Rate Constant and Orders of Reaction
Rate of reaction and orders
The rate of reaction tells us how fast a reaction is happening. We can calculate the rate of reaction by dividing the change in reactant or product by the time taken.
You can work out the rate from a graph by drawing a tangent and calculating its gradient. If you wanted to work out the initial rate of reaction (the rate right at the beginning of the reaction), you need to draw a tangent at t=0.
Orders tell us how much the concentration of a reactant affects its rate.
A reactant is zero-order if doubling its concentration has no effect on the reaction rate
A reactant is first-order if doubling its concentration also doubles the reaction rate – the concentration of the reactant and the rate of reaction are directly proportional
A reactant is second-order if doubling its concentration quadruples (2 x 2) the reaction rate. Tripling the concentration would increase the reaction rate nine times (3 x 3).
Worked example – deducing orders of reaction
Determine the orders of the reactants for the reaction between ozone and ethene.
Comparing experiments 1 and 2, the concentration of ozone quadruples, while the concentration of ethene stays the same. The reaction rate also quadruples, so the reaction is first order with respect to ozone.
Comparing experiments 2 and 3, the concentrations of both ozone and ethene doubles, causing the rate of reaction to quadruple. We already know that ozone is first order, so doubling this reactant should cause the rate to also double. This means that the reaction is also first order with respect to ethene since the reaction rate has doubled twice.
Overall orders and rate-concentration graphs
Reactions have overall orders – this is the sum of all the individual orders of the reactants. In the worked example above, where the two reactants were both first-order, this gives us an overall order of 2. In another reaction, where one reactant is second-order, another is first-order and the final reactant is zero-order, there would be an overall order of 2 + 1 + 0 = 3.
You can use rate-concentration graphs to work out the order of individual reactants. So long as you remember the different shapes of the lines produced in these graphs, you’ll know the order of a particular reactant just by looking at it.
A horizontal line means that increasing the concentration of the reactant results in no change to the reaction rate. The reaction is zero-order.
A straight line through the origin indicates a first-order reaction since the concentration of the reactant and reaction rate are directly proportional.
A curved upwards line indicates that the reaction is second-order since the rate is proportional to the squared product of the reactant concentration.
Rate equation
Once we know the orders of each reactant, we can write the rate equation. The rate equation tells us how the rate is affected by the concentrations of the reactants.
For the general reaction: A + B -> C + D, the rate equation is:
Rate = k [A]x[B]x
Where k is the rate constant and x is the order of reaction with respect to reactants A and B.
Let’s consider the example we saw before, where ozone and ethane react to form methanal and oxygen. Both of the reactants are first order, so the rate equation would look like this:
Rate = k [O3] [C2H4]
You could put a little number 1 after each of these reactants but it’s not needed. Anything written as [X] is equal to [X] x 1.
Let’s consider a more difficult example. Let’s say you have a reaction between iodine, hydrogen peroxide and hydrogen ions. You have worked out that the reaction is first order for iodine, second order for hydrogen peroxide and zero order for hydrogen ions. Our rate equation would be:
Rate = k [I2] [H2O2]2
We don’t need to include a number 1 after the iodine because its presence in the rate equation will indicate that it is first order. We write a little 2 after the hydrogen peroxide to show that it is second order. Anything that is zero order, in this case the hydrogen ions, are left out of the rate equation completely.
Concentration-time graph and half-life
First order reactants can be identified by concentration-time graphs because they have a constant half-life.
Half-life is the time taken for half of the reactant to be used up. For a reactant that is first order, its half-life is independent of the concentration, making each half-life the same length. In other words, the half-life is constant.
Look at the graph below to see how you can read off a concentration-time graph to find out whether the reactant is first order.
The concentration of bromine halves every 100 seconds – i.e. it takes 100 s to drop from 5 mol dm-3 to 2.5 mol dm-3, then another 100 s to drop from 2.5 mol dm-3 to 1.25 mol dm-3.
If you know the half-life of a first order reaction, you can use the equation below to calculate the rate constant, k.
In this instance, the units for k are s-1.
Rate-determining step
The rate-determining step refers to the slowest step in a reaction mechanism — it determines the rate of the reaction.
Anything that is involved in the rate-determining step will be present in the rate equation. This means that if we know the rate equation for a reaction, we can predict the rate-determining step.
Consider the reaction between hydrogen and iodine monochloride to form iodine and hydrogen chloride, as shown in the equation below.
Suppose you have carried out experiments to determine that the reaction is first order with respect to both hydrogen and iodine monochloride. That gives us the following rate equation:
Rate = k [H2] [ICl]
The rate equation tells us that one molecule of hydrogen and one molecule of ICl are involved in the slow (rate-determining step), since they are both first order. If ICl was second order it would mean that 2 molecules of ICl are present in the slow step.
We can predict the slow step as:
From looking at the overall equation, we know that another ICl molecule still needs to react. We also need to form another molecule of HCl and we need to get rid of HI, as this is not present in the overall equation. That means the fast step must be:
We can also go the other way and predict the rate equation from the rate-determining step. Consider the reaction between nitrogen dioxide and carbon monoxide, as shown in the equation below.
We know that anything in the slow step is present in our rate equation – in this case it is two molecules of nitrogen dioxide. Since there are two molecules involved, the reaction must be second order with respect to NO2. Therefore, the rate equation is:
Rate = k [NO2]2
Effect of temperature on rate constants
The rate constant, k, gives a numerical value to indicate how fast a reaction is happening and is only true for a reaction happening at a particular temperature. As soon as you change the temperature, you change the rate constant because you have changed the rate of reaction.
Remember that increasing temperature speeds up the rate of reaction by increasing the frequency of collisions (by providing molecules with more kinetic energy) and also increases the proportion of successful collisions (by increasing the number of particles that exceed the activation energy). This means that increasing temperature increases the rate constant, k.
The Arrhenius equation
The Arrhenius equation links the rate constant, k, with activation energy (Ea) and temperature (T).
The Arrhenius equation shows that as activation energy increases, the rate constant decreases. This makes sense – the larger the activation energy, the fewer successful collisions and the slower the rate of reaction. The equation also shows that as temperature increases, the rate constant also increases. This also makes sense, since we know that reactions happen faster at higher temperatures.
We can use the Arrhenius equation to calculate activation energy from a graph. First, we convert the equation into a logarithmic form, as shown below:
The graph should have ln K on the y-axis and 1/T on the x-axis. The gradient of the line is equal to –Ea/R and the intercept on the y-axis is equal to ln A.
Once we know the gradient and the y-axis intercept, we can pop these into the equation to work out the activation energy, as shown in the image below: