Question 1

Given below are two statements: R=8.314 J K^ -1 mol^ -1 and 1 cal =4.2 J Statement I: When E_a=12.6 kcal/mol, the room temperature rate constant is doubled by a 10\,^ increase in temperature (298 K to

Accepted Answer

For Statement I: According to the Arrhenius equation: ( k_2 k_1 ) = E_a 2.303 R ( T_2 - T_1 T_1 T_2 ) Given that the rate constant doubles, k_2 = 2k_1. The temperatures are T_1 = 298 K and T_2 = 308 K. The gas constant R = 8.314 J K^ -1 mol^ -1 = 8.314 4.2 cal K^ -1 mol^ -1 1.98 cal K^ -1 mol^ -1 . Substituting the values: (2) = E_a 2.303 1.98 ( 308 - 298 298 308 ) 0.301 = E_a 4.56 ( 10 91784 ) E_a = 0.301 4.56 91784 10 12598 cal/mol 12.6 kcal/mol. Thus, Statement I is true. For Statement II: Fo

Question 2

Decomposition of a hydrocarbon follows the equation k = (5.5 10^ 11 \, s ^ -1 )\,e^ -28000\, K T . The activation energy of reaction is __________ kJ mol^ -1 . (Nearest Integer) Given : R = 8.3 J K^ -

Accepted Answer

According to the Arrhenius equation, k = A e^ - E_a RT . The given equation is k = (5.5 10^ 11 s ^ -1 ) e^ - 28000 T . Comparing the exponent of e in both equations: E_a RT = 28000 T E_a = 28000 R Substituting the value of R = 8.3 J K ^ -1 mol ^ -1 : E_a = 28000 8.3 J mol ^ -1 E_a = 232400 J mol ^ -1 E_a = 232.4 kJ mol ^ -1 Rounding to the nearest integer, the activation energy is 232. Answer: 232

Question 3

For a reaction A P at T K , the half life (t_ 1/2 ) is plotted as a function of initial concentration [A]_0 of A as given below. The value of x in the given figure is _______ s (Nearest integer)

Accepted Answer

From the given graph, the plot of half-life t_ 1/2 versus initial concentration [A]_0 is a straight line passing through the origin. This indicates that t_ 1/2 is directly proportional to [A]_0, which means t_ 1/2 [A]_0. Therefore, the ratio t_ 1/2 [A]_0 is constant. Using the coordinates from the graph, we get: 240 4 10^ -3 = x 1.5 10^ -3 Solving for x: x = 240 1.5 10^ -3 4 10^ -3 x = 60 1.5 = 90 The value of x is 90. Answer: 90

Question 4

First order gas phase reaction A B + C p_i = initial pressure of gas A, p_t = total pressure of the reaction mixture at time t Expression of rate constant (k) is

Accepted Answer

The given reaction is A(g) B(g) + C(g) At t = 0, the initial pressure of A is p_i and the pressures of B and C are 0. At time t, let the decrease in pressure of A be x. Pressure of A = p_i - x Pressure of B = x Pressure of C = x Total pressure at time t is p_t = (p_i - x) + x + x = p_i + x From this, we get x = p_t - p_i The partial pressure of A at time t is P_A = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t For a first order reaction, the rate constant k is given by k = 1 t P_ A,0 P_A Substituting

Question 5

If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be _____ minutes. (Given : 2 = 0.3010)

Accepted Answer

For a first-order reaction, the rate constant k is given by: k = 2.303 2 t_ 1/2 Using the standard approximation 2.303 2 0.693: k = 0.693 6.93 = 0.1 min ^ -1 The time required for 99\% completion of the reaction is given by the integrated rate law: t = 2.303 k ( a a - x ) Here, the initial concentration a = 100 and the amount reacted x = 99, so a - x = 1. t = 2.303 0.1 ( 100 1 ) t = 23.03 (10^2) t = 23.03 2 = 46.06 minutes Answer: 46.06

Question 6

Consider the given graph showing variation of reactant concentration with time. Three different reactions were started with identical initial concentration of reactants. Which of the following stateme

