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Question
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?
Solution
The reaction \[\ce{X -> Y}\] follows second order kinetics.
Therefore, the rate equation for this reaction will be:
Rate = k[X]2
If the concentration of X is increased to three times, then [r'] = k[3X]2
`("r'")/"r" = ("k"[3"X"]^2)/("k"["X"]^2)` = 9
Hence, the rate of formation will increase by 9 times.
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