Question

Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF4 via the reaction

2COF2(g)⇌CO2(g)+CF4(g),    Kc=8.00

A.If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?

Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.810

B. 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

2COF₂(g) ⇌ CO₂(g) + CF₄(g),    Kc=8.00

IC: 2.0 M 0 0

EC: (2-m) m m

Kc =    ([CO₂] [CF₄]) / [COF₂]²

8 = (m x m) / (2 - m)²

7m² - 32m + 32 = 0

Solving the equation m = 3.09 and m = 1.48.

Since the concentration of the reactant can't be negative, so m= 3.09 is neglected. So, m = 1.48.

Hence, at equilibrium concentration of [COF₂] = (2 - 1.48) M = 0.52 M

(b)

CO(g) + NH₃(g) ⇌ HCONH₂(g),    Kc=0.810

IC: 1.00 M   2.00 M   0

EC: 1-m 2-m m

Now,

Kc = [HCONH₂] / ( [CO] [NH₃] )

0.81 = m / [ (1 - m) (2 - m)]

0.81 = m / [ m² - 3m + 2]

0.81m² - 3.43m + 1.62 = 0

Solving the quadratic equation, m = 3.69 and m = 0.54. Neglecting m = 3.69, since it will lead to negative concentration.

So, m = 0.54

At equilibrium concentration of [HCONH₂] = m=  0.54