4.1 Rates of Reactions

Reaction Rate = change in amount of reactants/products per unit time (units: mol dm^-3s^-1)

Following a reaction;

gas volume produced (gas syringe)
mass lost (balance)
colour change (colorimeter)
clock reaction (sudden change at particular time means specific concentration of product has been reached – the shorter the time taken, the faster the rate)
electrical conductivity (number of ions will change as reaction occurs)

Concentration-Time Graph

Rate at any point can be found by drawing a tangent at that point on the graph and finding the gradient.

Orders of Reaction

The order of reaction = how the reactants concentration affects the rate

INCREASE REACTANT – RATE STAYS THE SAME – ORDER OF 0

INCREASE REACTANT – RATE INCREASES BY 1 FACTOR – ORDER OF 1

INCREASE REACTANT – RATE INCREASES BY 2 FACTORS – ORDER OF 2

You can only find the order of a reaction *experimentally* – there is NO theoretical order system.

*square brackets indicate concentration. For example [X] = concentration of X.

Half-life = time taken for half the reactant to react

If the half life is constant = first order

If the half life is doubling = second order

You can also calculate the half life using reaction rates. For example, if you’re given the rate constant (see below) and the order you can work out half life (you don’t need to know how, just to be aware of it)

Rate Equations

Rate equation = tell you how the rate is affected by the concentrations of reactants.

E.G. Rate = k[A]^m[B]ⁿ

Where:

m = order of A

n = order of B

n+m = overall order

k = rate constant (always the same for a reaction at specific temp and pressure, increase temp = increase k = bigger value of k = faster reaction)

EXAMPLE

Propanone + Iodine —> Iodopropanone + H⁺ + I^– (reaction occurs in acid)

Info: First order with respect to propanone and H⁺ and zero order with respect to iodine

Rate equation = k[propanone]¹[H⁺]¹[iodine]⁰

Simplify to;

Rate equation = k[propanone][H⁺] (because anything to the power of 0 is 1)

How to calculate rate constant from the orders and rate?

Rearrange to make k the subject and calculate.

Units of k can be found as you know concentration is moldm^-3 and rate is moldm^-3s^-1 using a normal “cancelling” method.

Using data to deduce the order

The experiment: titrate sample solutions against sodium thiosulfate and starch to work out the concentration of the iodine. Repeat experiment, changing only the concentration for ONE REACTANT at a time.

experiment	1	2	3	4	5	6	7
[propanone]	0.4	0.8	1.2	0.4	0.4	0.4	0.4
[iodine]	0.002	0.002	0.002	0.004	0.006	0.002	0.002
[H⁺]	0.4	0.4	0.4	0.4	0.4	0.8	1.2

Here, first we changed the concentration of propanone for experiments 1, 2 and 3.

Then, we changed the concentration of iodine in experiments 4 and 5.

Lastly, we changed concentration of H⁺ in experiments 6 and 7.

From this table we can plot 7 Concentration-Time graphs. Finding the gradient at time zero for each of these plots will give us the INITIAL rate of each.
Compare the results e.g.

Experiment	Change compared to experiment 1	Rate of reaction	Change
1	—	0.033	—
2	[propanone] doubled	0.062	Rate doubled
3	[propanone] trebled	0.092	Rate trebled
4	[iodine] doubled	0.034	No change
5	[iodine] trebled	0.032	No change
6	[H⁺] doubled	0.058	Rate doubled
7	[H⁺] doubled	0.094	Rate trebled

*Reaction rates won’t be exactly double or treble due to experimental errors etc.

4) Now we can work out the rate equation:

Rate is proportional to [propanone] so the reaction is of order 1 with respect to propanone.
Rate does not change/is independent of [iodine] so the reaction is of order 0 with respect to iodine.
Rate is proportional to [H⁺] so the reaction is of order 1 with respect to [H⁺].

Rate determining step = slowest step in a multi-step reaction

(if a reactant appears in the rate equation it MUST be a rate determining step including catalysts which may appear in a rate equation)

PREDICITIONS

The order of a reaction with respect to a reactant shows the number of molecules that the reactant is involved in with regard to the rate-determining step. EXAMPLE: rate = k[X][Y]². Here, one molecule of X and 2 molecules of Y will be involved in the rate determining step.

