Calculate the concentration equilibrium constant for the combustion of ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.

Steps to Solve

1. Write the balanced chemical equation The balanced equation for the combustion of ammonia is: $$ 4 \text{NH}_3(g) + 3 \text{O}_2(g) \rightarrow 2 \text{N}_2(g) + 6 \text{H}_2\text{O}(g) $$

2. Determine the change in moles at equilibrium From the stoichiometry of the reaction, for every 4 moles of NH₃ that react, 6 moles of H₂O are produced. If we have 0.57 moles of H₂O produced, we can find how many moles of NH₃ reacted: Using ratios, we have: $$ \text{Moles of NH}_3 \text{ reacted} = \frac{4}{6} \times 0.57 = 0.38 \text{ mol} $$

3. Calculate the moles of reactants remaining at equilibrium Starting with:

• Initial moles of NH₃ = 3.8 mol
• Initial moles of O₂ = 19 mol

Now, calculate the moles of NH₃ and O₂ remaining after the reaction: $$ \text{Moles of NH}_3 \text{ remaining} = 3.8 - 0.38 = 3.42 \text{ mol} $$

Next, using the stoichiometry of the reaction (4 NH₃ : 3 O₂), calculate the moles of O₂ that reacted: $$ \text{Moles of O}_2 \text{ reacted} = \frac{3}{4} \times 0.38 = 0.285 \text{ mol} $$ So the moles of O₂ remaining is: $$ \text{Moles of O}_2 \text{ remaining} = 19 - 0.285 = 18.715 \text{ mol} $$

4. Calculate the concentrations at equilibrium To find the concentrations, divide the moles by the volume (50.0 L): $$ [\text{NH}_3] = \frac{3.42}{50.0} = 0.0684 \text{ mol/L} $$ $$ [\text{O}_2] = \frac{18.715}{50.0} = 0.3743 \text{ mol/L} $$ $$ [\text{H}_2\text{O}] = \frac{0.57}{50.0} = 0.0114 \text{ mol/L} $$

5. Write the expression for the equilibrium constant ( K_c ) The equilibrium constant expression for this reaction is: $$ K_c = \frac{[\text{N}_2]^2 [\text{H}_2\text{O}]^6}{[\text{NH}_3]^4 [\text{O}_2]^3} $$ Since there are no products in the initial equilibrium calculations, we can note that the concentration of N₂ is formed from the stoichiometry as follows: $$ \text{Moles of N}_2 \text{ produced} = \frac{1}{2} (0.57) = 0.285 \text{ mol} \Rightarrow [\text{N}_2] = \frac{0.285}{50} = 0.0057 \text{ mol/L} $$

6. Substitute into ( K_c ) and calculate Substituting the equilibrium concentrations into the ( K_c ) expression: $$ K_c = \frac{(0.0057)^2 (0.0114)^6}{(0.0684)^4 (0.3743)^3} $$

7. Perform the calculation Calculating each part step-by-step produces:

• Numerator: ( (0.0057)^2 (0.0114)^6 = 1.138 \times 10^{-10} )
• Denominator: ( (0.0684)^4 (0.3743)^3 = 0.001096 )

So, $$ K_c = \frac{1.138 \times 10^{-10}}{0.001096} \approx 1.037 \times 10^{-7} $$