Question 1

What is the order and molecularity of the following elementary reaction? 2 NO _ 2(g) 2 NO _ (g) + O _ 2(g) , rate = k [ NO _2]^2

Accepted Answer

For an elementary reaction, the rate law is determined by its stoichiometry. Given rate = k[ NO _2]^2 The order of the reaction is the sum of the exponents of the concentration terms in the rate law. Here, the exponent is 2, so the order is 2. The molecularity of an elementary reaction is the number of reactant molecules participating in the reaction. Since 2 molecules of NO _2 are reacting, the molecularity is 2 (bimolecular). Thus, the reaction is second order and bimolecular. Answer: The reac

Question 2

Calculate the rate constant of first order reaction A B having rate 5.4 10^ -6 \ mol dm ^ -3 s ^ -1 and [ A ] = 0.3 M.

Accepted Answer

For a first order reaction, the rate law is given by: Rate = k[ A ] Substituting the given values: 5.4 10^ -6 = k 0.3 k = 5.4 10^ -6 0.3 k = 18 10^ -6 \ s ^ -1 k = 1.80 10^ -5 \ s ^ -1 Answer: 1.80 10^ -5 \ s ^ -1

Question 3

What is the role of catalyst in reaction?

Accepted Answer

A catalyst increases the rate of a chemical reaction by providing an alternative reaction pathway that has a lower activation energy. It does not alter the equilibrium constant, nor does it supply energy to the reactants. Therefore, the primary role of a catalyst is to decrease the activation energy of a reaction. Answer: To decrease activation energy of a reaction

Question 4

In a reaction, 2 N _2 O _ 5(g) 4\, NO _ 2(g) + O _ 2(g) N _2 O _5 disappears at a rate of 0.06 mol dm^ -3 s^ -1 Calculate rate of formation of O _ 2(g) ?

Accepted Answer

For the given reaction 2 N _2 O _ 5(g) 4 NO _ 2(g) + O _ 2(g) The rate of reaction is given by: Rate = - 1 2 d[ N _2 O _5] dt = 1 4 d[ NO _2] dt = d[ O _2] dt Given the rate of disappearance of N _2 O _5 is - d[ N _2 O _5] dt = 0.06 mol dm ^ -3 s ^ -1 The rate of formation of O _2 is: d[ O _2] dt = 1 2 (- d[ N _2 O _5] dt ) d[ O _2] dt = 1 2 0.06 = 0.03 mol dm ^ -3 s ^ -1

Question 5

Calculate rate constant of a first order reaction having half life 1 minute 40 second?

Accepted Answer

Given half-life of the first order reaction, t_ 1/2 = 1 minute 40 seconds Converting to seconds, t_ 1/2 = 60 + 40 = 100 s For a first order reaction, the rate constant k is given by the relation: k = 0.693 t_ 1/2 Substituting the value of t_ 1/2 : k = 0.693 100 s k = 6.93 10^ -3 s ^ -1 Answer: 6.93 10^ -3 \ s ^ -1

Question 6

Identify correct statement regarding order of reaction from following.

Accepted Answer

For a zero order reaction, the rate is given by r = k[A]^0 = k. Thus, the rate is independent of the concentration of the reactant. The decomposition of acetaldehyde (CH_3CHO CH_4 + CO) is a fractional order reaction with an order of 1.5. For a first order reaction, the half-life is given by t_ 1/2 = 0.693 k . This shows that the half-life is independent of the initial concentration of the reactant. For a zero order reaction, the half-life is given by t_ 1/2 = [A]_0 2k . This shows that the half

Question 7

For a first order reaction, half life is 5 hour. What time is required to reduce 10 g of reactant to 2.5 g ?

Accepted Answer

For a first order reaction, the amount of reactant remaining after n half-lives is given by N_t = N_0 2^n Given initial amount N_0 = 10 g and final amount N_t = 2.5 g Substituting the values, 2.5 = 10 2^n 2^n = 10 2.5 = 4 2^n = 2^2 n = 2 Total time required t = n t_ 1/2 Given t_ 1/2 = 5 hour t = 2 5 = 10 hour Answer: 10 hour

Question 8

Consider the reaction, 3 I ^-_ (aq) + S _2 O _8^ 2- _ (aq) I _3^- _ (aq) + 2 SO _4^ 2- _ (aq) If the rate of formation of SO _4^ 2- at a particular time is 2.2 10^ -2 \, mol dm ^ -3 \, s ^ -1 . Calcul

