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17-1
(a) reversible reaction: a reaction that can occur in either direction and does not go to completion. Such reactions reach equilibrium with significant amount of reactants and products remaining.
(b) static equilibrium: a state of balance that exists when all acting influences are canceled, so there are no micro changes.
(c) equilibrium constant, K: a quantity that describes the position of equilibrium for a reversible reaction. It is equal to the product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of the products, each raised to the power that corresponds to its coefficient in the balanced chemical equation.
17-3
Equilibrium exists when all acting influences are canceled by others, resulting in a balanced system. In a “static equilibrium” there are no micro changes caused by the influences. In a “dynamic equilibrium” there are micro changes but the changes occur in such a way that the net result is that the system remains unchanged. Chemical equilibria are dynamic equilibria; individual molecules are continually reacting, even though the overall composition of the reaction mixture does not change.
17-5
(a) A very large value of K indicates that the reaction proceeds far toward completion (or that the forward reaction is highly favored over the reverse reaction). The equilibrium mixture has much higher concentrations of products than of reactants.
(b) A very small value of K indicates that the reaction proceeds only very slightly (or that the reverse reaction is highly favored over the forward reaction). The equilibrium mixture has much higher concentrations of reactants than of products.
(c) A value of K of about 1 indicates that the forward and reverse reactions are about equally favored. The equilibrium mixture contains about equal concentrations of products and reactants, if they are raised to the same power.
17-7 At equilibrium the rates of opposing processes are equal. Putting it another way: when the rates of opposing processes become equal, the system arrives at a state of equilibrium.
17-9 The equilibrium constant is the product of the equilibrium concentrations of the products divided by the product of the equilibrium concentrations of reactants, each raised to the power that corresponds to its coefficient in the balanced equation. When using the value of an equilibrium constant, it is necessary that we also know the balanced chemical equation since the powers of the terms in the equilibrium constant are tied to the coefficients in the balanced equation. For example, the following two equations have different values for their equilibrium constants.
17-11
(a) Your sketch should be a curve that is like that of Figure 17-2(b) except your curve has 2 reactants (the blue line) and 2 products (the red lines). After time tc, the line for D should be twice the value for C, as they are in Figure 17-2(b) for O2 and SO2. The blue lines for A and B should also be, after time tc, in the ratio of 3 to 1 if they started with equal concentrations.
(b) For K much less than 1, your sketch should be a curve much more like Figure 17-2(a) with the changes suggested in answer (a) above.
17-13
(a) Balanced equation: N2 + 3H2 2NH3 Kc = = 1
The statement given is false. The coefficients in the balanced equation represent the relative number of moles of reactants that are consumed or products that are formed, They have no bearing on the relative concentrations of the species in the equilibrium condition. If we assume [N2] = 1.0 M, [H2] = 3.0 M and [NH3] = 6.0 M, the corresponding Kc value is 1.5 instead of 1.
(b) This statement is also false. If the reaction started with only NH3 and went to equilibrium, then [H2] is three times that of [N2], but there are many other possibilities.
(c) This statement can be true. Let x = [N2], 3x = [H2] and 2x = [NH3]
17-15
(d) This statement can be true. Kc = = = 1
(e) This statement is false. The equilibrium expression is not satisfied if the concentrations of all reactants and products are equal and have any value other than 1M
(f) An equilibrium mixture does not require all reactants and products to have the same concentrations. It only requires that the concentrations of the components fulfill the Kc expression. This statement is false.
Of those species listed, the following would not be included in an equilibrium expression because they are solids: H2O(s), NaHCO3(s), and Fe2O3(s). These would not be included because they are liquids: H2 CH3 and NH3
17-17 (a) Kc = (b) Kc = (c) Kc = (d) Kc = (e) Kc =
17-3
The system is NOT at equilibrium. The concentration of NO2 must increase to achieve equilibrium. Thus the reverse reaction must proceed to a greater extent to increase the amount of the reactant and to bring the reaction to equilibrium (Q = K).
17-37 For the hypothetical reaction: aA + bB cC + dD
the reaction quotient Q is defined as Q
The difference between Q and K is that the concentration values in K must be equilibrium concentrations, while those in Q may be any values, including equilibrium ones.
