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:

3.466 × 10^–3 mol litre^–1 min^–1

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 given below:

\(k =  \frac{2.303}{60} log8\)

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, \(\frac{dx}{dt} = 3.466 \times 10^{-3}\, \ mol\, \ L^{-1}min^{-1}\)