Question

The rate of a first order reaction is \( 5.6 \times 10^{-4} \, mol \, L^{-1} \, s^{-1} \), when the concentration of reactant is \( 0.2 \, mol \, L^{-1} \). The rate constant \( 'k' \) is :

• A. \( 5.6 \times 10^{-3} \, s^{-1} \)
• B. \( 2.8 \times 10^{-4} \, s^{-1} \)
• C. \( 2.8 \times 10^{-5} \, s^{-1} \)
• D. \( 2.8 \times 10^{-3} \, s^{-1} \)

Hint:

For first-order reactions, the rate constant \( k \) is independent of the initial concentration.
Always be careful with powers of 10 during division in competitive exams.

Answer 1

The Correct Option is D

Concept:

• For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant raised to the power of one.

• The rate law expression is: \[ \text{Rate} = k[A] \] where \( k \) is the rate constant and \( [A] \) is the molar concentration of the reactant.

• The units of \( k \) for a first-order reaction are always \( \text{time}^{-1} \).
Step 1: Identify the given data
Rate of reaction (\( R \)) = \( 5.6 \times 10^{-4} \, mol \, L^{-1} \, s^{-1} \)
Concentration of reactant (\( [A] \)) = \( 0.2 \, mol \, L^{-1} \)

Step 2: Set up the equation for \( k \)
From the rate law: \[ k = \frac{\text{Rate}}{[A]} \]

Step 3: Perform the calculation
\[ k = \frac{5.6 \times 10^{-4}}{0.2} \] To simplify, rewrite \( 0.2 \) as \( 2 \times 10^{-1} \): \[ k = \frac{5.6 \times 10^{-4}}{2 \times 10^{-1}} \] \[ k = 2.8 \times 10^{-4 - (-1)} \] \[ k = 2.8 \times 10^{-3} \, s^{-1} \]

Step 4: Verify units
The units of rate are \( M \cdot s^{-1} \) and concentration is \( M \).
\[ k = \frac{M \cdot s^{-1}}{M} = s^{-1} \] The calculation and units are consistent with a first-order reaction.