Question

In a reaction,
\(2\text{N}_2\text{O}_{5(g)}\rightarrow 4\,\text{NO}_{2(g)}+\text{O}_{2(g)}\)
\(\text{N}_2\text{O}_5\) disappears at a rate of \(0.06\) mol dm\(^{-3}\) s\(^{-1}\)
Calculate rate of formation of \(\text{O}_{2(g)}\) ?

• A. \(0.02\) mol dm\(^{-3}\) s\(^{-1}\)
• B. \(0.03\) mol dm\(^{-3}\) s\(^{-1}\)
• C. \(0.04\) mol dm\(^{-3}\) s\(^{-1}\)
• D. \(0.06\) mol dm\(^{-3}\) s\(^{-1}\)

Hint:

Divide each rate by its stoichiometric coefficient to get the same reaction rate.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a reaction \(aA \to bB + cC\), the rate of reaction is \(-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt}\).

Step 2: Setting up:
For \(2\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2\):
\[ -\frac{1}{2}\frac{d[\text{N}_2\text{O}_5]}{dt} = \frac{1}{4}\frac{d[\text{NO}_2]}{dt} = \frac{d[\text{O}_2]}{dt} \]

Step 3: Detailed Explanation:
\(\text{N}_2\text{O}_5\) disappears at 0.06 mol dm\(^{-3}\) s\(^{-1}\), so
\[ \frac{d[\text{O}_2]}{dt} = \frac{1}{2} \times 0.06 = 0.03 \text{ mol dm}^{-3}\text{s}^{-1} \]

Step 4: Why the other options are wrong.
0.06 (D) would hold only if the coefficients of \(\text{N}_2\text{O}_5\) and \(\text{O}_2\) were equal. 0.02 and 0.04 do not match the 2:1 ratio of \(\text{N}_2\text{O}_5\) used to \(\text{O}_2\) formed.

Final Answer:
Rate of formation of \(\text{O}_2\) is \(0.03\) mol dm\(^{-3}\) s\(^{-1}\), option (B). \[ \boxed{0.03 \text{ mol dm}^{-3}\text{s}^{-1}} \]