A reactor processing a reaction is fed with a gas mixture containing 3 moles nitrogen, 6moles hydrogen, and 2-mole ammonia. The following reaction is occurring in the reactor:N2+ 3H2  2NH3If the extent of reaction for the above reaction is 0.5, find the mole fraction of the gases in theproduct mixture.

The extent of reaction is defined as the amount of reactant consumed or product formed divided by its stoichiometric coefficient. In this case, the extent of reaction is 0.5. This means that 0.5 moles of nitrogen and 1.5 moles of hydrogen have reacted to form 1 mole of ammonia.

Step 1: Calculate the final moles of each component after the reaction.

For Nitrogen (N2): Initial moles were 3. As per the reaction, 1 mole of nitrogen gives 2 moles of ammonia. So, 0.5 moles of nitrogen have reacted. Therefore, final moles of nitrogen = 3 - 0.5 = 2.5 moles.

For Hydrogen (H2): Initial moles were 6. As per the reaction, 3 moles of hydrogen gives 2 moles of ammonia. So, 1.5 moles of hydrogen have reacted. Therefore, final moles of hydrogen = 6 - 1.5 = 4.5 moles.

For Ammonia (NH3): Initial moles were 2. As per the reaction, 1 mole of nitrogen and 3 moles of hydrogen gives 2 moles of ammonia. So, 1 mole of ammonia has been formed. Therefore, final moles of ammonia = 2 + 1 = 3 moles.

Step 2: Calculate the total moles in the mixture after the reaction.

Total moles = moles of nitrogen + moles of hydrogen + moles of ammonia = 2.5 + 4.5 + 3 = 10 moles.

Step 3: Calculate the mole fraction of each component.

Mole fraction of Nitrogen = moles of nitrogen / total moles = 2.5 / 10 = 0.25 Mole fraction of Hydrogen = moles of hydrogen / total moles = 4.5 / 10 = 0.45 Mole fraction of Ammonia = moles of ammonia / total moles = 3 / 10 = 0.3

So, the mole fraction of nitrogen, hydrogen, and ammonia in the product mixture are 0.25, 0.45, and 0.3 respectively.