A → Products
For a first order reaction,
\(t_{1/2} = \frac{ln2}{k} = \frac{0.693 }{ k}\)
Time for \(90\% \) conversion,
\(t_{ 90\%} = \frac{1}{k} \;In \;\frac{100}{10 }= \frac{ln10}{k} = \frac{2.303}{k}\)
\(t_{90\%} = \frac{2.303}{0.693} \;t_{1/2} = 3.32 \;t_{1/2}\)
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.