Problem: Ammonia can be synthesized according to the following reaction: N 2(g) + 3H2(g) ⇌ 2NH3(g); Kp = 5.3 x 10–5 at 725 KA 200.0-L reaction container initially contains 1.27 kg of N2 and 0.310 kg of H2 at 725 K. Assuming ideal gas behavior, calculate the mass of NH3 (in g) present in the reaction mixture at equilibrium.

Step 1: Initial Partial Pressure of N₂, H₂

$\boxed{\mathbf{PV}\mathbf{=}\mathbf{nRT}}$

molar mass N₂: 2 x 14.00 g/mol = 28.00 g/mol

molar mass H₂: 2 x 1.01 g/mol = 2.02 g/mol

$$\begin{gathered} \boxed{\mathbf{P}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{V}}} \\ {\mathbf{P}}_{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\frac{\left(\mathbf{1}\mathbf{.}\mathbf{27}\mathbf{ }\cancel{\mathbf{kg}}\mathbf{\times }{\displaystyle \frac{{\mathbf{10}}^{\mathbf{3}}\cancel{\mathbf{g}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{kg}}}}\mathbf{\times }{\displaystyle \frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}}}{\mathbf{28}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }{\mathbf{N}}_{\mathbf{2}}}}}\right)\mathbf{ }\left(\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\cancel{\mathbf{L}}\mathbf{·}\mathbf{atm}}{\cancel{\mathbf{mol}}\mathbf{·}\cancel{\mathbf{K}}}}\right)\left(\mathbf{725}\mathbf{ }\cancel{\mathbf{K}}\right)}{\left(\mathbf{200}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{L}}\right)} \\ {\mathbf{P}}_{{\mathbf{N}}_{\mathbf{2}}}\mathbf{=}\frac{(0.310 \cancel{\mathrm{kg}}\times {\displaystyle \frac{{10}^3\cancel{g H_2}}{1 \cancel{\mathrm{kg}}}}\times {\displaystyle \frac{1 \cancel{\mathrm{mol} H_2}}{2.02 \cancel{g H_2}}}) (0.08206{\displaystyle \frac{\cancel{L}·\mathrm{atm}}{\cancel{\mathrm{mol}}·\cancel{K}}})(725 \cancel{K})}{(200.0 \cancel{L})} \end{gathered}$$

PN₂ = 13.49 atm

PH₂ = 45.65 atm

Step 2: ICE Chart

Step 3: K_p expression

$$\begin{gathered} \boxed{{\mathbf{K}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{products}}{\mathbf{reactants}}\mathbf{=}\frac{{\mathbf{\left[}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{N}}_{\mathbf{2}}\mathbf{\right]}{\mathbf{\left[}{\mathbf{H}}_{\mathbf{2}}\mathbf{\right]}}^{\mathbf{3}}}} \\ \mathbf{5}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{=}\frac{{\left[2x\right]}^2}{[13.49-x]{[45.65-3x]}^3} \end{gathered}$$