Understanding the rate law and orders of reaction

The reaction between hydroxide ions and bromoethane involves a collision between a molecule of each of the reactants, which means that the rate of the collisions is directly proportional to their concentration.

CH₃CH₂Br + OH^– ⇾ CH₃CH₂OH + Br^–

rate ∝ [OH^–] [CH₃CH₂Br]

We can write a rate law for the reaction:

rate = k [OH^–] [CH₃CH₂Br]

where k is the rate constant.

The value of the rate constant, k, is found experimentally and its value depends on the size of the molecules, their speed and the fraction of collisions which have both the correct orientation and sufficient energy to overcome the activation enthalpy, E_a, for the reaction. You can remind yourself about all these factors here.

We can determine the rate of the reaction by measuring how the concentration of either the reactants (CH₃CH₂Br and OH^–) or products (CH₃CH₂OH and Br^–) changes over time.

We can write a rate equation for any chemical reaction once we have found out experimentally how the rate depends on the concentration of the reactants – you can’t predict how the rate depends on the concentration of a reactant from the balanced / stoichiometric equation.

e.g. CH₃Br + OH^– ⇾ CH₃OH + Br^–

experiments show that the rate depends on the concentration of both reactants, [CH₃Br] and [OH^–]

the reaction is said to be first order with respect to CH₃Br, first order with respect to OH^– and second order overall
although each concentration is raised to the power of the order of that reactant in the rate equation, we don’t usually put in the ‘1’ in so this rate equation would look like

rate = k [CH₃Br] [OH^–]

e.g. BrO₃^– + 5Br^– + 6H⁺ ⇾ 3Br₂ + 3H₂O

experiments show that the rate equation for this reaction is

rate = k [BrO₃^–] [Br^–] [H⁺]²

the reaction is first order with respect to the bromate(V) ion, first order with respect to the bromide ion and second order with respect to the hydrogen ion. It is fourth order overall.

e.g. CH₃CH₂I + OH^– ⇾ C₂H₄ + I^– + H₂O

experiments show that the rate equation for this reaction is

rate = k [CH₃CH₂I]

the reaction is first order with respect to iodoethane, zero order with respect to the hydroxide ion and first order overall
this tells us that the rate of the reaction does not depend on the [OH^–]. The mechanism for this reaction has several steps and although the hydroxide ion is involved at some stage (it is a reactant after all), it is not involved in the key rate determining step – more on this later!

Essentially, if the reaction is first order with respect to a reactant, then doubling the concentration of that reactant will double the rate of the reaction.

The gradient of the line of best fit (Δ initial rate / Δ concentration) gives us a value for the rate constant, k.

What exactly do we have to do to get the experimental results that will allow us to plot a graph of concentration of reactant against initial rate of reaction?

react hydroxide ions with bromomethane and measure the decrease in [OH^–] as the reaction proceeds (see the post on experimental methods for more details)
plot a graph of [OH^–] vs. time
determine the initial rate of reaction from the gradient of a tangent drawn at t=0
repeat the experiment four more times, each time changing the starting [OH^–], plotting a graph for each experiment and determining the initial rate of reaction for each [OH^–]
plot a graph of initial rate of reaction (mol dm^-3 s^-1) against [OH^–] (mol dm^-3)

Kinetics experiments are never quick!

If a reaction is second order with respect to a reactant, doubling the concentration of that reactant leads to a quadrupling of the rate of reaction.

If a reaction is zero order with respect to a reactant, then doubling the concentration of that reactant has NO effect on the rate of the reaction

The following practice questions allow you to work through all of this material using a set of experimental results (the best way to get your head around it) and are an excellent opportunity to perfect your graph-drawing skills 😀.

Practice questions

2,4,6-trinitrobenzoic acid in solution loses carbon dioxide when heated:

The solution of 2,4,6-trinitrobenzoic acid was heated to 90°C. At certain time intervals, a sample was removed and added to a large volume of iced water to quench (stop) the reaction. The sample was then titrated with barium hydroxide using bromothymol blue as an indicator, and the concentration of 2,4,6-trinitrobenzoic acid remaining was determined.

Time / min	Concentration of 2,4,6-trinitrobenzoic acid / mol dm^-3
0	2.77 x 10^-4
18	2.32 x 10^-4
31	2.05 x 10^-4
55	1.59 x 10^-4
79	1.26 x 10^-4
157	0.58 x 10^-4

Plot a graph of concentration of 2,4,6-trinitrobenzoic acid against time. Use a large piece of graph paper as you will need to draw tangents to the curve later and you will need plenty of working space for this.
Using this graph, draw tangents to the curve at 10, 50, 100 and 150 minutes. Calculate their gradients and complete the table below:

Time / min	Concentration / mol dm^-3	Gradient / mol dm^-3 min^-1	Rate / mol dm^-3 min^-1
10
50
100
150

Plot a graph of rate of reaction against concentration of 2,4,6-trinitrobenzoic acid using your values from above.

(a) Explain why the line of best fit passes through the origin.

(b) Use your graph to figure out the relationship between the rate of reaction and the concentration of reactant – you can express this as the rate equation for the reaction.

Determine the gradient of the graph of rate of reaction vs. concentration of 2,4,6-trinitrobenzoic acid – this is the value of k, the rate constant.

Answers

To make life simpler the y-axis scale is 0-3 and the label indicates that each point is actually (x 10^-4) mol dm^-3. Just remember this when calculating gradients later on! You should always draw the graph so that it fills the graph paper given and a smooth line of best fit is drawn through the points.

2. Your values for gradient / rate will vary slightly compared with mine.

Time / min	Concentration / mol dm^-3	Gradient / mol dm^-3 min^-1	Rate / mol dm^-3 min^-1
10	2.5 x 10^-4	– 2.44 x 10^-6	2.44 x 10^-6
50	1.68 x 10^-4	– 1.64 x 10^-6	1.64 x 10^-6
100	1.04 x 10^-4	– 1.02 x 10^-6	1.02 x 10^-6
150	0.63 x 10^-4	– 0.635 x 10^-6	0.635 x 10^-6