Question

(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?

Answer 1

Moles I = 3.45 x 10^-2 g/ 126.904 g/mol=0.000272 
initial concentration of I = 0.000272 mol/ 10.0 L= 0.0000272 M 

I2 < => 2 I 
start 
. .. . . 2.72 x 10^-5 
change 
+x .. . . -2x 
at equilibrium 
x .. . . . .2.72 x 10^-5 - 2x 

3.1 x 10^-5 = ( 2.72 x 10^-5-2x)^2 / x 

solve for x = [I2] 
moles I2 = 10.0 x 
mass I2 = moles I2 x 253.808 g/mol 


moles SO3 = 1.30 g/80.066 g/mol=0.0162 
[SO3] = 0.0162 / 2.00 L=0.00810 M 
moles O2 = 0.106 g/ 32 g/mol=0.00331 
[O2]= 0.00331/ 2.00 L=0.00166 M 

Kp = Kc (RT)^delta n 
delta n = 2 - 2 - 1 = -1 
3.0 x 10^4 = Kc ( 0.08206 x 700)^-1 = Kc 0.0174 

Kc = 1.72 x 10^6 = (0.00810)^2 / [SO2]^2 x 0.00166 = 0.0395/ [SO2]^2 

[SO2]= 0.000152 M 

moles SO2 = 0.000152 x 2.00 L=0.000304 
mass SO2 = 0.000304 mol x 64.066 g/mol=0.0195 g