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'?
A. 4
B. $$\frac{1}{4}$$
C. 16
D. 2
Answer: Option A
Solution (By Examveda Team)
Step 1: Understanding the given information.The reaction given is X → Y, and it follows the rate equation:
Rate = k * Cxn
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 Cx. The initial rate is:
Rate1 = k * (Cx)n
When the concentration is doubled, the new rate is:
Rate2 = k * (2Cx)n
Step 3: Establishing the relationship.
Since it is given that the reaction rate quadruples:
Rate2 = 4 * Rate1
Substituting the expressions:
k * (2Cx)n = 4 * (k * (Cx)n)
Step 4: Solving for 'n'.
Dividing both sides by k * (Cx)n:
(2Cx)n / (Cx)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.
This Question Belongs to Chemical Engineering >> Chemical Reaction Engineering
n should be 2