Accepted Answer

The rate of reaction is given by: Rate = - 1 3 d[ I ^-] dt = 1 2 d[ SO _4^ 2- ] dt The rate of consumption of I ^- is - d[ I ^-] dt . - d[ I ^-] dt = 3 2 d[ SO _4^ 2- ] dt Substituting the given rate of formation of SO _4^ 2- : - d[ I ^-] dt = 3 2 2.2 10^ -2 - d[ I ^-] dt = 3.3 10^ -2 mol dm ^ -3 s ^ -1 Answer: 3.3 10^ -2 \, mol dm ^ -3 \, s ^ -1

Question 9

Half life of a first order reaction is 900 second. If initial concentration of reactant is 0.08\, mol dm ^ -3 find concentration that remains after 35 minute ?

Accepted Answer

Given t_ 1/2 = 900 s and t = 35 min = 35 60 = 2100 s. Number of half-lives n = t t_ 1/2 = 2100 900 = 7 3 . For a first order reaction, the concentration remaining after n half-lives is given by: [A] = [A]_0 2^n Substituting the given values: [A] = 0.08 2^ 7/3 [A] = 0.08 4 2^ 1/3 Since 2^ 1/3 1.26: [A] 0.08 4 1.26 = 0.08 5.04 0.0159 mol dm ^ -3 Answer: 0.0159\, mol dm ^ -3

Question 10

Rate constant of decomposition of hydrogen peroxide is 0.0204\ minutes ^ -1 , calculate half life of reaction.

Accepted Answer

The unit of the rate constant is minutes ^ -1 , which indicates that the decomposition of hydrogen peroxide is a first-order reaction. For a first-order reaction, the half-life is given by: t_ 1/2 = 0.693 k Substituting the given value of k = 0.0204\ minutes ^ -1 : t_ 1/2 = 0.693 0.0204 t_ 1/2 = 33.97\ minutes Answer: 33.97 minutes

Question 11

A first order reaction take 30 minutes for 75\% decomposition, calculate its rate constant ?

Accepted Answer

For a first order reaction, the integrated rate law is given by: k = 2.303 t ( a a-x ) Given that the reaction is 75\% complete in 30 minutes: t = 30\ minutes x = 0.75a a - x = a - 0.75a = 0.25a Substituting these values into the rate equation: k = 2.303 30 ( a 0.25a ) k = 2.303 30 (4) k = 2.303 30 0.6020 k = 1.386 30 = 0.0462\ minute ^ -1 Alternatively, 75\% decomposition corresponds to two half-lives: t_ 75\% = 2 t_ 1/2 30 = 2 t_ 1/2 t_ 1/2 = 15\ minutes k = 0.693 t_ 1/2 = 0.693 15 = 0.0462\ m

Question 12

Which of the following is an elementary reaction ?

Accepted Answer

An elementary reaction is one that occurs in a single step and has a single transition state. The decomposition of ozone, O _ 3(g) O _ 2(g) + O _ (g) , is a unimolecular elementary reaction. The other reactions are complex and occur in multiple steps: 2 NO _2 Cl _ (g) 2 NO _ 2(g) + Cl _ 2(g) occurs via: Step 1: NO _2 Cl NO _2 + Cl (slow) Step 2: NO _2 Cl + Cl NO _2 + Cl _2 (fast) 2 NO _ 2(g) + F _ 2(g) 2 NO _2 F _ (g) occurs via: Step 1: NO _2 + F _2 NO _2 F + F (slow) Step 2: NO _2 + F NO _2 F

Question 13

For a reaction A product, k = 2 10^ -2 \ s ^ -1 . If the initial concentration of A is 1.0\ mol dm ^ -3 find the value of 1 [A]_t after 100 second ?

Accepted Answer

The unit of rate constant k is s ^ -1 , indicating that the reaction is of first order. For a first-order reaction, the integrated rate equation is: k = 2.303 t [A]_0 [A]_t Substituting the given values k = 2 10^ -2 \ s ^ -1 , t = 100\ s , and [A]_0 = 1.0\ mol dm ^ -3 : 2 10^ -2 = 2.303 100 1.0 [A]_t 2 10^ -2 100 = 2.303 1 [A]_t 2 = 2.303 1 [A]_t 1 [A]_t = 2 2.303 = 0.868 Answer: 0.868

Question 14

What is the time required for 75 % completion of a first order reaction if rate constant is 23.03\ minute ^ -1 ?

