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

Answer:

The reaction rate will increase by a factor of 6

Rate2 = 6(Rate1)

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 = $[CH3COOC2H5] + \frac{1}{2}[CH3COOC2H5] = 1.5 [CH3COOC2H5]$

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

Therefore, the new rate is:

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

$\frac{Rate2}{Rate1} = (1.5)(4)= 6\\\\Rate2 = 6(Rate1)$

Answer 2

This problem involves the rate law of a chemical reaction relates the concentrations or pressures of the reactants with the rate of reaction together with some constant parameters. In this problem, we turn this into a mathematical problem by first assigning variables to the reactants.

Let:

x = concentration of ethyl acetate, [CH3COOC2H5]
y = concentration of sodium hydroxide, [NaOH]

The rate law is then equal to rate = xy since the reaction is first order in both reactants. It was then said that the concentration of ethyl acetate was increased by half and the concentration of NaOH was quadrupled. In mathematical form, this translates to the following expression:

rate = 0.5x*4y = 2xy

Thus, the reaction rate increased by a factor of 2.