The reaction X rightarrow text{products} is a | vedclass

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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 $[X]_0 = 1.0 	ext{ M}$,$[X]_t

Vedclass · Accepted Answer

$3.47 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$

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 $[X]_0 = 1.0 	ext{ M}$,$[X]_t = 0.25 	ext{ M}$,and $t = 40 	ext{ min}$.$k = \frac{2.303}{40} \log \frac{1.0}{0.25} = \frac{2.303}{40} \log 4$.Using $\log 4 = 0.60$,$k = \frac{2.303 	imes 0.60}{40} = \frac{1.3818}{40} = 0.034545 	ext{ min}^{-1}$.The rate of reaction is given by $	ext{Rate} = k[X]$.At $[X] = 0.1 	ext{ M}$,$	ext{Rate} = 0.034545 	imes 0.1 = 0.0034545 	ext{ mol L}^{-1} 	ext{ min}^{-1} \approx 3.45 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$.Rounding to the nearest option,the correct answer is $3.47 	imes 10^{-3} 	ext{ mol L}^{-1} 	ext{ min}^{-1}$.

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

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