Chapter 13 ChemicalEquilibrium

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–2 QUESTION Which of the comments given here accurately apply to the concentration versus time graph for the H 2 O + CO reaction? The changes in concentrations with time for the reaction H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) are graphed below when equimolar quantities of H 2 O(g) and CO(g) are mixed.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–3 QUESTION (continued) 1.The point where the two curves cross shows the concentrations of reactants and products at equilibrium. 2.The slopes of tangent lines to each curve at a specific time prior to reaching equilibrium are equal, but opposite, because the stoichiometry between the products and reactants is all 1:1. 3.Equilibrium is reached when the H 2 O and CO curves (green) gets to the same level as the CO 2 and H 2 curves (red) begin. 4.The slopes of tangent lines to both curves at a particular time indicate that the reaction is fast and reaches equilibrium.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–4 ANSWER Choice 2 is the only statement that accurately comments on this graph. When the stoichiometry is not 1:1 the slopes of tangent lines to both curves at a particular time reflect the stoichiometry ratio of the reactants. Section 13.1: The Equilibrium Condition

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–5 QUESTION The figure shown here represents the concentration versus time relationship for the synthesis of ammonia (NH 3 ). Which of the following correctly interprets an observation of the system? This is a concentration profile for the reaction N 2 (g) + 3H 2 (g)  2NH 3 (g) when only N 2 (g) and H 2 (g) are mixed initially.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–6 QUESTION (continued) 1.At equilibrium, the concentration of NH 3 remains constant even though some is also forming N 2 and H 2, because some N 2 and H 2 continues to form NH 3. 2.The NH 3 curve (red) crosses the N 2 curve (blue) before reaching equilibrium because it is formed at a slower rate than N 2 rate of use. 3.All slopes of tangent lines become equal at equilibrium because the reaction began with no product (NH 3 ). 4.If the initial N 2 and H 2 concentrations were doubled from what is shown here, the final positions of those curves would be twice as high, but the NH 3 curve would be the same.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–7 ANSWER Choice 1 properly states the relationship between reactants and products in a system that has reached a dynamic equilibrium. At equilibrium, the rate of the forward reaction continues at a pace that is equal to the continued pace of the reverse reaction. Section 13.1: The Equilibrium Condition

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–8 QUESTION One of the environmentally important reactions involved in acid rain production has the following equilibrium expression. From the expression, what would be the balanced chemical reaction? Note: all components are in the gas phase. K = [SO 3 ]/([SO 2 ][O 2 ] 1/2 ) 1.SO 3 (g)  SO 2 (g) + 2O 2 (g) 2.SO 3 (g)  SO 2 (g) + 1/2O 2 (g) 3.SO 2 (g) + 2O 2 (g)  SO 3 (g) 4.SO 2 (g) + 1/2O 2 (g)  SO 3 (g)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–9 ANSWER Choice 4 properly shows the product SO 3 on the right and incorporates the previous exponents from the equilibrium expression as coefficients in the chemical equation. Section 13.2: The Equilibrium Constant

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–10 QUESTION Starting with the initial concentrations of: [NH 3 ] = 2.00 M; [N 2 ] = 2.00 M; [H 2 ] = 2.00 M, what would you calculate as the equilibrium ratio once the equilibrium position is reached for the ammonia synthesis reaction? N 2 + 3H 2  2NH This cannot be done from the information provided.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–11 ANSWER Choice 4 is the correct response. The concentrations given are for the INITIAL position. In order to calculate the K ratio of products to reactants the equilibrium position concentrations must be observed. Section 13.2: The Equilibrium Constant

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–12 QUESTION One of the primary components in the aroma of rotten eggs is H 2 S. At a certain temperature, it will decompose via the following reaction. 2H 2 S(g)  2H 2 (g) + S 2 (g) If an equilibrium mixture of the gases contained the following pressures of the components, what would be the value of K p ? P H 2 S = 1.19 atm; P H 2 = 0.25 atm; P S 2 = 0.25 atm

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–13 ANSWER Choice 1 provides the correct K p value. Whether the equilibrium constant is based on pressure or molarity, the ratio is always set to show products divided by reactants. In addition, the pressure must be raised to the power that corresponds to the coefficients in the balanced equation. Section 13.3: Equilibrium Expressions Involving Pressures

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–14 QUESTION The balanced equation shown here has a K p value of What would be the value for K c ?(at approximately 1,100°C) 2H 2 S(g)  2H 2 (g) + S 2 (g)

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–15 ANSWER Choice 1 is obtained from K p = K c (RT)  n when T is expressed in K and the expression is solved for K c Section 13.3: Equilibrium Expressions Involving Pressures

