Answer:
Explanation:
From the graphical diagram attached below; we can see the relationship between the concentration of [tex]H_2O_2[/tex] which declines exponentially in relation to the time and it obeys the equation: [tex]\mathtt{y = 0.9951 e^{-8\times 10^{-4}x}}[/tex]
This relates to the 1st order reaction rate, whereby:
The integrated rate law[tex]\mathtt{ [A] = [A]_o e^{-kt}}[/tex]
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 [tex]\mathtt{k = 8\times 10^{-4} {s^{-1}}}[/tex]
Rate law [tex]\mathtt{= k[H_2O_2]}[/tex]
The integrated rate law [tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-(8*10^{-4})t}}[/tex]
From the given table:
the initial concentration of [tex]H_2O_2[/tex] = 1.00 M
∴
We can determine the concentration of the reactant at 4000s by using the formula:
[tex]\mathtt{[H_2O_2] = [H_2O_2]_oe^{-8*10^{-4}(t)}}[/tex]
[tex]\mathtt{[H_2O_2] = (1.00\ M)*e^{-8*10^{-4}(4000)\ s}}[/tex]
[tex]\mathtt{[H_2O_2] =0.0407 \ M}[/tex]
Finally, at 4000s: the average rate is:
[tex]\mathtt{= (8*10^{-4} \ s^{-1})(4000 \ s) }\\ \\ \mathtt{ = 3.256 \times 10^{-5} \ M/s}[/tex]
