Calculating Equilibrium Concentrations Part A Carbonyl fluoride, COF2, is an imp

Question

Calculating Equilibrium Concentrations

Part 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=7.10

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

Part B

Consider the reaction

CO(g)+NH3(g)HCONH2(g),    Kc=0.700

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

Part A:

                                                       2COF2(g)   CO2(g) + CF4(g)    

Initial concentration                     2 M                   0               0

Conc. At eqm.                                 (2-2X) M             X               X

And give that Kc=7.10

Equilibrium constant expression for this transformation is,

Kc = [CF4][CO2]/[COF2]2

Let us put all the known values,

7.10 = [(X) x (X)] /(2-2X)2

X2 / (2-2X)2 =7.1

Taking square roots of both sides,

X /(2-2X) = (7.1)1/2.

X/(2-2X) =2.66

X = (2-2x) x 2.66

X = 5.32 – 5.32X

X+5.32X = 5.32

6.32X = 5.32

X = 5.32/6.32

X = 0.842 M

Hence at equilibrium,

[COF2] = 2-0.842 = 2 – 0.842 = 1.158 M

[COF2] = 1.158 M

1.158 M COF2 will remains at equilibrium.

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Part B

Consider the reaction

                                                  CO(g)+         NH3(g)            HCONH2(g),    

Initial conc.                              1 M              2M                        0

Conc. At eqm.                           1-X                2-X                        ‘X’ M

And given Kc=0.700

Expression for equilibrium constant will be,

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

Let us put all known values,

0.7 = X /[(1-X) x (2-X)]

X/(2-3X+X2) = 0.7

X = 0.7(2-3X+X2)

X = 1.4-2.1X +0.7X2.

1.4-2.1X +0.7X2= X

0.7X2 - 3.1X + 1.4 = 0

Let us multiply the whole equation by 10,

7X2 - 31X + 14 = 0

Let us solve this quadratic equation by formula method

Comparing this quadratic equation with its standard form Ax2 + Bx + C = 0 we get,

A = 7, B= -31, C = 14

Then formula for X

X = [-B ± (B2-4AC)1/2] / 2A

X = [31 ± (-312 – 4x7x14)1/2] / (2x7)

X = (31 ± 23.85) / 14

X = (31- 23.85) / 14           or              X = (31+23.85) /14           

X = (7.15)/14                                           X = 54.85 / 14

X = 0.511 M                          or              X = 3.91 (Not possible as total moles of reactant are only 3 and product mole cannot exceed hence rejected)

So,

[HCONH2] = 0.511 M. at equilibrium.

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