Part a: Carbonyl fluoride, COF2, is an important intermediate used in the produc
ID: 873549 • Letter: P
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=6.90
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium? [COF2] =
Part B
Consider the reaction
CO(g)+NH3(g)?HCONH2(g), Kc=0.850
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? [HCONH2] =
Explanation / Answer
Construct an ICE table:
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Chemical Equation --------- 2COF2(g) =========> CO2(g) -------- + --------- CF4(g)
Intial Concentration ----------- 2 ------------------------------ 0 ------------------------------- 0
Change in Conc. ----------- -- - 2x --------------------------- x ------------------------------- x
Equilibrium Conc. ----------- 2 - 2x ------------------------- x ------------------------------- x
Kc = [CO2][CF4] / [COF2]^2
Kc = (x)(x) / (2 - 2x)^2
Kc = x^2 / (2 - 2x)^2 = 4.50 (solve for x algebraically)
x = [CO2] = [CF2] = 0.8093
[COF2] = 2 - 2x = 2 - (2)(0.8093)
[COF2] = 0.3814 (ANSWER)
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Part B
You have not specified the Temp. of reaction and I am assuming 400K.
The equilibrium constant values you have given are subject to change and I am assuming values given in literature.
So here is the sample calculation and please put different values and make it a practice problem
Thanks.
Equilibrium reaction: CO(g) + NH3(g) ? HCONH2(g)
Initial concentrations: 2.00 M .. 2.75 M ......... -
Change in concentr.: ... -x M .... -x M ............. +x M
At equilibrium: ....... 2.00-x M .. 2.75-x M ...... +x M
Kc = [HCONH2] / [CO][NH3]
Kc = (x) / (2.00-x)(2.75-x) = 0.207
0.207x^2 - 1.983x + 1.138 = 0
Solving this quadratic equation, one of the roots will be 0.61 which satisfies the conditions.
Equilibrium concentrations:
[HCONH2] = x = 0.61 M
[CO] = 2.00-x = 2.00 - 0.61 = 1.39 M
[NH3] = 2.75-x = 2.75 - 0.61 = 2.14 M
Sources Yahoo.Questions and answers.