Question

The rate constant for the reaction is 0.180 M -1.S-1 at 200 C.
A-->products
If the initial concentration of A is 0.00180 M, what will be the concentration after 535 s?

Answer 1

Given that k = 0.18  M^-1s^-1

It is rate constant of 2nd order reaction because units of first order reaction rate constant = M^-1s^-1

For first order recation,

k = [1/t] { 1/ [A]_t -1/[A]_o} -----Eq (1)

Given that

time t = 535 s

Initial concentration [A]_o = 0.0018 M

Final concentration [A]_t = ? M

Substitute ll the values in eq (1) ,

k = [1/t] { 1/ [A]_t -1/[A]_o}

0.18 = [1/535] { 1/[A]_t - 1/0.0018 }

On solving,

[A]_t = 0.00153 M

Therefore,

concentration after 535 s = 0.00153 M