The reversible decomposition of dinitrogen tetroxide, N2O4, to nitrogen dioxide, NO2, is shown. For this reaction, Keq= 0.15. N204 dinitrogen tetroxideitrogen dioxide 2 NO2 At equilibrium, is the concentration of reactants or products greater? The concentration of reactants equals the concentration of products at equilibrium. The concentration of reactants is greater than the concentration of products at equilibrium. The concentration of products is greater than the concentration of reactants at equilibrium. Does the reaction favor reactants or products? reactants neither products nor reactants O products

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Chapter1: Chemical Foundations
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The reversible decomposition of dinitrogen tetroxide, N204, to nitrogen dioxide, NO2, is shown. For this reaction,
Keq= 0.15.
N204
dinitrogen tetroxideitrogen dioxide
2 ΝΟ)
At equilibrium, is the concentration of reactants or products greater?
The concentration of reactants equals the concentration of products at equilibrium.
The concentration of reactants is greater than the concentration of products at equilibrium.
The concentration of products is greater than the concentration of reactants at equilibrium.
Does the reaction favor reactants or products?
reactants
neither products nor reactants
O products
Transcribed Image Text:The reversible decomposition of dinitrogen tetroxide, N204, to nitrogen dioxide, NO2, is shown. For this reaction, Keq= 0.15. N204 dinitrogen tetroxideitrogen dioxide 2 ΝΟ) At equilibrium, is the concentration of reactants or products greater? The concentration of reactants equals the concentration of products at equilibrium. The concentration of reactants is greater than the concentration of products at equilibrium. The concentration of products is greater than the concentration of reactants at equilibrium. Does the reaction favor reactants or products? reactants neither products nor reactants O products
At a certain temperature, 0.311 mol CH4 and 0.813 mol H,S are placed in a 3.50 L container.
CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g)
At equilibrium, 11.0 g CS, is present. Calculate Kc.
Kc =
Transcribed Image Text:At a certain temperature, 0.311 mol CH4 and 0.813 mol H,S are placed in a 3.50 L container. CH4(g) + 2 H2S(g) = CS2(g) + 4 H2(g) At equilibrium, 11.0 g CS, is present. Calculate Kc. Kc =
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