At a fixed temperature, equal moles of N2 (g) and H2 (g) are mixed in a constant

Question

At a fixed temperature, equal moles of N2 (g) and H2 (g) are mixed in a constant pressure container (the volume of the container changes in order to keep the pressure at a constant value). The N2 (g) and H2 (g) are allowed to react, producing NH3 (g): N2 (g) + 3 H2 (g) ? 2 NH3 (g) If the initial volume of the container, before any reaction takes place, is 2.50 L, determine the volume of the container after the N2 (g) and H2 (g) have reacted to completion. Hint: Because P and T are constant, Avogadro's Law applies (V is proportional to n). For this problem, assume that n moles of N2 (g) react with n moles of H2 (g) (equal moles are reacted). The initial moles of gas present are n + n = 2n. Next, determine the change in the number of moles of each reactant and product that occur when the reaction goes to completion. From the initial moles of reactants and product present along with the change in moles that occurs when the reaction goes to completion, determine the final moles of each reactant and product present. Finally, determine the total moles of gases present after the reaction, then use the relationship: nfinal/ninitial = Vfinal / Vinitial.

Explanation / Answer

n_initial = n+n = 2n n moles of H2 reacts with n/3 moles of N2 to give 2n/3 moles of NH3. So n_final = n-(n/3)+(2n/3) = n+(n/3) = 4n/3 n_initial/n_final = V_initial/V_final =>2n/(4n/3) = 2.5/V_final =>V_final = (2.5*2)/3 = 1.667 L

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