In: Chemistry
Calculating Equilibrium Concentrations
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=7.10
If only COF2 is present initially at a concentration of 2.00 M, what concentration of COF2 remains at equilibrium?
Part B
Consider the reaction
CO(g)+NH3(g)⇌HCONH2(g), Kc=0.700
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?
Part A:
2COF2(g) ⇌ CO2(g) + CF4(g)
Initial concentration 2 M 0 0
Conc. At eqm. (2-2X) M X X
And give that Kc=7.10
Equilibrium constant expression for this transformation is,
Kc = [CF4][CO2]/[COF2]2
Let us put all the known values,
7.10 = [(X) x (X)] /(2-2X)2
X2 / (2-2X)2 =7.1
Taking square roots of both sides,
X /(2-2X) = (7.1)1/2.
X/(2-2X) =2.66
X = (2-2x) x 2.66
X = 5.32 – 5.32X
X+5.32X = 5.32
6.32X = 5.32
X = 5.32/6.32
X = 0.842 M
Hence at equilibrium,
[COF2] = 2-0.842 = 2 – 0.842 = 1.158 M
[COF2] = 1.158 M
1.158 M COF2 will remains at equilibrium.
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Part B
Consider the reaction
CO(g)+ NH3(g) ⇌ HCONH2(g),
Initial conc. 1 M 2M 0
Conc. At eqm. 1-X 2-X ‘X’ M
And given Kc=0.700
Expression for equilibrium constant will be,
Kc = [HCONH2]/[CO][NH3]
Let us put all known values,
0.7 = X /[(1-X) x (2-X)]
X/(2-3X+X2) = 0.7
X = 0.7(2-3X+X2)
X = 1.4-2.1X +0.7X2.
1.4-2.1X +0.7X2= X
0.7X2 - 3.1X + 1.4 = 0
Let us multiply the whole equation by 10,
7X2 - 31X + 14 = 0
Let us solve this quadratic equation by formula method
Comparing this quadratic equation with its standard form Ax2 + Bx + C = 0 we get,
A = 7, B= -31, C = 14
Then formula for X
X = [-B ± (B2-4AC)1/2] / 2A
X = [31 ± (-312 – 4x7x14)1/2] / (2x7)
X = (31 ± 23.85) / 14
X = (31- 23.85) / 14 or X = (31+23.85) /14
X = (7.15)/14 X = 54.85 / 14
X = 0.511 M or X = 3.91 (Not possible as total moles of reactant are only 3 and product mole cannot exceed hence rejected)
So,
[HCONH2] = 0.511 M. at equilibrium.
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