\(Rate [X]^1[Y]^0\)
\(Rate = k[X]\)
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.
The following data were obtained for the reaction: \[ 2NO(g) + O_2(g) \rightarrow 2N_2O(g) \] at different concentrations:
The rate law of this reaction is:
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 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.