The reaction A + B —–> C + D has an initial rate of 0.0790 M/s.
rate = k [A] [B] ^2
a) What will the initial rate (in m/s ) be if [A] is halved and [B] is tripled?
b) What will the initial rate (in m/s ) be if [A] is tripled and [B] is halved?
2 Answers
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Let’s try this. Assume that the initial concentrations of each reactant are both 0.5 M. Then, from the rate law and the initial rate, you can calculate k as:
rate = k(0.5)(0.5)^2 = 0.0790
k = 0.632 M^-2s^-1
Now, if you have [A], it would be 0.25M and triple [B] to 1.5M. The rate then would be:
rate = 0.632 (0.25)(1.5)^2 = 0.356 M/s
Just to make sure this works, let’s assume that the initial concentrations of both were 0.25 M. In that instance, k would be:
0.079 = k (0.25)(0.25)^2
k = 5.056
And if you halve A and triple B, you would have
rate = 5.056 (0.125)(0.75)^2 = 0.356 M/s
So, under the first set of concentrations, if you triple [A] and halve [B],
rate = 0.632 (1.5)(.25)2 = 0.0593 M/s
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0.0790 M/s = k * [A][B]^2
a):
rate = k * 0.5[A] * (3*[B])^2
rate = k * 0.5[A] * 9[B]^2
rate = 4.5 * k * [A] * [B]^2
rate = 4.5 * 0.0790 = 3.56 M/s (3 sig figs)
b):
rate = k * 3[A] * (0.5[B])^2
rate = k * 3[A] * 0.25[B]^2
rate = 0.75 * k * [A] *[B]^2
rate = 0.75 * 0.790 = 0.0593 M/s (3 sig figs)
Source(s): ChemE
