The rate law for the following reactions:
`Ester + H^(o+) rarr "Acid" + "Alcohol"`, is
`dx//dt = k(ester)[H_(3)O^(o+)]^(0)`
What would be the effect on the rate if
(a) concentration of ester is doubled.
(b) concentration of `H^(o+)` ion is doubled.

Explanation

`:' Rate =K [ester]^(1) [H_(3)O^(+)]^(0)`
(a) `r_(1)=K [ester]^(1) [H_(3)O^(+)]^(0)`
Let, Initial conc. Of ester `=a`, initial conc. of `H_(3)O^(+)=b`
`:. r_(1)=K[a]^(1)[b]^(0)`
If conc. of ester is doubled, i.e., `[ester]=2a`, then
`r_(2)=K[2xxa]^(1) [b]^(0)`
`:. r_(1)/r_(2)=1/2`
or `r_(2)=2r_(1)`
(b) `r_(1)=K[a]^(1)[b]^(0)`
If conc. `H_(3)O^(+)` is doubled i.e., `[H_(3)^(+)O]=2b`
`r_(3)=[a]^(1) [2b]^(0)`
`:. r_(1)/r_(3)=1` or `r_(1)=r_(3)`