Question

Consider the reaction of ethyl acetate with sodium hydroxide: CH3COOC2H5(aq)+NaOH(aq)⇌CH3COONa(aq)+C2H5OH(aq) The reaction is first order in NaOH and second order overall. What is the rate law? View Available Hint(s) Consider the reaction of ethyl acetate with sodium hydroxide: The reaction is first order in and second order overall. What is the rate law? rate=k[CH3COOC2H5]2[NaOH]2 rate=k[CH3COOC2H5][NaOH]2 rate=k[NaOH] rate=k[CH3COOC2H5] rate=k[CH3COOC2H5][NaOH] rate=k[CH3COOC2H5]2[NaOH]

Answer 1

Answer: The rate law is
`rate=k[CH_3COOC_2H_5]^1[NaOH]^1`

Step-by-step explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

For the given reaction:

`CH_3COOC_2H_5(aq)+NaOH(aq)\rightleftharpoons CH_3COONa(aq)+C_2H_5OH(aq)`

k= rate constant

Rate law:
`rate=k[CH_3COOC_2H_5]^x[NaOH]^y`

For the given rate law:

y =1 = order with respect to
`NaOH`

n = total order = 2

2= (x+y)

2= (x+1)

x= 1

Thus order with respect to
`CH_3COOC_2H_5` is 1 and rate law is :
`rate=k[CH_3COOC_2H_5]^1[NaOH]^1`