The rate constant for a reaction \(P\longrightarrow Q\) is \(0.8 × 10^{-3}\, \ mol L^{-1}s^{-1}\) If the concentration of \(P\) is \(10 M\), then the concentration of \(Q\) after \(40\) minutes is
Answer & explanation
Correct answer: option 1
The correct answer is option 1. \(1.92\, \ mol L^{-1}\).
The given reaction is \(P \longrightarrow Q\)
Given,
Rate constant, \(k = 0.8 × 10^{-3}\, \ mol L^{-1}s^{-1}\)
From the unit of rate constant, it is clear that the reaction is a zero order reaction.
Hence, concentration of \(Q\) after \(40\, \ min\) \(= kT\)
or, concentration of \(Q\) after \(40\, \ min\) \(= 0.8 × 10^{-3} × 40 × 60 = 1.92\, \ mol L^{-1}\)