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Suppose you are studying the kinetics of the iodine-catalyzed decomposition of hydrogen peroxide. 2 H 2 O 2 ⟶ 2 H 2 O + O 2 If you determine the initial rate is 7.50 × 10 − 4 M/s when [ H 2 O 2 ] = 0.546 M and [ K I ] = 0.212 M , what is the rate constant? Assume that the order of both reactants is 1.

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Answer:

[tex]k=6.48x10^{-3}M^{-1}s^{-1}[/tex]

Explanation:

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In this case, based on the given information, we can write the rate law as shown below:

[tex]r=k[H_2O_2][KI][/tex]

Since the overall order of reaction is 2 being 1 for reach reactant. Thus, by knowing the initial rate and concentrations, the rate constant turns out:

[tex]k=\frac{r}{[H_2O_2][KI]}=\frac{7.50x10^{-4}M/s}{0.546M*0.212M}\\ \\k=6.48x10^{-3}M^{-1}s^{-1}[/tex]

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The rate constant, k is [tex]6.48x10^{-3}M^{-1}s^{-1}[/tex]

Rate law:

The rate law should be

[tex]r = k[H_2O_2]{KI}[/tex]

Since the total order of the reaction is 2 being 1 for reach reactant. So, by knowing the beginning rate and concentrations, the rate constant turns out:

[tex]k = \frac{r}{k[H_2O_2]{KI}} = \frac{7.50x10^{-4}M/s}{0.546M\times 0.212M}[/tex]

k = [tex]6.48x10^{-3}M^{-1}s^{-1}[/tex]

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