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

Effect of Increased Concentration on the Rate of Formation of Y

The reaction exhibits second-order kinetics. The rate law for such a reaction is expressed as: \[ \text{Rate} = k [X]^2 \] where \( \text{Rate} \) denotes the reaction rate, \( k \) is the rate constant, and \( [X] \) represents the concentration of reactant X.

Step 1: Initial Rate

Assuming the initial concentration of X is \( [X]_0 \), the initial reaction rate is: \[ \text{Rate}_0 = k [X]_0^2 \]

Step 2: Increased Concentration

Upon tripling the concentration of X, the new concentration becomes \( 3[X]_0 \). The resultant new rate is calculated as: \[ \text{Rate}_{\text{new}} = k (3[X]_0)^2 = k \times 9 [X]_0^2 \]

Step 3: Comparing the Rates

The ratio of the new rate to the initial rate is determined as follows: \[ \frac{\text{Rate}_{\text{new}}}{\text{Rate}_0} = \frac{k \times 9 [X]_0^2}{k [X]_0^2} = 9 \]

Conclusion:

A threefold increase in the concentration of X results in a ninefold increase in the rate of formation of Y.