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(D) For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t}$ Given $[X]_0 = 1.0 \ M$,$[X]_t = 0.25

Vedclass · Accepted Answer

$3.45 	imes 10^{-3} \ mol \ L^{-1} \ min^{-1}$

Vedclass · Answer

(D) For a first order reaction,the rate constant $k$ is given by: $k = \frac{2.303}{t} \log \frac{[X]_0}{[X]_t}$ Given $[X]_0 = 1.0 \ M$,$[X]_t = 0.25 \ M$,and $t = 40 \ min$: $k = \frac{2.303}{40} \log \frac{1.0}{0.25} = \frac{2.303}{40} \log 4$ $k = \frac{2.303 	imes 0.60}{40} = 0.034545 \ min^{-1}$ Now,the rate of reaction when $[X] = 0.1 \ M$ is: $Rate = k[X] = 0.034545 	imes 0.1$ $Rate = 0.0034545 \ mol \ L^{-1} \ min^{-1} = 3.45 	imes 10^{-3} \ mol \ L^{-1} \ min^{-1}$

The reaction $X \rightarrow$ products is a first order reaction. In $40 \ min$,the concentration of $X$ changes from $1.0 \ M$ to $0.25 \ M$. What is the rate of reaction when $[X] = 0.1 \ M$? $(\log 4 = 0.60)$

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