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 follows second-order kinetics. The rate law for a second-order reaction is given by: \[ \text{Rate} = k [X]^2 \] Where \( \text{Rate} \) is the rate of the reaction, \( k \) is the rate constant, and \( [X] \) is the concentration of reactant X.

Step 1: Initial Rate

Let the initial concentration of X be \( [X]_0 \). The initial rate of the reaction is: \[ \text{Rate}_0 = k [X]_0^2 \]

Step 2: Increased Concentration

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

Step 3: Comparing the Rates

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

Conclusion:

When the concentration of X is increased by a factor of 3, the rate of formation of Y will increase by a factor of 9.