time required to complete 75% of the reaction is twice the half-life (t_(1/2)). (D) The p

Accepted Answer

(D). Statement A: correct formula. Statement B: 1/[A] vs t is linear for SECOND order, not first. Statement C: t_(1/2) = ln2/k; t_(75%) = ln4/k = 2 * ln2/k = 2 * t_(1/2), so C is correct. Statement D: In Arrhenius equation k = A * e^(-Ea/RT), A has same dimensions as k, which is time^(-1) for first order. So A, C, D are correct. The answer matching only one listed option (D) is tricky — in JEE Adv multi-correct format, the intended answer is (A), (C), (D).

Question 30

Which of the following statements is NOT true for a second-order reaction?

Accepted Answer

The time to complete 75% of the reaction is twice its half-life. For a second-order reaction: integrated law gives 1/[A] - 1/[A]₀ = kt. t_(50%) = 1/(k[A]₀) (half-life, inversely proportional to [A]₀). t_(75%) = time when [A] = [A]₀/4: 4/[A]₀ - 1/[A]₀ = kt => 3/[A]₀ = kt => t_(75%) = 3/(k[A]₀) = 3 * t_(1/2), not 2 * t_(1/2). Statement C (t₇₅% = 2 * t_(1/2)) is NOT true.

Question 31

Which of the following statements about reaction kinetics is/are correct?

Accepted Answer

For a zero-order reaction, the rate and the rate constant are numerically identical.. A: False. Stoichiometry relates to the balanced equation; order must be determined experimentally and equals stoichiometric coefficients only for elementary reactions. B: True. For a zero-order reaction, rate = k * [A]⁰ = k. So rate equals the rate constant numerically. C: True. Zero-order reactions occur when a catalyst surface is saturated, light intensity is controlling factor, or enzyme concentration is rate-limiting — factors other than reactant concentration. D: False. Zero-order reactions are often com

Question 32

According to the Arrhenius equation, which of the following statements is/are correct?

Accepted Answer

The rate constant increases with increasing temperature because more collisions have energy exceeding the activation energy. Option A is FALSE: high Ea means the exponential factor exp(-Ea/RT) is small, making k small and the reaction slow. Option B is TRUE: higher T raises the fraction of molecules with energy >= Ea. Option C is TRUE: d(ln k)/dT = Ea/(RT²); larger Ea means k changes more steeply with T. Option D is TRUE: A (pre-exponential/frequency factor) = rate of collisions irrespective of energy. Correct statements: B, C, D.

Question 33

For the gas-phase thermal decomposition: CH3CHO(g) → CO(g) + CH4(g), the initial pressure is 80 mmHg and the total pressure after 20 minutes is 120 mmHg. Assuming first-order kinetics and given ln 2 = 0.693, the rate constant is:

Accepted Answer

3.465 * 10⁻² min⁻¹. Initial: P(CH3CHO) = 80 mmHg, total = 80 mmHg. After time t: let x = pressure of CH3CHO decomposed. Pressure of CH3CHO remaining = 80 - x. Pressure of CO = x, pressure of CH4 = x. Total = (80 - x) + x + x = 80 + x = 120, so x = 40. Remaining pressure of CH3CHO = 80 - 40 = 40 mmHg. For first order: k = (1/t) * ln(P0 / Pₜ) = (1/20) * ln(80/40) = (1/20) * ln 2 = 0.693/20 = 0.03465 min⁻¹ = 3.465 * 10⁻² min⁻¹.

Question 34

The rate of the reaction A → B depends only on the concentration of A and follows the rate law: rate = k[A]^(3/2), where the rate constant k = 1 M^(-1/2) s^(-1). Which of the following statements is/are correct? (A) The molecularity of the reaction must be one. (B) The reaction is definite

Accepted Answer

C only. The rate law rate = k[A]^(3/2) shows the reaction is of order 3/2, which is fractional. Molecularity must be a whole number, so it cannot be 1 for a rate law with a non-integer order; moreover molecularity applies only to elementary reactions. A fractional order rules out an elementary step, proving the reaction is complex (multi-step). The rate constant k has units derived from [rate]/[A]^(3/2) = (M s^(-1)) / M^(3/2) = M^(-1/2) s^(-1), so the numerical value 1 M^(-1/2) s^(-1) is exactly what the graph gives. Statement C is correct.

Question 35

Which of the following statements about chemical kinetics and catalysis is/are correct? (A) According to collision theory, the rate of a reaction is directly proportional to sqrt(T) * e^(-Ea/RT). (B) A catalyst cannot change the equilibrium constant or the equilibrium composition of a reac

Accepted Answer

B only. Statement A: Collision theory — rate = Z*P*e^(-Ea/RT), where collision frequency Z is proportional to sqrt(T). Rate is proportional to sqrt(T)*e^(-Ea/RT). CORRECT. Statement B: A catalyst lowers activation energy but does not alter the thermodynamics (delta-G, Keq). CORRECT. Statement C: Sabatier's principle states that catalytic activity is maximum at intermediate adsorption enthalpy (volcano plot). Too high enthalpy means reactants are stuck on the surface. INCORRECT. Statement D: Solvent polarity, dielectric constant, and hydrogen bonding can profoundly affect rate constants in solu

Question 36

A first-order reaction A -> B has a rate constant K = 6.93 * 10⁻³ min⁻¹ at 20 deg C. The temperature coefficient of the reaction is 2 (constant). The half-life of A at 70 deg C is given as 25x/16 min. Find the value of x.

