Question

A) 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=5.60
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Express your answer with the appropriate units.

B.) Consider the reaction
CO(g)+NH3(g)⇌HCONH2(g), Kc=0.750
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?
Express your answer with the appropriate units.

Answer 1

A) 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=5.60
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Express your answer with the appropriate units.

Ans:

Given reaction equation:
2 COF₂(g) ⇌ CO₂(g) + CF₄(g), Kc = 5.60
Kc equation for this reaction
Kc = ( [CO₂]∙[CF₄] ) / [COF₂]²
Kc = 5.60

ICE-Table
........... [COF₂]......... [CO₂].......... [CF₄]
I............. 2.0............... 0................ 0
C........... - 2∙x............ +x...............+x
E.......... 2 - 2∙x............ x................ x
I means initial concentration, C- Change in concentration at equilibrium, E- Concentration at equilibrium
Substitute the equilibrium concentrations values from the last row of the table in the equilibrium equation (Kc)
5.6 = x∙x / (2 - 2∙x)²
<=>
5.6 = ( x / (2 - 2∙x) )²
<=>
√5.6 = x / (2 - 2∙x)

2.36 = x / (2 - 2∙x)
<=>
(2 - 2∙x)∙2.36 = x

4.72 – 4.72 x = x

4.72 = 5.72 x

x = 0.825 M

So the equilibrium concentrations are:
[COF₂] = 2.0 M - 2∙x = 2.0M - 2 ∙ 0.825 M = 0.35 M
[CO₂] = [CF₄] = x = 0.825 M