Question

Hydrogen peroxide decays into water and oxygen in a

second-order process,
H2O2(aq) ---> H2O(l) + 1/2 O2(g)
What is the beginning concentration of H2O2 if after 320 seconds,
the peroxide concentration falls to 0.442 M, with K being 5.32 x 10-3
/sec ? What is the half-life?

Answer 1

The half-life of the decay process is approximately 119 seconds.

To determine the beginning concentration of H2O2 and the half-life of its decay, we can use the second-order rate equation:

Rate = k[H2O2]^2

Given:

- Rate constant (k) = 5.32 x
`10^-3 sec^-1`

- Concentration of H2O2 after 320 seconds = 0.442 M

1. Calculate the initial concentration of H2O2 ( [H2O2]0 ):

To find the initial concentration of H2O2, we need to rearrange the rate equation:

Rate =
`k[H2O2]^2`

k = Rate /
`[H2O2]^2`

[H2O2]0 = (
`√(Rate / k)`)

[H2O2]0 = √(0.442 M / (5.32 x
`10^-3 sec^-1`))

[H2O2]0 = √(0.442 M / 5.32 x
`10^-3 sec^-1`)

[H2O2]0 ≈ 16.62 M

Therefore, the beginning concentration of H2O2 is approximately 16.62 M.

2. Calculate the half-life:

The half-life (t1/2) is the time it takes for the concentration of H2O2 to reduce by half. In a second-order reaction, the relationship between half-life and initial concentration is given by:

t1/2 = 1 / (k[H2O2]0)

t1/2 = 1 / ((5.32 x
`10^-3 sec^-1`) * (16.62 M))

t1/2 ≈ 1.19 x
`10^2` seconds

Therefore, the half-life of the decay process is approximately 119 seconds.