Question

The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature.Time (s) [H2O2] (mol/L)0 1.00120 ± 1 0.91300 ± 1 0.78600 ± 1 0.591200 ± 1 0.371800 ± 1 0.222400 ± 1 0.133000 ± 1 0.0823600 ± 1 0.050 Assuming that the rate= -delta [H2O2]/delta t determine the rate law, integrated rate law, and the value of the rate constant. Calculate [H2O2] at 4000. s after the start of the reaction.

Answer 1

Answer:

Explanation:

From the graphical diagram attached below; we can see the relationship between the concentration of $H_2O_2$ which declines exponentially in relation to the time and it obeys the equation: $\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}$

This relates to the 1st order reaction rate, whereby:

The integrated rate law$\mathtt{ [A] = [A]_o e^{-kt}}$

here:

[A] = reactant concentration at time (t)

[A]_o = initial concentration for the reactant

k = rate constant

As such, the order of the reaction is the first order

Rate constant $\mathtt{k = 8\times 10^{-4} {s^{-1}}}$

Rate law $\mathtt{= k[H_2O_2]}$

The integrated rate law $\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}$

From the given table:

the initial concentration of $H_2O_2$ = 1.00 M

∴

We can determine the concentration of the reactant at 4000s by using the formula:

$\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}$

$\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}$

$\mathtt{[H_2O_2] =0.0407 \ M}$

Finally, at 4000s: the average rate is:

$\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}$