Answer:
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
= 6.02x10⁻⁶