Question

A 5.0 L reactor at 130 degrees celcius contains H2 at a partial pressure of 3.00 atm and O2 at a partial pressure of 3.00 atm. A spark ignites the reaction, and it goes to completion. What is the final pressure in the flask after it cools back down to 130 degrees celcius?

2H2(g) + O2(g) -> 2H2O(g)

Answer 1

moles of Hydrogen and oxygen in the reactor can be calculated from gas law equation

n= PV/RT

for both gases, P= 3 atm, V=5L, T= 130+273= 403K and R=0.0821 L.atm/mole.K

n= 3*5/(0.0821*403)= 0.4534, moles of oxygen = moles of hydrogen.

The reaction is 2H2(g) + O2(g)--------->2H2O(g), theoretical molar ratio of H2:O2= 2:1

actual ratio of H2: O2= 1:1. So Hydrogen is limiting reactant. Oxygen is excess

0.4534 moles of hydrogen consumes 0.4534/2= 0.2267 moles of O2

moles of H2O formed =0.4534 and moles of oxygen remaining = 0.2267 ( since the reaction goes to completion)

total moles pf products= 0.4534+0.2267=0.6801

This is n for calculating the final pressure, n= 0.6801, V= 5L, R =0.0821 L.atm/mole.K, T= 130+273= 403K

P= nRT/V= 0.6801*0.0821*403/5 =4.5 atm