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The reaction $X \rightarrow$ products is a first order reaction. In 40 minutes, the concentration of $X$…
The reaction $X \rightarrow$ products is a first order reaction. In 40 minutes, the concentration of $X$ changes from $1.0 \mathrm{M}$ to $0.25 \mathrm{M}$. What is the initial rate of reaction when $[X]=0.1 \mathrm{M}$ ? $(\log 4=0.60)$
$1.73 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ $3.47 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ $1.73 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ $3.45 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ Solution
Given, (concentration) $a=1.0 \mathrm{M}, a-x=0.25 \mathrm{M}, t=40$ minutes
For a first order reaction,
$
\begin{aligned}
k & =\frac{2.303}{t} \log \frac{a}{a-x} \\
& =\frac{2.303}{40} \log \frac{1}{0.25}=\frac{2.303}{40} \log 4 \\
& =\frac{2.303 \times 2 \times 0.3010}{40} \\
& =0.03466 \mathrm{~min}^{-1}
\end{aligned}
$
Now, Initial rate
$
\begin{aligned}
& =k[A] \\
& =0.03466(0.1) \\
& =0.003466 \mathrm{~mol} / \mathrm{L} / \mathrm{min} \\
& =3.466 \times 10^{-3} \mathrm{~mol} / \mathrm{L} / \mathrm{min} \\
& =3.47 \times 10^{-3} \mathrm{~mol} / \mathrm{L} / \mathrm{min}
\end{aligned}
$
According to options given, any option cannot match with answer. But (d) can be the correct answer
Asked in: AP EAMCET 2019 (20 Apr Shift 2)
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