Question

The initial rates for a reaction A and B were studied using the following initial concentrations in mol/L. Expt 1. [A]=0.300, [B]=0.200, rate=7.17x10-5 M/s Expt 2. [A]=0.664, [B]=0.200, rate=1.59x10-4 M/s Expt 3. [A]=0.664, [B]=0.221, rate=1.76x10-4 M/s What is the value of the rate constant?

Answer 1

k = 1.20 x 10⁻⁵

Step-by-step explanation:

The rate of reaction is given by

rate = k x ( molar conc A raised to x + molar conc B raided to y)

rate= k A^x B^y

taking the first and second equations given in the problem and dividing side by side we have

7.17 x 10⁻⁵ / 1.59 x 10 x ⁻⁴ = k 0.300^x 0.200 ^y/ k 0.664 ^x 0.200 ^y

0.45 = (0.300/0.664) = .45

( note: the concentration of A was almost doubled and the rate almost double hence the order is 1)

there fore the coefficient for the molar concentration A is one

taking the second and third equations and dividing again side by side we have:

1.59 x 10⁻⁴/ 1.76 x 10⁻⁴ = k 0.664 x 0.200/ 0.664 x 0.221

0.90 = 0.90

again the coefficient is one and we have then

rate = k x molar conc A x molar conc B

( note: here the concentrations varied by a proportion of 1.10 and the rate rates varied by the same amount. The rate dependence is one)

we can slve for k from any of the equations given. For example taking the first one

k = 7.17 x 10⁻⁵/(0.300 x 0.200) = 1.20 x 10⁻⁵