Correct question:
An 8.10-g sample of SO3 was placed in an evacuated container, where it decomposed at 590°C according to the following reaction:
SO3(g) <-----> SO2(g) + 1/2 O2 (g)
At equilibrium the total pressure and the density of the gaseous mixture were 1.83 atm and 1.57 g/L respectively. Calculate Kp for this reaction
Answer:
Kp for this reaction is 0.149atm
Step-by-step explanation:
Given details
The state reaction
SO3(g) <-----> SO2(g) + 1/2 O2 (g)
Density = 1.57 g/L
Temperature = 590°C = 863K
The given mass of SO3 is 8.10g
The molar mass of SO3 is
S + 3O = {(32) + 3(16)} = 80g/mol
Numbers of mole =
Given mass/molar mass = 8.10/80
Numbers of mole of SO3 = 0.1013mol
From density = mass/volume
Volume V = 8.10/1.56 = 5.2L
Initial pressure from PV = nRT
R = Universal gas constant
R = 0.0821 atm/K/mol
P = (0.1013*0.0821*863)/5.2
P = 1.38 atm
At equilibrium
moles SO3 = 0.10 - X
moles SO2 = X
moles O2 = X/2
moles total = (0.10 - X) + X + X/2
Total mole = 0.10 + X/2
Ptot = (0.10 + X/2)*0.0821*863/5.2 = 1.83
(0.10 + X/2)* 70.8523 = 9.516
X/2 = 0.1343 -0.10 = 0.0343
X = 0.0686
At equilibrium
moles SO3 = 0.10 - X = 0.10 - 0.0686 = 0.0314
moles SO2 = X = 0.0686
moles O2 = X/2 = 0.0343
moles total = 0.10 + X/2 = 0.10 + 0.0343 = 0.1343
P(SO3) = Ptot*X(SO3) = 1.83*0.0314/0.1343 = 0.428atm
P(SO2) = 1.83*0.0686/0.1343 = 0.935atm
P(O2) = 1.83*0.0343/0.1343 = 0.467atm
Kp for this reaction is
Kp = [P(SO2)*P(O2)^1/2]/P(SO3)
Kp = {0.935*(0.467)^0.5}/0.428
Kp = 0.149atm