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.
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Comments (1)
A. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\text{k}} + {{\text{k}}_{\text{g}}}$$
B. $${{\text{k}}_{\text{e}}}{\text{ff}} = \frac{{{\text{k}} + {{\text{k}}_{\text{g}}}}}{2}$$
C. $${{\text{k}}_{\text{e}}}{\text{ff}} = {\left( {{\text{k}}{{\text{k}}_{\text{g}}}} \right)^{\frac{1}{2}}}$$
D. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$
The half life period of a first order reaction is given by (where, K = rate constant. )
A. 1.5 K
B. 2.5 K
C. $$\frac{{0.693}}{{\text{K}}}$$
D. 6.93 K
Catalyst is a substance, which __________ chemical reaction.
A. Increases the speed of a
B. Decreases the speed of a
C. Can either increase or decrease the speed of a
D. Alters the value of equilibrium constant in a reversible
A. $$ \propto {\text{CA}}$$
B. $$ \propto \frac{1}{{{\text{CA}}}}$$
C. Independent of temperature
D. None of these

n should be 2