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 Reaction \mathrm{X+Y \rightarrow P+Q} follows rate law, \mathrm{r=k[X]^{1 / 2}[Y]^{1 / 2}}. Starting with \mathrm{1M} of \mathrm{X} and \mathrm{Y} each, what is the time taken for conc. of \mathrm{A} to become \mathrm{0.2 \mathrm{M}} ? (Given \mathrm{k=\frac{\ln 10}{100} \mathrm{~s}^{-1}})

Option: 1

70 s


Option: 2

100 s


Option: 3

60 s


Option: 4

140 s


Answers (1)

best_answer

         \mathrm{X+Y \rightarrow P+Q}

\mathrm{t=0 \quad 1 \quad 1 \quad x}
\mathrm{t=t \quad 1-x \quad 1-x \quad x \quad x}

\mathrm{r=k[x]^{1 / 2}[y]^{1 / 2}}
\mathrm{ \frac{d x}{d t}=r=k(1-x)^{1 / 2}(1-x)^{1 / 2}}
\mathrm{\frac{d x}{d t}=k(1-x) \Rightarrow t=\frac{1}{k} \ln \left(\frac{1}{1-x}\right) }

\mathrm{t=\frac{2.303 \times 100}{\ln 10} \log \left(\frac{1}{0.2}\right)=\frac{2.303 \times 100}{2.303} \log 5}
\mathrm{ t=69.898 \approx 70 \, s}

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