17-39 The forms are the same but Q uses instantaneous concentrations, while K is valid only with equilibrium concentrations. Q varies with changes in concentration at constant T whereas K does not. 17-41 Q = [ClF]2 [Cl2][F
] = (7.3)2 (0.5)(0.2) = 5.3 x 102 > K = 19.9
The reaction will favor the reactants so that ClF concentration will decrease and both Cl2 and F2 concentrations will increase until equilibrium is established.
2 17-43 C(graphite) + CO2(g)
(3.5 mol/1.5 L)2 (3.5 mol/1.5 L) = 2.3; Q >> Kc; reverse rxn is favored
17-53 If a change of condition (stress) is applied to a system at equilibrium, the system responds in a way that tends to reduce the stress and to reach a new state of equilibrium. Factors that affect a system at equilibrium are changes in concentration, temperature and total pressure (by changing volume). A catalyst has no effect on the value of K or on the equilibrium position; it changes the time required to reach equilibrium. A catalyst affects both forward and reverse reactions the same; hence, it has no net effect on K
17-55
(a) Decreasing the temperature of an exothermic reaction favors products, so the amount of B present at the new equilibrium will be less than before the temperature was decreased.
(b) Adding more A will favor the forward reaction so the amount of B will decrease.
(c) Adding more C will favor the reactants so the amount of B will increase.
(d) Removing small amounts of D favors the forward reaction, decreasing the amount of B.
(e) Increasing the pressure will favor the side of the equation that has the fewer number of moles of gaseous species. In this reaction, the increased pressure will favor the reactants so the amount of B will increase.
17-57 If the total pressure were decreased, the reaction that forms more moles of gas will be favored (forward reaction is favored).
17-59 Pink color indicates “Moist” air the presence of a high concentration of H2O will shift the reaction toward the more hydrated pink compound.
17-61 Increasing the temperature causes the reaction to re-establish equilibrium by producing less heat. (We note that this involves a change in the value of the equilibrium constant; for an exothermic reaction, increasing T decreases K, whereas for an endothermic reaction, increasing T increases K.) So we expect increasing the temperature will cause these reactions to shift as follows:
(a) to the left (b) to the right (c) to the left (d) to the right (e) to the right
Change in the concentration of NH3
(1) removing O2(g) increase- reverse reaction favored
(2) adding N2 (g) increase- reverse reaction favored
(3) adding water increase- reverse reaction favored
(4) expanding the container at constant pressure increase ( 7 moles of reactant gas to 2 moles of product gas favors the reverse reaction)
(5) increasing the temperature increase- reverse reaction favored
17-65 Note: Volume is not given. Because there are equal numbers of moles on both sides of the equation, the volumes would cancel and we can work with the reaction summary in numbers of moles (a)
+ B C + D
(b) Let x = moles of A that react to reach equilibrium. (See note in part (a) about neglect of volume.)
x = = 10.3
= = = 65 – 10. =
17-77 CaCO3(s) CaO(s) + CO2(g)
(a) Kc = Kp(RT) n = 1.16(RT)-1 = = =
(b) Kp = ; Kp = 1.16 atm = Alternatively, since Kc =[CO2] and the equilibrium concentration of CO2 is x when 22.5 g of CaCO3 are placed in a 9.56-L container, then Kc =[CO2] = 0.0132 M. To convert to pressure, we will find the number of moles of CO2 in the 9.56 L container and use the ideal gas law.
9.56L x = 0.128 mol CO2
P = = = 1.18 atm
(c) While beginning with 22.5 g of CaCO3, the amount left undecomposed can be calculated by finding the beginning moles and subtracting the moles reacted.
? mol CaCO3 = 22.5 g CaCO3 x = 0.224 mol CaCO3 originally
? mol CaCO3 at equilibrium = 0.224 mol originally − 0.128 mol reacted = 0.0968 mol CaCO3
? g CaCO3 = 0.0968 mol unreacted x =
PH2 is very nearly zero assuming there is an excess of CuO(s).
