Calculate Equilibrium Constant of 2BrCl <=> Br2 + Cl2

[markosheehan]

bromine monochloride dissociated on heating the following equation

2BrCl <=> Br2 + Cl2

.9 moles of BrCl were heated in a 5 liter vessel until equilibrium was established . the amount of free chlorine in the mixture was found to be .4 moles

calculate the equilibrium constant for the reaction.

i thought this would be the answer kc=(.4*.4)/(.8)^2

if there are .4 moles of cl2 there must be .4 moles of Br2

there are 2 moles of 2BrCl for every mole of cl2 so there must be .8 mole of BrCl

sadly this does not give the right answer which is kc=16

 

[I like Serena]

if there are .4 moles of cl2 there must be .4 moles of Br2

there are 2 moles of 2BrCl for every mole of cl2 so there must be .8 mole of BrCl

Careful here.

We are indeed dealing with 0.8 moles of $\ce{BrCl}$.
But those 0.8 moles of $\ce{BrCl}$ were dissociated in the reaction $\ce{2BrCl <=> Br2 + Cl2}$.
It means that from the original 0.9 moles, there are 0.1 moles left.
So in equilibrium we have 0.1 moles of $\ce{BrCl}$, 0.4 moles of $\ce{Br2}$, and 0.4 moles of $\ce{Cl2}$.

That makes the equilibrium constant:
$$K_c = \frac{0.4\cdot 0.4}{0.1^2} = 16$$

 

[markosheehan]

thanks.

isnt the concentration always supposed to be in moles per liter so should we divide by 5 as its in a 5 litre vessel. the answer at the back of the book is 16 though.

the next part of the question is explain why it is not necessary to know the volume of the solution when calculating the equilibrium constant.

i am not sure what i should say.

 

[I like Serena]

thanks.

isnt the concentration always supposed to be in moles per liter so should we divide by 5 as its in a 5 litre vessel. the answer at the back of the book is 16 though.

the next part of the question is explain why it is not necessary to know the volume of the solution when calculating the equilibrium constant.

i am not sure what i should say.

The divisions by the volume cancel in this case, because we have the same number of molecules on both sides of the equation.
With concentrations in mole per liter the equilibrium constant is:
$$K_c = \frac{\frac{0.4\text{ mol}}{\cancel{5\text{ L}}}\cdot \frac{0.4\text{ mol}}{\cancel{5\text{ L}}}}
{\left(\frac{0.1\text{ mol}}{\cancel{5\text{ L}}}\right)^2} = 16$$