Method of intial rates: Given the following data, determine the rate law expression and the value of the rate constant for the reaction. `2X+Y+ZrarrU+V`

Strategy: The rate law is of the from
`"Rate" = k[X]^(a)[Y]^(b)[Z]^(c )`
We must evaluate a,b,c, and k by using the reasoning outlined earlier. Thus, to determine order with respect to X, we must select those experiments in which concentration of X changes but the concentrations of Y and Z are kept onstant and so on. Note that the coefficient of U in the balenced equation is 1, so the rate of reaction is equal to the rate of formation of U.

Explanation

Dependence on [Y] : In experiments 1 and 3, the intial concentrations of X and Z are unchanged. Thus, any change in the rate would be due to the change in concentration of Y. But we observethat the rate is the same in experiments 1 and 3, even though the concentration of Y is different. Thus, the reaction rate is independendent of `[Y]`, thus `b = 0([Y]^(@) =1)`. The rate law must be
`"Rate" = k[X]^(a)[Z]^(c )`
We can neglet changes in [Y] in the subsequent reasoning. Dependence on [Z]: Experiments 1 and 4 involve the same initial concentration of X, thus the observed [Z]. Thus, we must compare experiments 1 and 4 to find Z. [Z] has been multiplied by a factor of `(0.60M)/(0.20M) = 3.0 = [Z]` ratio
Consequently, the rate changes by a factor of
`(7.2xx10^(-6)Ma^(-1))/(2.4xx10^(-6)Ms^(-1)) = 3.0` = rate ratio
The exponent C can be reduced from
rate ratio `= ([Z] ratio)^(c )`
`(3.0) = (3.0)^(c )`
thus `c=1`
i.e., the reaction is first order in Z. Now we know that the rate law is of the form .
Rate `= k[X]^(a)[Z]`
Dependence on [X]: We use experiments 1 and 2 to evaluate a, because [X] is changed, [Y] does not matter, and [Z] is unaltered. The observed rate change is due only to changed [X]. [X] has been multiplied by a factor of
`(0.40M)/(0.20M)= 2.0= [X]` ratio
Consequently, the rate changes by a factor of
`(9.6xx10^(-6)Ms^(-1))/(2.4xx10^(-6)Ms(-1))= 4.0` = rate ratio
The exponent a can be deduced from
rate ratio `= ([X]ratio)^(a)`
`4.0 = (2.0)^(a)`
`(2.0)^(2)=(2.)^(a)`
thus `a=2`
i.e., the reaction is second order in X
From these results, we can write the complete rate - law expression
Rate `= k[X]^(2)[Y]^(@)[Z]^(1)`
or Rate `= k[X]^(2)[Z]^(1)`
To evalute the specific reaction rate , k, we can substitute any of the four sets of data into the rate-law expression we have just derived. Data from experiment 4 give
`Rate_(4)= k[X]_(4)^(2)[Z]_(4)`
`k = (Rate_(4))/([X]_(4)^(2)[Z]_(4))`
`= (7.2x10^(-6)Ms(-1))/((0.20M)^(2)(0.60M))`
`= 300xx10^(-6)M^(-2)s^(-1)`
`= 3.0xx10^(-4)M^(-2)s^(-1)`
The rate-law expression can also be written with the value of k incorporated.
`Rate = 3.0xx10^(-4)M^(-2)s^(-1)[X]^(2)[Z]`
This expression allows us to calculate the reaction rate at which this reaction occurs with any known concentrations of X and Z (provided some Y is present).