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25 May, 15:22

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.

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  1. 25 May, 19:15
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    0.65, 1.30, 0.55

    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
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