A-Level Chemistry Equilibrium Question

Question

N2O4(g) <==> 2NO2(g) Kp = 0.115 atm at 298K If the equilibrium partial pressure of N2O4 is 0.500 atmosphere, calculate the equilibrium partial pressure of NO2, and the total equilibrium pressure.

Answer 1

N2O4(g) <==> 2 NO2(g), Kp = 0.115 atm at 298 K, with p(N2O4) = 0.500 atm.

This one is a direct substitution - no ICE table needed, because you are handed an equilibrium pressure rather than a starting one.

Step 1 - write Kp

Kp = p(NO2)^2 / p(N2O4)

The NO2 term is squared because its coefficient in the balanced equation is 2. That squaring is the single most common slip on this problem, and it is also why Kp comes out with units of atm here: pressure squared divided by pressure leaves pressure.

Step 2 - rearrange and substitute

p(NO2)^2 = Kp x p(N2O4) = (0.115 atm)(0.500 atm) = 0.0575 atm^2

p(NO2) = sqrt(0.0575) = 0.240 atm

Step 3 - total pressure

By Dalton's law the total is just the sum of the partial pressures:

P(total) = 0.500 + 0.240 = 0.740 atm

Check it

(0.240)^2 / 0.500 = 0.0576 / 0.500 = 0.115 atm, which is Kp. Substituting your answer back into the expression takes five seconds and catches a forgotten square root immediately.

Worth noticing about the answer

NO2 is present at less than half the pressure of N2O4, so at room temperature this equilibrium sits well to the left - the dimer is favoured. That is consistent with Kp being smaller than 1.

This is the classic sealed-tube demonstration your course is probably building toward. N2O4 is colourless and NO2 is dark brown, so the equilibrium position is visible to the eye. Warm the tube and it darkens; cool it in ice and it fades toward colourless. The forward reaction breaks an N-N bond and is endothermic, so heating shifts it right, exactly as Le Chatelier predicts - and unlike most equilibria you can simply watch it happen.