CO(g)+NH3(g)⇌HCONH2(g), Kc=0.830

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?

*Posted: August 14, 2013*

ICE, ICE, baby ...

CO(g) + NH3(g) ⇌ HCONH2(g),

Kc = (products)/(reactants)

Kc = [HCONH2]/([CO][NH3])

Set up an ICE table

Species.....Initial...Change...Equil...

CO............1.00 M....-x........1.00-x.

NH3...........2.00 M....-x.......2.00-x.

HCONH2.......0.........+x.........x..

Kc = [HCONH2]/([CO][NH3])

Kc = 0.830 = (x)/((1.00 - x)*(2.00 - x))

Multiply this out to get a quadratic equation:

(0.830)*((1.00 - x)*(2.00 - x)) = x

(0.830)*(2 - 3x + x^2) = x

1.66 - 2.49x + x^2 = x

x^2 - 3.49x + 1.66 = 0

Use the quadratic formula to solve for the roots of this equation.

negative root = 0.5681

positive root = 2.9219

We reject the positive root (it is greater than the starting amount of NH3, so it is unrealistic), so

x = 0.5681

We plug that back into the ICE table to determine the equilibrium quantities.

Species.......Equil...

CO..............1.00-x = (1.00 - 0.5681) = 0.4319 M

NH3............2.00-x = (2.00 - 0.5681) = 1.4319 M

HCONH2......x = 0.5681 M

So at equilibrium, [HCONH2] = 0.5681 M