Practicing Success
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 |
\(1.92\, \ mol L^{-1}\) \(1.46\, \ mol L^{-1}\) \(1.23\, \ mol L^{-1}\) \(0.24\, \ mol L^{-1}\) |
\(1.92\, \ mol L^{-1}\) |
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 \(40\, \ min\) \(= 0.8 × 10^{-3} × 40 × 60 = 1.92\, \ mol L^{-1}\) |