Examveda

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)

  1. Mukunda Madhab
    Mukunda Madhab:
    8 months ago

    n should be 2

Related Questions on Chemical Reaction Engineering

A first order gaseous phase reaction is catalysed by a non-porous solid. The kinetic rate constant and the external mass transfer co-efficients are k and $${{\text{k}}_{\text{g}}}$$ respectively. The effective rate constant (keff) is given by

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. $$\frac{1}{{{{\text{k}}_{\text{e}}}{\text{ff}}}} = \frac{1}{{\text{k}}} + \frac{1}{{{{\text{k}}_{\text{g}}}}}$$