Question

Air and hydrogen are mixed in the ratio 1/4 and are heated to 1200 K. Calculate the partial pressures of H2 and O2 in the equilibrium gas at 1 atm pressure and at 10 atm pressure. Air contains 21 volume percent O2.

Answer 1

Here the volume of oxygen in 1 liter of air = .210 litre

And 22.4 litres of any gas weighs one mole.

So, 22.4 litres of oxygen = 1 mole

So, .21 litre of oxygen will have = .21/22.4 = .009375 mole.

Likewise hydrogen which is present in air contains 0.000005lts

So, 0.000005ltr of hydrogen will have = 0.000005/22.4=2.2*10-7 moles

We have mole fraction of a gas = {partial pressure of gas}/{total pressure of a the mixture of gas}

Also we have mole fraction of hydrogen = {2.2*10^-^7}/{0.009375+2.2*10^-^7}=2.33*10^{-5}

At 1 atm, Partial pressure of a hydrogen = 2.33*10-5 * 1atm= 2.33*10-5 atm

Mole fraction of oxygen = 0.0975

At 1 atm, partial pressure of oxygen = 0.0975atm

At 10 atm pressure, partial pressure of oxygen is 0.0975*10atm=0.975atm