InterviewSolution
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InterviewSolution
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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. |
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Answer» 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). |
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