Determination of Rate Equations & the Rate Determining Step
Determination of Rate Equations & the Rate Determining Step Revision
Determination of Rate Equations & the Rate Determining Step
Rate equations are determined experimentally by investigating the effect of changing concentrations and then deducing the order.
Using Initial Rate to Calculate Reaction Variables
![](https://mmerevise.co.uk/app/uploads/2022/10/log-rate-graph-1012x1024.png)
![](https://mmerevise.co.uk/app/uploads/2022/10/log-rate-graph-1012x1024.png)
Rate equations have a general form:
\text{Rate} =k[X]^n
To work out the value of n graphically, we need to take logarithms of both sides of the equation to convert it to the form y = mx + c.
\text{log Rate}=\text{log k}+\text{n log [X]}
A graph of this equation would produce a straight line, where the \text{y-intercept} is equal to \text{log k} allowing us to calculate the rate constant \text{K}.
\text{K}=10^{\text{ y-intercept}}
The gradient of the line is equal to the order of reaction, \text{n}, with respect to X.
The Rate Determining Step
The slowest step in a reaction is known as the rate-determining step (RDS). It is rate-determining because it controls the overall rate of a reaction.
The rate equation contains species up to and including the rate-determining step. Therefore, we can use the rate equation to determine the rate-determining step. Take the reaction
\text{A} + 3\text{B} + 2\text{C} \rarr\text{D}
A suggested mechanism for this reaction is
Step 1: \text{A} + 2\text{B}\rarr\text{X}
Step 2: \text{B} + 2\text{C}\rarr\text{Y}
Step 3: \text{X} + \text{Y}\rarr\text{D}
The rate equation for the reaction is
\text{Rate} = \text{k[A][B]}^2
The rate equation tells us that there is 1 mole of A and 2 moles of B up to and including the rate-determining step. Step 1 contains both 1 mole of A, 2 moles of B, Step 2 contains only 1 mole of B and none of A, and Step 3 contains no moles of either species.
Step 3 cannot be the RDS as it does not contain either of the species contained in the rate equation. step 2 contains 1 mole of B, but taking both steps 1 and 2 together gives 3 moles of B in total. As the rate equation contains only 2 moles of B step 3 can’t be the RDS. Step 1 contains the correct number of both moles according to the rate equation and so must be the rate determining step.
Determining Reaction Orders From Experimental Data.
By carrying out a series of experiments using different concentrations of reactants, we can determine the rate equation of a reaction. To do this we need to compare the effects of the different concentrations on the initial rate.
Experiment | Initial Concentration of X /\text{mol dm}^{-3} | Initial Concentration of Y /\text{mol dm}^{-3} | Initial Rate /\text{mol dm}^{-3}\text{ s}^{-1} |
1 | 0.12 | 0.26 | 2.10\times10^{-4} |
2 | 0.36 | 0.26 | 1.98\times10^{-3} |
3 | 0.72 | 0.13 | 3.78\times10^{-3} |
[5 marks]
We can start by comparing the rates of experiments 1 and 2. Although the concentration of Y remains constant, the initial rate changes. That immediately tells us that the order of X is greater than 0 and is part of the rate equation.
From experiments 1 to 2, the concentration of X triples while the initial rate increases by a factor of 9. This means that the order of X must be 2 (you can work out the amount the initial rate increases if you divide the rates).
\text{Rate} = \text{k[X]}^2
We can now look at experiments 2 and 3 to find the effect of Y. Firstly, we have to consider the effect of X. Between experiments 2 and 3, the concentration of X doubles. We know that X is second order so from experiments 2 to 3, the initial rate would be expected to quadruple from 1.8\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} to 7.56\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}.
Since that is not the initial rate of the reaction, Y must have an order greater than 0. From experiments 2 to 3, the concentration of Y halves, meanwhile, the rate is half the expected value of 7.56\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1}, so Y must be first order.
\text{Rate} = \text{k[X]}^2\text{[Y]}
To calculate the value for \text{k}, the rate equation needs to be rearranged to make \text{k} the subject, and then the known values can be substituted into the equation.
For example, if we use the rate equation above, \text{Rate} = k [X]^2[Y], we can find \text{k} using the equation k=\frac{\text{Rate}}{[X]^2[Y]}.
Required Practical
Initial Rate Experiments – The Iodine Clock
The initial rate of a reaction can be calculated by carrying out a clock reaction. In clock reaction, a system is designed so that it can be monitored over a set period of time or until one product has been produced in a specific concentration.
