f325 rate of reaction

rate = change in concentration of reactant or product time taken for the change to take place

the UNITS of rate are therefore moles per dm 3 second

which we would write as mol dm-3 s-1

F325 Kinetics – Rate of reaction

More about reaction rates

We want to know about the rate of reactions- in order to be able to control them- because rate change gives us insights into how reactions actually work

We have already met collision theory- reactants must collide with more energy than the activation energy, EA, for a reaction to take place- the proportion of molecules that exceed EA can be changed by altering the temperature, or the presence of a catalyst

We will define the rate of a reaction as:

Rate graphs and orders explain and use the terms: rate of reaction, order, rate constant, half-life, rate-determining step deduce, from a concentration–time graph, the rate of a reaction and the half-life of a first-order

reaction (concentration–time can be plotted from continuous measurements taken during the course of a reaction – this is continuous monitoring)

state that the half-life of a first-order reaction is independent of the concentration deduce, from a rate–concentration graph, the order (0, 1 or 2) with respect to a reactant determine, using the initial rates method, the order (0, 1 or 2) with respect to a reactant (initial

rates require separate experiments using different concentrations of one of the reactants, clock reactions are an approximation of this method)

Rate equations, rate constants deduce, from orders, a rate equation of the form: rate = k[A]m[B]n, for which m and n are 0, 1 or 2

(integrated forms of rate equations are not required) calculate the rate constant, k, from a rate equation explain qualitatively the effect of temperature change on a rate constant and hence the rate of a

reaction

Rate-determining step for a multi-step reaction:

(i) propose a rate equation that is consistent with the rate-determining step(ii) propose steps in a reaction mechanism from the rate equation and the

balanced equation for the overall reaction


We may sometimes measure another physical quantity (such as volume of a gas) and use that instead of concentration – so we could for example end up using units like cm3 s-1

, but the common factor is that rates are always measured in something per unit time.

Determining the rate of a reactionTypically we'd plot a graph of concentration of a reactant or product (y-axis) vs. time (x-axis). On such a graph the gradient of the curve at any time, t, is the rate of the reaction at that time.

We can calculate (approximate) the gradient by drawing a tangent to the curve and calculating change in y-axis divided by change in x-axis. In particular, this method allows us to determine the rate at time = 0, which we call the initial rate. Note: we don't worry about the sign ! The "changes in" are both taken as positive.

Constructing a graph of this kind requires us to take regular measurements as a reaction proceeds. We could:

- measure pH with a probe/by titrating samples during acid/base reactions- measure volume of gas produced or loss in mass of reactants for a reaction which produces a gas- measure changes in colour (e.g. absorption of a specific wavelength as a coloured species is used or formed) or opacity (if a precipitate is formed).

Rate with respect to what ?Depending on which reactant or product we choose to measure, we may get a different value for the rate for the same reaction. This is to do with MOLE RATIOS.

Consider H2 + I2 2HILets say we measure the rate of disappearance of I2 and find that it is 1.0 mol dm-3 s-1. The equation tells us that one mole of I2 makes 2 moles of HI, so the rate of appearance of HI will be 2.0 mol dm-3 s-1.


- the rate of disappearance of a reactant will not necessarily be the same as the rate of appearance of a product; you've got to check the mole ratios

- when we talk about "the rate of a reaction" we need to be specific; the rate with respect to which reactant or product ? We should talk about the rate of formation of HI or the rate of disappearance of I2 or whatever.

How each reactant affects the rateEach individual reacting species (including any catalysts) contributes to overall rate of the reaction - the way in which each affects the rate gives us valuable information about how the reaction works. We determine the effect each reactant has on the rate experimentally. It cannot be deduced from the chemical equation.

We call the effect the concentration of a reactant has on the rate of a reaction the ORDER with respect to that reactant.

For a reactant A: rate [A]x

where x is 0, 1 or 2, called the ORDER with respect to A(if you haven't met it before, means "proportional to")

Zero orderIf the order is 0 with respect to A, then rate is independent of (not affected by) the concentration of A. As A is used up the rate doesn't change.

