Part A Carbonyl fluoride, COF 2 , is an important intermediate used in the produ

Question

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=5.60

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

Express your answer with the appropriate units.

[COF2] =_____________________________

Part B

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?

Express your answer with the appropriate units.

[HCONH2] = __________________________

Explanation / Answer

Part A Ans:
the equilibrium concentrations equation in M are related as:
Kc = ( [CO][CF] ) / [COF]²
Kc = 5.60

........... [COF]......... [CO].......... [CF]
Initial....... 2.0............... 0................ 0
Change.. - 2x............ +x...............+x
Equilib..... 2 - 2x............ x................ x

When you substitute the expressions for the equilibrium concentrations from the last row of the table to the equilibrium equation you get:
Kc = ( [CO][CF] ) / [COF]²
5.6 = xx / (2 - 2x)²
5.6 = ( x / (2 - 2x) )²

5.6 = x / (2 - 2x)  
2.36 = x / (2 - 2x)  
(2 - 2x)*2.36 = x  
4.72 – 4.72 x = x

4.72 = 5.72 x

x = 0.825 M

So the equilibrium concentrations are:
[COF] = 2.0 M - 2x = 2.0M - 2 0.825 M = 0.35 M
[CO] = [CF] = x = 0.825 M

Part B Ans:
the equilibrium concentrations equation in M are related as:
Kc = [HCONH2] / ( [CO][NH3] )

Kc = 0.63

........... [CO]......... [NH3].......... [HCONH2]
Initial....... 1.0............2.0................ 0
Change.. x............ x...............+x
Equilib..... 1- x...........2-x................ x

When you substitute the expressions for the equilibrium concentrations from the last row of the table to the equilibrium equation you get:
Kc = [HCONH2] / ( [CO][NH3] )

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

0.63[2-2x-x+x2] = x

0.63[2-3x+x2] = x

1.26-1.89x+0.63x2 = x

0.63x2 -2.89 x+1.26 = 0

ax2 + bx + c = 0 The solution to the quadratic equation is given by the quadratic formula:

Solution x = 1.815 or x = -6.4

Negative values is not possible so x = 1.815

So at equilibrium [HCONH2] = 1.815 M

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