. Kc = 2.19*10-10 @ 100 oC for the rxn: COCl2(g)  CO(g) + Cl2(g) the following mixtures may or may not be at equilibrium. Identify which mixtures are at equilibrium, if not, indicate which direction the reaction will go to obtain equilibrium (don’t guess, no credit given without supporting evidence). a) [CO] = 1.0*10-3, [Cl2] = 1.0*10-3, [COCl2]= 2.19*10-1 b) [CO] = 3.31*10-6, [Cl2] = 3.31*10-6, [COCl2]= 5.00*10-2 c) [CO] = 4.5*10-7, [Cl2] = 5.73*10-6, [COCl2]= 8.57*10-2

Respuesta :

Answer: a) Not in equilibrium and the reaction will go to the right; b) NOt in equilibrium with reaction going to the left; c) Not in equillibrium with reaction also going to the left.

Explanation: Kc is an equilibrium constant which is calculated using the molar concentrations of each element, which are in gaseous form, of the reaction.

For the reaction in question: COCl2(g) ⇄ CO(g) + Cl2(g),

the Kc is: Kc = [tex]\frac{[CO][Cl2]}{[COCl2]}[/tex]

a) Kc1 = [tex]\frac{1.10^{-3}. 1.10^{-3} }{2.19.10^{-1} }[/tex] = 0.456.[tex]10^{-5}[/tex]

Comparing  Kc1 with Kc in equilibrium (Kc=2.19.[tex]10^{-10}[/tex]), it can be deducted that it is not in equilibrium, since Kc1≠Kc and Kc1 is larger than Kc. When that occurs, it means the reaction is favoring the products, producing more of it. So, the equilibrium is going to the right.

b) Kc2 = [tex]\frac{3.31.10^{-6} . 3.31.10^{-6} }{5.10^{-2} }[/tex] = 0.662.[tex]10^{-10}[/tex]

Again, Kc2 and Kc are different but, this time, Kc2 is smaller than in equilibrium, so it means the reaction is going towards the reactants, in other words, it is to the left.

c) Kc3 = [tex]\frac{4.5.10^{-7}.5.73.10^{-6} }{8.57.10^{-2} }[/tex] = 3.01.[tex]10^{-11}[/tex]

With those concentrations, this reaction is not in equilibrium and, as Kc3 is smaller than Kc, the equilibrium is towards the reactants or to the right.