The reaction between ethyl bromide (c2h5br) and hydroxide ion in ethyl alcohol is shown below at 330 k. c2h5br(alc) + oh −(alc) → c2h5oh(l) + br−(alc) the reaction is first order each in ethyl bromide and hydroxide ion. when [c2h5br] is 0.0478 m and [oh − ] is 0.162 m, the rate of disappearance of ethyl bromide is 2.8 10-7 m/s.

what is the value of the rate constant?

3.62 x 10⁻⁵ M⁻¹s⁻¹.

Explanation:

where, k is the rate constant,

[C₂H₅Br] = 0.0478 M, and [OH⁻] = 0.162 M.

∴ The rate constant (k) = The rate of disappearance of ethyl bromide / [C₂H₅Br][OH⁻] = (2.8 x 10⁻⁷ M/s) / (0.0478 M)(0.162 M) = 3.62 x 10⁻⁵ M⁻¹s⁻¹.

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'

2b2t

explanation:

2b2t