Question

The rate constant of a first order reaction is \(231\times10^{-5}\ s^{-1}\). How long will 4 g of this reactant reduce to 2 g?

• A. 310 s
• B. 300 s
• C. 210 s
• D. 30.1 s
• E. 230.3 s

Hint:

For first order reactions, the time required to reduce a reactant to half is always \(t_{1/2}=\frac{0.693}{k}\).

Answer 1

The Correct Option is B

Concept: For a first order reaction, half-life is: \[ t_{1/2}=\frac{0.693}{k} \]

Step 1: Identify change in amount.
The reactant decreases from 4 g to 2 g. This is exactly half of the original amount.

Step 2: Use half-life formula.
\[ k=231\times10^{-5}=2.31\times10^{-3}\ s^{-1} \] \[ t_{1/2}=\frac{0.693}{2.31\times10^{-3}} \]

Step 3: Calculate time.
\[ t_{1/2}=300\ s \]