Question

A first order reaction has a specific reaction rate of 10^–2 s^–1. How much time will it take for 20 g of the reactant to reduce to 5 g?

• A. 238.6 second
• B. 138.6 second
• C. 346.5 second
• D. 693.0 second

Answer 1

The Correct Option is B

To solve this problem, we need to find out how much time it takes for the concentration or amount of a reactant in a first-order reaction to decrease from 20 g to 5 g. The given problem states that the specific reaction rate constant, \( k \), is \( 10^{-2} \, \text{s}^{-1} \).

For a first-order reaction, the formula relating the initial concentration or amount \( [A]_0 \), the final concentration or amount \( [A] \), the rate constant \( k \), and the time \( t \) is given by:

t = \frac{1}{k} \ln \left(\frac{[A]_0}{[A]}\right)

In this problem:

• Initial amount, \( [A]_0 = 20 \, \text{g} \)
• Final amount, \( [A] = 5 \, \text{g} \)
• Rate constant, \( k = 10^{-2} \, \text{s}^{-1} \)

Let's substitute these values into the formula to calculate the time:

t = \frac{1}{10^{-2}} \ln \left(\frac{20}{5}\right)

t = \frac{1}{0.01} \ln (4)

t = 100 \ln (4)

We know:

\ln(4) \approx 1.386

Substitute this value back into the equation:

t = 100 \times 1.386 = 138.6 \, \text{seconds}

Therefore, the correct answer is 138.6 seconds.