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The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular temperature: Time (s) (mol/L) 1.00 120 ± 1 0.91 300 ± 1 0.78 600 ± 1 0.59 1200 ± 1 0.37 1800 ± 1 0.22 2400 ± 1 0.13 3000 ± 1 0.082 3600 ± 1 Assuming that determine the rate law, the integrated rate law, and the value of the rate constant. Calculate at s after the start of the reaction.

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Answer:

Part a: The rate of the equation for 1st order reaction is given as  [tex]Rate=k[H_2O_2][/tex]

Part b: The integrated Rate Law is given as [tex][H_2O_2]=[H_2O_2]_0 e^{-kt}[/tex]

Part c: The value of rate constant is [tex]7.8592 \times 10^{-4} s^{-1}[/tex]

Part d: Concentration after 4000 s is 0.043 M.

Explanation:

By plotting the relation between the natural log of concentration of [tex]H_2O_2[/tex], the graph forms a straight line as indicated in the figure attached. This indicates that the reaction is of 1st order.

Part a

Rate Law

The rate of the equation for 1st order reaction is given as

[tex]Rate=k[H_2O_2][/tex]

Part b

Integrated Rate Law

The integrated Rate Law is given as

[tex][H_2O_2]=[H_2O_2]_0 e^{-kt}[/tex]

Part c

Value of the Rate Constant

Value of the rate constant is given by using the relation between 1st two observations i.e.

t1=0, M1=1.00

t2=120 s , M2=0.91

So k is calculated as

[tex]-k(t_2-t_1)=ln{\frac{M_2}{M_1}}\\-k(120-0)=ln{\frac{0.91}{1.00}}\\k=\frac{-0.09431}{-120}\\k=7.8592 \times 10^{-4} s^{-1}[/tex]

The value of rate constant is [tex]7.8592 \times 10^{-4} s^{-1}[/tex]

Part d

Concentration after 4000 s is given as

[tex]-k(t_2-t_1)=ln{\frac{M_2}{1.0}}\\-7.8592 \times 10^{-4}(4000-0)=ln{\frac{M_2}{1.00}}\\-3.1437=ln{\frac{M_2}{1.00}}\\M_2=e^{-3.1437}\\M_2=0.043 M[/tex]

Concentration after 4000 s is 0.043 M.

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