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–16 QUESTION The liquid metal mercury can be obtained from its ore cinnabar via the following reaction: HgS(s) + O 2 (g)  Hg(l) + SO 2 (g) Which of the following shows the proper expression for K c ? 1.K c = [Hg][SO 2 ]/[HgS][O 2 ] 2.K c = [SO 2 ]/[O 2 ] 3.K c = [Hg][SO 2 ]/[O 2 ] 4.K c = [O 2 ]/[SO 2 ]

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–17 ANSWER Choice 2 correctly presents the product to reactant ratio. Recall that pure liquids and solids are not shown in the equilibrium constant expression. Section 13.4: Heterogeneous Equilibria

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–18 QUESTION Consider the reaction: CO(g) + 3H 2 (g)  CH 4 (g) + H 2 O(g). At a certain temperature, the value of K is equal to 185. If the equilibrium concentrations of CO, H 2 and H 2 O respectively are M, M and M, what would you calculate as the equilibrium concentration of CH 4 ? M M M M

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–19 ANSWER Choice 2 correctly reports the resulting equilibrium concentration of CH 4. The equilibrium expression includes raising [H 2 ] to the third power: K = [H 2 O][CH 4 ]/[CO][H 2 ] 3 Section 13.5: Applications of the Equilibrium Constant

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–20 QUESTION At a certain temperature, FeO can react with CO to form Fe and CO 2. If the K p value at that temperature was 0.242, what would you calculate as the pressure of CO 2 at equilibrium if a sample of FeO was initially in a container with CO at a pressure of 0.95 atm? FeO(s) + CO(g)  Fe(s) + CO 2 (g) atm atm atm atm

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–21 ANSWER Choice 3 provides the correct pressure in this equilibrium system. Solids do not appear in the equilibrium expression, so K p = P CO 2 /P CO. Also the reaction indicates a 1:1 ratio between the change in CO and CO 2. Therefore K p = X/(0.95 – X) Section 13.5: Applications of the Equilibrium Constant

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–22 QUESTION The weak acid HC 2 H 3 O 2, acetic acid, is a key component in vinegar. As an acid the aqueous dissociation equilibrium could be represented as HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 – (aq). At room temperature the K c value, at approximately 1.8  10 –5, is not large. What would be the equilibrium concentration of H + starting from 1.0 M acetic acid solution?  10 –5 M  10 –3 M  10 –5 M 4.More information is needed to complete this calculation.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–23 ANSWER Choice 2 is correct assuming that 1.0 – X can be approximated to 1.0. The relatively small value for K indicates that, compared to 1.0, X would not be large enough to include in the calculation. The equilibrium expression could be simplified to K = X 2 /1.0. A quick test of this hypothesis could be made by using the 4.2  10 –3 value as X and checking the right side of the expression to see if it was the same as 1.8  10 –5. Section 13.6: Solving Equilibrium Problems

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–24 QUESTION The boulder in (a) has what in common with the reactants in (b)?

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–25 QUESTION (continued) 1.The position of both after the “hill” is less stable than the top of the “hill” position. 2.The position of the boulder will not change, nor will the position of the H 2 and O 2 in the H 2 + O 2  2H 2 O chemical system because the activation energy of the forward reaction is not large enough to equal the activation of the reverse reaction. 3.The position of the boulder will not change, nor will the position of the H 2 and O 2 in the H 2 + O 2  2H 2 O chemical system because the activation energy is too large. 4.The low energy position on the right shows an overall stable position, but the K value is not large enough.

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–26 ANSWER Choice 3 makes an accurate comparison within the analogy. Some chemical changes may have large values for K (as a boulder may be near a lower energy position) but the rate of the reaction, greatly influenced by the activation energy, may prevent the change or cause it to be too slow to measure. Section 13.5: Applications of the Equilibrium Constant

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–27 QUESTION The following table shows the relation between the value of K and temperature of the system: At 25°C; K = 45; at 50°C; K = 145; at 110°C; K = 467 (a) Would this data indicate that the reaction was endothermic or exothermic? (b) Would heating the system at equilibrium cause more or less product to form? 1.Exothermic; less product 2.Exothermic; more product 3.Endothermic; less product 4.Endothermic; more product

Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 13–28 ANSWER Choice 4 makes correct assumptions using Le Châtelier’s principle. The increase in K with temperature indicates that the reaction uses energy to produce a higher ratio of product to reactant at equilibrium. The stress of heat for an endothermic reaction causes more product to form. Section 13.7: Le Châtelier’s Principle