The conversion of molecules 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

Let the reaction is X →Y
This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]²                                                             … (1)
Let initial concentration is x mol L⁻¹,
Plug the value in equation (1)
Rate, R₁ = k .(a)²
= ka²
Given that concentration is increasing by 3 times so new concentration will 3a mol L⁻¹
Plug the value in equation (1) we get
Rate, R₂ = k (3a)²
= 9ka²
We have already get that R₁ = ka₂ plus this value we get
R₂ = 9 R₁
So that, the rate of formation will increase by 9 times.
Do you know any other method of solving this problem ?
The conversion of molecules 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?
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