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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? - Chemistry

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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?

Numerical
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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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Chapter 3: Chemical Kinetics - Intext Questions [Page 71]

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NCERT Chemistry Part 1 and 2 [English] Class 12
Chapter 3 Chemical Kinetics
Intext Questions | Q 3.4 | Page 71

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