[tex]\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}[/tex]
[tex]\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}[/tex]
[tex]\text{and in the liquid form it is easily transported. An industrial chemist studying this}[/tex]
[tex]\text{reaction fills a} \ \mathbf{100 \ L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \ \text{of oxygen gas, }[/tex]
[tex]\text{to be} \ \mathbf{2.6\ mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}[/tex]
[tex]\text{ammonia at the final temperature of the mixture. Round your answer to 2 significant digits.}[/tex]
Answer:
Explanation:
From the correct question above:
The reaction can be represented as:
[tex]\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }[/tex]
From the above reaction; the ICE table can be represented as:
[tex]\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }[/tex]
I (mol/L) 0.086 0.28 0 0
C -4x -3x +2x +6x
E 0.086 - 4x 0.28 - 3x +2x +6x
At equilibrium;
The water vapor = [tex]\dfrac{2.6 \ mol}{100 \ L} = 6x[/tex]
[tex]x = \dfrac{2.6}{100} \times \dfrac{1}{6}[/tex]
[tex]x = 0.00433[/tex]
[tex]\text{equilibrium constant} ({k_c}) = \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }[/tex]
[tex]\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\[/tex]
Replacing the value of x, we have:
[tex]K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}[/tex]
[tex]K_c = \mathbf{5.5 \times 10^{-8} \ to \ 2 \ significant \ figures}[/tex]