Accepted Answer

Analysis of the graph: The graph shows reactant concentration [R] vs time t for three reactions starting from the same initial concentration. Curve 1 is a straight line with negative slope. For a zero-order reaction, the integrated rate law is [R] = [R]_0 - kt, which is a straight line in [R] vs t. Hence, reaction 1 is a zero-order reaction. Curves 2 and 3 are non-linear (exponential-type decays), so they correspond to reactions of order greater than zero (e.g., first or second order). Evaluatin

Question 7

For a first order reaction A B t/ min [A]/M 0 0.6500 x 0.0650 20 0.00065 x= ______ min. (Nearest integer)

Accepted Answer

For a first order reaction, the rate constant k is given by: k = 1 t [A]_0 [A]_t Using the data at t = 20 min: k = 1 20 0.6500 0.00065 = 1 20 1000 = 3 20 10 Using the data at t = x min: k = 1 x 0.6500 0.0650 = 1 x 10 Equating the two expressions for k: 3 20 10 = 1 x 10 3 20 = 1 x x = 20 3 = 6.66... Rounding to the nearest integer, we get x = 7. Answer: 7

Question 8

Consider the first order reaction R P. The fraction of molecules decomposed in the given first order reaction can be expressed as

Accepted Answer

For a first order reaction, the concentration of reactant at time t is given by [R]_t = [R]_0 e^ -k_1 t The amount of reactant decomposed is [R]_0 - [R]_t = [R]_0 - [R]_0 e^ -k_1 t = [R]_0 (1 - e^ -k_1 t ) The fraction of molecules decomposed is [R]_0 - [R]_t [R]_0 = [R]_0 (1 - e^ -k_1 t ) [R]_0 = 1 - e^ -k_1 t Answer: 1 - e^ -k_1 t

Question 9

Consider the reaction aX bY, for which the rate constant at 30°C is 1 10^ -3 mol^ -1 L s^ -1 . Which of the following statements are true? A. When concentration of 'X' is increased to four times, the

Accepted Answer

The unit of the rate constant k is mol^ -1 L s^ -1 . The general unit of the rate constant for an n-th order reaction is (mol L^ -1 )^ 1-n s^ -1 . Equating the powers of the concentration units, we get 1-n = -1 n = 2. Thus, the reaction is a second-order reaction. Statement B is correct. For a second-order reaction, Rate = k[X]^2. When the concentration of X is increased to 4 times, the new rate is k(4[X])^2 = 16k[X]^2, which is 16 times the original rate. Statement A is correct. The half-life o

Question 10

Consider the following gas phase reaction being carried out in a closed vessel at 25°C. 2A(g) 4B(g) + C(g) time (min) total pressure of the system (mm Hg) 30 300 600 The pressure of C(g) at 30 minutes

Accepted Answer

For the given gas phase reaction: 2A(g) 4B(g) + C(g) Let the initial pressure of A(g) be P_0. At t = 0: P_A = P_0, P_B = 0, P_C = 0 At t = t: P_A = P_0 - 2p, P_B = 4p, P_C = p Total pressure at time t, P_t = P_0 - 2p + 4p + p = P_0 + 3p At t = , the reaction is complete, so P_A = 0 2p = P_0 p = P_0 2 Total pressure at t = , P_ = 4 ( P_0 2 ) + P_0 2 = 5P_0 2 Given P_ = 600 mm Hg: 5P_0 2 = 600 P_0 = 240 mm Hg At t = 30 min, P_t = 300 mm Hg: P_0 + 3p = 300 240 + 3p = 300 3p = 60 p = 20 mm Hg The pr

Question 11

t_ 100\% is the time required for the 100\% completion of the reaction while t_ 1/2 is the time required for 50\% of the reaction to be completed. Which of the following option correctly represents th

