posted by Hannah .
(a) At 800 K the equilibrium constant for I2(g) 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.25 10-2 g of I(g), how many grams of I2 are in the mixture?
(b) For 2 SO2(g) + O2(g) 2 SO3(g), Kp = 3.0 104 at 700 K. In a 2.00-L vessel the equilibrium mixture contains 1.46 g of SO3 and 0.130 g of O2. How many grams of SO2 are in the vessel?
how do you do these?
3.25E-2/10L = 2.56E-4 mol and
2.56E-4/10 = 2.56E-5 M
........I2 ==> 2I
Kc = (I)^2/(I2)
You know Kc and I, solve for I2 in molarity, then convert to grams in the 10 L container.
1.46/80 = about 0.01825 mols SO3.
0.130/32 = about 0.0041 mols O2.
............2SO2 + O2 ==> 2SO3
Why can't you use PV = nRT to solve for pSO2 and again for pO2. Substitute these numbers in Kp expression and solve for pSO2. Finally, use PV = nRT and convert pSO2 to n and to grams. Post your work if you get stuck.
You have to convert the concentrations to molarity
3.25 x10^-2 g/126.9g*mol-1=moles of I
moles of I/10L=molarity
2I=(moles of I2/10L)^2
Since Kc =3.1 x10^-5=products/reactants,
Kc=3.1 x10^-5=(moles of 2I/10L)^2/[x]
Solve for x
to get grams of I2,
Molarity*10L=moles of I2
moles of I2*253.8g*mol-1= g of I2
2SO2(g)----> O2(g) + 2SO3(g)
You have to convert everything to molarity. I showed you how to do this in A, and you have to find the molecular weights of the compounds in question.
Kp = 3.0 x 10^4 =products/reactants=(SO3)^2(O2)/[2x]^2
solve for x
Multiply x by 2 to get molarity of SO2 and use the same setup that I gave you in A to get grams of SO2
Didn't know that you were answering it.
Note that in a you must convert to M as I did.
Note that in b you do not need to convert to M because part b is Kp and not Kc. Pressure is what you want in part b and molarity won't do it unless you wish to use molarity to find pressure.
Kp=is pressure, I apologize about that one. So, you have to use the setup that DrBob222 gave you in his initial post.