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

(a) `r = k[Ester]^(1) [H_(3)O^(o+)]^(0)`
`r_(1) = k[a]^(1) [b]^(0)`
`r_(2) = k[2a]^(1)[b]^(0)`
`(r_(1))/(r_(2)) = (1)/(2), r_(2) = 2r_(1)`
(b) `r_(3) = k[a]^(1) [2b]^(0)`
`(r_(1))/(r_(3)) = 1, r_(1) = r_(3)`