Accepted Answer

For a zero-order reaction: The integrated rate law is given by t = [A]_0 - [A]_t k . For 50\% completion (t_ 1/2 ), [A]_t = [A]_0 2 : t_ 1/2 = [A]_0 - [A]_0 2 k = [A]_0 2k For 100\% completion (t_ 100\% ), [A]_t = 0: t_ 100\% = [A]_0 - 0 k = [A]_0 k Comparing the two, we get: t_ 100\% = 2t_ 1/2 For a first-order reaction: The integrated rate law is given by t = 1 k ( [A]_0 [A]_t ). For 100\% completion (t_ 100\% ), [A]_t = 0: t_ 100\% = 1 k ( [A]_0 0 ) = A first-order reaction takes infinite tim

Question 12

For reaction A P, rate constant k = 1.5 10^3 s^ -1 at 27°C. If activation energy for the above reaction is 60 kJ mol^ -1 , then the temperature (in °C) at which rate constant, k = 4.5 10^3 s^ -1 is __

Accepted Answer

Using the Arrhenius equation: ( k_2 k_1 ) = E_a 2.3 R ( 1 T_1 - 1 T_2 ) Given: k_1 = 1.5 10^3 s ^ -1 k_2 = 4.5 10^3 s ^ -1 T_1 = 27^ C = 300 K E_a = 60 kJ mol ^ -1 = 60000 J mol ^ -1 R = 8.3 J K ^ -1 mol ^ -1 Substituting the values into the equation: ( 4.5 10^3 1.5 10^3 ) = 60000 2.3 8.3 ( 1 300 - 1 T_2 ) 3 = 60000 19.09 ( 1 300 - 1 T_2 ) 0.48 = 60000 19.09 ( 1 300 - 1 T_2 ) 1 300 - 1 T_2 = 0.48 19.09 60000 1 300 - 1 T_2 = 9.1632 60000 = 1.5272 10^ -4 1 T_2 = 1 300 - 1.5272 10^ -4 1 T_2 = 3.333

Question 13

A B (first reaction) C D (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K. At 500 ~K , 50 \% of the reaction

Accepted Answer

For A K_1 B (2) = E_ a_1 R [ 1 300 - 1 500 ] E_ a_1 = 2 R 1500 2 E_ a_2 = E_ a_1 2 = 2 R 1500 4 (K_1)_ at 500 K = 2 2 (K_2)_ at 500 K = 2 Now for C K_2 D [ (K_2)_ at 500K (K_2)_ at 300K ] = ( 2 R 1500 4 ) 1 R [ 1 300 - 1 500 ] (K_2)_ at 300 K = 2 2 = 0.49 (K_2)_ at 300 K = 4.9 10^ -1 . So, answer is 5.

Question 14

An organic compound undergoes first order decomposition. The time taken for decomposition to ( 1 8 )^ th and ( 1 10 )^ th of its initial concentration are t _ 1 / 8 and t _ 1 / 10 respectively. What i

Accepted Answer

For first-order decomposition with rate equation [A] = [A]_0 e^ -kt , the time for concentration to reduce to fraction f is: t = 1 k 1 f = (1/f) k For reduction to 1/8: t_ 1/8 = 8 k = 3 2 k For reduction to 1/10: t_ 1/10 = 10 k Therefore: t_ 1/8 t_ 1/10 = 8 10 = 3 2 10 = 3 0.3 1 = 0.9 Thus: t_ 1/8 t_ 1/10 10 = 0.9 10 = 9

Question 15

The half-life of ^ 65 Zn is 245 days. After x days, 75 \% of original activity remained. The value of x in days is \_\_\_\_. (Nearest integer) (Given: 3=0.4771 and 2=0.3010)

Accepted Answer

Using radioactive decay: A = A_0 ( 1 2 )^ t/t_ 1/2 . With 75% activity remaining: 0.75 = ( 1 2 )^ x/245 . Taking log base 10: (0.75) = x 245 (0.5). Calculate: (0.75) = 3 - 4 = 0.4771 - 2(0.3010) = -0.1249 and (0.5) = -0.3010. Thus -0.1249 = x 245 (-0.3010), giving x = 0.1249 245 0.3010 = 101.66 102 days (nearest integer).

