The following reaction: X + Y --> Z was studied by measuring the [X] as a functi
ID: 1049842 • Letter: T
Question
The following reaction: X + Y --> Z was studied by measuring the [X] as a function of time. Reaction Y was always present in excess and it was assumed that the [Y] remained essentially constant during the reaction. Shown below is a table of the [X] as a function of time, in seconds. In the first experiment, [Y] = 2.0 M and in the second experiment, [Y] = 4.0 M. Also shown are the plots of these data which gave linear relationships, ln[X] versus time.
[Y] = 2.0 M [Y] = 4.0 M
t (s) [X] (M) [X] (M)
2 1.000 0.837
3 0.916 0.700
4 0.837 0.548
5 0.765 0.488
6 0.700 0.407
Use these data to determine the form of the rate equation for this reaction.
Explanation / Answer
Assume
rate= k[X]a[Y]b
rate1 = -delta[X]/dt =-(0.837-1)/(4-2) =8.15*10^-2 M/s at [Y]= 2.0M
rate2= -delta[X]/dt =-(0.548-0.837)/(4-2) = 14.45*10^-2 at [Y]= 4.0M
rate1/rate2 = ([Y1]/[Y2])^b
8.15/14.45 = (2/4)^b
2^b = 1.77
b=0.823 , we can approximate it as 1.
order w.r.t X is 1 , from the data we can observe that , rate doesnt remain constant. and decreasing .
comparing rates of reaction for [Y]=2.0M at t= 2 s and t= 3 and t=4 and relating to concentration it gives a=1
ratelaw is k[X][Y].
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