Question

The rate constant for a zero order reaction \( A \rightarrow B + C \) is \( 6.0 \times 10^{-3} \, \text{mol L}^{-1} \, \text{s}^{-1} \). What would be the time taken for the initial concentration of A to decrease from 0.2 M to 0.024 M?

• A. 15.83 s
• B. 37.34 s
• C. 31.90 s
• D. 29.33 s

Hint:

For zero-order reactions, the concentration decreases linearly with time. The time required for a specific decrease in concentration can be found using the integrated rate law.

Answer 1

The Correct Option is D

Step 1: Zero-order reaction rate law.
For a zero-order reaction, the rate law is given by:
\[ \text{Rate} = k \]
where \( k \) is the rate constant, and the concentration of the reactant decreases linearly with time. The integrated rate law for a zero-order reaction is:
\[ [A] = [A]_0 - kt \]
where:
- \( [A]_0 \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time taken.

Step 2: Rearranging the equation.
Rearranging the equation to solve for \( t \), we get:
\[ t = \frac{[A]_0 - [A]}{k} \]
where:
- \( [A]_0 = 0.2 \, \text{M} \) (initial concentration),
- \( [A] = 0.024 \, \text{M} \) (final concentration),
- \( k = 6.0 \times 10^{-3} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

Step 3: Substitute the given values.
Substitute the given values into the equation: \[ t = \frac{0.2 \, \text{M} - 0.024 \, \text{M}}{6.0 \times 10^{-3} \, \text{mol L}^{-1} \, \text{s}^{-1}} \]

Step 4: Calculate the time.
Perform the calculation: \[ t = \frac{0.176 \, \text{M}}{6.0 \times 10^{-3} \, \text{mol L}^{-1} \, \text{s}^{-1}} = 29.33 \, \text{s} \]

Step 5: Conclusion.
Therefore, the time taken for the concentration of A to decrease from 0.2 M to 0.024 M is 29.33 seconds. The correct answer is option (D).