Final answer:
To increase the N2 concentration to 1.1 M in the reaction vessel, 0.15 mol of H2 needs to be removed due to the 1:3 stoichiometric ratio between N2 and H2 in the equilibrium reaction. However, this answer is not listed among the provided options, indicating there may be a mistake in the interpretation of the question or the provided answer choices.
Step-by-step explanation:
To determine how many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 M, we first need to understand the stoichiometry of the equilibrium reaction: N2(g) + 3H2(g) ⇌ 2NH3(g). In this reaction, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.
Due to this 1:3 ratio, removing nitrogen would require an equivalent adjustment in hydrogen concentration to maintain equilibrium.
To change the concentration of N2 from 1.00 M to 1.1 M, an extra 0.1 mol of N2 is required in a volume of 1.00 L. This means 0.3 mol of H2 (since 1 mol of N2 reacts with 3 mol of H2) must be removed to shift the equilibrium to produce this additional amount of N2.
However, because the removal of H2 also shifts the equilibrium towards N2 production, only half of this amount needs to be physically removed (0.3 mol / 2 = 0.15 mol). The remaining adjustment comes from the shift in equilibrium.
Hence, 0.15 mol of H2 must be removed to achieve the desired increase in N2 concentration to 1.1 M, which is not one of the answer options provided, suggesting a potential misinterpretation of the question or options given.