The rate constant for the decomposition of H_ | vedclass

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(A) The decomposition of $H_2O_2$ follows first-order kinetics.The integrated rate equation for a first-order reaction is: $k = \frac{2.303}{t} \log \

Vedclass · Accepted Answer

$105$

Vedclass · Answer

(A) The decomposition of $H_2O_2$ follows first-order kinetics.The integrated rate equation for a first-order reaction is: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:$k = 3.66 	imes 10^{-3} \ s^{-1}$$[A]_0 = 0.882 \ M$$[A]_t = 0.600 \ M$Substituting the values:$3.66 	imes 10^{-3} = \frac{2.303}{t} \log \frac{0.882}{0.600}$$3.66 	imes 10^{-3} = \frac{2.303}{t} \log(1.47)$$3.66 	imes 10^{-3} = \frac{2.303}{t} 	imes 0.1673$$t = \frac{2.303 	imes 0.1673}{3.66 	imes 10^{-3}}$$t = \frac{0.3853}{0.00366} \approx 105.27 \ s$Thus,the time required is approximately $105 \ s$.

The rate constant for the decomposition of $H_2O_2$ is $3.66 \times 10^{-3} \ s^{-1}$. If the initial concentration of $H_2O_2$ is $0.882 \ M$,then in how many seconds will its concentration become $0.600 \ M$?

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