Question

Be sure to answer all parts. The reaction 2A rightarrow B is second order in A with a rate constant of 32.1 M^-1 s^-1 at 25 degree C. Starting with [A]0 = 0.00779 M, how long will it take for the concentration of A to drop to 0.001803M? s Calculate the half-life of the reaction for [A]_0 = 0.00779 M. s Calculate the half-life of the reaction for [A)0 = 0.00269 M.

Answer 1

Part A

second order equation

1/[A_t] = kt + [A₀]

where [A₀] = intial concentration = 0.00779 M

[A_t] = concentrationat certain time = 0.0018

k = rate constant = 32.1 M-1 s-1

put all the values in the above equation

1/0.0018 = 32.1 x t + 1/0.00779

555.56 = 32.1 x t + 128.37

2.1 x t = 555.56 - 128.37

2.1 x t = 427.19

t = 427.19 / 2.1

= 203.424 sec

Part B

t1/2 = 1/k [A₀]

= 1/ 32.1 M-1 s-1x 0.00779M

= 1/0.25

= 4 s

part C

t1/2 = 1/k [A₀]

= 1/ 32.1 M-1 s-1x 0.00269M

= 1/0.0863

= 11.58 s