45
Step 1: Using First-Order Reaction Formula
The time required for the concentration to become \( \frac{1}{n} \) of its initial value in a first-order reaction is: \[ t_{1/n} = \frac{2.303}{k} \log n \] For \( t_{1/4} \): \[ t_{1/4} = \frac{2.303}{k} \log 4 = \frac{2.303}{k} (2 \times \log 2) \] Using \( \log 2 = 0.3 \), \[ t_{1/4} = \frac{2.303}{k} (2 \times 0.3) = \frac{2.303 \times 0.6}{k} \] For \( t_{1/10} \): \[ t_{1/10} = \frac{2.303}{k} \log 10 = \frac{2.303}{k} (1) \] Step 2: Calculating Ratio
\[ \frac{t_{1/4}}{t_{1/10}} = \frac{2.303 \times 0.6}{2.303 \times 1} = 0.6 \] \[ \frac{t_{1/4}}{t_{1/10}} \times 100 = 0.6 \times 100 = 60 \] Thus, the correct answer is 60.
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 | \(5 \times 10^{-3}\) |
Match the following: