Solution for this Half life's Question

Question

The half-life for the decomposition of hydrogen peroxide, H2O2, at 25^0C is 990 s.

a. Determine the first order rate constant.

b. A solution of H2O2 is initially 0.155 M. Determine the concentration of H2O2 after five (5) half-lives have elapsed.

Answer 1

Hello Divine

 

a)  For a 1st order reaction,

 

t_1/2 = 0.693 / k

 

Rearrange to solve for k

 

k = 0.693 / 990 = 7.0 x 10^-4 s^-1

 

 

b)  Each half-life is 990 s, therefore, 5 half-lives = 990 x 5 = 4950 s

 

Equation to use is:

 

ln[A_t] = ln[A_o] – kt

 

where [A_t] is the concentration of the substance at a time "t", which is what we need to calculate.

[A_o] = initial concentration = 0.155 M

k = rate constant = 7.0 x 10^-4 s^-1

t = time = 4950 s

 

ln[A_t] = ln[A_o] – kt

 

ln[A_t] = ln(0.155) – (7.0 x 10^-4 x 4950)

 

ln[A_t] = –1.86 – (3.47)

 

ln[A_t] = –5.33

 

Take the antilog (e^x) of both sides of the equation

 

[A_t] = e^–5.33 = 0.00484 M