We show this as rate [A]0

The shape of a concentration vs. time graph for a reactant where the rate is zero-order with respect to that reactant will be a straight line – the concentration of A will decrease steadily; the rate (gradient) not changing as A is used up.

Since rate does not depend on A, if the concentration of A is doubled, the rate would be the same.

First orderIf the order is 1 with respect to A, then rate [A]1 i.e. rate [A]As time progresses the concentration of A decreases rapidly at first, then more slowly as A is used up.Since rate is proportional to the concentration of A, if the concentration of A is doubled, the rate increases x2.


Second orderIf the order is 2 with respect to A, then rate [A]2. As time progresses the concentration of A decreases very rapidly at first, and the change in rate is more dramatic than for first order.

Since rate depends on the concentration of A squared, if concentration of A increases x2, the rate increases x4.

Half-lifeIt is easy to spot when a reaction is zero-order with respect to the reactant you are following, but a little harder to tell first order from second order. Measuring the half-life allows them to be told apart.

The half life is the time taken for the concentration to reach half of it previous value. This can be read off the graph.

For a first-order reaction, successive half-lives will be equal – e.g. if the concentration halved after 30 seconds, it would have halved again after a further 30 seconds. If a reaction has a constant half-life with respect to one reactant it is first order with respect to that reactant.

For a second-order reaction the half-life increases as the reaction progresses. For a first order reaction the half-life is constant as the reaction progresses. For a zero order reaction the half-life decreases as the reaction progresses.

Overall order of a reactionThe overall order of a reaction is the sum of the orders for each of the reactants.

For example, a reaction is zero-order with respect to A, first order with respect to B and second order with respect to C.

The overall order of the reaction = 0 + 1 + 2 = 3. This reaction is third-order overall.

Rate constant and Rate EquationRather than keep using signs all the time we can write a rate equation:

For the example above:rate [A]0 rate [B]1 rate [C]2

The rate equation is rate = k [A]0[B]1[C]2


Because anything to the power 0 = 1, and anything to the power 1 = itself, we can simplify this:

The rate equation is rate = k [B][C]2

Here k is called the rate constant and the overall expression is called the rate equation. The rate constant is different for every reaction, and varies with temperature, so the temperature must be stated.

Value and Units of the rate constantOnce we know the rate equation we can measure the initial rate of reaction that we get with known concentrations of the reactants. We can then put these concentrations into the rate equation to determine the value of the rate constant.

e.g. The reaction between H2 and NO has the following rate equation:

rate = k [H2][NO]2

When 6.0x10-3 mol dm-3 H2 was reacted with 3.0 x 10-3 mol dm-3 NO, the initial rate was measured to be 4.5x10-3 moldm-3s-1. Calculate the value of the rate constant.

rearranging, k = rate / [H2][NO]2

= 4.5 x 10-3 / (6.0x10-3 x (3.0x10-3)2)

= 8.3 x 10-4

Units of the rate constantWe will be expected to write the correct units for the rate constant as well as calculating a value for it. To get the units we need to remember that units of rate are mol dm-3 s-1 and the units of concentrations are mol dm-3. We can put the units into the rate equation and cancel units just like simplifying an expression in maths:

e.g. for first order overall:rate = k[A] so k = rate/[A] so units of k = mol dm-3 s-1

/ mol dm-3 = s-1

zero order overall: k = rate / [A]0 Units: mol dm-3 s-1

second order overall: e.g. k = rate / [A][B] Units: dm3 mol-1 s-1

third order overall: k = rate / [A][B]2 Units: dm3 mol-2 s-1

So in the example above, k = rate / [H2][NO]2


k = mol dm -3 s -1 = mol-2dm6s-1

[mol dm-3][mol dm-3]2

Initial Rates MethodIt isn't always possible to measure the concentration of every reacting species and track how it changes as the reaction proceeds. So long as we can measure the change in concentration of ONE reactant or product over time, we can determine the order for ALL reacting species by a series of experiments, and hence determine the rate equation and value of the rate constant.

