![The equilibrium constant, Kc. for the following reaction is 1.29×102 at 600 K. COC12(g) = CO(g) + Ch(g) Calculate the equilibrium concentrations of reactant and products when 0.305 moles of COCL() are introduced into a 1.00 L vessel a 600 K CoC] Co]](http://img.homeworklib.com/questions/ff931a00-a39a-11eb-b9c9-8fce2c4f07da.png?x-oss-process=image/resize,w_560)
Homework Answers
concentration of CoCl2= moles/Volume in L=0.305moles/1L=0.305M
the reaction is CoCl2(g)<------>Co(g)+Cl2(g)
preparing the ICE table and noting that =x concentration of CO ( or Cl2) at equilibrium
initial (M) change equilibrium
CoCl2 0.305 -x 0.305-x
Co 0 x x
Cl2 0 x x
K= Equilibrium constant = x2/(0.305-x)= 1.29*10-2,when solved using excel, x=0.0566
hence at Equilibrium, [Co]=[Cl2]=0.0566 and [CoCl2]=0.305-0.0566=0.2484M
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