[dhoom3]

Here it is, **** IF ANYTHING, PLEASE 'TRY' TO HELP ME WITH #2 PROBLEM ***

1) (Im sure i have it correct, but it wont hurt to post it now will it?)
At 527 C the quilibrium constant for the reaction COCL₂  ::equil::CO + CL₂ has the value Kc = 4.63x10^-3. The Initial Partial pressure of COCL2 is 0.760atm. Calculate the equilibrium partial pressures for each component.
   What i Got -> Kp = Kc(RT)^delta N
                           delta N = 2-1 = 1
                           temp = 800.15
                           kp =  4.63 x 10^-3 x 0.0821 x 800.15 = 0.304
                           
                           0.304 = P(CO) * P (Cl2) / P (COCl2)
                           0.304 = x^2 / 0.760 - x
                           0.231 - 0.304x = x^2
                           0 = x^2 + 0.304x - 0.231
                           used quadratic equation and got x = .352 or x= -656
                           so the one that is positive is right correct? <----
                           .352 pressure is for CO and CL2
                           .760 - .352 = .354 for COCL2 (correct?)

2) In the reaction 2ICI  ::equil::I₂ + Cl₂ at 682 K, a .682g sample of ICI is placed in a 625 mL reaction vessel. When equilibrium is reached, 0.0383g of I2 is found in the mixture. What is the Kc for this reaction?

Help needed at this spot.

[leaftye]

On #2. 

Remember that PV=nRT.  Reorganize to n/V = P/RT.  You should see that n/V is a concentration.

Convert 0.0383g of I2 to moles.  There will be the same number of moles of Cl2, and twice as many moles of ICl.  Now divide each by volume to get m/v.

Kp=([m/v of I2 * RT][m/v of Cl2 * RT]/[m/v of ICl * RT]^2) x (RT)

Which is the same as:
Kp=([m/v of I2][m/v of Cl2]/[m/v of ICl]^2) x (RT)

Where:
Kc=[m/v of I2][m/v of Cl2]/[m/v of ICl]^2