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 decomposition of a compound A follows first-order kinetics. The concentration of A at time t = 0 is 1.0 mol L-1. After 60 minutes, it reduces to 0.25 mol L-1. What is the initial rate of the reaction at t = 0? (Take ln 2 = 0.693)
The following graph indicates the system containing 1 mole of gas involving various steps. When it moves from Z to X, the type of undergoing process is: