In a first-order reaction A P, the ratio of a/(a – x) was found to be 8 after 60 minutes. If the concentration is 0.1 Μ then the rate of reaction in moles of A reacted per minute is:
Answer & explanation
Correct answer: option 2
The correct answer is option 2. 3.466 × 10–3 mol litre–1 min–1.
The given ratio of \(\frac{a}{a-x}\) is 8 after 60 minutes, where \(a\) is the initial concentration of A and \(x\) is the concentration of A at time \(t\).
We know, for a first order reaction,
\(k = \frac{2.303}{t} log\frac{a}{a - x}\)
Applying the above data:
\(k = \frac{2.303}{60} log8\)
or, \(k = 3.466 \times 10^{-2}\, \ min^{-1}\)
The given concentration of \([A] = 0.1\). Thus,
\(\frac{dx}{dt} = k[A]\)
or, \(\frac{dx}{dt} = 3.466 \times 10^{-2} \times 0.1\)
or, \(\frac{dx}{dt} = 3.466 \times 10^{-3}\, \ mol\, \ L^{-1}min^{-1}\)