Conversion of molecule X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the rate will increase 9 times
For a second-order reaction, the rate law is:
$\text{Rate} = k [X]^2$
- If [X] is tripled ([X]→3[X]), then:
$\text{New rate} = k (3[X])^2 = 9 k [X]^2$
The rate increases 9 times