An 8.10-g sample of was placed in an evacuated container, where it decomposed at 590°C according to the following reaction: At equilibrium the total pressure and the density of the gaseous mixture were 1.83 atm and 1.57 g/L respectively. Calculate for this reaction.

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

Kp for this reaction is 0.149atm

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

geosphere and hydrosphere.

this is because the geosphere has to do with geological events, while the hydrosphere controls all water on earth.