Rate Laws Chapter 14 part II Rate Laws Chemical reactions are reversible. So far we have only considered the forward reaction in our rates. Eventually.

Rate Laws

Chapter 14 part II

Rate Laws Chemical reactions

are reversible. So far we have only

considered the forward reaction in our rates.

Eventually the rate is the difference between the forward & reverse rates.

2NO2(g) -> 2NO(g) + O2(g)

2NO(g) + O2(g) -> 2NO2(g)

(-∆[NO2]/∆t) -(∆[NO]/∆t

Rate1 = -∆[NO2]/∆t

Rate2 = ∆[NO]/∆t

How to deal with that? For our purposes, we

will concentrate on reactions that in which the reverse rate will be insignificant. Specifically when the [products] is very low to zero.

Chemical Kinetics

0.0000

0.0020

0.0040

0.0060

0.0080

0.0100

0.0120

0 200 400 600

Time (seconds)

Con

cn

etr

ati

on

(m

ol/

Lit

er)

Concentration NO2Concentration NOConcentration O2

Integrated Rate Law

• Rate1 = -∆[NO2]/∆t

Rate2 = ∆[NO]/∆t

The Integrated Rate Law expresses how the concentration depends on time.

Differential Rate Law Differential rate law

expresses how the rate depends on concentration.

The differential rate law is also known as just Rate Law.

Rate = k[NO2]n

k is the rate constant

n is called the order of the reaction

The order of a reaction must be determined experimentally.

Rate = k[NO2]n= -∆[NO2]/∆t

Rate may be defined in different ways.

Defining rates 2NO2(g) -> 2NO(g)+O2(g)

In the integrated rate law, the rate can be defined using any of the species in the equation, product or reactant.

However, one must be careful. In this case it takes 2 NO2 to make one O2.

Defining Rate Laws

If Rate1 = -∆[NO2]/∆t = k[NO2]n

Then Rate2 = ∆[O2]/∆t = k’ [NO2]n

Since 2NO2 molecules are consumed for every O2 molecule produced,

Rate = 2 x Rate’ Or k[NO2]n = 2 k’ [NO2]n

Therefore k = 2 x k’ The value of the rate constant depends on how the

rate is defined.

Two important points The value of the

exponent n must be determined experimentally; it cannot be written from the balanced equation.

The concentration of the products do not appear in the rate law because the reaction is being studied under conditions where the reverse reaction does not significantly contribute to the overall rate.

Summary There are two types of rate laws:

Intergrated & differential

We consider rates at where the reverse reaction is insignificant so rate laws involve only [reactants].

Because both laws for a reaction are well defined, the experimental determination of either rate law is sufficient.

Finally: From the rate law, one may infer the mechanism

of a reaction.

Determining the Form First one needs to

determine the form of the rate law

We need to look at how the rate changes as the concentration changes.

Consider dinitrogen pentaoxide in carbon tetrachloride:

2N2O5(soln) -> 4NO2(soln) + O2(g)

2N2O5(soln) -> 4NO2(soln) + O2(g)

time (s) N2O5 (M)0 1.00

200 0.88400 0.78600 0.69800 0.61

1000 0.541200 0.481400 0.431600 0.381800 0.342000 0.30

Concentration vs Time

0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 500 1000 1500 2000 2500

Time (s)

[N2

O5

] (M

)

Rate Laws: First Order First determine how

the rate changes as the concentration changes.

We find that as the concentration is halved, the rate is halved.

This is the definition of a First Order RXN.

[N2O5] Rate M/s

0.90 M 5.4 x 10-4

0.45 2.7 x 10-4

Rate =( - ∆[N2O5]/∆t) =

k[N2O5]1 = k[N2O5]

First Order Reactions

Rate = - ∆[A]/∆t = k[A] Where A is the concentration of the reactant.

Method of Initial Rates

The initial rate is where the reaction has just started, just after t=0.

Several experiments are carried out using different concentrations and the initial rate is determined for each run.

The rates are then compared to the concentrations to see how the rate depends on the initial concentration.

For Example: NH4

+(aq)+NO2-(aq)-> N2(g)+2H2O(l)

# [NH4+]M [NO2

-]M Rate M/s

1. 0.100 0.0050 1.35x10-7

2. 0.100 0.010 2.70x10-7

3. 0.200 0.010 5.40x10-7

The general for of the rate law is…

Rate=-∆[NH4+]/∆t

=k[NH4+]n[NO2

-]m We can determine

the values of m & n experimentally

See: Rate2/Rate1

Rate2 =2.70x10-7M/s=k(0.100M)n(0.010M)m

Rate1 1.35x10-7M/s k(0.100M)n(0.005M)m

Determining a Rate Law

# [NH4+]M [NO2

-]M Rate M/s

1. 0.100 0.0050 1.35x10-7

2. 0.100 0.010 2.70x10-7

3. 0.200 0.010 5.40x10-7

Rate2 =2.70x10-7M/s=k(0.100M)n(0.010M)m

Rate1 1.35x10-7M/s k(0.100M)n(0.005M)m

2 =(0.010M)m = (2.0)m m=1

(0.0050M)m

NH4+(aq)+NO2

-(aq)-> N2(g)+2H2O(l)

Since m=1 that means the rate law is first order for [NO2

-] Now find the value of

n. Since both n & m are

1, the rate law is written

k [NH4+][NO2

-]

# [NH4+]M [NO2

-]M Rate M/s

1. 0.100 0.0050 1.35x10-7

2. 0.100 0.010 2.70x10-7

3. 0.200 0.010 5.40x10-7

Overall Rate Since the sum of the

orders in this rate law is two, n&m=2, then the rate law is said to have an overall reaction order of 2

How do you find the value for k?

Chose one experiment and add in your known values to solve for k.

What is k? 2.7x10-4 l/mol•s

Examples . Calculator problem - Cyclopropane rearranges to form

propene: CH2CH2CH2 --> CH2=CHCH3 by first-order kinetics. The rate constant is k = 2.74 x

10-3 s-1. The initial concentration of cyclopropane is 0.290 M. What will be the concentration of cyclopropane after 100 seconds?

a) 0.220 Mb) 0.760 Mc) 2.74 x 10-1 Md) 7.94 x 10-2 M

Answer: A