Equlibrium problem 2?

Question

1 mole of COCl2(g) was placed in a 500ml reaction flask. At equ'm the concentration of the product Cl2(g) was 0.7M. The other product is CO(g). Calculate K.

Please show your work! Thanks :D

Answer 1

Initial concentration COCl2 = 1 mole / 0.500 L = 2 M
Concentration at equilibrium = 0.7 M

The equation is :
COCl2 (g) <-----> CO ( g) + Cl2 (g)
Initial concentration
2 M
change
- 0.7. . .. . . . .. .. +0.7 . . . ..+ 0.7
at equilibrium
1.3.. . . . . .. .. .. . . 0.7 . .. . . .0.7
K = [CO] [Cl2] / [ COCl2] = 0.7 x 0.7 / 1.3 = 0.377

Answer 2

Initial conc of phosgene = 1/0.5 = 2M.
The Cl2 has gone up from 0 to 0.7, so the phosgene has gone down from 2 to 1.3. The CO has also gone up to 0.7. So we now have the 3 eq concs.

Kc = 0.7 x 0.7/1.3