Question

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (mol/L) 1.00 120 ± 1 0.91 300 ± 1 0.78 600 ± 1 0.59 1200 ± 1 0.37 1800 ± 1 0.22 2400 ± 1 0.13 3000 ± 1 0.082 3600 ± 1 Assuming that determine the rate law, the integrated rate law, and the value of the rate constant. Calculate at s after the start of the reaction.

Answer 1

Part a: The rate of the equation for 1st order reaction is given as
`Rate=k[H_2O_2]`

Part b: The integrated Rate Law is given as
`[H_2O_2]=[H_2O_2]_0 e^(-kt)`

Part c: The value of rate constant is
`7.8592 * 10^(-4) s^(-1)`

Part d: Concentration after 4000 s is 0.043 M.

Step-by-step explanation:

By plotting the relation between the natural log of concentration of
`H_2O_2`, the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

`Rate=k[H_2O_2]`

Part b

Integrated Rate Law

The integrated Rate Law is given as

`[H_2O_2]=[H_2O_2]_0 e^(-kt)`

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

`-k(t_2-t_1)=ln{(M_2)/(M_1)}\\-k(120-0)=ln{(0.91)/(1.00)}\\k=(-0.09431)/(-120)\\k=7.8592 * 10^(-4) s^(-1)`

The value of rate constant is
`7.8592 * 10^(-4) s^(-1)`

Part d

Concentration after 4000 s is given as

`-k(t_2-t_1)=ln{(M_2)/(1.0)}\\-7.8592 * 10^(-4)(4000-0)=ln{(M_2)/(1.00)}\\-3.1437=ln{(M_2)/(1.00)}\\M_2=e^(-3.1437)\\M_2=0.043 M`

Concentration after 4000 s is 0.043 M.