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

Answer:

Concentration of $H_{2}O_{2}$ after 7.00 days is $1.2\times 10^{-5}$ M

Explanation:

Integrated rate equation for decomposition of $H_{2}O_{2}$ is-

             $[H_{2}O_{2}]=[H_{2}O_{2}]_{0}\times (0.5)^{(\frac{t}{t_{0.5}})}$

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-

$[H_{2}O_{2}]= 0.52\times (0.5)^{\frac{604800}{39200}}$

or, $[H_{2}O_{2}]$ = $1.2\times 10^{-5}$

So concentration of $H_{2}O_{2}$ after 7.00 days is $1.2\times 10^{-5}$ M