For first order reactions:
\[ K_1 t_1 = \ln\left(\frac{1}{1 - \frac{2}{3}}\right) = \ln 3 \]
\[ K_2 t_2 = \ln\left(\frac{1}{1 - \frac{4}{5}}\right) = \ln 5 \]
\[ \therefore K_1 t_1 = 5 \quad \text{and} \quad K_2 t_2 = 2 \]
\[ \frac{K_1}{K_2} = \frac{\ln 3}{\ln 5} \]
\[ \frac{t_1}{t_2} = \frac{0.477}{0.699} \times 5 = 1.7 \times 10^{-1} \]
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:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: