COOP IV ________________________
Chem 1403-40 (name)
Dr. Robert E. Belford
1. Does the following reaction favor reactants or products?
2SO3(g) -->
2SO2(g) + O2(g) K = 7.9 x 10-9
Reactants, K is a very small fraction
2. Calculate the equilibrium constant Kc for the reaction
PCl3(g) + Cl2(g) --> PCl5(g)
when a mixtures of the gasses has the following equilibrium concentrations
[PCl3] = 2.4 x 10-2, [Cl2] = 8.9 x 10-2, [PCl5] = 3.7 x 10-2
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3. Using the results from question 2, determine Kc for the following reaction.
PCl5(g) --> PCl3(g) + Cl2(g)
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4. Using the results from question 2, determine Kp for the following reaction at 25 deg C.
PCl3(g) + Cl2(g) --> PCl5(g)
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5. Consider the following equilibrium:
2HI(g) <==> H2(g) + I2(g)
At equilibrium a 5.00 L vessel contains 4.50 mol I2, 4.50 mol H2 and an unknown amount of HI. Kc for this equilibrium is 0.0183. What is the equilibrium concentration of HI?

6. For the reaction H2(g) + I2(g) --> 2HI(g), DHo = -9.4kJ, for each of the following changes, will the equilibrium product concentration increase or decrease?
a. Addition of hydrogen Increase
b. Removal of Iodine Decrease
c. Increase in Pressure No Effect
d. An increase in Temperature Decrease
7. For the reaction 2NF3 (g) à N2(g) + 3F2(g) , DHo = +4.3kJ/mol, for each of the following changes, will the equilibrium product concentration increase or decrease?
a. Addition of
Fluorine Decrease
b. Removal of Nitrogen Increase
c. Increase in Pressure Decrease
d. An increase in
Temperature Increase
8. Consider the reaction
2SiO(g) <==> 2Si(l) + O2(g); Kc = 9.62 x 10-1
If 1.00 mol SiO is placed into a 1.00 L container, what are the equilibrium concentrations of all species?
2SiO(g) <==> 2Si(l) + O2(g)
Initial moles: 1 0 0
Change: -2x +x +x
Equil. Conc. [1-2x] (constant) [x]

The first answer is nonsense, so [O2] = 0.247M & SiO=0.506M
9. Calculate the value of the equilbrium constant for the reaction
2NO(g) + O2(g) <==> N2O4(g) K = ?
Knowing the following equilibrium constants at the same Temperature and Pressure
2NO(g) + O2(g) <==> NO2(g) K1 = 2.0 x 102
N2O4(g) <==> 2 NO2(g) K2 = 7.2 x 10-4
Writing the second one backward makes it the reciprocal of the forward, then adding coupled equations makes the coupled equilibrium constant multiplicative

10. A mixture of 0.659 g of H2 and 26.7 g Br2 is heated in a 3 L container. These substances react as follows
H2(g) + Br2(g) <==> 2HBr(g)
At equilibrium the vessel is found to contain 10.4 g of Br2. Calculate Kc for this reaction

H2(g) + Br2(g) <==> 2HBr(g)
[Initial]: .110 0.0557 0
Change: -x -x 2x
[Equilibrium]: .11-x .0557-x 2x
[0.0217]
From Bromine data: 0.0217 = 0.0557-x => x = 0.036
So [H2]eq = 0.11 – 0.026 = 0.084 and [HBr]eq = 2(0.036) = 0.072

11. Given the reaction:
Cl2(g) + 2HBr(g) çè Br2(g) + 2HCl(g)
70.9g of chlorine is mixed with 161.8g of HBr in a 10.0 L vessel. What are the equilibrium concentrations of all species if Kc = 1.5 x 101?
The initial molarity of chlorine is 70.9g(70.9g/mol)/10L=0.100M
The initial molarity of hydrogen bromide is 161.8g(80.9g/mol)/10L=0.200M
Cl2(g) + 2HBr(g) çè Br2(g) <==> 2HCl(g)
Initial moles: .1 .2 0 0
Change: -x -2x +x +2x
Equil. Conc. [.1-x] [.2-2x] [x] [2x]

[Cl2]= .1-.0712 = 0.0288M
[HBr] = .2-2(0.0712) = 0.0576M
[Br2]=.0712
[HCl] = 2(0.0712) = 0.142M
12. Calculate the equilibrium constant if one mole of pure ethyl alcohol is mixed with one mole of pure acetic acid at room temperature. The products are ethyl-methyl ester and the equilibrium mixture contains 2/3 moles of the ester.
C2H5OH + CH3OOH ç è H2O + H3CCOOCH2CH3
Initial moles: 1 1 0 0
Change: -x -x +x +x
Equil. moles [1-x] [1-x] [x] [x]
X=2/3, 1-X=1/3
EquilConc: [1/3]/V [1/3]/V [2/3]/V [2/3]/V
