Read the passage carefully and answer the questions based on the passage:
The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. In the rate equation Rate = $k [A]^x [B]^y$
$x$ and $y$ indicate how sensitive the rate is to the change in concentration of A and B, respectively. Sum of these exponents, i.e., $x + y$ gives the overall order of a reaction where $x$ and $y$ represent the order with respect to the reactants A and B, respectively. Hence, the sum of powers of the concentration of the reactants in the rate law expression is called the order of that chemical reaction. For a first order reaction, the concentration of the reactant varies as $[R] = [R]_0e^{-kt}$
The conversion of molecules X to Y follows second order kinetics. If the concentration of X is increased to four times, how does it affect the rate of formation of Y?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The rate will increase by 16 times
The reaction X → Y follows second-order kinetics.
For a second-order reaction: Rate = k [X]²
Let initial concentration of X = [X] Initial rate = k [X]²
Now, concentration is increased 4 times: New concentration = 4[X]
New rate = k (4[X])² = k × 16 [X]² = 16 × initial rate
∴ The rate of formation of Y becomes 16 times the original rate.