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14 March, 02:56

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 3.10 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?

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  1. 14 March, 03:09
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    PCO2 = 0.6 25 atm

    PSO2 = 1.2 75 atm

    PO2 = 0.6 atm

    Explanation:

    Step 1: Data given

    Volume = 10.0 L

    Temperature = 100.0 °C

    Pressure = 3.10 °C

    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

    Step 2: The balanced equation

    CS2 (g) + 3O2 (g) →CO2 (g) + 2SO2 (g)

    Step 3: Name the reactants and products

    a = CS2

    b = O2 before reaction

    c = CO2

    d = SO2

    e = nS O2 after reaction with n = the number of moles

    Step 4: Calculate moles before reaction

    PV = nRT

    n = PV / (RT)

    (na + nb) = (3.10atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

    (na + nb) = 1.0124

    Step 5: Calculate moles after reaction

    PV = nRT

    n = PV / (RT)

    nc + nd + ne) = PV / (RT) = (2.50 atm) * (10.0L) / ((0.08206 Latm/moleK) * (373.15K))

    (nc + nd + ne) = 0.816 moles

    Step 6: Calculate mol fraction

    For 1 mole CS2 we need 3 moles O2 to produce 1 mole of CO2 and 2 moles of SO2

    moles O2 remaining = ne = nb - 3na

    moles CO2 produced = nc = na

    moles SO2 producted = nd = 2na

    (nc + nd + ne) = 0.816 moles = nb - 3na + na + 2na = 0.816

    nb = 0.816

    . (na + nb) = 1.0124

    na = 1.0124 moles - 0.816 moles = 0.208

    which leads to

    nc = na = 0.208

    nd = 2na = 2*0.208 = 0.416

    ne = 0.816 - 3*0.208 = 0.192

    mole fraction CO2 = 0.208 / (0.208 + 0.416 + 0.192) = 0.25

    mole fraction SO2 = 0.416 / (0.208 + 0.416 + 0.192) = 0.5 1

    mole fraction O2 = 0.192 / (0.208 + 0.416 + 0.192) = 0.24

    Step 6: Calculate partial pressure

    PCO2 = 0.25 * 2.50 atm = 0.6 25 atm

    PSO2 = 0.51 * 2.50 atm = 1.2 75 atm

    PO2 = 0.24 * 2.50 atm = 0.6 atm

    Step 7: Control results

    now let's verify a couple of things

    PV = nRT

    P = nRT/V

    before rxn

    P = (0.208 + 0.816) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 3.10 atm

    after rxn

    P = ((0.208 + 0.416+0.192) * (0.08206 L*atm/mole*K) * (373.15K) / (10.0L) ≈ 2.50 atm
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