A reaction is of second order with respect to its reactant. How will its reaction rate be affected if the concentration of the reactant is (i) doubled (ii) reduced to half ?

Explanation

Let the concetration of the reactant , [A] =a
Order of reaction =2 so that
Rate fo reactioin will Rate =`k[A]^(2)" "…(i)`
Plug the values we get Rate `=ka^(2)`
(i) Given that concentration of the reactant of the reactant is double
So that `[A]=2a,`
Plug the value in equation (1) we get
New Rate of reaction, `R_(1)=k(2a)^(2)=4ka^(2)`
We have already calculate that initial rate `R=Ka^(2)`
Plud the value we get `R_(1)=4R`
Hence rate of reaction will increased to 4 times
(ii) Given that concentration of the reactant is reduced to half
So that [A]=(1/2)a
Plug the values in equation (1) we get
New Rate of reaction , `R_(2)=k((1//2)a)^(2)`
`=(1//4)ka^(2)`
We have already calculate that initial rate `R=ka^(2)`
Plug the value we get `R_(1)=(1//4)R`
Hence rate of reaction will reduced to 1/4