when a 3.00 mol dm-3 sample of PCL5 is heated to 250 degrees the decomposition

PCL5 (g) <------>  PCL3 (g) + CL2  (g)

occurs and reaches equilibrium with Kc = 1.80.  What is the composition of the eqilibirum mixture? what percentage of  PCL5 has decomposed at 250 degrees??

can any1 help me with this question??

i have no idea of how to even start this problem!!!!

when a 3.00 mol dm-3 sample of PCL5 is heated to 250 degrees the decomposition

PCL5 (g) <------>  PCL3 (g) + CL2  (g)

occurs and reaches equilibrium with Kc = 1.80.  What is the composition of the eqilibirum mixture? what percentage of  PCL5 has decomposed at 250 degrees??

can any1 help me with this question??

i have no idea of how to even start this problem!!!!

the same question i answered on another forum. Here's the answer:

Kc = [PCl3][Cl2] / [PCl5]

we start with 3,00 M PCl5 and due to the decomposition x mol PCl5 reacted and resulted into the forming of x mol Cl2 and PCl3. So we get the following equation (hence all molecules are in a same volume so concentration can be replaced by mols):

1,8 = x^2 / (3 - x) which we need to solve for x ( mol reacted PCl5 and thus mols Cl2 and PCl3)

--> 5,4 - 1,8x = x^2 --> x^2 + 1,8x - 5,4 = 0

now we get the values x as x(1) = (-1,8 + sqrt 24,84) / 2 = 1,591987159

and x(2) = (-1,8 - sqrt 24,84) / 2 = -3,391987159 which does not satisfie, because there can't be formed any PCl5.

That's why 1,591987159 mol PCl5 reacted resulting in the following mols molecules in equilibrium:

1,308012841 mol PCl5 and 1,591987159 mol Cl2 and PCl3. That means that 1,591987159 /3 * 100 = 53,1 % PCl5 is decomposed.