Question

For a first-order reaction \( A \rightarrow B \), the rate constant is \( 0.1 \text{ s}^{-1} \). What is the time required for \( 50\% \) completion?

• A. \( 6.93 \text{ s} \)
• B. \( 5 \text{ s} \)
• C. \( 10 \text{ s} \)
• D. \( 0.693 \text{ s} \)

Hint:

For first-order reactions: \[ t_{1/2}=\frac{0.693}{k} \] The half-life is independent of the initial concentration.

Answer 1

The Correct Option is A

Step 1 : Understanding the Question:
The question requires us to determine the time necessary for a first-order chemical reaction to reach 50% completion. This specific duration of time is defined as the half-life of the reaction, where the concentration of reactant decreases to half of its initial value.
Step 2 : Key Formulas and Approach:
For a first-order chemical reaction, the rate law and integration lead to the expression for half-life:
\[ t_{1/2} = \frac{\ln(2)}{k} \approx \frac{0.693}{k} \] Here, \( k \) represents the reaction rate constant, and \( t_{1/2} \) is the half-life. We will substitute the given rate constant into this standard formula.
Step 3 : Detailed Solution:

Identify that 50% completion of a reaction refers precisely to its half-life parameter, \( t_{1/2} \).

Write down the first-order half-life formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Retrieve the given value of the rate constant: \( k = 0.1 \text{ s}^{-1} \).

Substitute \( k \) into the equation and perform the division:
\[ t_{1/2} = \frac{0.693}{0.1} = 6.93 \text{ s} \]
Step 4 : Final Answer:
The time required for 50% completion is \( 6.93 \text{ s} \), which corresponds to option (A).
\[ \boxed{6.93\text{ s}} \]