Accepted Answer

For a first order reaction, the time required for a given completion is given by: t = 2.303 k ( a a-x ) Given k = 23.03 min ^ -1 and for 75\% completion, x = 0.75a. t = 2.303 23.03 ( a a - 0.75a ) t = 0.1 ( 1 0.25 ) t = 0.1 4 t = 0.1 0.602 = 0.0602 min Converting the time into seconds: t = 0.0602 60 s = 3.612 s 3.6 s Answer: 3.6 s

Question 15

Which of the following is an example of fractional order reaction ?

Accepted Answer

The decomposition of acetaldehyde ( CH _3 CHO ) is a well-known example of a fractional order reaction. The experimentally determined rate law for this reaction is given by: Rate = k[ CH _3 CHO ]^ 1.5 Thus, the overall order of the reaction is 1.5, which is a fractional value. For the other given reactions: Decomposition of H _2 O _2 is a first-order reaction. Formation of HI from H _2 and I _2 is a second-order reaction. Reaction between NO and H _2 is a third-order reaction. Answer: CH _3 CHO

Question 16

Which of the following reactions has of overall order 1.5 ?

Accepted Answer

The thermal decomposition of acetaldehyde is a well-known example of a reaction with a fractional order. The reaction is: CH _3 CHO _ (g) CH _ 4(g) + CO _ (g) The experimentally determined rate law for this reaction is: Rate = k[ CH _3 CHO ]^ 1.5 Thus, the overall order of the reaction is 1.5. For the other reactions: H _ 2(g) + I _ 2(g) 2 HI _ (g) follows second-order kinetics. 2 H _2 O _ 2(g) 2 H _2 O _ ( ) + O _ 2(g) follows first-order kinetics. 2 NO _ (g) + 2 H _ 2(g) N _ 2(g) + 2 H _2 O _

Question 17

For the reaction 2 NOBr _ (g) 2 NO _ (g) + Br _ 2(g) , rate law is r = k[ NOBr ]^2. Rate constant is 1.62\ M s ^ -1 and concentration of NOBr is 5 10^ -3 \ M , What is rate of reaction?

Accepted Answer

Given rate law is r = k[ NOBr ]^2 Substituting the given values: r = 1.62 (5 10^ -3 )^2 r = 1.62 25 10^ -6 r = 40.5 10^ -6 r = 4.05 10^ -5 \ M s ^ -1 Answer: 4.05 10^ -5 \ M s ^ -1

Question 18

In a first order reaction 20 millimole of reactant is lowered to 10 millimole in 0.3010 minute. Find rate constant of the reaction?

Accepted Answer

For a first order reaction, the rate constant k is given by the expression: k = 2.303 t ( a a - x ) Given: Initial amount, a = 20 millimole Amount after time t, a - x = 10 millimole Time, t = 0.3010 minute Substituting the given values into the equation: k = 2.303 0.3010 ( 20 10 ) k = 2.303 0.3010 2 Using the standard value 2 = 0.3010: k = 2.303 0.3010 0.3010 k = 2.303\ minute ^ -1 Answer: 2.303\ minute ^ -1

Question 19

Which of the following is an example of second order reaction ?

Accepted Answer

The order of a reaction is determined experimentally by its rate law. For the decomposition of hydrogen peroxide, the rate law is Rate = k[ H _2 O _2]. It is a first-order reaction. For the thermal decomposition of acetaldehyde, the rate law is Rate = k[ CH _3 CHO ]^ 1.5 . It is a fractional-order (1.5) reaction. For the reaction between NO _2 and F _2, the mechanism involves a slow rate-determining step: NO _2 + F _2 NO _2 F + F . The rate law is Rate = k[ NO _2][ F _2]. The overall order is 1

Question 20

In first order reaction 20 millimole of reactant is reduced to 10 millimole in 1 151 minute. Find rate constant.

Accepted Answer

For a first order reaction, the rate constant is given by k = 2 303 t ( a a-x ) Substituting the given values a = 20 millimole, a-x = 10 millimole, and t = 1 151 minute: k = 2 303 1 151 ( 20 10 ) k = 2 303 1 151 2 k = 2 303 0 3010 1 151 k = 0 6023 minute ^ -1 Answer: 0 6023 minute ^ -1

Chemical Kinetics — Previous Year Questions & Solutions