For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:Trial[A] (M)[B] (M)[C] (M)Initial rate(M/s)10.500.500.501.5×10−420.500.501.504.5×10−431.000.500.506.0×10−441.001.000.506.0×10−4What is the value of the rate constant k for this reaction?

Asked Jun 4, 2019

For the reaction A+B+C→D+E, the initial reaction rate was measured for various initial concentrations of reactants. The following data were collected:

Trial [A] 
Initial rate
1 0.50 0.50 0.50 1.5×10−4
2 0.50 0.50 1.50 4.5×10−4
3 1.00 0.50 0.50 6.0×10−4
4 1.00 1.00 0.50 6.0×10−4

What is the value of the rate constant k for this reaction?


Expert Answer

Step 1

The order of product [A] we first look for two trails where [A] and the inital rate changes, and [B] and [C] remain the same: these will be trial one and three.

 This is zero order reaction based on the units.

Then we calculate by how much one concentration of product [A] is bigger than the other:

Trial          A (M)

  1            0.50
  3            1.00

Step 2

Here, we observe that Trial 3 is twice as big than Trial 1. 

we repeat this procedure for the initial rates of Trial 1 and 3:

Trial        Initial Rate (M/s)
  1           1.5 x 10-4
  3           6.0 x 10-4

Here, we see that Trial 3 is four times bigger than Trial 1. With these results we divide four by two and we obtain the final order of [A] which is 2. This happens because every time we double the concentration of A the initial rate will be four times bigger than the original.

Step 3

To find the rest of the orders we repeat this process where in order to find [B], [A] and [C] would not have to change and for [C] the concentrations of [A] and [B] must be the same.

These operations yield the following order:

r = K [A]2 [C]

Where [B] is omitted since it has an order of zero because the initi...

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