Question

Consider the reaction CO(g)+NH3(g)⇌HCONH2(g), Kc=0.790 If a reaction vessel initially contains only CO and NH3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH2 be at equilibrium?

Answer 1

Kc = [HCONH2] / [CO][NH3]
In going to equilibrium, the balanced equation says that CO and NH3 will both lose x in concentration, while HCONH2 will gain x. Substitute all we know into the equation:
0.790 = x/(1.00-x)(2.00-x),
rearranges to:
0.790x² - 3.370x + 1.580 = 0.
Using the quadratic formula, we get two answers, x = 0.536 and x = 3.7296. Since [CO] cannot lose more than 1.00 M in concentration, the answer is x = [HCONH2] = 0.536 M.
To check, calculate Q:
Q = 0.536/(1.00-0.536)(2.00-0.536) = 0.789. Since Q = Kc, the reaction is at equilibrium.