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Question

(C) For a first order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \f

Vedclass · Accepted Answer

$4.606 	imes 10^{-3} \; \sec^{-1}$

Vedclass · Answer

(C) For a first order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \frac{1}{10} \; M = 0.1 \; M$Final concentration $[A]_t = \frac{1}{100} \; M = 0.01 \; M$Time $t = 8 \; 	ext{minutes} + 20 \; 	ext{seconds} = (8 	imes 60) + 20 = 500 \; \sec$Substituting the values:$K = \frac{2.303}{500} \log \frac{0.1}{0.01}$$K = \frac{2.303}{500} \log 10$Since $\log 10 = 1$:$K = \frac{2.303}{500} = 0.004606 \; \sec^{-1}$$K = 4.606 	imes 10^{-3} \; \sec^{-1}$

$A$ first order reaction was started with a decimolar solution of the reactant. $8$ minutes and $20$ seconds later,its concentration was found to be $M/100$. The rate constant of the reaction is:

Similar Questions

If a $75\%$ first-order reaction is completed in $32 \text{ minutes}$,how many minutes will it take for the same reaction to be $50\%$ completed?

Derive the equation showing the relation between the concentration $[R]_1$ and $[R]_2$ at time $t_1$ and $t_2$ for a first-order reaction.

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For the reaction $2A + B \to \text{Product}$,the rate law is given as $\frac{-d[A]}{dt} = K[A]$. At a time when $t = \frac{1}{K}$,the concentration of the reactant $A$ is ($Co =$ initial concentration).

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