Question

The rate constant for the reaction \(2\text{N}_2\text{O}_5\rightarrow 4\text{NO}_2+\text{O}_2\) is \(3.0\times 10^{-5} \text{s}^{-1}\). If the rate is \(2.4\times 10^{-5} \text{mole L}^{-1} \text{s}^{-1}\) then the concentration of \(\text{N}_2\text{O}_5\,(\text{mol L}^{-1})\) is

• A. \(1.4\)
• B. \(1.2\)
• C. \(0.04\)
• D. \(0.8\)

Hint:

The unit of k shows a first order reaction, so rate = k[N2O5].

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The unit of \(k\) is \(\text{s}^{-1}\), which tells us the reaction is first order in \(\text{N}_2\text{O}_5\). So rate = \(k[\text{N}_2\text{O}_5]\).

Step 2: Detailed Explanation:
\[ 2.4\times10^{-5} = 3.0\times10^{-5}\times[\text{N}_2\text{O}_5] \]
\[ [\text{N}_2\text{O}_5] = \frac{2.4\times10^{-5}}{3.0\times10^{-5}} = 0.8\ \text{mol L}^{-1} \]
Options (A) and (B) are greater than 0.8 and would give a larger rate. Option (C) \(0.04\) would give a rate of only \(1.2\times10^{-6}\) mol L\(^{-1}\)s\(^{-1}\).

Step 3: Final Answer:
\([\text{N}_2\text{O}_5] = 0.8\) mol L\(^{-1}\), option (D).

Final Answer:
Units of k show first order: rate = k times concentration. \[ \boxed{\text{(D) }0.8\ \text{mol L}^{-1}} \]