1. Carbonyl fluoride, COF2, is an important intermediate used in the production

Question

1. 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=7.40

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

2.Consider the reaction

CO(g)+NH3(g)?HCONH2(g),    Kc=0.630

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?

The reversible chemical reaction

A+B?C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=6.2

3.Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached? 2 sig fig and correct units

4.What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00M and[B] = 2.00M ?

2 sig fig and correct units

Explanation / Answer

1) 2COF2(g) ---> CO2(g)+CF4(g)

initial conc of   C0F2 , C02 and CF4 are 2 , 0 , 0

equilibrium conc of   C0F2 , C02 and CF4 are 2-2x , x , x

Kc = [C02] [CF4] / [C0F2]^2

7.40 = x * x / ( 2-2x )^2

solving we get

x = 0.8447

so [CoF2] at equilibrium = 2 - 2*0.8447 = 0.31 M

2)

CO(g)+NH3(g)----> HCONH2(g)

initial conc of   C0 , NH3 and HCONH2 are 1 , 2 , 0

equilibrium conc of   C0 , NH3 and HCONH2 are 1-x , 2-x , x

Kc = [HCONH2] / [C0] [NH3]

0.63 = x / ( 2-x )* (1-x)

solving we get

x = 0.488

so [HCONH2] at equilibrium = x = 0.488 M

3) A+B ---> C+D

initial conc of   A, B, C and D are 2 , 2 , 0 , 0

equilibrium conc of   A, B , C and D are 2-x , 2- x , x , x

Kc =[ C] [D] / [A] [B]

6.2 = x2 / ( 2-x )^2

solving we get

x = 1.4268

so [A] at equilibrium = 2 -1.4268 = 0.57 M

4) initial conc of   A, B, C and D are 1 , 2 , 0 , 0

equilibrium conc of   A, B , C and D are 1-x , 2- x , x , x

Kc =[ C] [D] / [A] [B]

6.2 = x2 / ( 1-x ) ( 2-x)

solving we get

x = 0.886

so [D] at equilibrium = x = 0.89 M

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