Question

A second order reaction of the type A+B>P was carried out in a solution that was initially .075 mol dm-3 in A and .03 mol dm-3 in B. After 1 hour, the concentration of A had fallen to .02 mol dm-3. a. Calculate the rate constant. b. What is the half life of the reactant? Answers: a. 16.2 dm3mol-hr-, 4.5E-3 dm3mol-s- b. 5.1E3s, 2.1E3 s

Answer 1

The rate constant of the reaction is
`1.0185* 10^(-2) dm^3/ mol s`.

`1.31* 10^3 s` is the half life of the reactant.

Step-by-step explanation:

A+B → P

Integrated rate law for second order kinetics is given by:

`(1)/(A)=kt+(1)/(A_0)`

A = concentration left after time t =
`0.02 mol /dm^3`

`A_o` = Initial concentration =
`0.075 mol /dm^3`

t = 1 hour = 3600 seconds

`(1)/(0.02 mol /dm^3)=k* 3600 s+(1)/(0.075 mol /dm^3)`

`k=(1)/(3600 s)* ((1)/(0.02 mol /dm^3)-(1)/(0.075 mol /dm^3))`

`k = 1.0185* 10^(-2) dm^3/ mol s`

Half life for second order kinetics is given by:

`t_{(1)/(2)}=(1)/(k* a_0)`

`t_{(1)/(2)}=(1)/(1.0185* 10^(-2) dm^3/ mol s* 0.075 mol /dm^3)=1.31* 10^3 s`

The rate constant of the reaction is
`1.0185* 10^(-2) dm^3/ mol s`.

`1.31* 10^3 s` is the half life of the reactant.