Answer:
k = 1.20 x 10⁻⁵
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
