The equilibrium constant, Kc, for the following reaction is 77.5 at 600 K. CO(g) Cl2(g) COCl2(g) Calculate the equilibrium concentrations of reactant and products when 0.555 moles of CO and 0.555 moles of Cl2 are introduced into a 1.00 L vessel at 600 K.

[CO] = 0.078M

[Cl2] = 0.078M

[COCl2] = 0.477M

Explanation:

Based on the reaction:

CO(g) Cl2(g) ⇄ COCl2(g)

Where equilibrium constant, kc, is:

kc = 77.5 = [COCl2] / [CO] [Cl2]

[] represents the equilibrium concentration of each gas. The initial concentration of each gas is:

[CO] = 0.555mol/1.00L = 0.555M

[Cl2] = 0.555M

And equilibrium concentrations are:

[CO] = 0.555M - x

[Cl2] = 0.555M - x

[COCl2] = x

Where x is reaction coordinate

Replacing in kc expression:

77.5 = [x] / [0.555M - x] [0.555M - x]

77.5 = x / 0.308025 - 1.11 x + x²

23.8719 - 86.025 x + 77.5 x² = x

23.8719 - 87.025 x + 77.5 x² = 0

x = 0.477M. Right answer

x = 0.646M. False answer. Produce negative concentrations

Replacing:

And those concentrations are the equilibrium concentrations

answer; protein;

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.