Consider the following reaction: A ( g ) + B ( g ) → C ( g ) + D ( g ) and the following experimental initial rate data: [ A ( g ) ] / mol dm - 3 [ B ( g ) ] / mol dm - 3 Initial rate/ mol dm - 3 s - 1 Experiment 1 1 . 50 × 10 - 2 1 . 50 × 10 - 2 2 . 32 × 10 - 3 Experiment 2 1 . 50 × 10 - 2 3 . 00 × 10 - 2 4 . 64 × 10 - 3 Experiment 3 3 . 00 × 10 - 2 1 . 50 × 10 - 2 4 . 64 × 10 - 3 Deduce the orders with respect to each reactant and the overall reaction order.

A ( g ) + B ( g ) → C ( g ) + D ( g ) The rate law for the above reaction can be written as, rate = [ A ] x [ B ] y . In order to solve this question we can use the working method to deduce the rate equation from the method of initial rates: rate 1 rate 2 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 1 . 50 × 10 - 2 ) x ( 3 . 00 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 y = 1 2 So, y = 1 rate 1 rate 3 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 3 . 00 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 x = 1 2 So, x = 1 The reaction is first-order with respect to both A and B ; the overall reaction order is second order.

Consider the following reaction: A ( g ) + B ( g ) → C ( g ) + D ( g ) and the following experimental initial rate data: [ A ( g ) ] / mol dm - 3 [ B ( g ) ] / mol dm - 3 Initial rate/ mol dm - 3 s - 1 Experiment 1 1 . 50 × 10 - 2 1 . 50 × 10 - 2 2 . 32 × 10 - 3 Experiment 2 1 . 50 × 10 - 2 3 . 00 × 10 - 2 4 . 64 × 10 - 3 Experiment 3 3 . 00 × 10 - 2 1 . 50 × 10 - 2 4 . 64 × 10 - 3 Deduce the rate equation.

A ( g ) + B ( g ) → C ( g ) + D ( g ) The rate law for the above reaction can be written as, rate = [ A ] x [ B ] y . In order to solve this question we can use the working method to deduce the rate equation from the method of initial rates: rate 1 rate 2 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 1 . 50 × 10 - 2 ) x ( 3 . 00 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 y = 1 2 So, y = 1 rate 1 rate 3 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 3 . 00 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 x = 1 2 So, x = 1 The reaction is first-order with respect to both A and B . The rate equation is, rate = [ A ] [ B ] .

Consider the following reaction: A ( g ) + B ( g ) → C ( g ) + D ( g ) and the following experimental initial rate data: [ A ( g ) ] / mol dm - 3 [ B ( g ) ] / mol dm - 3 Initial rate/ mol dm - 3 s - 1 Experiment 1 1 . 50 × 10 - 2 1 . 50 × 10 - 2 2 . 32 × 10 - 3 Experiment 2 1 . 50 × 10 - 2 3 . 00 × 10 - 2 4 . 64 × 10 - 3 Experiment 3 3 . 00 × 10 - 2 1 . 50 × 10 - 2 4 . 64 × 10 - 3 Calculate the value of the rate constant, k , for the reaction from experiment 2 and state its units.

A ( g ) + B ( g ) → C ( g ) + D ( g ) The rate law for the above reaction can be written as, rate = [ A ] x [ B ] y . In order to solve this question we can use the working method to deduce the rate equation from the method of initial rates: rate 1 rate 2 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 1 . 50 × 10 - 2 ) x ( 3 . 00 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 y = 1 2 So, y = 1 rate 1 rate 3 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 3 . 00 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 x = 1 2 So, x = 1 The reaction is first-order with respect to both A and B . The rate law for the reaction is, rate = k [ A ] [ B ] . ∴ k = rate A B = 4 . 64 × 10 - 3 mol dm - 3 s - 1 1 . 50 × 10 - 2 mol dm - 3 × 3 . 00 × 10 - 2 mol dm - 3 = 1 . 03 × 10 1 mol - 1 dm 3 s - 1

