ANSWERS to Worksheet on Equilibrium Chapter 18
1) Explain the following statements with a sentence
a) The concentrations of reactants and products in a system at dynamic equilibrium don’t change. In a chemical system at equilibrium the forward and reverse reaction rates occur at the same rate. Therefore, even though the reactions are taking place the products are being formed at the same rate that they are being used up. Thus the concentrations don't change.
b) A change in the pressure on a gaseous system can cause a shift in the equilibrium position. A change in pressure on an equilibrium system that contains gases can disturb the equilibrium if there are unequal moles of gases on the reactant side and product side of the equation. An increase of pressure will cause an equilibrium system to shift to the side with fewer moles of gas. (This is vice versa for a decrease in pressure.)
c) When a chemical equilibrium is established, the amount of reactants and products is the same. This is usually not the case. When a system is at equilibrium the amounts of reactants and products don't change, but they don't necessarily have to be equal.
d) The Keq for a certain reaction was 2 x 10-7. This reaction is said to be reactant favored. This is correct since the Keq < 1, which means that at equilibrium there are more reactants than products.
e) When a common ion is added to a system at equilibrium the system shifts to …. use up the added common ion. The effect of this is that the presence of a common ion makes a salt less soluble.
2) Calculate Keq for this reaction: 2SO3(g) à 2SO2(g) + O2(g), if the equilibrium concentrations are: [SO2] = 0.42 M, [O2] = 0.21 M, [SO3] = 0.072 M. Keq = 7.1
3) At 25oC, the following reaction has an equilibrium constant of 9.0 x 10-4. Find the concentration of A(g) in moles per liter, when the concentration of B(g) is 0.030 mol/L and the concentration of C(g) is 0.060 mol/L. [A] = 0.12 mol/L
A(g) ßà B(g) + 2C(g)
4)
Indicate how the equilibrium position of this reaction is shifted by the
following changes:
heat + 4HCl(g) + O2(g)
ßà
2Cl2(g) + 2H2O(g)
a) add Cl2 This will cause a shift in reverse or to the left to use up some of Cl2 added and restore equilibrium. The value of Keq will not change.
b) remove O2 This will cause a shift in reverse or to the left in order to form more O2, replacing what was removed. Equilibrium will be restored and the value of Keq will not change.
c) increase pressure This will cause a shift in the forward direction since the product side of the equation has 4 moles of gas and the reactant side has more, at 5 moles. This shift will restore equilibrium and there will not be a change in the value of Keq.
d) lower the temperature This will cause a change in reverse, the exothermic direction in order that some of the removed heat will be released. This shift will cause a change in the Keq; it will be lower.
For each of the changes above state whether the value of Keq would change or not. see above.
Explain one of the adjustments in the equilibrium position above in terms of Le Chatelier’s Principle.
see above
5) At a certain temperature, an equilibrium is established for this reaction in a 3.0 L container: Cl2(g) + Br2(g) ßà 2BrCl(g). Given 0.85 mol Cl2, 0.85 mol Br2, and 0.30 mol BrCl, what is Keq? Keq = 0.12
6) Aluminum hydroxide has a Ksp of 3.0 x 10-34. Calculate the equilibrium concentrations of aluminum and hydroxide ions. [Al3+] = 1.8 x 10-9 M, [OH-] = 5.5 x 10-9 M
7) Explain what happens when HCl is added to a saturated solution of AgCl? Since the Cl- in HCl is common to AgCl, an insoluble salt, adding HCl will be adding Cl- and will cause some more AgCl to precipitate. This is an example of the "common ion effect" and we could also say that AgCl is less soluble in the HCl.
8) The solubility of SrCrO4 in water is 1.2 grams per liter of solution. Calculate Ksp. Ksp = 3.5 x 10-5
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