Question

At 200°C, Kc​ = 1.6*10−10 for: N2​O(g) + NO2(g) ⇌ 3 NO(g). If 350 mL of NO measured at 797 torr and 25.0°C is placed into a 3.00 L container, what will be the molarity of N2​O at equilibrium when the reaction is heated to 200°C? what is the total equilibrium pressure, in units of torr at 200°C?

Answer 1

From Ideal Gas law we have, PV = nRT

For NO, P = 797torr, V = 0.350L, T = 298K, R = 62.364 L torr mol^-1K^-1. Substituting above values, we will get number of moles of NO.

n = PV/RT = 797*0.350/(62.634*298) = 0.015moles

Molarity of NO = moles/Volume of solution = 0.015/3 = 0.005M

N2​O(g) + NO2(g) ⇌    3 NO(g)

at t = 0 0 0    0.005M

at t = eq    x x    0.005-3x

K_c = (0.005-3x)³/ x² = 1.6*10^-10

solving for x we have, x = 0.00167M (other two roots are imaginary, hence rejected)

Molarity of N2O = 0.00167M

Total equilibrium pressure = Sum of partial pressure of all gases

P = CRT, C = concentration, T = 273+200K = 473K

P_N2O = 0.00167*62.364*473 = 49.261 torr = P_NO2

As Kc<<1 it means reactants exceed product. the reaction almost goes to completion in reverse direction. So P_NO = 0

This can also be verified as conc of NO at equilibrium = 0.005-3(0.00167) = 0

Total Pressure = 49.461+49.261+0 torr = 98.522torr