Consider the following reaction: A ( g ) + B ( g ) → C ( g ) + D ( g ) and the following experimental initial rate data: [ A ( g ) ] / mol dm - 3 [ B ( g ) ] / mol dm - 3 Initial rate/ mol dm - 3 s - 1 Experiment 1 1 . 50 × 10 - 2 1 . 50 × 10 - 2 2 . 32 × 10 - 3 Experiment 2 1 . 50 × 10 - 2 3 . 00 × 10 - 2 4 . 64 × 10 - 3 Experiment 3 3 . 00 × 10 - 2 1 . 50 × 10 - 2 4 . 64 × 10 - 3 Determine the rate of the reaction when [ A ( g ) ] = 2 . 00 × 10 - 2 mol dm - 3 and [ B ( g ) ] = 4 . 00 × 10 - 2 mol dm - 3 .

A ( g ) + B ( g ) → C ( g ) + D ( g ) The rate law for the above reaction can be written as, rate = [ A ] x [ B ] y . In order to solve this question we can use the working method to deduce the rate equation from the method of initial rates: rate 1 rate 2 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 1 . 50 × 10 - 2 ) x ( 3 . 00 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 y = 1 2 So, y = 1 rate 1 rate 3 = ( 1 . 50 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y ( 3 . 00 × 10 - 2 ) x ( 1 . 50 × 10 - 2 ) y = 2 . 32 × 10 - 3 4 . 64 × 10 - 3 ⇒ 1 2 x = 1 2 So, x = 1 The reaction is first-order with respect to both A and B . The rate law for the reaction is, rate = k [ A ] [ B ] . ∴ k = rate A B (Using the experiment 2 data) = 4 . 64 × 10 - 3 mol dm - 3 s - 1 1 . 50 × 10 - 2 mol dm - 3 × 3 . 00 × 10 - 2 mol dm - 3 = 1 . 03 × 10 1 mol - 1 dm 3 s - 1 Therefore, rate for the given concentrations of [ A ( g ) ] and B g is, Rate = k [ A ] [ B ] = ( 1 . 03 × 10 1 mol - 1 dm 3 s - 1 ) × ( 2 . 00 × 10 - 2 mol dm - 3 ) × ( 4 . 00 × 10 - 2 mol dm - 3 ) = 8 . 25 × 10 - 3 mol dm - 3 s - 1

The rate constant, k 1 , of a first-order reaction is 6 . 30 × 10 3 s - 1 at 32 ° C and the corresponding rate constant, k 2 , is 2 . 25 × 10 5 s - 1 at 83 ° C . Deduce the activation energy, E a in kJ mol - 1 , correct to two significant figures.

Given, for a first-order reaction, The rate constant k 1 is 6 . 30 × 10 3 s - 1 And, the rate constant k 2 is 2 . 25 × 10 5 s - 1 T 1 = 32 ° C = ( 32 + 273 ) = 305 K T 2 = 83 ° C = ( 83 + 273 ) = 356 K The gas constant, R = 8 . 314 J K - 1 mol - 1 The rate constant of a reaction is related to activation energy as, k = Ae - E a RT ∴ E a = ln k 1 k 2 × R 1 T 2 - 1 T 1 = 2 . 303 × log 6 . 30 × 10 3 2 . 25 × 10 5 × 8 . 314 1 356 - 1 305 = 63 . 3 kJ mol - 1

Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions

Author:Sergey Bylikin, Gary Horner & Brian Murphy

Sergey Bylikin Chemistry Solutions for Exercise - Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions

Attempt the practice questions on Chapter 16: Chemical Kinetics [AHL], Exercise 4: Questions with hints and solutions to strengthen your understanding. Oxford IB Diploma Programme Chemistry Course Companion solutions are prepared by Experienced Embibe Experts.

Questions from Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions with Hints & Solutions

EASY

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IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 8

What are the units of the frequency factor in the Arrhenius equation?

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Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 9

Ozone is considered to decompose according to the following two-step mechanism:

step $1$ : $O_{3} (g) ⇌_{k_{- 1}}^{k_{1}} O_{2} (g) + O (g)$ fast

step $2$ : $O (g) + O_{3} (g) \overset{k_{2}}{\to} 2 O_{2} (g)$ slow

Which of the following are correct?
I. The overall reaction is $2 O_{3} (g) \to 3 O_{2} (g)$ .

