Question

In: Chemistry

Part A 1.Carbonyl fluoride, COF2, is an important intermediate used in the production of fluorine-containing compounds....

Part A

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=5.50

If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2remains at equilibrium?

Express your answer with the appropriate units.

Hints

[COF2] =

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Part B

Consider the reaction

CO(g)+NH3(g)⇌HCONH2(g),    Kc=0.710

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.

Hints

[HCONH2] =

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Part A

What is the equilibrium concentration of CO at 1000 K?

Express your answer in molarity to three significant figures.

[CO] =   M  

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Part B

What is the equilibrium concentration of Cl2 at 1000 K?

Express your answer in molarity to three significant figures.

[Cl2] =   M  

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Part C

What is the equilibrium concentration of COCl2 at 1000 K?

Express your answer in molarity to three significant figures.

[COCl2] =   M  

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Solutions

Expert Solution

         2COF2(g)⇌CO2(g)+CF4(g),   

I          2                 0             0

C       -2x                x            x

E      2-2x               +x          +x

Kc   = [CO2][CF4]/[COF2]^2

5.5   = x*x/(2-2x)^2

5.5*(2-2x)^2 = x^2

     x = 0.82

[COF2] = 2-2x = 2-2*0.82   = 0.36M

[CO2]    = x      = 0.82M

[CF4]    = x     = 0.82M

part-B

           CO(g)+NH3(g)⇌HCONH2(g),

I            1          2                  0

C         -x        -x                  +x

E       1-x        2-x                  +x

      Kc   = [HCONH2]/[CO][Nh3]

     0.71   = x/(1-x)(2-x)

   0.71*(1-x)(2-x) = x

     x   = 0.513

[CO] = 1-x = 1-0.513   = 0.487M

[NH3]   = 2-x = 2-0.513    = 1.487M

[HCONH2] = x= 0.513M


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