Question

The conversion of molecule X to Y follows second-order kinetics. If the concentration of X is increased 3 times, how will it affect the rate of formation of Y? 

Hint:

In a second-order reaction, the rate of the reaction is proportional to the square of the concentration of the reactant.

Answer 1

For a second-order reaction, the rate of the reaction is proportional to the square of the concentration of reactant \(X\): \[ \text{Rate} = k \cdot [X]^2 \] If the concentration of \(X\) is increased 3 times, the rate of formation of \(Y\) will increase by a factor of: \[ \left( 3 \right)^2 = 9 \] Thus, the rate of formation of \(Y\) will increase by a factor of 9 when the concentration of \(X\) is increased 3 times. \vspace{10pt}