Using Initial Rate Data (office mix)
By Will Paddock
Created 3 years ago
An example we would have done in class Friday, if not for the snow day.
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Slide 1 - SAMPLE EXERCISE 14.6 Determining a Rate Law from Initial Rate Data
- The initial rate of a reaction was measured for several different starting concentrations of A and B, and the results are as follows:
- Using these data, determine (a) the rate law for the reaction, (b) the magnitude of the rate constant, (c) the rate of the reaction when [A] = 0.050 M and [B] = 0.100 M.
- Analyze: We are given a table of data that relates concentrations of reactants with initial rates of reaction and asked to determine
- (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table.
- Plan: (a) We assume that the rate law has the following form: Rate = k[A]m[B]n, so we must use the given data to deduce the reaction orders m and n. We do so by determining how changes in the concentration change the rate. (b) Once we know m and n, we can use the rate law and one of the sets of data to determine the rate constant k. (c) Now that we know both the rate constant and the reaction orders, we can use the rate law with the given concentrations to calculate rate.
- Solve: (a) As we move from experiment 1 to experiment 2, [A] is held constant and [B] is doubled. Thus, this pair of experiments shows how [B] affects the rate, allowing us to deduce the order of the rate law with respect to B. Because the rate remains the same when [B] is doubled, the concentration of B has no effect on the reaction rate. The rate law is therefore zero order in B (that is, n = 0).
Slide 2 - SAMPLE EXERCISE 14.6 continued
- In experiments 1 and 3, [B] is held constant so these data show how [A] affects rate. Holding [B] constant while doubling [A] increases the rate fourfold. This result indicates that rate is proportional to [A]2 (that is, the reaction is second order in A). Hence, the rate law is
- This rate law could be reached in a more formal way by taking the ratio of the rates from two experiments:
- Using the rate law, we have
- 2n equals 1 under only one condition:
- We can deduce the value of m in a similar fashion
- Using the rate law gives
Slide 3 - SAMPLE EXERCISE 14.6 continued
- Because 2m = 4, we conclude that
- (b) Using the rate law and the data from experiment 1, we have
- (c) Using the rate law from part (a) and the rate constant from part (b), we have
- Because [B] is not part of the rate law, it is irrelevant to the rate, provided that there is at least some B present to react with A.
- Check: A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see
- if we can correctly calculate the rate. Using data from experiment 3, we have
- Thus, the rate law correctly reproduces the data, giving both the correct number and the correct
- units for the rate.
Slide 4 - SAMPLE EXERCISE 14.6 continued
- PRACTICE EXERCISE
- The following data were measured for the reaction of nitric oxide with hydrogen:
- Answers: (a) rate = k[NO]2[H2]; (b) k =1.2 M–2s–1; (c) rate = 4.5 10–4 M/s
- (a) Determine the rate law for this reaction. (b) Calculate the rate constant. (c) Calculate the rate when [NO] = 0.050 M and [H2] = 0.150 M.