Rates of Reactions

Learning GoalsStudents will:understand the Rate Law Equationdetermine the Rate Law Equation given

experimental data

Success CriteriaStudents will:Be able to write Rate Law expressions for

chemical reactions

PurposeWe understand:How the rate of a reaction changes during a

reaction (the rate decreases as reactants are used up)

How to calculate the rate of reaction based on data from a graph. (r = ∆c/∆t)

which factors affect the rates of a reaction.What we want to do:Is develop a mathematical equation for rates

of reaction (r) that incorporates some rate factors.

The Rate Law or Law of Mass ActionThe rate of a chemical reaction is proportional to

the product of the concentrations of the reactantsOR The rate, r, will always be proportional to the

product of the initial concentrations of the reactants, where these concentrations are raised to some exponential values.

For a typical chemical reaction:

a A + b B (products)This can be expressed as r α [A]m[B]n

Where the exponents m and n are “orders of reaction”These exponents are NOT related to the co-efficients

Rate Law EquationThe relationship r α [A]m[B]n can be

converted to the Rate Law equation:r = k [A]m[B]n

Where:[A] and [B] represent the concentrations of

substance A and substance Bk = rate law constant (which is not affected

by the concentration)m and n are the “orders of reaction”

Determining Orders of ReactionThe values m and n MUST be determined

empirically (by conducting an experiment and mathematically analyzing the data)

The exponents, m and n may be zero, fractions or integers.

Some “experimental” data is provided below for the following equation:

2 X + 2 Y + 3 Z → productsThis means the rate law equation will be:

r = k[X]m[Y]n[Z]p Let’s determine m, n and p.

2 X + 2 Y + 3 Z → productsDetermine how the rate of reaction changes

due to changes in concentration of X ([X]).To do this, we must keep the concentrations of

Y and Z constant – they are controlled variables. Any changes in r must be entirely due to X.

Experimental data showed that:As the concentration of X is doubled (x2), the

rate of reaction also doubles (x2)As the concentration of X is tripled (x3), the

rate of reaction also triples (x3)

Determining mWe have discovered a

linear relationship.r [X]1 (y = x1)

Linear relationships are known as first order relationships because the exponent is 1.

Since the exponent is 1, then m=1

2 X + 2 Y + 3 Z → productsDetermine how the rate of reaction changes

due to changes in concentration of Y ([Y]).To do this, we must keep the concentrations of

X and Z constant – they are controlled variables. Any changes in r must be entirely due to Y.

Experimental data showed that:As the concentration of Y is doubled (x2), the

rate of reaction multiplies by 4 (x4) (4 = 22)As the concentration of Y is tripled (x3), the

rate of reaction multiplies by 9 (x9) (9 = 32)

Determining nWe have discovered an

exponential relationship.r [Y]2 (y = x2)

Linear relationships are known as second order relationships because the exponent is 2.

Since the exponent is 2, then n=2

2 X + 2 Y + 3 Z → productsDetermine how the rate of reaction changes

due to changes in concentration of Z ([Z]).To do this, we must keep the concentrations of

X and Y constant – they are controlled variables. Any changes in r must be entirely due to Z.

Experimental data showed that:As the concentration of Z is doubled (x2), the

rate of reaction does not change (x1) (1 = 20)As the concentration of Z is tripled (x3), the

rate of reaction does not change (x1) (1 = 30)

Determining pWe have discovered an

line relationship.r [Z]0 (y = xo)

Line relationships are known as zeroth order relationships because the exponent is 0.

Since the exponent is 0, then p=0

Rewrite the Rate Lawr = k[X]1[Y]2[Z]0

The overall order of reaction is determined by adding the individual orders of reaction.

In this case 1 + 2 + 0 = 3

Let’s use some real data!

Start by writing the Rate Law equation r = k[BrO3

-(aq)]m[HSO3

-(aq)]n

Determine how the rate of reaction changes due to changes in concentration of BrO3

- ([BrO3-(aq)]).

To do this, we must keep the concentration of HSO3- constant

([HSO3-(aq) ]) – it is a controlled variable. Any changes in r must be

entirely due to BrO3-(aq).

