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

Answer 2

The concentration of hydrogen peroxide after seven days is $1.2 \times 10^{-5} \;\rm M$.

The decomposition of the hydrogen peroxide is the first order reaction as the concentration of the reactants and the rate is directly proportional.

How to calculate the concentration?

Given,

• Initial concentration of $[\rm H_{2}O_{2}]_{0}$ = 0.52 M
• The half-life for reaction $(t_{0.5})$ = 39200 seconds

The equation for the decomposition of hydrogen peroxide can be given as,

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

Substituting values in the above equation:

$\begin{aligned}\rm [H_{2}O_{2}] &= [0.52]\times (0.5)^{(\frac{604800}{{39200}})}\\\\&= 1.2 \times 10^{-5}\;\rm M\end{aligned}$

Therefore, the concentration of hydrogen peroxide after seven days is $1.2 \times 10^{-5} \;\rm M$.

Learn more about half-life here:

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