Question

A mixture of H_2(g) and O_{2 (g)} at 1.50 atm in a rigid container at temperature T is ignited. The remaining gas, which is entirely H_{2 (g)} exerts a pressure of 0.30 atm at tempertaure T. Assume that the gases are ideal and the H₂O_(l) formed in the combustion reaction takes up negligible volue in the container. Determine the mole fraction of O_2(g) in the original mixture.

Answer 1

Ans. Let the volume of vessel be V liters.

# Step 1: Calculate initial moles in the reaction vessel:

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in equation 1-

            1.50 atm x V L = n x (0.0821 atm L mol^-1K^-1) x T K

            Or, n = 1.50V atm L / (0.0821T atm L mol^-1)

            Hence, n = 18.270V/T mol

So, initial number of moles of gases (H₂ and O₂) = 18.240 V/T mol

# Step 2: Calculate moles of H_-`in the reaction vessel:

Putting the values in equation 1-

            0.30 atm x V L = n x (0.0821 atm L mol^-1K^-1) x T K

            Or, n = 0.30V atm L / (0.0821T atm L mol^-1)

            Hence, n = 3.6541V/T mol

Therefore, moles of H₂ remaining in the vessel after combustion = 3.6541 mol

# Step 3:        Balanced Reaction: 2 H₂(g) + O₂(g) ------------> H₂O

Let the initial moles of H₂ = X mol

We have, Moles of H₂ remaining after combustion = 3.6541V/T mol

# For all O₂ to be consumed up, O₂ must be the limiting reactant.

In balanced reaction, 1 mol O_­2 reacts with 2 mol H₂.

Now,

            Initial moles of O₂ = (1/2) x initial moles of H₂ = (1/2)X = 0.50X mol

And, Moles of H₂ remaining after combustion =

Initial moles of H₂ – moles of H₂ consumed

Or, 3.6541V/T mol = X mol – 0.50X mol = 0.50X mol

            Or, X = (3.6541V/T mol) / 0.50 = 7.3082V/T mol

Hence, initial moles of H₂, X = 7.3082V/T mol

# Step 4: Calculating mole fraction of O₂

Initial moles of O₂ = 0.50X = 0.50 x (7.3082 V/T) mol = 3.6541V/T mol

Now,

Mole fraction of O₂ = Initial moles of O₂ / Total initial moles in reaction mixture

                                    = (3.6541V/T mol) / (18.270V/T mol)

                                    = 0.200