Chlorine free radicals in the ozone consist of 2 steps:

Cl•(g) + O₃(g) —> ClO•(g) + O₂(g) slow rate determining step

ClO•(g) + O•(g) —> Cl•(g) + O₂(g) fast reaction

Therefore, Cl• and O₃ must be in the rate equation as they are the reactants from the slowest step.

RATE = k[Cl•][O₃]

Predicting Mechanisms:

Once you know what the rate determining reactants are, you can think about what reaction mechanism it follows.

EXAMPLE:

If the rate equation is: rate = k[X][Y]

And the two different mechanisms are:

X + Y —> Z

X —> Y + Z

From the rate equation, we know that X and Y MUST be in the rate determining step, therefore, it’s mechanism 1 which is the right one.

Halogenoalkanes – Nucleophilic Substitution (S_N)

Halogenoalkanes can be hydrolysed by OH^–ions by nucleophilic substitution. This is where a nucleophile (e.g. :OH^–) attacks a molecule and is swapped/substituted for one of the attached groups (e.g. Br^δ-). In this case the Carbon (C^δ+) to Halogen (X^δ-) bond is POLAR as halogens are much more electronegative than the carbon so they draw in electrons making the Carbon slightly/delta positive. The bond looks like this:

C^δ+— X^δ-

Thus, the carbon can be easily attacked by a nucleophile who likes positive areas. This mechanism occurs:

*C-Br bond breaks heterolytically (unevenly)

Primary – react by SN2 where 2 molecules/ions are involved in the rate determining step
Secondary – react by SN1 and SN2
Tertiary – react by SN1 where 1 molecule/ion is involved in the rate determining step

You can see by the rate equation if there are 1 or 2 molecules in the rate determining step, which in turn, tells you if the mechanism is SN1 or SN2.

EXAMPLE:

Rate = k[X][Y] = 2 molecules in rate determining step = SN2 = primary/secondary halogenoalkane

Rate = k[X] = 1 molecule in rate determining step = SN1 = tertiary/secondary halogenoalkane

Activation Energy

We can calculate the activation energy using the Arrhenius equation:

Where; (you don’t have to learn this, just understand the relationship)

k = rate constant E_A = activation energy (J)

T = temperature (K) R = gas constant (8.31 JK^-1mol^-1)

A = another constant

Some relationships to note:

As E_A increases, k will get smaller. Therefore large activation energy, means a slow rate – this makes sense!
As T increases, k increases. Therefore at high temperatures, rate will be quicker – this makes sense too!

If we “ln” both sides of Arrhenius’ equation, we get;

ln k = – E_A/RT + ln A

(don’t forget, ln A is just a constant, a number)

This looks a bit like:

y = mx + c

If we plot ln k (y) against 1/T (x), the gradient we produce will be –E_A/R (m). Then R is just a number that we know (8.31 JK^-1mol^-1) we can rearrange and find the activation energy.

EXAMPLE:

Iodine clock reaction

S₂O₈^2- (aq) + 2I^– (aq) —> 2SO₄^2-_(aq)+ I₂ (aq)

Rate of reaction is inversely proportional to the time taken for the solution to change colour

i.e. increased rate = decreased time taken

k α 1/t

We can say that 1/t is the same as k (rate constant) and we can substitute 1/t instead of k in Arrhenius’ equation and find the gradient again to find a value for E_A.

Catalysts

Catalyst = increases rate of a reaction by providing an alternative reaction pathway with a LOWER activation energy (E_A). A catalyst will be chemically unchanged at the end of a reaction.

Adv: Small amount needed to catalyse a lot of reactions, also they are remade, thus reusable.

Disadv: High specificity to the reactions they catalyse.

There are two types of catalysts:

HOMOGENOUS CATALYSTS

HETEROGENOUS CATALYSTS

These are catalysts in the same state as the reactants.

E.G. when enzymes catalyse reactions in your body, all reactants are aqueous, this is a homogenous catalysis.

These are catalysts in different physical states to the reactants.

They are easily separated from products – GOOD

Can be poisoned (i.e. a substance clings to a catalyst stronger than the reactant would, preventing reaction speeding up) example: sulphur in the Haber process is a “poison”– BAD

Solid catalysts provide a large surface area for the reaction to occur e.g. mesh/powder

E.G. vanadium pentoxide in the contact process to make sulphuric acid