Q1: One mole of methane undergoes complete combustion in a stoichiometric amount of air. The reaction proceeds as CH₄ + 2O₂ → CO₂ + 2H₂O. Both the reactants and products are in gas phase. ΔH°₂₉₈ = - 730 kJ/mole of methane. Mole fraction of water vapour in the product gases is about

ANS:A - 0.19

To find the mole fraction of water vapor (𝑋H2OXH2​O​) in the product gases, we need to consider the stoichiometry of the reaction and the enthalpy change (Δ𝐻298°ΔH298°​). Given:

1. One mole of methane (𝐶𝐻4CH4​) undergoes complete combustion with 2 moles of oxygen (𝑂2O2​).
2. The reaction produces carbon dioxide (𝐶𝑂2CO2​) and water (𝐻2𝑂H2​O).
3. The enthalpy change (Δ𝐻298°ΔH298°​) for the combustion of one mole of methane is -730 kJ.

From the balanced equation: 𝐶𝐻4+2𝑂2→𝐶𝑂2+2𝐻2𝑂CH4​+2O2​→CO2​+2H2​O We can see that for every mole of methane consumed, 2 moles of water vapor are produced. So, the mole fraction of water vapor (𝑋H2OXH2​O​) is equal to the mole fraction of water vapor produced in the product gases. Now, let's calculate 𝑋H2OXH2​O​: The enthalpy change (Δ𝐻298°ΔH298°​) for the combustion of one mole of methane is -730 kJ. This means that 730 kJ of heat is released when one mole of methane is combusted. Since the reaction produces 2 moles of water vapor for every mole of methane consumed, the amount of heat released per mole of water vapor produced is half of the enthalpy change for the combustion of methane. So, the heat released per mole of water vapor produced = −730 kJ2=−365 kJ−2730kJ​=−365kJ. Now, to find the mole fraction of water vapor, we need to consider the total moles of products formed. For every mole of methane consumed, 1 mole of carbon dioxide and 2 moles of water vapor are produced. So, the total moles of products per mole of methane consumed is 1+2=31+2=3 moles. Therefore, the mole fraction of water vapor (𝑋H2OXH2​O​) in the product gases is: 𝑋H2O=moles of water vaportotal moles of products=23=0.67XH2​O​=total moles of productsmoles of water vapor​=32​=0.67 So, the mole fraction of water vapor in the product gases is approximately 0.67.