4. The rate of formation of C was measured in trials with varying concentrations
ID: 1011255 • Letter: 4
Question
4. The rate of formation of C was measured in trials with varying concentrations of reactants A and B. Date Data was taken at 25°C. All rates given are initial rates. Trial LAL (M) Reaction 2A, B 3C 0.150 0.150 6.0 X 10 2 0.370 1.5 x 10 0.150 0.370 0.250 1.5 x 10 a) Determine the rate law for this reaction. What is the overall order for this reaction? b) Determine the rate constant for this reaction at 25°C-be sure to include units. c) Calculate the rate of formation of C when [A] 0.170 M and [B] 0.100 M. d) What was the initial rate of consumption of A in trial 3? 1Explanation / Answer
1.a) Rate law for the reaction is:
rate is proportional to [A]x [B]y , where x,y are the coefficients of A&B respectively.
Calculation of the value of x:
consider trail1 and trail2:
concentration of A is becoming 0.37/0.15 = 2.46 that is approximately 2.5 times
rate of the reaction is becoming 1.5*10^-3/ 6*10^-4 = 2.5
so the concentration of A and rate are changing equally.
therefore the value of x= 1
Finding the value of y :
consider trail2 and trail 3:
The concentration of B is changing from 0.15M to 0.25M i.e., 0.25/0.15 = 1.67 times
rate of the reaction is remaining the same. i.e., 1.5*10^-3/ 1.5*10^-3 = 1
that means rate of the reaction is independent of the concentration of B.
therefore the rate law is : rate = K [A]1
HERE K is rate constant of the reaction.
Overall order of the reaction is one.
b) calculation of rate constant of reaction: AT 250C
rate = K [A]1
6* 10^-4M/s = K * 0.15M
therefore K= 6* 10^-4M/s / 0.15 M
= 4 * 10^ -3 s-1
c) Rate of formation of C is:
rate = K * [ A ]
= 4* 10^-3 s-1 * 0.17 M
= 6.8 * 10^-4 Ms-1
d) Initial rate of consumption of A in trail 3:
= concentration of the reactant * rate constant
= 0.37 M * 4 * 10^ -3 s-1
= 1.48* 10^ -3 M/s
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