Topic #24:Topic #24:The Rate ExpressionThe Rate Expression

EQ: How can we tell the difference between a EQ: How can we tell the difference between a second order reaction and a third order second order reaction and a third order reaction?reaction?

Rate LawRate Law Math expression to show how rate depends Math expression to show how rate depends

on concentrationon concentration

Rate = k[reactant 1]Rate = k[reactant 1]mm[reactant 2][reactant 2]nn…… m m and and nn are called are called reaction ordersreaction orders. Their . Their

sumsum is called the is called the overall reaction order.overall reaction order. k k is the is the rate constantrate constant. It is specific to a . It is specific to a

reaction at a certain temperature.reaction at a certain temperature.

Rate = k [A]m [B]n

• The exponents in a rate law must be determined by experiment. They are not derived from the stoichiometric coefficients in a chemical equation

• The values of exponents establish the order of a reaction for each species and overall order for the reaction

• The proportionality constant, k, is the rate constant and its value depends on the reaction, the temperature, and the presence or absence of a catalyst.

How do we measure the rate of a reaction?

Remember: the rate of disappearance equals the rate of appearance

So for: CH2O + O2 CO2 + H2O

Rate = - Δ[CH2O] = - Δ[O2] = Δ [CO2] = Δ [H2O] Δt Δt Δt Δt

(g)2(l)2(aq)22 OOH2OH2

dt

O]d[H

2

1

dt

]d[O

dt

]Od[H

2

1rate 2222

122 ]O[Hkrate

Chemical Equation showing stoichiometry:

General expressions for the rate:

Actual experimentally determined form of the rate law:

Sample Exercise 1:Sample Exercise 1:2NO + 2H2NO + 2H22 N N22 + 2H + 2H22O O is the reaction we’re is the reaction we’re

studying, this is the data found during our studying, this is the data found during our experimentationexperimentation

ExperimentNumber

Conc. of NO(M)

Conc. of H2

(M)Rate of N2 forming (Ms-1)

1 0.210 0.122 0.0339

2 0.210 0.244 0.0678

30.420 0.122 0.1356

2NO + 2H2NO + 2H22 N N22 + 2H + 2H22OO

Exp #1 and #2, [NO] is unchanged, [HExp #1 and #2, [NO] is unchanged, [H22] is doubled ] is doubled and this causes the rate to double (0.0678/0.0339 and this causes the rate to double (0.0678/0.0339 =2), the H=2), the H22’s rate order is 1.’s rate order is 1.

Exp #1 and #3, [HExp #1 and #3, [H22] is unchanged, [NO] is doubled ] is unchanged, [NO] is doubled and this causes the rate to quadruple and this causes the rate to quadruple (0.1356/0.0339 =4), NO’s rate order is 2.(0.1356/0.0339 =4), NO’s rate order is 2.

Rate = k [NO]Rate = k [NO]22[H[H22]]

The overall The overall rate orderrate order for this reaction is for this reaction is 33 (1+2)(1+2)

[NO] [H2] Rate

1 0.210 0.122 0.0339

2 0.210 0.244 0.0678

30.420 0.122 0.1356

Suppose…Suppose…

If when we ran exp #1 and #3 , the rate If when we ran exp #1 and #3 , the rate didn’t change, what rate law would we didn’t change, what rate law would we expect?expect?

Since it didn’t change when we doubled Since it didn’t change when we doubled [NO], the rate order is 0, meaning the [NO], the rate order is 0, meaning the rate doesn’t depend on the concentration rate doesn’t depend on the concentration of NO at all, so rate= k[Hof NO at all, so rate= k[H22] and the overall ] and the overall

reaction order is 1.reaction order is 1.