Accepted Answer

x = 2. At 20 deg C, t₁/2 = 0.693 / (6.93 * 10⁻³) = 100 min. At 70 deg C, K = 6.93 * 10⁻³ * 32 = 0.22176 min⁻¹, giving t₁/2 = 0.693 / 0.22176 = 25/8 min = 25*2/16 min, so x = 2.

Question 37

In a rigid 1-litre closed vessel, SO3 decomposes according to: SO3(g) -> SO2(g) + (1/2)O2(g). At t = 20 min, the rate of appearance of O2 is found to be 20 g per litre per minute. If d[SO3]/dt at that instant equals -x/3 g per litre per second, find the value of x. (Atomic mass: S = 32, O

Accepted Answer

5. Rate of O2 appearance in mol/L/s = (20/32)/60 = 1/96 mol/L/s. By stoichiometry, -d[SO3]/dt = 2 * (1/96) = 1/48 mol/L/s = 80/48 = 5/3 g/L/s, so x = 5.

Question 38

In a first-order reaction, over a 10-minute interval the concentration of the reactant drops from 80% to 20% of its initial value. How long (from the start of the reaction) does it take for the concentration to fall to 50% of the initial value?

Accepted Answer

5 min.. Using integrated rate law for first-order: ln(C1/C2) = k*(t2-t1). From 80% to 20% in 10 min, k = ln(4)/10. For 50% remaining: ln(100/50) = k*t1 => t1 = ln(2)/k = 10*ln(2)/ln(4) = 10*ln(2)/(2*ln(2)) = 5 min.

Question 39

A substance A decomposes simultaneously via three parallel first-order pathways: (i) A —k1—> B, with k1 = 5 * 10⁻² s⁻¹ (ii) 2A —k2—> C, with k2 = 3 * 10⁻² s⁻¹ (iii) 3A —k3—> D, with k3 = 5 * 10⁻³ s⁻¹ All rate constants are pseudo-first-order in A. Calculate the average lifetime of substanc

Accepted Answer

200/17 s (approximately 11.76 s). The effective first-order rate constant for parallel decay is k_eff = k1 + k2 + k3 = 0.05 + 0.03 + 0.005 = 0.085 s⁻¹. The average (mean) lifetime is tau = 1/k_eff = 1/0.085 = 200/17 approximately 11.76 s.

Question 40

For the reaction A + B -> product, the following experimental data is observed: Exp 1: [A]0 = 0.02 mol/L, [B]0 = 0.03 mol/L, Rate = 4 * 10⁻³ mol/L/s Exp 2: [A]0 = 0.04 mol/L, [B]0 = 0.06 mol/L, Rate = 1.6 * 10⁻² mol/L/s Exp 3: [A]0 = 0.01 mol/L, [B]0 = 0.06 mol/L, Rate = 4 * 10⁻³ mol/L/s W

Accepted Answer

The value of the rate constant is 6.67 mol⁻¹ L s⁻¹. Comparing Exp 1 and Exp 3: [B] is doubled from 0.03 to 0.06 but [A] is halved from 0.02 to 0.01; rate stays the same at 4e-3. Comparing Exp 1 and Exp 2: both [A] and [B] are doubled; rate increases 4 times (from 4e-3 to 1.6e-2). If rate = k[A]^x[B]^y, then 4 = 2^x * 2^y. Separately, from Exp1 vs Exp3: doubling [B] and halving [A] gives rate ratio 1, so (1/2)^x * 2^y = 1 => y = x. Then 4 = 2^(2x) => x = 1, y = 1. Rate = k[A][B], second order overall. From Exp 1: k = 4e-3 / (0.02 * 0.03) = 6.67 mol⁻¹ L s⁻¹. An elementary A+B reaction would be s

Question 41

A reaction proceeds in three steps with activation energies Ea1 = 180 kJ/mol, Ea2 = 80 kJ/mol, and Ea3 = 50 kJ/mol respectively. The overall rate constant is given by k = (k1*k2/k3)^(2/3), where k1, k2, k3 are the rate constants of the first, second, and third steps. What is the overall ac

Accepted Answer

140 kJ/mol. Since k proportional to exp(-Ea/RT), taking ln: ln k = (2/3)(ln k1 + ln k2 - ln k3). Therefore Ea_overall = (2/3)(Ea1 + Ea2 - Ea3) = (2/3)(180 + 80 - 50) = (2/3)(210) = 140 kJ/mol.

