Question

Hydrogen peroxide decomposes into water and oxygen in a first-order process.H2O2(aq) --> H2O(l) + 1/2 O2(g)At 20.0 °C, the half-life for the reaction is 3.92 x 104 seconds. If the initial concentration of hydrogen peroxide is 0.52 M, what is the concentration after 7.00 days?1.2 x 10-5 M0.034 M0.074 M0.22 M0.52 M

Answer 1

Concentration of
`H_(2)O_(2)` after 7.00 days is
`1.2* 10^(-5)` M

Step-by-step explanation:

Integrated rate equation for decomposition of
`H_(2)O_(2)` is-

`[H_(2)O_(2)]=[H_(2)O_(2)]_(0)* (0.5)^{((t)/(t_(0.5)))}`

where [
`H_(2)O_(2)`] is concentration of
`H_(2)O_(2)` after "t" time,
`[H_(2)O_(2)]_(0)` is initial concentration of
`H_(2)O_(2)` and
`t_(0.5)` is half-life

Here
`[H_(2)O_(2)]_(0)` is 0.52 M, t is 604800 second (7 days) and
`t_(0.5)` is 39200 seconds

Plug in all the values in the above equation-

`[H_(2)O_(2)]= 0.52* (0.5)^{(604800)/(39200)}`

or,
`[H_(2)O_(2)]` =
`1.2* 10^(-5)`

So concentration of
`H_(2)O_(2)` after 7.00 days is
`1.2* 10^(-5)` M