What we do is choose suitable concentrations for all the reacting species, determine the initial rate of the reaction (gradient at t = 0), then repeat the experiment changing the concentration of one reacting species at a time and keeping the others fixed, and see what that does to the rate.

e.g. 2NO(g) + O2(g) 2NO2(g)

Experiment [NO] in mol dm-3 [O2] in mol dm-3 Initial rate in mol dm-3 s-1

1 0.00100 0.00100 1.82 x 10-6

2 0.00100 0.00200 3.64 x 10-6 3 0.0020 0.0010 7.28 x 10-6

Comparing experiment 1 and 2 we can see that doubling the concentration of O2 and keeping the concentration of NO the same has doubled the rate of the reaction. This means that the reaction is first order with respect to O2.

Comparing experiment 1 and 3 we can see that doubling the concentration of NO while keeping the concentration of O2 the same has increased the rate x 4. This means the reaction is second order with respect to NO.

The rate equation is therefore rate = k [O2][NO]2

The value of the rate constant can now be found by rearranging the equation and putting any one set of experiment data into the equation. Here we've used Experiment 2 data:

k = 3.64 x 10-6 / (0.002 x 0.0012) = 1,820 mol-2dm6s-1

Using graphical dataUsing a comparison of just two rates to decide on the order can be unreliable, so we fix the concentrations of other reactants and then try a range of concentrations for each in turn, determining the initial rate in each case. We would then plot a graph of how the initial rate (y-axis) varied as we changed the concentration. The shape of this graph tells us the order with respect to that reactant:


Imagine a reaction with rate equation rate = k[A]0[B]1[C]2

Zero order:

As the concentration of A changes, the rate is unaffected.

First order:

When the concentration of B is doubled, the rate also doubles.

Second order:

When the concentration of C is doubled, the rate is quadrupled.

More about the rate constantThe value of k in a rate equation will be different under different reaction conditions such as temperature. The rate constant tells us how fast the reaction will be: high values of k mean fast reactions, low values of k mean slow reactions.

We should be able to relate this to our AS ideas about temperature and the Boltzmann distribution:


The rate equation depends on k and on the concentrations of reactants. If the concentration of the reactants is kept the same but the temperature increases, then the rate increases. This means that the value of k must have increased.

The higher the temperature, the larger the value of k becomes

For many reactions the value of k doubles for every 10°C temp rise.It is worth noting that the value of k also depends on the activation energy for the reaction. Lowering the activation energy increases the value of k. The Arrhenius Equation relates all of these factors:

k = Ae(-Ea/RT) (A and R constants)

Determining Reaction MechanismsReactions occur in a sequence of steps (a mechanism). The products from one step are often the reactants for subsequent steps.

Some steps in the mechanism will happen slowly and others will be fast. One step that is slower than all the others will determine the overall rate of reaction – the overall rate can't be faster than the slowest step. We call this step the RATE DETERMINING STEP (RDS).

Any reactants which are involved after the rate determining step won’t have any effect on the overall rate, so won't be part of the rate equation. We can propose that what may be reacting in the rate determining step are the substances which are shown in the rate equation. The order with respect to each of those reactants tells us how many molecules of that substance are involved:

Example 1: When we know the rate equationNO2(g) + CO(g) NO(g) + CO2(g)

Seems straight forward – collision of NO2 with CO… until we do the experiments

…and find that the rate equation is rate = [NO2]2

This tells us two NO2 collide in the Rate Determining Step.

EVIDENCE for the reaction proceeding by more than one step:1) The rate equation has different numbers of reacting particles (two NO2 molecules) compared to the overall chemical equation (one CO and one NO2) so


there must be other subsequent reaction steps. We say the overall equation and the rate equation have DIFFERENT STOCHIOMETRY.

