Final answer:
To increase the concentration of nitrogen to 1.1 M in the equilibrium mixture, we need to remove 0.30 moles of hydrogen from the vessel.
Step-by-step explanation:
To increase the concentration of nitrogen (N2) to 1.1 M in the equilibrium mixture, we need to calculate how many moles of hydrogen (H2) must be removed from the vessel. The balanced equation for the reaction is N2 + 3H2 ⇌ 2NH3. From the given equilibrium concentrations, we can calculate the moles of nitrogen and hydrogen present:
Moles of N2 = concentration of N2 × volume of vessel = 1.00 M × 1.00 L = 1.00 mol
Moles of H2 = concentration of H2 × volume of vessel = 0.50 M × 1.00 L = 0.50 mol
To increase the concentration of nitrogen to 1.1 M, we need to find the difference in the moles of nitrogen before and after the change. The change in moles of nitrogen is given by Δn = moles of nitrogen final - moles of nitrogen initial = 1.1 mol - 1.00 mol = 0.10 mol
According to the balanced equation, the stoichiometry between nitrogen and hydrogen is 1:3. Therefore, for every mole of nitrogen that is formed or consumed, three moles of hydrogen are formed or consumed. In this case, since the concentration of nitrogen is increasing, the number of moles of hydrogen that needs to be removed is the difference in moles of nitrogen multiplied by the stoichiometric ratio:
Change in moles of hydrogen = Δn × (3 mol H2 / 1 mol N2) = 0.10 mol × 3 = 0.30 mol
Therefore, to increase the concentration of nitrogen to 1.1 M, 0.30 moles of hydrogen must be removed from the vessel.