Sample Exercise 2:Sample Exercise 2: Use the kinetics data to write the rate law for the Use the kinetics data to write the rate law for the

reaction. What overall reaction order is this?reaction. What overall reaction order is this? 2NO + O2NO + O22 2NO2NO22

Exp # [NO] [O2] Rate forming NO2 (M/s)

1 0.015 0.015 0.048

2 0.030 0.015 0.192

3 0.015 0.030 0.096

4 0.030 0.030 0.384

exp #1 and exp #2, [Oexp #1 and exp #2, [O22] remained constant, ] remained constant,

where [NO] is doubled. The rate is where [NO] is doubled. The rate is quadrupled. The rate order for [NO] is 2.quadrupled. The rate order for [NO] is 2.

Exp #1 and exp #3, [NO] remained constant, Exp #1 and exp #3, [NO] remained constant, where [O2] is doubled. The rate is doubled. where [O2] is doubled. The rate is doubled. The rate order for [O2] is 1.The rate order for [O2] is 1.

Overall reaction orderOverall reaction order (1+2) = (1+2) = 33

Sample Exercise 3Sample Exercise 3Rate data for the reaction: Rate data for the reaction:

CHCH33Br + OHBr + OH-- CH CH33OH + BrOH + Br--

Exp # CH3Br OH- Rate of forming CH3OH

1 0.200 0.200 0.015

2 0.400 0.200 0.030

3 0.400 0.400 0.060

Use the data to find the experimental rate law.

Distinctions Between Rate And Distinctions Between Rate And The Rate Constant The Rate Constant kk

The The raterate of a reaction is the change in concentration with of a reaction is the change in concentration with time, whereas the time, whereas the rate constantrate constant is the proportionality constant is the proportionality constant relating reaction rate to the concentrations of reactants.relating reaction rate to the concentrations of reactants.

The rate constant remains The rate constant remains constantconstant throughout a reaction, throughout a reaction, regardless of the initial concentrations of the reactants.regardless of the initial concentrations of the reactants.

The rate and the rate constant have the same numerical The rate and the rate constant have the same numerical values values andand units only in units only in zerozero-order reactions.-order reactions.

For reaction orders other than zero, the rate and rate constant For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.reactants are 1 M. Even then, their units are different.

Method Of Initial Rates Summary

The effects of doubling one initial concentration:– For zero-order reactions, no effect on rate.– For first-order reactions, the rate doubles.– For second-order reactions, the rate quadruples.– For third-order reactions, the rate increases

eightfold.

The value of k for the reaction can be calculated.

Graphing conc. Vs timeGraphing conc. Vs time

Shows the effect of Shows the effect of reactants being used reactants being used up on the rate of up on the rate of reactionreaction

If constant = zero If constant = zero orderorder Decreasing Decreasing

concentration is not concentration is not affecting the rateaffecting the rate

Graphing first orderGraphing first order

If the reaction rate is If the reaction rate is halved when the halved when the concentration is concentration is halved, then the halved, then the reaction is first orderreaction is first order

Graphing second orderGraphing second order

If halving the If halving the concentration causes concentration causes the rate to decrease the rate to decrease by a factor of 4, the by a factor of 4, the reaction is second reaction is second orderorder

(1/2 )(1/2 )22 = ¼ = ¼

Rate vs. conc graphsRate vs. conc graphs

First orderFirst order

Conc. shows an exponential decrease, Conc. shows an exponential decrease, that is the time for the conc. to fall from that is the time for the conc. to fall from its initial value to half its initial value, is its initial value to half its initial value, is equal to the time required for it to fall equal to the time required for it to fall from half to one quarter to one eighth, from half to one quarter to one eighth, etc…etc…

This is known as half-life, t This is known as half-life, t ½½ Look at next slide for graphical Look at next slide for graphical

representationrepresentation

Radioactive decayRadioactive decay

First order exponential decayFirst order exponential decay Half life is important and can be found from a Half life is important and can be found from a

graph or the equationgraph or the equation t t ½ ½ = = ln2ln2

kk

Example: if the rate constant of a first order reaction Example: if the rate constant of a first order reaction is 0.005 s-1, then the half life will beis 0.005 s-1, then the half life will be

t t ½ ½ = = ln2 ln2 = 139 s= 139 s

0.0050.005Now you can try to do #7 on handout