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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0180 mol gaseous NH3 and 0.0100 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 1.50×10-3 M. Calculate Keq for the reaction at this temperature.
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at eq mol N2 = 0.0015 mol NH3 = 0.018 - 0.0015 = 0.00165 mol O2 = 0.01 - 0.0015 = 0.0085 \[K _{eq} = \frac{ [N _{2}]^{2}[H _{2}O]^{6} }{ [NH _{3}]^{4}[O _{2}]^{3} }\]
\[K _{eq} = \frac{ (0.0015)^{2}(0.0015) ^{6}}{ (0.0165)^{4}(0.0085)^{3} }\] * mol of NH3 = 0.0165
it says that is wrong..
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