For the nitrogen fixation reaction 3H₂(g) + N₂(g) ⇌ 2NH₃(g), K_c = 6.0 * 10^-2 at 500^$\circ$C. If 0.250 M H₂ and 0.050 M NH₃ are present at equilibrium, what is the equilibrium concentration of N₂

Question

Quizplus · Accepted Answer

Given the equilibrium constant $K_c = 6.0 \times 10^{-2}$ and the concentrations of $H_2$ and $NH_3$, we can set up the expression for $K_c$ as $\frac{[NH_3]^2}{[H_2]^3[N_2]}$. Solving for $[N_2]$ using the given concentrations and $K_c$ value, we find that $[N_2] = 2.7 M$.

For the Nitrogen Fixation Reaction 3H2(g) + N2(g) ⇌ 2NH3(g) ∘\circ∘

For the Nitrogen Fixation Reaction 3H₂(g) + N₂(g) ⇌ 2NH₃(g) $\circ$