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 ?

Let the reaction be X→Y

As, this reaction follows second order kinetics, the rate of the reaction will be,

Rate = k(X)^{2} → Equation 1

Since the concentration of X is increased three times, Equation 1

will become,

Rate = k(3X)^{2}

= k × 9(X)^{2}

∴ The rate of formation of Y will become 9 times.

__Thus, the rate of formation of Y will become 9 times__

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