Question

The reaction between ethyl bromide (C₂H₅Br) and hydroxide ion in ethyl alcohol at 330 K, C₂H₅BR(alc) + OH-(alc) --> C₂H₅OH(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 x 10⁻⁷ M/s.What is the value of the rate constant?

k=?

Answer 1

The value of the rate constant (k) for the reaction between ethyl bromide and hydroxide ion is calculated to be 3.56 x 10^-5 s^-1 by rearranging the rate equation k = rate / ([C2H5Br][OH-]).

Step-by-step explanation:

The question at hand is calculating the rate constant (k) for the reaction between ethyl bromide (C2H5Br) and hydroxide ion that is first order in both reactants.

To find the rate constant for a reaction where rate = k [C2H5Br][OH-], we rearrange the rate equation to solve for k: k = rate / ([C2H5Br][OH-]). Given that the rate of disappearance of ethyl bromide is 1.7 x 10-7 M/s when [C2H5Br] is 0.0477 M and [OH-] is 0.100 M, we can calculate the rate constant (k) accordingly.

k = (1.7 x 10-7 M/s) / (0.0477 M x 0.100 M) = 3.56 x 10-5 s-1

Thus, the value of the rate constant is 3.56 x 10-5 s-1.