Method
In this common clock reaction, hydrogen peroxide (\text{H}_2\text{O}_2), iodide ions (\text{I}^-), thiosulfate ions (\text{S}_2\text{O}_3^{2-}) and starch are mixed together. The following reactions take place:
\text{H}_2\text{O}_{\text{2(l)}} + 2\text{H}^+_{\text{(aq)}} + 2\text{I}^{-}_{\text{(aq)}} \rarr \text{I}_{\text{2 (aq)}}
2\text{S}_2\text{O}_3^{2-}\text{}_{\text{(aq)}}+\text{I}_{\text{2(aq)}}\rarr2\text{I}^-_{\text{(aq)}}+\text{S}_4\text{O}_6^{2-}\text{}_{\text{(aq)}}
- Fill a burette (50\text{ cm}^3) with potassium iodide solution. Rinse the 50\text{ cm}^3 burette with potassium iodide beforehand to improve accuracy.
- Put 10.0\text{ cm}^3 of hydrogen peroxide into a clean and dry measuring cylinder.
- Use a measuring cylinder to add 25\text{ cm}^3 of sulfuric acid to a second 250\text{ cm}^3 beaker. Then, add 20\text{ cm}^3 of deionised water into the beaker.
- Use a plastic dropping pipette to add about 1\text{ cm}^3 of starch solution to the 250\text{ cm}^3 beaker.
- Use the burette to add 5.0\text{ cm}^3 of potassium iodide solution to the mixture in the 250\text{ cm}^3 beaker.
- Finally add 5.0\text{ cm}^3 of sodium thiosulfate solution to the mixture in the 250\text{ cm}^3 beaker.
- Stir the mixture in the 250\text{ cm}^3 beaker and pour the 10.0\text{ cm}^3 hydrogen peroxide solution from the measuring cylinder into the 250\text{ cm}^3 beaker and immediately start the timer.
- Stop the timer when the mixture in the 250\text{ cm}^3 beaker turns blue-black and record the time.
- Repeat steps 1-8 at least two more times, changing the potassium iodide concentration each time. This is to allow the order of reaction to be determined.
- Plot a graph of initial rate versus concentration to determine the order.
![](https://mmerevise.co.uk/app/uploads/2022/10/Iodine-Clock-e1665761761633.png)
Determination of Rate Equations & the Rate Determining Step Example Questions
Question 1: The overall equation for a reaction is as follows.
\text{NO}_{2\text{(g)}} + \text{CO}_{\text{(g)}} \rarr \text{NO}_{\text{(g)}} + \text{CO}_{2\text{(g)}}
Below is the mechanism for the reaction.
Step 1:
Step 2:
\text{NO}_3+\text{CO}\rarr\text{NO}_2+\text{CO}_2
Step 1 is the slowest step in the reaction. Determine the rate equation for the reaction. Explain your answer.
[2 marks]
\text{Rate} = \text{k[NO}_2\text{]}^2.
\text{NO}_2 appears twice in the rate-determining step while \text{CO} does not appear at all.
Question 2: When placed in a dilute alkaline solution, ethanal undergoes the following reaction.
2\text{CH}_3\text{CHO}\rarr \text{CH}_3\text{CH(OH)CH}_2\text{CHO}
Below is the rate equation of the reaction.
\text{Rate} = \text{k}\left[\text{CH}_3\text{CHO}\right]\left[\text{OH}^\text{-}\right]
The initial rate of the reaction is found to be 2.5\times10^{-3}\text{ mol dm}^{-3}\text{ s}^{-1} when the initial concentration of ethanal is 0.15\text{ mol dm}^{-3}, and the initial concentration of sodium hydroxide is 0.030\text{ mol dm}^{-3}.
Calculate the value for the rate constant at this temperature and give its units.
[3 marks]
\begin{aligned}\text{k}&=\frac{2.5\times10^{-3}}{0.15\times0.03}\\ &=0.555\end{aligned}
Units:\text{ mol}^{-1}\text{dm}^{3}\text{ s}^{-1}
Question 3: The data in the following table was obtained by carrying out a series of experiments to find the initial rate of reaction between substances X and Y.
Experiment | Initial Concentration of X /\text{mol dm}^{-3} | Initial Concentration of Y /\text{mol dm}^{-3} | Initial Rate /\text{mol dm}^{-3}\text{ s}^{-1} |
1 | 0.14 | 0.24 | 2.60\times10^{-4} |
2 | 0.42 | 0.24 | 2.34\times10^{-3} |
3 | 0.84 | 0.48 | 9.36\times10^{-3} |
Deduce the orders of the reaction with respect to X and Y.
[2 marks]
Order with respect to X: 2 or \text{[X]}^2
Order with respect to Y: 0 or \text{[Y]}^0
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