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

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Hydrogen gas, $H_{2} (g)$ , reacts with iodine gas, $I_{2} (g)$ , to form hydrogen iodide, $HI (g)$ :

$H_{2} (g) + I_{2} (g) \to 2 HI (g)$

The mechanism of the two-step reaction is considered to be:

step $1$ : $I_{2} (g) ⇋_{k_{- 1}}^{k_{1}} 2 I (g)$ fast

step $2$ : $2 I (g) + H_{2} (g) \overset{k_{2}}{\to} 2 HI (g)$ slow

What is the rate equation for the overall reaction?

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Important Questions on Chemical Kinetics [AHL]

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

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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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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.

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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 $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

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

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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 $3$	$3.00 \times 10^{- 2}$	$1.50 \times 10^{- 2}$	$4.64 \times 10^{- 3}$

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

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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.

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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 $20 ° C$ .

Hydrogen gas, H2(g), reacts with iodine gas, I2(g), to form hydrogen iodide, HI(g): H2(g) + I2(g) → 2HI(g) The mechanism of the two-step reaction is considered to be: step 1: I2(g) ⇋k-1k1 2I(g) fast step 2: 2I(g) + H2(g) →k2 2HI(g) slow What is the rate equation for the overall reaction?

Important Questions on Chemical Kinetics [AHL]

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