Question

The reaction of ethyl acetate with sodium hydroxide, ch3cooc2h5(aq)+naoh(aq)⇌ch3coona(aq)+c2h5oh(aq) is first order in ch3cooc2h5 and first order in naoh. if the concentration of ch3cooc2h5 was increased by half and the concentration of naoh was quadrupled, by what factor would the reaction rate increase? express your answer numerically.

Answer 1

The reaction rate will increase by a factor of 6

Rate2 = 6(Rate1)

Step-by-step explanation:

The given reaction is:

`CH3COOC2H5(aq)+NaOH(aq)\rightleftharpoons CH3COONa(aq)+C2H5OH(aq)`

`Rate = k[CH3COOC2H5]^(m)[NaOH]^(n)`

k = rate constant

m = order w.r.t to
`CH3COOC2H5` = 1

n = order w.r.t to
`NaOH` = 1

Therefore, the initial rate is:

`Rate1 = k[CH3COOC2H5][NaOH]`

It is given that
`[CH3COOC2H5]` is increased by 1/2. Therefore the new concentration =
`[CH3COOC2H5] + (1)/(2)[CH3COOC2H5] = 1.5 [CH3COOC2H5]`

It is given that the concentration of
`NaOH` was quadrupled. Similarly, new concentration of
`NaOH` is =
`4[NaOH]`

Therefore, the new rate is:

`Rate2 = k[1.5CH3COOC2H5][4NaOH]`

`(Rate2)/(Rate1) = (1.5)(4)= 6\\\\Rate2 = 6(Rate1)`