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?
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:
Rate1 = k .(a)2
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
Hence, the rate of formation will increase by 9 times.
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