\(Rate [X]^1[Y]^0\)
\(Rate = k[X]\)
From Exp I and II,
\(\frac{4 \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{L}{0.1}\right)^1 \left(\frac{0.2}{0.1}\right)^0\)
\(2 = (10 L)^1.\)
Hence L = 0.2 mol/L
From Exp III and IV,
\(\frac{M \times 10^{-3}}{2 \times 10^{-3}} = \left(\frac{0.4}{0.1}\right) \left(\frac{0.4}{0.2}\right)^0\)
\(\frac{M}{2} = 4\)
\(M = 8\)
\(\frac{M}{L} = \frac{8}{0.2}\)
\(\frac{M}{L} = 40\)
So, the answer is 40.
For a first-order reaction, the concentration of reactant was reduced from 0.03 mol L\(^{-1}\) to 0.02 mol L\(^{-1}\) in 25 min. What is its rate (in mol L\(^{-1}\) s\(^{-1}\))?
The Order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. In order to obtain the reaction order, the rate equation of the reaction will given in the question.