When the reaction N 2 (g) + O 2 (g) 2NO(g) is run at 2000 C,

Question:

When the reaction N₂(g) + O₂(g) ⇆ 2NO(g) is run at 2000 °C, appreciable amounts of reactants and product are present at equilibrium.

(a) A sealed 2.00-L container at 2000 °C is filled with 1.00 mole of NO(g) and nothing else. At that moment, which reaction is faster, forward or reverse? Justify your answer.

(b) At equilibrium, the concentration of NO(g) is 0.0683 M and the concentration of N₂(g) is 0.2159 M. What is the value of K_eq at 2000 °C?

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a In a sealed 200L container filled with 100 mole of NOg and nothing else the initial concentrations of N2g and O2g are zero The forward reaction N2g O2g 2NOg cannot occur without any reactant present N2 and O2 Therefore there are no reactants available for the forward reaction to proceed making the forward reaction slower or essentially nonexistent On the other hand there is 100 mole of product present NO This means that the reverse reaction 2NOg N2g O2g can occur as the NO molecules have the potential to collide and combine to form N2 and O2 Thus at that moment the reverse reaction is faster than the forward reaction b The equilibrium expression for the reaction N2g O2g 2NOg is Keq NO2 N2 O2 Given that the concentration ...View the full answer

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