Question 16

At 27^ C in presence of a catalyst, activation energy of a reaction is lowered by 10 ~kJ ~mol ^ -1 . The logarithm of ratio of k (catalysed) k (uncatalysed) is.... (Consider that the frequency factor

Accepted Answer

Using the Arrhenius equation: k = A e^ -E_a/RT The ratio of rate constants is: k_ catalysed k_ uncatalysed = A e^ -E_a^c/RT A e^ -E_a^u/RT = e^ (E_a^u - E_a^c)/RT = e^ E_a/RT Since the catalyst lowers activation energy by 10 kJ/mol: E_a = 10 kJ/mol = 10000 J/mol At T = 27°C = 300 K, R = 8.314 J/(mol·K): k_c k_u = e^ 10000/(8.314 300) = e^ 10000/2494.2 = e^ 4.009 Taking logarithm base 10: _ 10 k_c k_u = 4.009 _ 10 (e) = 4.009 0.4343 = 1.741

Question 17

Given above is the concentration vs time plot for a dissociation reaction : A nB . Based on the data of the initial phase of the reaction (initial 10 min), the value of n is \_\_\_\_.

Accepted Answer

A nB At t = 0: [A] = 0.05 M, [B] = 0 At t = 10 min: [A] = 0.04 M, [B] = 0.01 n From the graph, [B] at t = 10 min = 0.03 M 0.01 n = 0.03 n = 3

Question 18

Observe the following reactions at T ( K ). I. A products. II. 5 Br ^ - ( aq )+ BrO _ 3 ^ - ( aq )+6 H ^ + ( aq ) 3 Br _ 2 ( aq )+3 H _ 2 O ( l ) Both the reactions are started at 10.00 am. The rates

Accepted Answer

For reaction II: 5Br^- + BrO_3^- + 6H^+ 3Br_2 + 3H_2O, The overall reaction rate is: r = - 1 5 [Br^-] t = - 1 3 [Br_2] t . Given [Br^-] t = 2 10^ -4 mol L^ -1 min^ -1 The reaction rate is r = 2 10^ -4 5 = 4 10^ -5 mol L^ -1 min^ -1 . Since the rates of both reactions are equal at 10.10 am and reaction I is first-order: r = k[A], so 4 10^ -5 = k 10^ -2 , giving k = 4 10^ -3 min^ -1 .

Question 19

For the thermal decomposition of reactant AB ( g ), the following plot is constructed. The half life of the reaction is ' x^ . x= \_\_\_\_ min. (Nearest integer)

Accepted Answer

From the concentration vs. time graph, the thermal decomposition of AB(g) shows a linear decrease in concentration, indicating zeroth-order kinetics. Initial concentration: [AB]₀ = 0.6 M at t = 0 s. Reading from the graph: [AB] = 0.5 M at t = 200 s. Rate of decomposition: Δ[AB]/Δt = (0.6 - 0.5) M / 200 s = 0.0005 M/s. For zeroth-order kinetics: [AB] = [AB]₀ - kt. Half-life occurs when [AB] = [AB]₀/2 = 0.3 M. 0.3 = 0.6 - 0.0005 × t₁/₂. t₁/₂ = 0.3 / 0.0005 = 600 s = 10 min.

Question 20

Correct statements regarding Arrhenius equation among the following are : A. Factor e^ -E a / R T corresponds to fraction of molecules having kinetic energy less than Ea. B. At a given temperature, lo

Accepted Answer

The Arrhenius equation is given by k = A e^ -E_a/RT . Statement A: The factor e^ -E_a/RT represents the fraction of molecules having kinetic energy equal to or greater than the activation energy E_a. Thus, statement A is incorrect. Statement B: From the equation, as E_a decreases, the value of k increases, leading to a faster reaction rate. Thus, statement B is correct. Statement C: For many reactions, the temperature coefficient is approximately 2, meaning the rate of reaction doubles for every

Chemical Kinetics — Previous Year Questions & Solutions