Question

A mixture of CS2(g) and excess O2(g) is placed in a 10 L reaction vessel at 100.0 ∘C and a pressure of 3.15 atm . A spark causes the CS2 to ignite, burning it completely, according to the equation: CS2(g)+3O2(g)→CO2(g)+2SO2(g) After reaction, the temperature returns to 100.0 ∘C, and the mixture of product gases (CO2, SO2, and unreacted O2) is found to have a pressure of 2.50 atm . What is the partial pressure of each gas in the product mixture? Enter your answers numerically separated by commas.

Answer 1

0.65, 1.30, 0.55

Step-by-step explanation:

By the Raoult's Law, the partial pressure of a gas in a gas mixture is its molar fraction multiplied by the total pressure of the mixture. Let's found out the total number of moles of the mixture using the ideal gas law:

PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm.L/mol.K), and T is the temperature (100°C + 273 = 373 K).

3.15*10 = n*0.082*373

30.586n = 31.5

n = 1.03 moles of the reactants

Let's call the total number of moles of CS₂ as x and the number of moles of O₂ is 1.03 - x, thus by the stoichiometry of the reaction, 1:3:1:2, the number of moles of O₂ that reacts is 3x, and the remaining number of moles of it, and the number of moles of each product is:

nO₂ = (1.03 - x) - 3x

nCO₂ = x

nSO₂ = 2x

After the reaction happened, the number of moles is:

2.50*10 = n*0.082*373

30.586n = 25.0

n = 0.8174 mol

Thus,

nO₂ + nCO₂ + nSO₂ = 0.8174

(1.03 - x) -3x +x + 2x = 0.8174

1.03 - x = 0.8174

x = 0.2126 mol

nO₂ = (1.03 - 0.2126) - 3*0.2126 = 0.1796 mol

nCO₂ = 0.2126 mol

nSO₂ = 2*0.2126 = 0.4252 mol

Using Roult's Law, the partial pressures will be:

pCO₂ = (nCO₂/n)*2.50 = (0.2126/0.8174)*2.50 = 0.65 atm

pSO₂ = (nSO₂/n)*2.50 = (0.4252/0.8174)*2.50 = 1.30 atm

pO₂ = (nO₂/n)*2.50 = (0.1796/0.8174)* 2.50 = 0.55 atm