Question

A first-order reaction has the rate constant of \(1.15 \times 10^{-3} \, \text{s}^{-1}\). The time required to reduce 10 g of reactant to 6 g is \(x \times 10^2\) sec. What is the approximate value of \(x\)?

• A. 2.2
• B. 3.3
• C. 4.4
• D. 1.1

Hint:

For first-order reactions, time can be calculated using the logarithmic formula.

Answer 1

The Correct Option is C

For a first-order reaction, the relationship between the concentration at any time and the initial concentration is given by: \[ \ln \left(\frac{[A_0]}{[A_t]}\right) = k t \] where: - \( [A_0] \) is the initial concentration, - \( [A_t] \) is the concentration at time \( t \), - \( k \) is the rate constant. We are given: - \( [A_0] = 10 \, \text{g} \), - \( [A_t] = 6 \, \text{g} \), - \( k = 1.15 \times 10^{-3} \, \text{s}^{-1} \). The time \( t \) is given by the formula: \[ t = \frac{1}{k} \ln \left(\frac{[A_0]}{[A_t]}\right) \] Substitute the values into the equation: \[ t = \frac{1}{1.15 \times 10^{-3}} \ln \left(\frac{10}{6}\right) \] \[ t = \frac{1}{1.15 \times 10^{-3}} \ln (1.6667) \approx \frac{1}{1.15 \times 10^{-3}} \times 0.5108 \] \[ t \approx 4.4 \times 10^2 \, \text{sec} \Rightarrow x = 4.4 \] Thus, the correct answer is \( \boxed{4.4} \).