Question

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?

Answer 1

3.62 x 10⁻⁵ M⁻¹s⁻¹.

Step-by-step explanation:

• Since the reaction is first order each in ethyl bromide and hydroxide ion.

• The rate of disappearance of ethyl bromide = k[C₂H₅Br][OH⁻],

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.