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Social ScienceFor 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?
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) |
| 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
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
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.
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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