Help! With Kc and Equilibrium?

Please consider this Reaction:
N2(g) + O2(g) ⇌ 2 NO(g);
Kc = 2.8 ✕ 10^−4 at 1968 K

a.)What is the value of Kc for the reaction 2 NO(g) ⇌ N2(g) + O2(g) at the same temperature?
b.)What is the value of Kc for the reaction 1/2 N2(g) + 1/2 O2(g) ⇌ NO(g) at the same temperature?
c.)Does the equilibrium in (a) favor the reactant or the products?
d.)Does the equilibrium in (b) favor the reactants or the product?
e.)Consider the original reaction. If N2 = 0.114 atm, O2 = 0.321 atm,
and NO = 2.30 atm, is the system at equilibrium? If not, will the system shift to the left or to the right in order to achieve equilibrium?

1 Answer
Feb 24, 2018

Consider,

#N_2(g) + O_2(g) rightleftharpoons 2NO(g)#, where #K_"c" = 2.8*10^-4# at #1968"K"#

In the first case, we are undergoing the reverse reaction, where,

#([N_2][O_2])/([NO]^2)#

is the reciprocal of the forward reaction's equilibrium expression.

Hence,

#1/K_"c" approx 3.6*10^4#

this reaction favors the products, because #K_"c" > 1#.

In the second case, we are halving the coefficients, thereby taking the square root of the rate expression,

#sqrt(([NO]^2)/([N_2][O_2])) = ([NO])/([N_2]^(1/2)[O_2]^(1/2))#

Hence,

#K_"c"^(1/2) approx 1.8*10^-2#

this reaction favors the reactants, because #K_"c" < 1#.

Since the constituents of the reaction are all gases, we can assume these pressures are indeed their concentrations, and solve for the current state of the reaction,

#Q_"c" = (2.30"atm")^2/(0.114"atm" * 0.321"atm") approx 150#

#Q_"c" > K_"c"#, so the reaction would reverse to achieve equilibrium.