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(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: Initial concentrat

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$0.023 \ minute^{-1}$

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(B) For a first order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: Initial concentration $[A]_0 = 20 \ mmol \ dm^{-3}$ Final concentration $[A]_t = 8 \ mmol \ dm^{-3}$ Time $t = 40 \ minute$ Substituting the values: $k = \frac{2.303}{40} \log \frac{20}{8}$ $k = \frac{2.303}{40} \log(2.5)$ Since $\log(2.5) \approx 0.3979$: $k = \frac{2.303 	imes 0.3979}{40} \approx 0.0229 \ minute^{-1}$ Rounding off,we get $k \approx 0.023 \ minute^{-1}$.

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.

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