The decomposition of XY is second order in XY and has a rate constant of 7.06×10

Question

The decomposition of XY is second order in XY and has a rate constant of 7.06×103 M1s1 at a certain temperature. a. What is the half-life for this reaction at an initial concentration of 0.100 M? b. How long will it take for the concentration of XY to decrease to 12.5% of its initial concentration when the initial concentration is 0.100 M? c. How long will it take for the concentration of XY to decrease to 12.5% of its initial concentration when the initial concentration is 0.200 M? d.If the initial concentration of XY is 0.140 M , how long will it take for the concentration to decrease to 6.60×102 M ? e. If the initial concentration of XY is 0.050 M, what is the concentration of XY after 55.0 s ? f.If the initial concentration of XY is 0.050 M, what is the concentration of XY after 500 s ?

Explanation / Answer

a) The governing equation for 2nd order reaction is :-

k*t = {1/[A]} - {1/[A0]} ; where k = rate constant, t = time after which the concentration reduces from [A0] to [A]

Now, t1/2 = (1/k)*{1/[A0]} = (1/7.06*10-3)*(1/0.1) = 1/7.06*10-4 = 1416.43 sec

b) [A] = (1/8)*[A0]

Thus, t = (1/k)*(7/[A0]} = 9915.01 sec

c) [A] = (1/8)*[A0]

Thus, t = (1/k)*(7/[A0]} = 4957.5 sec

e) [A0] = 0.05 M & t = 55 sec

Thus, 1/[A] = 20.39

or, [A] = concentration left after time 55 sec = 0.05 M

f) [A0] = 0.05 M & t = 500 sec

Thus, 1/[A] = 23.53

or, [A] = concentration left after time 55 sec = 0.0425 M

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