Question

For the reaction
\(\text{C}_{2}\text{H}_{5}\text{Br}_{(l)}\rightarrow \text{C}_{2}\text{H}_{4(g)}+\text{HBr}\), \(r=K(\text{C}_{2}\text{H}_{5}\text{Br})\)
What is order and molecularity of the reaction respectively ?

• A. \(1,2\)
• B. \(2,1\)
• C. \(2,2\)
• D. \(1,1\)

Hint:

Order is the sum of the powers in the rate law; molecularity counts molecules in the elementary step.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Order is the sum of the powers of the concentration terms in the experimental rate law. Molecularity is the number of molecules colliding in the elementary step.

Step 2: Order
The rate law is \(r=K[\text{C}_2\text{H}_5\text{Br}]\), so the order is 1.

Step 3: Molecularity
A single molecule of \(\text{C}_2\text{H}_5\text{Br}\) decomposes, so the step is unimolecular and the molecularity is 1.

Final Answer:
The order is 1 and the molecularity is 1. \[ \boxed{\text{(D) }1,\ 1} \]