The total pressure of a mixture of oxygen and hydrogen is 1.0 atm. The mixture is ignited and
the water is removed. The remaining gas is pure hydrogen and exerts a pressure of 0.40 atm
when measured at the same values of T and V as the original mixture. What was the
composition of the original mixture in mole per cent?
Answers
Answer:
The original mixture will have 80% of H₂ and 20% of O₂.
Explanation:
Given data:
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
To find: composition of original mixture in mole percent.
According to the question, we can write
2H₂(g) + O₂(g) → 2H₂O(l)
Let initial no.of moles of H₂ is “n1”, Initial no. 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) = χ₁, mole fraction of hydrogen and n2/(n1+n2) = χ₂, mole fraction of oxygen]
∴ P2 / P1 = χ₁ – 2(χ₂) …… (iii)
Or, P2/P1 = 1 – χ₂ – 2 (χ₂)
Or, P2/P1 = 1 – 3 (χ₂)
Or, 0.40 / 1 = 1 – 3(χ₂)
Or, 3 (χ₂) = 1 - 0.40 = 0.60
Or, χ₂ = 0.60 / 3 = 0.2
Putting value of χ₂ in eq. (iii)
∴ 0.4/1 = χ₁ – 2 * 0.2
Or, χ₁ = 0.4 + 0.4 = 0.8
∴ Mole % of oxygen = 20 % and Mole fraction of hydrogen = 80 %.
:
Answer:
The original mixture will have 80% of H₂ and 20% of O₂.
Explanation:
Given data:
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
To find: composition of original mixture in mole percent.
According to the question, we can write
2H₂(g) + O₂(g) → 2H₂O(l)
Let initial no.of moles of H₂ is “n1”, Initial no. 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) = χ₁, mole fraction of hydrogen and n2/(n1+n2) = χ₂, mole fraction of oxygen]
∴ P2 / P1 = χ₁ – 2(χ₂) …… (iii)
Or, P2/P1 = 1 – χ₂ – 2 (χ₂)
Or, P2/P1 = 1 – 3 (χ₂)
Or, 0.40 / 1 = 1 – 3(χ₂)
Or, 3 (χ₂) = 1 - 0.40 = 0.60
Or, χ₂ = 0.60 / 3 = 0.2
Putting value of χ₂ in eq. (iii)
∴ 0.4/1 = χ₁ – 2 * 0.2
Or, χ₁ = 0.4 + 0.4 = 0.8
∴ Mole % of oxygen = 20 % and Mole fraction of hydrogen = 80 %.