1) The following reaction was carried out in a 2.00 L reaction vessel at 1100 K

Question

1) The following reaction was carried out in a 2.00L  reaction vessel at 1100 K:

C(s)+H2O(g)?CO(g)+H2(g)

If during the course of the reaction, the vessel is found to contain 7.00mol  of C, 13.5mol  of H2O, 3.90mol  of CO, and 6.00mol  of H2, what is the reaction quotient Q?

2) 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?

3) Consider the reaction

CO(g)+NH3(g)?HCONH2(g), Kc=0.780

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?

Explanation / Answer

1)[CO]= 3.90/2.00=1.95

[H2]= 6.00/2.00=3.00 M
[H2O]= 13.5 / 2.00=6.75 M
Q = 1.95 x 3.00/6.75=0.867

2)   2 COF2 <=> CO2 + CF4

2.00 -2x <=> x & x

K = [CO2] [CF4] / [COF2)^2

6.90 = [x] [x] / [2 - 2x]^2

6.90 [2 - 2x]^2 = [x] [x]

6.90 [4 - 8x +4x2] = x2

27.6 - 55.2x + 27.6x2 = x2

27.6 - 55.2x + 26.6x2 = 0

http://www.math.com/students/calculators...

x = 0.84,....

what concentration of COF2 remains at equilibrium:
2.00 - 2x =

2.00 - 1.68 = 0.32 Molar

3)   Kc = [HCONH2] / [CO][NH3]

let X = moles/L of HCONH2 formed
[CO] = 1.00 M - X
[NH3] = 2.00 M - X
[HCONH2] = X
0.780 = X / (1.00-X)(2.00-X) = X / 2.00 - 3.00 X + X^2
0.780X^2 - 3.34X + 1.56 = 0

X = 0.5335
HCONH2 =0.5335  M

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