Answer to Question #115297 in General Chemistry for Sora

Question

An 8.00-gram sample of SO3 was placed in an evacuated container, where it decomposed at 600.8°C according to the following reaction:
SO3 (g)⇌SO2 (g)+1⁄2O2 (g)
At equilibrium the total pressure and the density of the gaseous mixture were 1.80 atm and 1.60
g/L, respectively. Calculate Kp for this reaction.

Answer 1

n(SO₃) = m(SO₃) / M_r(SO₃) = 8.00 g / (32 + 3x16) g/mol = 8.00 g / 80 g/mol = 0.1 mol

Density = m / V

V = m / Density = 8.00 g / 1.6 g/L = 5.0 L

T = 600.8 + 273 = 873.8 K

pV = nRT

p = nRT / V = 0.1 mol x 0.0821 L•atm/K•mol x 873.8 K / 5.0 L =1.435 atm

The initial pressure of SO₃ is 1.435 atm

K_p = (x)(1/2x)^1/2 / (1.435 - x)

x is the change in the partial pressure of SO₃

The equilibrium pressure are represented as

SO₃ (g)⇌SO₂ (g) + 1⁄2O₂ (g)

Initial(atm) 1.435

Change x

Equilibrium 1.80 atm

x = 0.365 atm

K_p = 0.365 x (1/2x0.365)^1/2 / (1.435 - 0.365) = 0.146