The rate law for the reaction,
Ester + H+ → Acid + Alcohol ; is given as
dx
dt = k [Ester] [H+
]0
What would be effect on the rate if
(i) Concentration of the ester is halved ?
(ii) Concentration of H+ is doubled ?
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Answer:
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Explanation:
υ=k[ester][H
3
O
+
]
(a) Conc. of ester is doubled rate also double that is 2υ because rate of the reaction depends upon ester concentration.
(b) Conc. of H
+
is doubled rate does not change that is υ because rate of the reaction does not depends on H
3
O
+
concentration.
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