For the chemical reaction X → Y, it is observed that, on doubling the concentration of 'X', the reaction rate quadruples. If the reaction rate is proportional to Cxn, then what is the value of 'n'?

Answer: Option A

Step 1: Understanding the given information.

The reaction given is X → Y, and it follows the rate equation:

Rate = k * C_xⁿ

It is observed that when the concentration of 'X' is doubled, the reaction rate quadruples.

Step 2: Setting up the rate equation.

Let the initial concentration of 'X' be C_x. The initial rate is:

Rate₁ = k * (C_x)ⁿ

When the concentration is doubled, the new rate is:

Rate₂ = k * (2C_x)ⁿ

Step 3: Establishing the relationship.

Since it is given that the reaction rate quadruples:

Rate₂ = 4 * Rate₁

Substituting the expressions:

k * (2C_x)ⁿ = 4 * (k * (C_x)ⁿ)

Step 4: Solving for 'n'.

Dividing both sides by k * (C_x)ⁿ:

(2C_x)ⁿ / (C_x)ⁿ = 4

2ⁿ = 4

Since 4 = 2², we get:

n = 2

Final Answer: The value of 'n' is 2.

The correct answer is Option D: 2.