II. $O (g)$ is a reaction intermediate.

III. The rate equation is:

rate $= k {[O_{3}]}^{2} {[O_{2}]}^{3}$

MEDIUM

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Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 10

Consider the following reaction:

$A (g) + B (g) \to C (g) + D (g)$

and the following experimental initial rate data:

	$[A (g)] / mol {dm}^{- 3}$	$[B (g)] / mol {dm}^{- 3}$	Initial rate/ $mol {dm}^{- 3} s^{- 1}$
Experiment $1$	$1.50 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$2.32 \times 10^{- 3}$
Experiment $2$	$1.50 \times 10^{- 2}$	$3.00 \times 10^{- 2}$	$4.64 \times 10^{- 3}$
Experiment $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

Deduce the orders with respect to each reactant and the overall reaction order.

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IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 10

Consider the following reaction:

$A (g) + B (g) \to C (g) + D (g)$

and the following experimental initial rate data:

	$[A (g)] / mol {dm}^{- 3}$	$[B (g)] / mol {dm}^{- 3}$	Initial rate/ $mol {dm}^{- 3} s^{- 1}$
Experiment $1$	$1.50 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$2.32 \times 10^{- 3}$
Experiment $2$	$1.50 \times 10^{- 2}$	$3.00 \times 10^{- 2}$	$4.64 \times 10^{- 3}$
Experiment $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

Deduce the rate equation.

MEDIUM

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IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 10

Consider the following reaction:

$A (g) + B (g) \to C (g) + D (g)$

and the following experimental initial rate data:

	$[A (g)] / mol {dm}^{- 3}$	$[B (g)] / mol {dm}^{- 3}$	Initial rate/ $mol {dm}^{- 3} s^{- 1}$
Experiment $1$	$1.50 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$2.32 \times 10^{- 3}$
Experiment $2$	$1.50 \times 10^{- 2}$	$3.00 \times 10^{- 2}$	$4.64 \times 10^{- 3}$
Experiment $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

Calculate the value of the rate constant, $k$ , for the reaction from experiment $2$ and state its units.

MEDIUM

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IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 10

Consider the following reaction:

$A (g) + B (g) \to C (g) + D (g)$

and the following experimental initial rate data:

	$[A (g)] / mol {dm}^{- 3}$	$[B (g)] / mol {dm}^{- 3}$	Initial rate/ $mol {dm}^{- 3} s^{- 1}$
Experiment $1$	$1.50 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$2.32 \times 10^{- 3}$
Experiment $2$	$1.50 \times 10^{- 2}$	$3.00 \times 10^{- 2}$	$4.64 \times 10^{- 3}$
Experiment $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

Determine the rate of the reaction when $[A (g)] = 2.00 \times 10^{- 2} mol {dm}^{- 3}$ and $[B (g)] = 4.00 \times 10^{- 2} mol {dm}^{- 3}$ .

MEDIUM

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IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 11

The rate constant, $k_{1}$ , of a first-order reaction is $6.30 \times 10^{3} s^{- 1}$ at $32 ° C$ and the corresponding rate constant, $k_{2}$ , is $2.25 \times 10^{5} s^{- 1}$ at $83 ° C$ . Deduce the activation energy, $E_{a}$ in $kJ {mol}^{- 1}$ , correct to two significant figures.

MEDIUM

Diploma

IMPORTANT

Oxford IB Diploma Programme Chemistry Course Companion>Chapter 16 - Chemical Kinetics [AHL]>Questions>Q 11

The rate constant, $k_{1}$ , of a first-order reaction is $6.30 \times 10^{3} s^{- 1}$ at $32 ° C$ and the corresponding rate constant, $k_{2}$ , is $2.25 \times 10^{5} s^{- 1}$ at $83 ° C$ . Calculate the rate constant, $k_{3}$ in $s^{- 1}$ , at $20 ° C$ .

Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions

Sergey Bylikin Chemistry Solutions for Exercise - Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions

Questions from Sergey Bylikin, Gary Horner and, Brian Murphy Solutions for Chapter: Chemical Kinetics [AHL], Exercise 4: Questions with Hints & Solutions

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