Please help me by showing the work and the answer, because I\'m not understandin

Question

Please help me by showing the work and the answer, because I'm not understanding the material. Thanks.

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=7.80 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.

Part B

Consider the reaction

CO(g)+NH3(g)?HCONH2(g),    Kc=0.890

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.

Explanation / Answer

Part A      

         2COF2(g)? CO2(g)+CF4(g)

initial.....................2.................0...............0

at equilibrium........2-x..............x...............x

KC=[CO2][CF4]/[COF2]2=x2/(2-x)2=7.8

x2=7.8(4+x2-4x)

0.1282x2=4+x2-4x

0=4+x2-0.1282x2-4x

0=4-0.87179x2-4x

(solve for x using quadratic formula) take positive value

x=0.844

hence equilibrium concentration of COF2=2.00-0.844=1.156M

Part B

                            CO(g)+NH3(g)?HCONH2(g),    Kc=0.890

initial.....................1...........2...............0

at equilibrium......1-x.......2-x...............x

Kc=x/(1-x)(2-x)=0.89(x2-3x+2)=x

x2-3x-1.12359x+2=0

x2-4.12359x+2=0 (solve for x using quadratic equation)

((((1-x)(2-x)=2-x-2x+x2=x2-3x+2)))

Hence concentration of HCONH2(g) at equilibrium will be

[HCONH2]=x=0.56

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