Question

The reaction \( \text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \text{O}_2 \) is first order in \( \text{N}_2\text{O}_5 \), having rate constant \( 6 \times 10^{-4} \, \text{L}^{-1} \, \text{s}^{-1} \). What is the value of rate of reaction when concentration of \( \text{N}_2\text{O}_5 \) is 1.25 mol L\(^{-1}\)?

• A. \( 7.75 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \)
• B. \( 8.15 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \)
• C. \( 4.96 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \)
• D. \( 2.01 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \)

Hint:

For first-order reactions, the rate is directly proportional to the concentration of the reactant. Use the rate law to calculate the rate of reaction.

Answer 1

The Correct Option is B

Step 1: First-order rate law.
For a first-order reaction, the rate law is given by: \[ \text{Rate} = k \times [\text{N}_2\text{O}_5] \] Where \( k \) is the rate constant and \( [\text{N}_2\text{O}_5] \) is the concentration of the reactant.
Step 2: Substituting values.
Substitute the given values: \[ \text{Rate} = (6 \times 10^{-4}) \times (1.25) = 8.15 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \] Step 3: Conclusion.
The correct answer is (B) \( 8.15 \times 10^{-4} \, \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \).