Intitial rate method: Use the following initial rate data determine the form of the rate law expressoin for the reaction `3X+2Yrarr2U+V`

Strategy: The rate law is of the form
`Rate = k[X]^(a)[Y]^(b)`
No two experiments have the same in initial [Y], thus, use the alternative method discussed earlier to evalaute a and b.

Explanation

The initial concentration of X is the same in expermental 1 and 3, thus we divide the third rate law expression by the corresponding terms in the first one:
`(Rate_(3))/(Rate_(1))= (k[X]_(3)^(a)[Y]_(3)^(b))/(k[X]_(1)^(a)[Y]_(1)^(b))`
The value k cancels, and so do the concentrations of X because they are equal. Consequently, the expression simplifies to
`(Rate_(3))/(Rate_(1))= (([Y]_(3))/([Y]_(1)))^(b)`
Subsituting known values of rates and [Y],
`(1.20xx10^(-2)Ms^(-1))/(6.00xx10^(-3)Ms^(-1))= ((2.00xx10^(-2)M)/(1.00xx10^(-2)M))^(b)`
`2.0 = (2.0)^(b)`
or `b = 1`
Thus, the reaction is first order in Y. No two of the expermental runs have the same concentrations of Y, so we must proceed somewhat differntly. Let us compare expermints 1 and 2. The observed change in rate must be due to the ombination of the change in [X] and [Y]. We can divide the second rate law expression by the correponding terms in the first one. Canceling the equal k values and collecting terms, we get
`(Rate_(2))/(Rate_(1)) = (k[X]_(2)^(a)[Y]_(2)^(b))/(k[X]_(1)^(a)[Y]_(1)^(b)) = (([X]_(2))/([X]_(1)))^(a) (([Y]_(2))/([Y]_(1)))^(b)`
Inserting the known the known values for rates and concentratoins and the known [Y] exponent of 1, we have.
`(1.44xx10^(-1)Ms^(-1))/(6.00xx10^(-3)Ms^(-1))= ((2.00xx10(-2)M)/(1.00xx10^(-2)M))^(a) ((3.00xx10^(-2)M)/(1.00xx10^(-2)M))^(1)`
`24.0= (2.00)^(a) (3.00)`
`800=(2.00)^(a)`
`(2.00)^(3)= (2.00)`
Thus, `a=3`
and the reaction is third order in X. The complete rate law expression has the from
`Rate = k[X]^(3)[Y]`