Let’s look at the data again – are there any 2 trials we can use for comparison in which [BrO3

-(aq)] changes and [HSO3

-] remains constant.

Reaction: 2 BrO3-(aq) + 5 HSO3

-(aq) = Br2(g) + 5 SO4

2-(aq) +

H2O(l) + 3 H+(aq)

trial Initial [BrO3-(aq) ]

(mmol/L)Initial

[HSO3-(aq)](mmol/L)

Initial rate of Br2(g) production (mmol/L∙s)

1 2.0 3.0 0.20

2 2.0 6.0 0.80

3 4.0 6.0 1.60

Determine m and n

Let’s choose trials 2 and 3. [BrO3-(aq)] changes

and [HSO3-] remains constant.

As the concentration of [BrO3-(aq)] is doubled

(x2), the rate of reaction also doubles (x2) (2 = 21)

This is a first-order reaction, m = 1

Determine m and n

Determine how the rate of reaction changes due to changes in concentration of HSO3

- ([HSO3-(aq) ]).

To do this, we must keep the concentration of BrO3-

constant ([BrO3-(aq)]) – it is a controlled variable. Any

changes in r must be entirely due to HSO3-(aq).

Let’s look at the data again – are there any 2 trials we can use for comparison in which [HSO3

-] changes and [BrO3-(aq)]

remains constant.

Determine m and n

Let’s choose trials 1 and 2. [HSO3-] changes and

[BrO3-(aq)] remains constant.

As the concentration of [HSO3-] is doubled (x2),

the rate of reaction multiplies by 4 (x4) (4 = 22)This is a second-order reaction, m = 2Rewrite the Rate Law Equation;

r = k[BrO3-(aq)]1[HSO3

-(aq)]2

Now let’s determine k(the Rate Law constant)Since r = k[BrO3

-(aq)]1[HSO3

-(aq)]2

Then k = r

[BrO3-(aq)]1[HSO3

-(aq)]2

Go back to the data and choose any trial. I will use trial 1. Input the r, [BrO3

-(aq)], and [HS.O3

-(aq)]

values.

Therefore:k = 0.20 mmol/L·s[2.0 mmol/L]1[3.0 mmol/L]2

k = 0.011 L2/mmol2·s Now: r = 0.011 L2/mmol2· s[BrO3

-(aq)]1[HSO3

-

(aq)]2 See page 376 for tips on the units for k

ApplicationsNow that we have a completed rate Law

Equation:

r = 0.011 L2/mmol2· s[BrO3-

(aq)]1[HSO3-(aq)]2

We can input any concentrations of the reactants and determine a rate of reaction.

Try This: What is the rate of reaction if: [BrO3

-(aq)] = 0.10 mmol/L and [HSO3

-(aq)]=

0.10 mmol/L?

Review of Order of Reaction

Review of Order of Reaction

Determine the Rate Law Equation for these sets of

dataReaction: 2 NO(g) + Br2(g) → 2 NOBr(g)

trial Initial [NO(g) ](mol/L) Initial [Br2(g)](mol/L) Initial rate of NOBr(g) production (mol/L∙s)

1 0.10 0.10 0.040

2 0.10 0.20 0.080

3 0.20 0.20 0.320

BrO3-(aq)  +  5Br-

(aq)  +  6H+(aq)

  →  3 Br2(g)  +  3H2O(l)

trial

[BrO3-] (mol/L) [Br-] (mol/L) [H+] (mol/L) Initial Rate

(mol/L·s)

1 0.10 0.10 0.10 8.03 x 10 -4

2 0.20 0.10 0.10 1.62 x 10 -3

3 0.20 0.20 0.10 3.17 x 10 -3

4 0.10 0.10 0.20 3.22 x 10 -3

Experiment

[BrO3-] (M) [Br-] (M) [H+] (M) Initial Rate

(M/s)

1 0.10 0.10 0.10 8.0 x 10 -4

2 0.20 0.10 0.10 1.6 x 10 -3

3 0.20 0.20 0.10 3.2 x 10 -3

4 0.10 0.10 0.20 3.2 x 10 -3