Question

ammonia has been studied as an alternative \clean\ fuel for internal combustion engines, since its reaction with oxygen produces only nitrogen and water vapor and in the liquid form it is easily transported. an industrial chemist studying this reaction fills a 100. l tank with 5.3 mol of ammonia gas and 6.3 mol of oxygen gas, and when the mixture has come to equilibrium measures the amount of water vapor to be 6.4 mol. calculate the concentration - equilibrium constant for the combustion of ammonia at the final temperature of the mixture. round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The combustion of ammonia is \(4NH_3(g)+3O_2(g)
ightleftharpoons 2N_2(g)+6H_2O(g)\)

Step2: Calculate the molar - concentrations

The volume of the tank \(V = 100.L\).
The molar - concentration of \(NH_3\), \(c_{NH_3}=\frac{n_{NH_3}}{V}=\frac{5.3mol}{100.L}=0.053mol/L\)
The molar - concentration of \(O_2\), \(c_{O_2}=\frac{n_{O_2}}{V}=\frac{6.3mol}{100.L}=0.063mol/L\)
The molar - concentration of \(N_2\), \(c_{N_2}=\frac{n_{N_2}}{V}\), since \(n_{N_2}\) is not given we assume it is formed from the reaction. But we can also use the given \(n_{H_2O}\) to calculate the equilibrium constant.
The molar - concentration of \(H_2O\), \(c_{H_2O}=\frac{n_{H_2O}}{V}=\frac{6.4mol}{100.L}=0.064mol/L\)

Step3: Write the expression for the equilibrium constant \(K_c\)

For the reaction \(4NH_3(g)+3O_2(g)
ightleftharpoons 2N_2(g)+6H_2O(g)\), the equilibrium - constant expression is \(K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}\)
Since we are not given the amount of \(N_2\) and we can assume that the reaction has reached equilibrium with the given amounts of reactants and products, and if we assume the reaction has gone to completion in the forward direction (based on the information of amounts at equilibrium), we can calculate \(K_c\) using the concentrations of \(NH_3\), \(O_2\) and \(H_2O\).
\[

$$\begin{align*} K_c&=\frac{(0)^2\times(0.064)^6}{(0.053)^4\times(0.063)^3}\\ &=\frac{(0.064)^6}{(0.053)^4\times(0.063)^3} \end{align*}$$

\]
\[

$$\begin{align*} (0.064)^6&=6.871947674\times10^{-10}\\ (0.053)^4&=7.890481\times10^{-6}\\ (0.063)^3&=2.50047\times10^{-4} \end{align*}$$

\]
\[

$$\begin{align*} K_c&=\frac{6.871947674\times 10^{-10}}{7.890481\times 10^{-6}\times2.50047\times 10^{-4}}\\ &=\frac{6.871947674\times 10^{-10}}{1.97307\times10^{-9}}\\ &\approx0.35 \end{align*}$$

\]

Answer:

0.35