Question

The total pressure of a mixture of oxygen and hydrogen is 1.00 atm. The mixture is ignited and the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.400 atm when measured at the same values of T and V as the original. What is the mole fraction of hydrogen in the original mixture

Answer 1

The original mixture will have 80% of H₂ and 20% of O₂.

Explanation:

Total pressure of the original mixture of oxygen and hydrogen, P₁ = 1 atm

Pressure of pure hydrogen, P₂= 0.40 atm at same values of T and V as the original mixture

According to the question, we can write

2H₂+ O₂ --> 2H₂O

Let initial number of moles of H₂ is “n₁”, Initial number of moles of O₂ be “n₂” and initial mole of 2H₂O is zero.

At equilibrium,  

Moles of 2H₂ = n₁- 2α

Moles of O₂ = n₂ - α

Moles of 2H₂O = 2α  

Since the mixture is ignited therefore the O2 is consumed ∴ n₂ – α = 0 <---> n2 = α and for the remaining pure Hydrogen the no. of mole will be n₁- 2α = n₁ – 2n₂.

By ideal gas law, we can write the equations as  

P1 = (n₁ + n₂)(RT/V) …… (i)

P2 = (n₁ – 2n₂)(RT/V) …. (ii)

On dividing the eq. (ii) by (i), we get

P₂ / P₁= [n₁/(n₁+n₂)] – [2n₂/(n₁+n2_​​​​​​)]

[Where, n1/(n1+n2) = X₁, mole fraction of hydrogen and n2/(n1+n2) = X₂ mole fraction of oxygen]

∴ P2 / P1 = X₁– 2(X₂) …… (iii)

Or, P2/P1 = 1 – X₂ – 2 (X₂)  

Or, P2/P1 = 1 – 3 (X₂)

Or, 0.40 / 1 = 1 – 3(X₂)

Or, 3 (X₂) = 1 - 0.40 = 0.60

Or, X₂ = 0.60 / 3 = 0.2

Putting value of X₂ in eq. (iii)

∴ 0.4/1 = X₁ – 2 * 0.2

Or, X₁ = 0.4 + 0.4 = 0.8

So X₁=0.8 and X₂=0.2

∴ Mole fraction of oxygen = 20 % and Mole fraction of hydrogen = 80 %.