A second order reaction is 41.5% complete in 500 sec.

Question

a) Calculate the rate constant? 

b) What is the value of the half-life? 



c) How long will it take for the reaction to go to 25%, 80% completion?

Answer 1

For a second order reaction, 1/[A] = 1/[A]_o + kt

a) If we start with 100 as [A]_o, when 41.5% complete, [A] = 58.5

1/58.5 = 1/100 + k(500 sec)

0.017 = 0.01 + 500k

0.007/500 = k

k = 1.4x10^-5 M^-1s^-1

b) t_1/2 = 1/k[A]_o = 1/(1.4x10^-5 *100) = 1/1.4x10^-3)

t_1/2 = 714 sec

c) For 25% completion, [A] = 75

1/75 = 1/100 + 1.4x10^-5 t

0.0033/1.4x10^-5 = t = 238 seconds

Do the same process for 80%