Question

What is the order and molecularity respectively for the elementary reaction given below?

\[ \mathrm{O}_{3(\mathrm{~g})}+\mathrm{O}_{(\mathrm{g})} \longrightarrow 2 \mathrm{O}_{2(\mathrm{~g})} \text { if } \mathrm{r}=\mathrm{k}\left[\mathrm{O}_3\right][\mathrm{O}] \]

• A. \(2^{\text{nd }}\) and 1
• B. \(2^{\text{nd }}\) and 2
• C. \(1^{\text{st }}\) and 2
• D. \(1^{\text{st }}\) and 1

Hint:

For an elementary reaction the order equals the sum of the powers in the rate law, and molecularity equals the number of reacting species.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For an elementary step, molecularity is the number of reacting species colliding.

Step 2: Molecularity
Reactants are \(\text{O}_3\) and \(\text{O}\), so the molecularity is 2.

Step 3: Order
The rate law is \(r=k[\text{O}_3][\text{O}]\), so the order is \(1+1=2\).

Final Answer:
The order is 2 and the molecularity is 2. \[ \boxed{\text{(B) }2^{\text{nd}}\text{ and }2} \]