17-81 The calculation from Go gives the value of Kp directly (Go = -RT ln Kp) for a gaseous reaction. This may then be converted to the value of Kc if desired.
17-83
(b) For an all gas reaction, Go is related to Kp, so we first calculate the value of Kp
The left side of the equation (ln term) is positive; the right side is negative so H is negative
17-87 (a) The reaction is exothermic as written, so the reverse reaction is favored by high temperature.
Thus, the value of Kp at 500°C is expected to be less than the value at 25°C.
(b)
Using the van’t Hoff equation,
17-89 Fish need O2(aq) to live. With heating has caused less O2(aq) to be present. This would be shifting the equation to the left or towards more reactants; therefore, the equation must be exothermic. With an increased temperature, the equilibrium constant would decrease since there are less products..
17-91 (a) Product favored; K>1
(b) Reactant favored; K<1
(c) Reactant favored; K<1
K
= ( )( )8 = (0.20)(0.00262)8 = (b)
Kp = ( ) = 0.20
In the second case, water is in a condensed phase (pure solid) and is taken as unity.
(b) Increasing volume with Br2 absent will have no effect on the equilibrium position since all of the partial pressures of all components will change in a similar manner and the total number of moles of gas are the same on both sides of the reaction.
Increasing volume with Br2 present will keep the PBr2 constant at 0.28 atm while allowing the other partial pressures of the other components to change (decrease momentarily). The net effect should be that of adding more Br2(g), which would shift the equilibrium toward products (to the right).
17-97 Cl2(g) + F2(g) 2ClF(g) [ClF]2
Q = [Cl2][F2] = = 15.0; Q < K = 19.9
The reaction will favor the product, so the ClF concentration will increase and both Cl2 and F2 concentrations will decrease until equilibrium is established.
17-99 WCl6(g) + 3H2(g) W(s) + 6HCl(g)
PHCl6 (0.10)6 21
Kp = PWCl6 PH23 = (0.012) PH 3 = 1.37 x 10
PH2 3 = = 6.1 x 10–26
PH2 = (6.1 x 10–26)1/3 =
17-101 In a chemical equilibrium, a balanced system exists in which the relative amounts of reactants and products remain constant or equal to some value. At the molecular level there are changes, but the changes occur in such a way that the system remains unchanged or equal to its stable proportions.
17-103 C6H5COOH(s) + H2 C6H5COO–(aq) + H3O+(aq)
As a weak acid, benzoic acid will ionize to a small extent as indicated in the above equilibrium reaction. If D2O is added to the water, it should be predicted that it will react like H2O and that some of it will become HD2O+. The HD2O+ can give up a D in the reverse reaction yielding HDO and C6H5COOD(s). Thus, with sufficient time there should be some transfer of the D from the D2O to the C6H5COOH.
17-105 Assuming a 1 liter container and that at equilibrium, [N2O4] = the concentration of [NO2] at 110 oC.
N2O4 (g) 2 NO2 (g) start 3.00M 0 change
x +2x equil 3.00 – x M 2 x M
We know that at equilibrium, the concentration of N2O4 = the concentration of NO2 , so: 3.00 – x = 2x x = 1M
[N2O4] = 3.00 – 1 = [NO2] = 2x =
Since the problem asks how much NO2 will be present, the answers could be in M, g, or mol. Alternatively: mol NO2 = 2.00 M x 1 L = 2.00 mol or 2.00 mol x 46.0 g/mol = 92 g NO2
17-107 From the equilibrium constant, you can get the following:
8HNO3 (aq) + 3Cu(NO3)2 (aq) + 2NO (g)
To balance the equation, you will have to add copper in a form that would NOT be in the equilibrium constant:
8HNO3 (aq) + 3Cu(s) 3Cu(NO3)2 (aq) + 2NO (g)
Then to balance the equation, we are missing 4 O and 8H, but it must be in a form that would NOT be in the equilibrium constant:
17-109 Mountain or space sickness must reflect the decreased partial pressure of O2 in these environments. With less O2 in the inspired air, less O2 will be taken up by the hemoglobin and less O2 will then circulate to the tissues. This leads to hypoxia at the tissues and could result in some damage to tissues.