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 3.10 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?

Answer 1

V= 10 L
T = 100 C = 373 K
P1 = 3.10 atm
P2 = 2.50 atm

n = moles

Before Rxn:
(n1 + n2) = PV/RT
(n1 + n2) = (3.10 atm)(10 L) / (0.0821 L *atm/mol * K)(373 K)
(n1 + n2) = 1.01 mol

After Rxn:
(n3 + n4 + n5) = PV/RT
(n3 + n4 + n5) = (2.50 atm)(10 L) / (0.0821 L *atm/mol * K)(373 K)
(n3 + n4 + n5) = 0.82 mol

Therefore

mol O2 (g) remaining: n5 = n2 - 3n1
mol CO2 (g) produced: n3 = n1
mol SO2 (g) produced: n4 = 2n1

(n3 + n4 + n5) = 0.82 mol

now

(n2 - 3n1 + 2n1) = 0.82 mol
n2 = 0.82 mol

(n1 + n2) = 1.01 mol
n1 = 0.9960 mol - 0.8000 mol = 0.19 mol

n3 = n1 = 0.19
n4 = 2n1 = (2 * 0.19) = 0.38 mol
n5 = 0.82 - (3 * 0.19) = 0.25 mol

Mole Fractions:

mol fraction CO2 (g): (0.19 mol) / (0.82 mol) = 0.23
mol fraction SO2 (g): (0.38 mol) / (0.82 mol) = 0.46
mol fraction O2 (g) : (0.25 mol) / (0.82 mol) = 0.30

Partial Pressures:

P CO2 (g) = (0.23 * 2.50 atm) = 0.575 atm
P SO2 (g) = (0.46 * 2.50 atm) = 1.15 atm
P O2 (g) = (0.30 * 2.50 atm) = 0.75 atm