The conversion of molecules `X` to `Y` follows second order kinetics. If the concenration of `X` is increased to three times, how will it affect the rate of formation of `Y` ?

Options

A. increased by three times
B. decreased by three times
C. increases by nine times
D. decrease by nine times

Answer

C

Explanation

For second order kinetics,
`"rate"=k[reactant]^(2)`
So make two equation first by taking initial rate and concentration, second by taking tripled concentration and unknown rate (r'). Compare them to find a relation between r and r'.
For a reaction, `xrarry`
`"Reaction rate"(r)=k[x]^(2)`
If the concentration of x is increased three times, then Reaction rate (r')`=k[3x]^(2)=kxx[9x^(2)]`
On dividing Eq. (ii) and (i), we get
`(r^(1))/(r)=(kxx[9x^(2)])/(kxx[x^(2)])=9`
It means that the rate of formation of y will increase by nine times.