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

The reaction is CS2(g)+3O2(g)→CO2(g)+2SO2(g)
1 mole of CS2 reacts with excess of oxygen to give 1 mole of CO2 and 2 mole of SO2.

Let n1= mole of CS2, n2= mole of oxygen

Since oxygen is excess and all the moles of CS2(g) is consumed, products are free from CS2.

Let n3= mole of O2 after the reaction, n4= mole of CO2 and n5= moles of SO2

Before the reaction,   n1+n2 can be calculated from gas law

(n1+n2)= PV/RT, P= 3.1 atm, V=10L, T= 100deg.c= 100+273= 373K, R=0.0821 L.atm/mole.K, Hence n1+n2= 3.1*10/(0.0821*373)= 1.01moles

After the reaction, n3+n4+n5= 2.5*10/(0.0821*373)= 0.82

But from the reaction, moles of CO2 formed, n4 = moles of CS2 consumed , n4=n1 and moles of SO2 formed, n5 = 2* moles of n1 = 2n1

Moles of O2 remaining , n3 =n2-3n1

But   n3+n4+n5= 0.82, replacing n3, n4 and n5 in terms of n1

, n1+ 2n1+ n2-3n1= 0.82, n2= 0.82 moles

But n1+n2=1.01, n1=1.01-0.82=0.19 moles

Hence n3= 0.82-3*0.19=0.25, n4=0.19 and n5=2*0.19=0.38

Mole fraction= moles of component/total moles

Mole fraction of products : O2=0.25/0.82=0.31, CS2=0.19/0.82=0.23, SO2=0.46

Partial pressures = mole fraction* total pressures

Partial pressure (atm): O2=0.31*2.5= 0.775 atm, C2=0.23*2.5= 0.575 and SO2=0.46*2.5= 1.15 atm