Module Question

[Emily_Stenzler_2H]

Nitrogen dioxide is produced by the reaction N2O4 (g) ->/<- 2NO2 (g) with K=1.7*10^3. When PNO2 = 0.5 atm and PN2O4 = 0.5 atm is the system at equilibrium? If not, is there a tendency to form more reactants or more products?

I am struggling with the above problem. Since we are given K, do we assume that is in M and we would therefore need to convert it to atm in order to compare the two constants? How would we do this without the temperature?

When I leave the units as they are, I get 0.5, meaning the reaction will proceed towards the right. Would this be correct?

[Jenny Lee 2L]

Hi Emily! I think for this question, K is in Kp because the units of NO2 and N2O4 are in partial pressures (atm). When asking for whether these conditions are at equilibrium, use Kp = (PNO2)^2/(PN2O4), plug in the numbers of PNO2 and PN2O4 (which in this case are 0.5 atm), and check to see if this is <Kp, = Kp, or >Kp. If <Kp (1.7 x 10^3), this means the equilibrium will form more products because at this point there are more reactants than there would be at equilibrium. If =Kp, the reaction is at equilibrium. Lastly, if >Kp, more reactants will be made because at this point there is more product than there would be at equilibrium.

Hope this helps!

[KatarinaReid_3H]

Yes, you are correct. The question is asking whether at the current partial pressure values the rxn is at equilibrium. Therefore, it gives you the Kp value, and asks you to find the Q value (Q is used in place of K if we are unsure that the rxn is at equilibrium).
The Q value is found: Q = P(NO2)^2/P(N2O4) = 0.5
Therefore, Q<K. Which means the products side must get larger so that the Q value can increase to K.
You don't need the temperature. The K value is affected by temperature, but since we are already given K, then it does not matter.