The concentrations of reactants and products for a chemical reaction can be calc

Question

The concentrations of reactants and products for a chemical reaction can be calculated if the equilibrium constant for the reaction and the starting concentrations of reactants and/or products are known. Carbonyl fluoride, COF_2, is an important intermediate used in the production of fluorine-containing compounds. For instance, it is used to make the refrigerant carbon tetrafluoride, CF_4 via the reaction 2COF_2(g) CO_2(g) + CF_4(g), K_c = 5.10 If only COF_2 is present initially at a concentration of 2.00 M, what concentration of COF_2 remains at equilibrium? Express your answer with the appropriate units. Consider the reaction CO(g) + NH_3(g) HCONH_2(g), K_c = 0.600 If a reaction vessel initially contains only CO and NH_3 at concentrations of 1.00 M and 2.00 M, respectively, what will the concentration of HCONH_2 be at equilibrium? Express your answer with the appropriate units.

Explanation / Answer

Answer 1:

According to the reaction equation:
2 COF(g) CO(g) + CF(g)
the equilibrium concentrations in M are related as:
Kc = ( [CO][CF] ) / [COF]²
with Kc = 8.30

ICE-Table
........... [COF]......... [CO].......... [CF]
I............. 2.0............... 0................ 0
C........... - 2x............ +x...............+x
E.......... 2 - 2x............ x................ x

When we substitute the expressions for the equilibrium concentrations from the last row of the table to the equilibrium equation you get:
<=> 8.3 = xx / (2 - 2x)²
<=> 8.3 = ( x / (2 - 2x) )²
<=> 8.3 = x / (2 - 2x)
<=> (2 - 2x)8.3 = x
<=> 28.3 = x(1 + 28.3)
x = 28.3 / (1 + 28.3) = 0.852 M

So the equilibrium concentrations are:
[COF] = 2.0 M - 2x = 2.0M - 2 0.852 M = 0.296 M
[CO] = [CF] = x = 0.852 M

Therefore concentration of COF2 is 0.852M.

Answer 2:

K=[HCONH2] / [CO][NH3]

K=[x]/[1.00-x][2.00-x]
0.6=x/[1.00-x][2.00-x]
0.6(1-x)(2-x)=x

x= 0.477412 OR 4.18925
Disregard 4.18925 because it is larger than 1.00

therefore the concentration of HCONH2 at equilibrium will be 0.477412 Molarity.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.