CH3Cl(aq)+OH(aq)=CH3OH(aq)+Cl Equilibrium constant Kc is 1*10^16 Calculate the equilibrium concentrations of CH3Cl, CH3OH, OH- and C1- in a solution prepared by mixing equal volumes of 0.1 M CH3Cl and 0.4 M NaOH. (Hint: In defining , assume that the reaction goes 100% to completion, and then take account of a small amount of the reverse reaction.)

[CH₃Cl] = 1,3×10⁻¹⁸

[CH₃OH] = 0,05

[Cl⁻] = 0,05

[OH⁻] = 0,2M

Explanation:

For the reaction:

CH₃Cl(aq) + OH⁻(aq) ⇄ CH₃OH(aq) + Cl⁻

Kc is defined as:

Kc = 1x10¹⁶ = [CH₃OH] [Cl⁻] / [CH₃Cl] [OH⁻] (1)

As the solution is mixed with equal volumes, concentration of reactants decreases in the half

Assuming the reaction goes 100% to completion, equilibrium concentrations are:

[CH₃OH] = 0,05-X

[Cl⁻] = 0,05-X

[CH₃Cl] = X

[OH⁻] = 0,2M - X

Replacing in (1):

1x10¹⁶ = [0,05-X] [0,05-X] / [X] [0,2-X]

1x10¹⁶ = [X² - 0,1X + 0,0025] / [0,2X - X²]

- 1x10¹⁶X² + 2x10¹⁵X - 0,0025= 0

Solving for X:

X = 1.3×10⁻¹⁸

That means equilibrium concentrations are:

[CH₃OH] = 0,05 - 1,3×10⁻¹⁸ ≈ 0,05M

[Cl⁻] = 0,05 - 1,3×10⁻¹⁸ ≈ 0,05M

[CH₃Cl] = 1,3×10⁻¹⁸

[OH⁻] = 0,2M - 1,3×10⁻¹⁸ ≈ 0,2M

I hope it helps!

your answer is a. the enthalpy of formation for a pure element in its standard state is always positive.