Question

The rate constant for a first order reaction is $4.606 \times 10^{-3} s^{-1}$. The time required to reduce $2.0\,g$ of the reactant to $0.2\,g$ is :

• A. 100 s
• B. 200 s
• C. 500 s
• D. 1000 s

Answer 1

The Correct Option is C

To find the time required for a first-order reaction where the reactant is reduced from 2.0 g to 0.2 g, we use the formula for first-order kinetics:

\(t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right)\)

Where:

• \(t\) is the time required.
• \(k\) is the rate constant, given as \(4.606 \times 10^{-3} \, s^{-1}\).
• \([A]_0\) is the initial concentration of the reactant, here \(2.0 \, g\).
• \([A]\) is the final concentration of the reactant, here \(0.2 \, g\).

Plugging the values into the formula:

\(t = \frac{2.303}{4.606 \times 10^{-3}} \log\left(\frac{2.0}{0.2}\right)\)

The log term calculates as:

\(\log\left(\frac{2.0}{0.2}\right) = \log(10) = 1\)

Therefore:

\(t = \frac{2.303}{4.606 \times 10^{-3}} \times 1\)

Calculating further:

\(t = \frac{2.303}{4.606 \times 10^{-3}} = 500 \, s\)

Hence, the time required to reduce the reactant from 2.0 g to 0.2 g is 500 seconds.