Chemistry 40S Unit 3: Chemical Kinetics Lesson 4.

Chemistry 40SUnit 3: Chemical Kinetics

Lesson 4

Learning OutcomesO C12-3-09 Explain the concept of a

reaction mechanism. Include: rate-determining step

O C12-3-10 Determine the rate law and order of a chemical reaction from experimental data. Include: zero-, first-, and second-order reactions and reaction rate versus concentration graphs

Reaction MechanismO A reaction mechanism

summarizes the individual steps a reaction follows

O Each individual step is called an elementary step or an elementary process

O The balanced equation of a reaction represents the net reaction mechanism

Reaction Mechanism

O Example 1: 2NO(g) + O2(g) 2NO2(g)

Step 1: 2NO(g) N2O2(g)

Step 2: N2O2(g) + O2(g) 2NO2(g)

Net reaction: 2NO(g) + O2(g) 2NO2(g)

O As the N2O2 appears in the reaction mechanism but not in the overall chemical equation, it is called an intermediate.

Reaction MechanismO Example 2: 2O3(g) 3O2(g)

Step 1: Cl2(g) + O3(g) ClO(g) + O2(g)

Step 2: O3(g) O2(g) + O(g)

Step 3: ClO(g) + O(g) Cl2(g) + O2(g)

Net reaction: 2O3(g) 3O2(g)

O Cl2(g) is a catalyst and the ClO(g) is an intermediateO Catalysts are unchanged & not in the overall

reactionO Intermediates are not in the overall reaction

because they are transformed

Rate-Determining StepO Not all steps in a mechanism have

the same speedO Rate-Determining Step - slowest

of the elementary processesO Determines the rate of the reaction

because it has a stronger affect than the others

Rate-Determining Step Problem

O Example 1: Given the following mechanism:P + Q X + T (slow)X + P Y + R (fast)

Y + S T (moderate)

a) What is the net reaction?b) What are the reaction intermediates?c) Which step is the rate determining step?

Rate-Determining Step Problem

d) What would be the effect of increasing the concentration of P?e) What would be the effect of decreasing the concentration of Q?f) What would be the effect of increasing the concentration of S?

Rate LawO Rate Law – an expression that

relates the rate of reaction to the concentration of reactantsO Helps us calculate rate based on the

concentrations of reactantsO Shows the quantitative effect of

concentration on reaction rate

Rate LawO Rate of consumption is directly proportional to

concentration O The faster the reactants are consumed, the lower its

concentrationO This can be represented by:

Rate = k[A]x

O k – specific rate constant O [A] – concentration of AO x – order of the reaction

O The rate constant is unique to each reaction at a specific temperature (its value depends on the speed/types of molecules)

O Therefore = changing T change in the rate constant

Order of ReactionO Order of reaction – indicates the

effect of concentration of reactants on reaction rate

O First Order Reaction (x = 1) – reaction rate is directly proportional to changes in reactant concentrationO i.e. doubling reactant concentration =

doubling rate

Order of ReactionO Second Order Reaction (x =2) –

reaction rate is proportional to the square of the change in concentrationO i.e. doubling concentration = 22 = 4x

increase in rateO Zero Order Reaction (x=0) – the

reaction rate does not depend on the concentration of reactants

Rate LawO When a reaction has two reactants

the rate depends on the concentration of both reactants

O Each reactant affects the rate differently therefore:

Rate = k[A]x[B]y

O The total order of the reaction is the sum of the order of all reactants (i.e. x + y)

Determining the Rate Law of a Reaction

O Rate law can only be determined experimentally (despite our knowledge of reaction stoichiometry)

O There are 3 ways to determine rate law, but all three work to determine the order of each reactant:O Differential Rate Law (calculus)O Integrated Rate Law (time vs.

concentration graphs)O Initial Rates

Rate Law using Initial Rates

Example 1: What is the rate law for the below reaction based on the provided experimental data?

H2O2 + 2 HI → 2 H2O + I2Trial [H2O2] (M) [HI] (M) Initial Rate

(mol/L·s)1 0.10 0.10 0.0076

2 0.10 0.20 0.0152

3 0.20 0.10 0.0152

Rate Law using Initial Rates

Example 2: For the reaction A + B → products, the following data was collected. Determine the rate law.

Trial [A] (M) [B] (M) Initial Rate (mol/L·s)

1 0.10 0.20 2.0

2 0.30 0.20 18.0

3 0.20 0.40 16.0

Determining the Specific Rate Constant

Example 1: Determine the value of the specific rate constant, k, based on the below information.

Rate = k[A]2[B]

Trial [A] (M) [B] (M) Initial Rate (mol/L·s)

1 0.10 0.20 2.0

2 0.30 0.20 18.0

3 0.20 0.40 16.0

Rate Law & Stoichiometry

O For single step (elementary reactions) the order of each reactant in the rate law is equal to the coefficient in the reaction's balanced equation

Example 1: aA + bB → cC + dDrate = k[A]a[B]b

Rate Law & Reaction Mechanisms

O Rate law corresponds to the stoichiometry of the rate determining step

O Example 1: The mechanism for the reaction: 3 M + N → P + 2 Q

O a) What is the rate law for this reaction?b) What would be the effect of tripling the [M]?c) What would be the effect of doubling the [N]?