A) Carbonyl fluoride, COF2, is an important intermediate used in the production

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=8.00

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

B)

Consider the reaction

CO(g)+NH3(g)?HCONH2(g),    Kc=0.660

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?

Explanation / Answer

First, let us define the equilibrium constant for any species:

The equilibrium constant will relate product and reactants distribution. It is similar to a ratio

The equilibrium is given by

rReactants -> pProducts

Keq = [products]^p / [reactants]^r

For a specific case:

aA + bB = cC + dD

Keq = [C]^c * [D]^d / ([A]^a * [B]^b)

For this specific case:

K = [CO2][CF4] / [COF2]^2

Kc = 8 initially

NOW;

initial conditions:

[COF2] = 2

[CO2] =0

[CF4] = 0

in equilbirium

[COF2] = 2 -2x

[CO2][ =0 +x

[CF4] = 0 +x

substitute in Kc:

K = [CO2][CF4] / [COF2]^2

8 = (x)(x)/(2-2x)^2

sqrt(8) = x/(2-2x)

2.8284(2-2x) = x

0.35355x = 2-2x

(0.35355+2)x = 2

x = 2/(0.35355+2) = 0.84978

substitute

[COF2] = 2 -2x = 2-2*0.84978 = 0.30044

[CO2][ =0 +x = 0.84978

[CF4] = 0 +x = 0.84978

answer is:

[COF2] = 2 -2x = 2-2*0.84978 = 0.30044 M left in equilbirium

B)

CO(g)+NH3(g) = HCONH2(g), Kc=0.660

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

initially [CO] = 1 [NH3] = 2 [HCONH2] = 0

in equilibrium:

[CO] = 1 -x

[NH3] = 2 - x

[HCONH2] = 0+x

substitute in equilbirium

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

0.66 = x / ((1-x)(2-x))

solve for x

0.66*(1-x)(2-x) = x

2-3x+x^2 = 1.51515151x

x^2 -4.5151x +2 = 0

x = 0.4978

[CO] = 1 -0.4978 = 0.5022 M

[NH3] = 2 - 0.4978 = 1.5022

[HCONH2] = 0+x = 0.4978 M

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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