Question

Consider the reaction, \(2\text{N}_{2}\text{O}_{5}\rightarrow 4\text{NO}_{2}+\text{O}_{2}\). In this reaction, \(\text{NO}_{2}\) is being formed at the rate of \(0.025\) mol \(L^{-1}s^{-1}\). What would be the rate of reaction?

• A. \(0.625\times 10^{-3}molL^{-1}S^{-1}\)
• B. \(0.0625\times 10^{-3}molL^{-1}S^{-1}\)
• C. \(6.25\times 10^{-3}molL^{-1}S^{-1}\)
• D. \(6.25\times 10^{-2}molL^{-1}S^{-1}\)

Hint:

Divide the rate of formation of a species by its stoichiometric coefficient to get the rate of reaction.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a reaction the rate is the rate of change of concentration of any species divided by its stoichiometric coefficient.

Step 2: Write the rate expression
For \(2\text{N}_{2}\text{O}_{5}\rightarrow 4\text{NO}_{2}+\text{O}_{2}\):
\(\text{Rate}=-\dfrac{1}{2}\dfrac{d[\text{N}_2\text{O}_5]}{dt}=\dfrac{1}{4}\dfrac{d[\text{NO}_2]}{dt}=\dfrac{d[\text{O}_2]}{dt}\)

Step 3: Calculation
\(\text{Rate}=\dfrac{1}{4}\times 0.025=6.25\times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}\)

Final Answer:
The rate of reaction is 6.25 times ten to the power minus three mol per litre per second. \[ \boxed{\text{(C) }6.25\times 10^{-3}\ \text{mol L}^{-1}\text{s}^{-1}} \]