The rate constant for the oxidation of hydrog | vedclass

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Question

(A) The reaction follows first-order kinetics as indicated by the unit of the rate constant $(s^{-1})$.For a first-order reaction,the formula is $k =

Vedclass · Accepted Answer

$1.326 	imes 10^4 \ s$

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(A) The reaction follows first-order kinetics as indicated by the unit of the rate constant $(s^{-1})$.For a first-order reaction,the formula is $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.Here,$[A]_0 = 20 \ mL$,$[A]_t = 8 \ mL$,and $k = 6.93 	imes 10^{-5} \ s^{-1}$.Substituting the values: $6.93 	imes 10^{-5} = \frac{2.303}{t} \log \frac{20}{8}$.$6.93 	imes 10^{-5} = \frac{2.303}{t} \log(2.5)$.Since $\log(2.5) \approx 0.3979$,we have $t = \frac{2.303 	imes 0.3979}{6.93 	imes 10^{-5}}$.$t = \frac{0.91636}{6.93 	imes 10^{-5}} \approx 1.322 	imes 10^4 \ s$.Thus,the time taken is approximately $1.326 	imes 10^4 \ s$.

The rate constant for the oxidation of hydrogen peroxide by $KMnO_4$ is $6.93 \times 10^{-5} \ s^{-1}$. How much time will it take for the volume of a standard $KMnO_4$ solution to decrease from $20 \ mL$ to $8 \ mL$?

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