chemical ideas 10.3 the effect of concentration on rate

Chemical Ideas 10.3

The Effect of Concentration on Rate

rate Listen: [ rāt ]n.

A quantity measured with respect to another measured quantity.

speed = rate of change of distance

inflation = rate of change of

prices

when taking about rate you MUST be clear about units being

used

m/s %/year

rate of reaction

rate at which products are converted to

reactants

0.0001 mol O2 formed

0.0002 mol H2O formed

0.0002 mol H2O2 used up

EACH SECOND!!!

Measuring the rate of a reaction.

1. Decide on a property of reactant or product that you can measure.

2. Measure the change in property over a certain time

3. Find the rate

measure the change in amount of a reactant or product in a certain

time

change of property

time

concentration of H2O2 at start / mol

dm3

Initial rate / (cm3 of O2(g))s-1

0.40 0.51

0.32 0.41

0.24 0.32

0.16 0.21

0.08 0.10

This graph shows us that rate is directly proportional to the

concentration of hydrogen peroxide

rate [H2O2(aq)]rate = constant x

[H2O2(aq)]

The concentration of the enzyme catalase also affects the rate of the reaction …

rate = constant x [catalase]We can combine the two equations to

get …rate = constant x [H2O2(aq)] x [catalase]

rate = k [H2O2(aq)] [catalase]

• This is the rate equation for the reaction

• the constant k is called the rate constant

•k varies with temperature, therefore you must always state the temperature at which measurements are made.

rate = k [H2O2(aq)] [catalase]

Order of a Reaction• For a reaction in which A & B are

reactants …–A + B products

• The general rate equation is…– rate = k [A]m [B]n

• m and n are powers to which the concentration must be raised. usually have values of 0, 1 or 2.

• m & n are called the order of the reaction

decomposition of hydrogen peroxide

rate = k [H2O2(aq)] [catalase]• The reaction is first order with respect to

H2O2

• The reaction is also first order with respect to catalase.

• The overall order of a reaction is given by (m + n).

• the reaction is overall second order

• For the reaction 2Br (g) Br2 (g)

• Rate equation is – rate = k [Br]2

• S2O82-

(aq) + 2I- (aq) SO4

2- (aq) + I2 (aq)

– rate = k [S2O82-

(aq) ] [I- (aq) ]

you cannot cannot predict the rate equation for a reaction from it’s balanced

equation

BrO3-(aq) + 5Br-

(aq) + 6H+ (aq)

5H2O (l) + 3Br2 (aq)

rate = k [BrO3-] [Br-] [H+]2

you cannot cannot predict the rate equation for a reaction from it’s balanced

equation

Half Lifes

0.0

20.0

40.0

60.0

80.0

100.0

120.0

140.0

160.0

180.0

200.0

220.0

240.0

0 20 40 60 80 100 120 140 160 180

time / s

Chemical Ideas 10.3 (again)

Knowing how concentration affects rate can tell us something about the way reactions occur.

Remember …The rate of any reaction can be

expressed in terms of the concentrations of its reactants

rate = k [A]x [B]y [C]z

x,y & z are the order of the reaction with respect to that reaction.

If they =1 the number is not shown

decomposition of hydrogen peroxide

rate = k [H2O2(aq)] [catalase]• The reaction is first order with respect to

H2O2

• The reaction is also first order with respect to catalase.

• The overall order of a reaction is given by (m + n).

• the reaction is overall second order

half-lives (t ½ )Reactions which are first

order will show a curve that is identical to radioactive decay!

Time taken for half of a reactant to get used up

in the reaction

For a first order reaction the half-life is always constant no matter what the starting amount!

zero order & second order reactions do not have this

feature

Finding the Order of a Reaction

• To find out the order of a reaction it is necessary to carry out practical experiments.

• The data can then be used to determine the order of the reaction.

you cannot cannot predict the rate equation for a reaction from it’s balanced equation

Progress Curve Method

• Rate is calculated by drawing tangents to the curve at various points

• Can then find the order with respect to a reactant/product

• tedious & inaccurate (unless using a PC?)

Initial Rates Method – drawing tangents

• most used• Several

experimental runs are completed (as in activity EP6.3).

0.00

0.10

0.20

0.30

0.40

0.50

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40

concentr ation

• Initial rate is calculated by drawing tangents at the origin.

• We then plot initial rate against concentration

producing graphs

[A]

[A] [A]2

[A]

rate

raterate

rate

zero order first order

second order second order

Initial Rates Method – reciprocal of time

• Measuring how long to produce a small fixed amount of one of the products.

• Time taken is called the reaction time.• Rate is high – reaction time small• Rate low – reaction time large.

• Average rate 1/t .

• Graph of 1/t against concentration.

half-lives method

• You can use the progress curve to determine half-lives for the reaction.

• If they are constant then the reaction is first order.

rate equations & rate mechanisms

when we know the rate equation we can link it to the reaction mechanism.We can then work out the rate determining step.

CH3

CH3

CH3

BrC + OH-

CH3

CH3

CH3

Br-C OH +

first order w.r.t. (CH3)3CBr

zero order w.r.t. OH-

rate = k[(CH3)3CBr]

CH3

CH3

CH3

BrC +

OH-CH3

CH3

CH3

C OH+

step one CH3

CH3

CH3

C+ Br-

CH3

CH3

CH3

C+step two

Ξfast

slow

mechanism of enzyme catalysed reactions …

• when the substrate concentration is low for the reaction – rate = k[E][S] ([E] is concentration of enzyme)

• we can deduce from this that the rate determining step involves one enzyme molecule & one substrate molecule.

• Following steps are faster.

• Substrate concentration high then– rate = k [E]

WHY? – why are some steps slow & others fast?

• One reason = different steps have different activation enthalpies.– Large activation enthalpy, only a small

number of molecules pass over it each second so rate of reaction is slow.

– Small activation enthalpy, greater proportion of molecules can pass each second, hence a faster rate.