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.

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.