The rate constant of a first-order reaction i | vedclass

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(A) For a first-order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ Given: $k = 6 \ min^{-1}$ $[R]_0 = 0.5 \ mo

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$0.38$

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(A) For a first-order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ Given: $k = 6 \ min^{-1}$ $[R]_0 = 0.5 \ mol \ L^{-1}$ $[R] = 0.05 \ mol \ L^{-1}$ Substituting the values in the equation: $6 = \frac{2.303}{t} \log \frac{0.5}{0.05}$ $6 = \frac{2.303}{t} \log(10)$ Since $\log(10) = 1$: $6 = \frac{2.303}{t} 	imes 1$ $t = \frac{2.303}{6} \ min$ $t \approx 0.3838 \ min$ Rounding to two decimal places,$t \approx 0.38 \ min$.

The rate constant of a first-order reaction is $6 \ min^{-1}$. If the initial concentration of the reactant is $0.5 \ mol \ L^{-1}$,after how many minutes will the concentration of the reactant become $0.05 \ mol \ L^{-1}$?

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