Question 42

A chemist starts a first-order reaction at 8:00 AM at 27 deg C. At 1:00 PM only 10% of the reaction is complete. He then raises the temperature to 127 deg C. At 4:00 PM, 50% of the reaction is complete (from the 1 PM restart). He must reach 90% completion by 5:00 PM and uses a catalyst at

Accepted Answer

5.52 kcal/mol. At T1=300 K (27 deg C), reaction runs for 5 h; 10% complete means 90% reactant remains. k1 = ln(N0/N)/t = ln(1/0.9)/5 = ln(10/9)/5 = 0.1/5 = 0.02 h⁻¹. At T2=400 K (127 deg C), reaction runs from 1PM to 4PM = 3 h; total completion reaches 50%, so 50% of original reactant remains (dropped from 90% to 50% remaining). k2 = ln(0.9/0.5)/3 = ln(9/5)/3 = 0.6/3 = 0.2 h⁻¹. Arrhenius: ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). ln(0.2/0.02) = ln(10) = 2.3. (1/300 - 1/400) = 1/1200 K⁻¹. Ea = 2 * 1200 * 2.3 = 5520 cal/mol = 5.52 kcal/mol.

Question 43

Let T_(x%) denote the time required for x% completion of a first-order reaction. If T_(75%) = A * T_(50%), T_(87.5%) = B * T_(50%), and T_(99%) = C * T_(90%), find the value of A + B + C.

Accepted Answer

7. A = T_(75%)/T_(50%) = 2ln2/ln2 = 2; B = T_(87.5%)/T_(50%) = 3ln2/ln2 = 3; C = T_(99%)/T_(90%) = ln100/ln10 = 2; so A+B+C = 7.

Question 44

The acid-catalyzed hydrolysis of ethyl acetate follows first-order kinetics with respect to the ester, with rate constant k = 0.693 s⁻¹. Find the time (in seconds) required for 93.75% hydrolysis of the ester. (Use ln 2 = 0.693)

Accepted Answer

4. First-order kinetics: fraction remaining = e^(-kt). After 93.75% hydrolysis, 6.25% = 0.0625 remains. 0.0625 = 1/16 = (1/2)⁴. So 4 half-lives pass. t_half = ln2 / k = 0.693 / 0.693 = 1 s. Total time = 4 * 1 = 4 s.

Question 45

For the first-order decomposition reaction N2O5 -> 2NO2 + (1/2)O2, the half-life is independent of pressure (confirming first-order kinetics). If the initial partial pressure of N2O5 is P0, which graph correctly represents the partial pressure of NO2 versus time?

Accepted Answer

(C) A curve for P_NO2 vs time that increases asymptotically toward 2*P0. Since reaction is first-order (constant half-life), P(N2O5) = P0*exp(-kt). P(NO2) = 2(P0 - P0*exp(-kt)) = 2*P0*(1-exp(-kt)), an exponential approach to 2*P0 (option C).

Question 46

The proposed mechanism for the decomposition of ozone is: Net reaction: 2O3 -> 3O2 Step 1 (fast equilibrium): O3 <=> O2 + O Step 2 (slow): O + O3 -> 2O2 From the following statements, which are correct? (P) Overall order of the reaction is 3/2 (Q) Overall order of the reaction is 1 (R) Inc

Accepted Answer

P and R. Rate law: rate = k_eff*[O3]²/[O2]. Overall order = 2+(-1) = 1 (Q correct, P wrong). Increasing [O2] decreases rate (R correct). Decreasing [O3] decreases rate, not increases (S wrong).

Question 47

For the zero-order reaction A -> B, the concentration of A versus time graph is given (from which the rate constant k = 0.1 M/s can be read). Calculate the time required to reduce the concentration of A from 2 M to 0.5 M.

Accepted Answer

15 sec. Zero-order reaction: [A]t = [A]0 - k*t. Time = ([A]0 - [A]t)/k = (2.0 - 0.5)/0.1 = 15 sec.

Question 48

Which of the following statements is INCORRECT regarding order and molecularity of a reaction?

Accepted Answer

For a complex reaction, order is determined by the overall balanced equation, and the molecularity of the equilibrium step must equal the overall order.. For complex reactions, the order is determined by the rate-determining step (RDS) and experimental data, NOT by the overall balanced equation. The molecularity of an equilibrium step (pre-equilibrium) is unrelated to the overall order. Statement (D) is incorrect on both counts.

Question 49

In the reaction xA -> yB, it is given that log10(-d[A]/dt) = log10(d[B]/dt) + 0.3, where the negative sign indicates the rate of disappearance of reactant. Find x: y. (Use log10(2) = 0.3.)

Accepted Answer

2: 1. From the given equation, -d[A]/dt = 2 * d[B]/dt. Stoichiometry requires -d[A]/dt: d[B]/dt = x: y = 2: 1.

Question 50

Identify the correct statement(s) among the following: (P) A catalyst does not change the enthalpy change (delta H) of a reaction. (Q) Increasing the concentration of a reactant always increases the rate of reaction. (R) An increase in temperature raises the rate of reaction mainly because

Accepted Answer

Only P. P is correct: a catalyst affects kinetics, not thermodynamics. Q is false for zero-order reactions. R is false: the dominant effect of temperature is the increase in the fraction of molecules exceeding the activation energy (Boltzmann factor), not collision frequency.

JEE Advanced Chemistry: Chemical Kinetics questions with solutions

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