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The Effect of Concentration on Rate

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1 The Effect of Concentration on Rate
Chemical Ideas 10.3 The Effect of Concentration on Rate

2 A quantity measured with respect to another measured quantity.
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 m/s %/year when taking about rate you MUST be clear about units being used

3 rate at which products are converted to reactants
rate of reaction

4 EACH SECOND!!! 0.0001 mol O2 formed 0.0002 mol H2O formed
mol H2O2 used up EACH SECOND!!!

5 Measuring the rate of a reaction.
measure the change in amount of a reactant or product in a certain time Decide on a property of reactant or product that you can measure. Measure the change in property over a certain time Find the rate change of property time

6

7 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 0.16 0.21 0.08 0.10

8 rate = constant x [H2O2(aq)]
This graph shows us that rate is directly proportional to the concentration of hydrogen peroxide rate = constant x [H2O2(aq)] rate  [H2O2(aq)]

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

10 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.

11 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

12 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

13 For the reaction 2Br (g)  Br2 (g)
Rate equation is rate = k [Br]2 S2O82-(aq) + 2I- (aq)  SO42- (aq) + I2 (aq) rate = k [S2O82-(aq) ] [I- (aq) ] you cannot predict the rate equation for a reaction from it’s balanced equation

14 rate = k [BrO3-] [Br-] [H+]2
BrO3-(aq) + 5Br-(aq) + 6H+ (aq)  5H2O (l) + 3Br2 (aq) rate = k [BrO3-] [Br-] [H+]2 you cannot predict the rate equation for a reaction from it’s balanced equation

15 Half Lifes

16

17 Chemical Ideas 10.3 (again)
Knowing how concentration affects rate can tell us something about the way reactions occur.

18 Remember … rate = k [A]x [B]y [C]z
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

19 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

20 Time taken for half of a reactant to get used up in the reaction
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

21 zero order & second order reactions do not have this feature
For a first order reaction the half-life is always constant no matter what the starting amount!

22 Finding the Order of a Reaction
you cannot predict the rate equation for a reaction from it’s balanced equation 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.

23 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?)

24 Initial Rates Method – drawing tangents
most used Several experimental runs are completed (as in activity EP6.3). Initial rate is calculated by drawing tangents at the origin. We then plot initial rate against concentration

25 producing graphs first order zero order second order second order [A]
rate rate [A] [A] second order second order rate rate [A] [A]2

26 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.

27 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.

28 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.

29 first order w.r.t. (CH3)3CBr
+ OH- first order w.r.t. (CH3)3CBr CH3 zero order w.r.t. OH- CH3 C OH + Br- CH3 rate = k[(CH3)3CBr]

30 slow Ξfast + + CH3 Br C CH3 C+ Br- CH3 C+ CH3 C OH OH- step one
step two OH- + Ξfast

31 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]

32 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.


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