The reaction between ethyl bromide (C₂H₅Br) and hydroxide ion in ethyl alcohol at 330 K, C_2H_5Br (alc)+OH^- (alc) \rightarrow C_2H_5OH(l)+Br^-(alc), is first order each in ethy…

Question

The reaction between ethyl bromide (C₂H₅Br) and hydroxide ion in ethyl alcohol at 330 K,
`C_2H_5Br (alc)+OH^- (alc) \rightarrow C_2H_5OH(l)+Br^-(alc)`, is first order each in ethyl bromide and hydroxide ion. When [C₂H₅Br] is 0.0477 M and [OH⁻] is 0.100 M, the rate of disappearance of ethyl bromide is 1.7×10⁺⁷M/s.

Answer 1

How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

If the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution, then thee rate of reaction will decrease by a factor of 4 to 4.25×10⁶ M/s

Step-by-step explanation:

For a first order reaction we have

Rate =
`k_(c)`[A] where [A] is the concentration of the reagent

Hence the reaction between (C₂H₅Br) and OH⁻ is given by

Rate = Kc[C₂H₅Br][OH⁻]

The addition of pure ethyl which only dilute the concentration of the reactants we have

Final concentration = (initial concentration)/ 2

Therefore the rate becomes

Rate =
`k_(c) ([C_(2) H_(5) Br])/(2) ([OH^(-) ])/(2)` =
`(1)/(4)`×Kc[C₂H₅Br][OH⁻]

Then the rate of reaction is divided by 4 thus

The rate of the reaction =
`(1.7*10^(+7) )/(4) = 4.25*10^(6) M/s`