Question

For the reaction, \(2\text{N}_{2}\text{O}_{5(g)}\rightarrow 4\text{NO}_{2(g)}+\text{O}_{2(g)}\) in liquid bromine, the rate of formation of \(\text{NO}_{2}\) is \(0.08\ \text{mol dm}^{-3}\text{min}^{-1}\). What is the rate of disappearance of \(\text{N}_{2}\text{O}_{5}\)?

• A. \(0.01\ \text{M min}^{-1}\).
• B. \(0.02\ \text{M min}^{-1}\)
• C. \(0.03\ \text{M min}^{-1}\).
• D. \(0.04\ \text{M min}^{-1}\).

Hint:

Relate the rates using the stoichiometric coefficients of the balanced equation.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For \(2\text{N}_{2}\text{O}_{5}\rightarrow4\text{NO}_{2}+\text{O}_{2}\), the reaction rate is \(-\dfrac{1}{2}\dfrac{d[\text{N}_{2}\text{O}_{5}]}{dt}=\dfrac{1}{4}\dfrac{d[\text{NO}_{2}]}{dt}\).

Step 2: Rearrange
\(-\dfrac{d[\text{N}_{2}\text{O}_{5}]}{dt}=\dfrac{2}{4}\dfrac{d[\text{NO}_{2}]}{dt}=\dfrac{1}{2}\dfrac{d[\text{NO}_{2}]}{dt}\).

Step 3: Calculation
\(=\dfrac{1}{2}\times0.08=0.04\ \text{M min}^{-1}\).

Final Answer:
N2O5 disappears at 0.04 M per minute. \[ \boxed{\text{(D) }0.04\ \text{M min}^{-1}} \]