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# 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 ? Sunil Kumar FP
7 years ago
X-------Y
The conversion follows second order equation
rate=k[X]^2
If the concentration of X is increased 3 times then the rate of formation
R=k(3X)2=9K[X]^2
Thus the final rate =9 times the initian rate
is increasedX^2 times that is 9 times.
3 years ago

The order of reaction is defined as the sum of the powers of concentrations In the rate law.

The rate of second order reaction can be expressed as     rate = k[A]2

The reaction X → Y follows second order kinetics.

Therefore, the rate equation for this reaction will be:

Rate = k[X]2                                                    ________________________   (1)

Let [X] = a mol L-1 , then equation (1) can be written as:

Rate = k [a]2

ka2

If the concentration of X is increased to three times, then [X] = 3a mol L-1

Now, the rate equation will be:

Rate = k [3a]2

= 9(ka2)

Hence, the rate of formation will increase by 9 times.