Chemistry Question!!
posted by Trixie .
When 1.00 g of gaseous I2 is heated to 1000. K in a 1.00 L sealed container, the resulting equilibrium mixture contains 0.830 g of I2. Calculate Kc for the dissociation equilibrium below.
I2 <> 2 I
What I did was converted the initial I2 into moles then molarity since it is just over 1 L (.00394 M) then I found the molarity of I by subtracting .830 from 1 and converting first to moles (.00134 M of I) I then set up Kc= [.00134 M of I]^2/ [.00394 M of I2] which i found to be .00045...but it's incorrect. I'm not sure what I am missing...

Your procedure appears to be ok down to the last step. The divisor should be the 0.830/253.8 = 0.00327 and not 0.00395 since the 0.830 g was the equilibrium amount. Also, watch the number of significant figures. That sometimes causes problems.

Realize there are 0.830 g of I2 at equilibrium... not I