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 unit of the rate constant, mol L−1 s−1, shows that the reaction is of zero order. For a zero order reaction the rate is constant and equal to k, so the product formed grows linearly with time:
\( [\mathrm{Q}] = kt \)
\( t = 40\ \text{min} = 40 \times 60 = 2400\ \text{s} \)
\( [\mathrm{Q}] = 0.8 \times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1} \times 2400\ \text{s} = 1.92\ \text{mol L}^{-1} \)
This is less than the 10 M of P present, so P is not used up within 40 minutes and the zero order law holds throughout.