2) There is a reactant in the chemical equation (CO) which is not in the rate equation, so it must be used in another step after the RDS

To suggest a sequence of steps in the mechanism, write all the information you have about each step. This will include the reactants in the RDS, any reactants which don't figure in the RDS, and the final products:

in the RDS, two molecules of NO2 are the reactantsNO2 + NO2 ? + ? STEP 1 SLOW (RDS)

CO2 is a product, and it can't be made in the first step so it must be made subsequently. CO is a reactant and we haven't used it in the first step so it must react here

? + CO ? + CO2 STEP 2 FAST

the overall equation only has one molecule of NO2 reacting, so one must be created as a product subsequently

NO2 + NO2 ? + ? STEP 1 SLOW (RDS) ? + CO NO2 + CO2 STEP 2 FAST

we figure the rest out by filling the gaps. - the overall equation creates NO, which isn't yet created in either of our steps- balancing the second equation suggests NO3 as the other reactant

so our proposed steps are:NO2 + NO2 NO + NO3 SLOW: RDSNO3 + CO NO2 + CO2 FAST

Example 2: Proposing a rate equation, given a mechanismOverall equation: 4HBr(g) + O2(g) 2H2O(g) + 2Br2(g)

Step 1: HBr + O2 HBrO2 SLOW: RDSSteps 2+: FAST

We would propose that the rate depends on the concentrations of the reactants in the RDS. In this case it is one HBr molecule reacting with one O2 molecule, so we'd suggest that the rate equation is rate = k[HBr][O2] (which we'd verify by experimentation).


Example 3: When unexpected species appear in the rate equation H+ catalyst

CH3COCH3(aq) + I2(aq) CH2ICOCH3(aq) + HI(aq)

But when we do the experimentation we find the rate equation is:

rate = k[CH3COCH3][H+] (zero-order with respect to I2)

Explanation:From the rate equation we can see that I2 is not involved in the RDS, so this is evidence that the mechanism has more than one step, and the one involving the iodine happens after the RDS.

We would propose a rate-determining step in which H+ reacts with the propanone. We need to remember that H+ is a catalyst, so we'll need to regenerate it in a subsequent step. We can have the catalyst in the rate equation. The rate can depend on the concentration of the catalyst because it IS involved in the reaction (i.e. it is a reactant which is regenerated). We can propose the following:

Step 1: CH3COCH3 + H+ ? SLOW, RDSSteps 2+: ? + I2 CH2ICOCH3 + HI + H+ FAST

and it is clear that [CH3COHCH3]+ must be the intermediate.

In fact the mechanism is as displayed here, showing it is consistent with the evidence we have obtained.


Writing an overall equationWe may be asked to combine the steps of a mechanism to show the overall equation. Combining the steps simply requires adding all the steps together and cancelling any species found on both sides of the reaction.

e.g. here's the mechanism for the oxidation of hydrogen bromide:Step 1: HBr + O2 HBrO2 SLOWStep 2: HBrO2 + HBr 2 HBrO FASTStep 3: HBrO + HBr H2O + Br2 FASTStep 4: HBrO + HBr H2O + Br2 FAST

Overall: 4HBr(g) + O2(g) 2H2O(g) + 2Br2(g)

Spotting a catalystGiven a sequence of equations for the steps in a mechanism, we may be asked to state which substance is the catalyst.

A catalyst will be used (be a reactant) in one step AND will be regenerated (be a product) in a later step. This means that overall it is "not used up". It will not show up in the overall equation as a reactant or a product.

e.g.CH3COCH3 + H+ [CH3COHCH3]+ SLOWCH3COHCH3]+ CH2=C(CH3)OH + H+ FASTCH2=C(CH3)OH + I2 CH2ICI(CH3)OH FASTCH2ICI(CH3)OH CH2ICOCH3 + HI FAST

Overall: CH3COCH3 + I2 CH2ICOCH3 + HI

Identify the catalyst: The catalyst is H+. Justify your answer: It is used in the first step, and regenerated in the second step. It does not appear in the overall equation.

f325 rate of reaction

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