For this reaction:
\[K = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\]
Given: \( k = 4.6 \times 10^{-2} \, \text{s}^{-1}, \, [A]_0 = 1 \, \text{M}, \, [A] = 0.1 \, \text{M} \)
\[4.6 \times 10^{-2} = \frac{2.303}{t} \log \frac{1}{0.1}\]
\[4.6 \times 10^{-2} = \frac{2.303}{t} \times 1\]
\[t = \frac{2.303}{4.6 \times 10^{-2}} \approx 50 \, \text{sec.}\]
The rate of a reaction:
A + B −→ product
is given below as a function of different initial concentrations of A and B.
Experiment | \([A]\) (mol L\(^{-1}\)) | \([B]\) (mol L\(^{-1}\)) | Initial Rate (mol L\(^{-1}\) min\(^{-1}\)) |
---|---|---|---|
1 | 0.01 | 0.01 | \(5 \times 10^{-3}\) |
2 | 0.02 | 0.01 | \(1 \times 10^{-2}\) |
3 | 0.01 | 0.02 | \(5 \times 10^{-3}\) |
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: