SET HOMEWORK WEEK 2 CHAPTER 14
4.32 a. Soluble; The ions present would be NH4+ and SO42- (in 2:1 ratio) b. Insoluble c. Insoluble d. Soluble; The ions present would be Ca2+ and NO3- (in 1:2 ratio) 4.34c. Total Ionic equation: Pb2+(aq) + 2 NO3-(aq) + 2 Na+(aq) + 2 Br-(aq) PbBr2(s) + 2 Na+(aq) + 2 NO3-(aq) Net Ionic equation: Pb2+(aq) + 2 Br-(aq) PbBr2(s) This is actually an equilibrium, i.e.: Pb2+(aq) + 2 Br-(aq) PbBr2(s) ; This type of equilibrium will be discussed further in Expt. #7 and in lecture near the end of term. 14.4 The addition of reactions 1 and 2 yields reaction 3:
(Reaction 1) HCN (Reaction 2)
+ OHR H2O R
H
R R+
CN +
OH-
+
H2O
Sum: HCN + OH- + H2O CN- + H2O + H+ + OHWe cancel OH- and H2O, which appear on both sides of the equation, to yield: R (Reaction 3) HCN R H+ + CNThe rule states that, if a given equation can be obtained from the sum of other equations, the equilibrium constant for the given equation equals the product of the other equilibrium constants. Thus, K for reaction 3 is K = K1 x K2 = (4.9 x 104) x (1.0 x 10-14) = 4.9 x 10-10
Note that K is small (<< 1), as expected for the ionization of a weak acid.
010. a0. To study the equilibrium, create a saturated solution. Dissolve enough Mg(OH)2 so that after mixing well, there is still excess solid that settles to the bottom of the container. As long as there is solid present in the flask, equilibrium is established. This solution could then be filtered and titrated with standardized HCl solution to determine [OH].
[Once we know about pH, we would see that Mg(OH)2 is a base, so its equilibrium could be studied by pH measurements.]
b0. Ksp = [Mg2+][OH]2 c0. If more solid Mg(OH)2 is added to a saturated solution, it will settle to the bottom of the container and not dissolve. Remember that pure solids do not appear in the Kc expression (see (b)). Therefore, adding more solid will have no effect on the concentrations of Mg2+ and OH ions in solution. d0. Yes, If a solution is prepared from solid Mg(OH)2, then [Mg2+] = 1/2[OH]. We know this from the stoichiometric coefficients in the balanced equation.
In addition, if we know the values of Ksp and [OH-], we can substitute into the K expression to determine [Mg2+]. Presumably, both methods must give the same value for [Mg2+].
14.20
The equilibrium is: H2O(l) H2O(g) and the Kp expression is:
Kp = p(H2O(g))
The important concept is the Kp expression: it contains the vapor, but not the liquid. The addition of a pure liquid (i.e. water) does not affect the equilibrium. (The pure liquid does not appear in the equilibrium constant). Thus, the amount of water vapor does not change appreciably. This behavior is analogous to the non-effect of adding solid Mg(OH)2 in Question 14.18c. (More precisely, the amount of vapor decreases slightly because the liquid takes up more room in the container.) If, instead, you add water vapor to the container, the partial pressure of the vapor has changed and the equilibrium has been disturbed. Thus vapor condenses until the original vapor pressure is restored. Thus, the amount of liquid water in the container increases.
*** If water vapor is allowed to escape from the container, some liquid has to vaporize in order to restore the equilibrium pressure of H2O(g). Given long enough, all the liquid will evaporate.
14.21 First consider this as a limiting reagent problem, in which the limiting reagent is completely consumed. Hydrogen, H2, is the limiting reactant, so the maximum amount of CH3OH that could form is one mol [i.e. 2 mol H2 x (1 mol CH3OH/2 mol H2)] For this situation, the ICE table would look like: I C E CO(g) 2.0 mol -1.0 1.0 + 2 H2(g) 2.0 mol -2.0 0 CH3OH(g) 0 mol +1.0 1.0
Concept: The changes (line C) are in the same proportion as the coefficients in the balanced equation. Now to the actual question: because the reaction comes to equilibrium before it can go to completion, less than one mol of CH3OH forms. The answer is a.
*** With the Excel spreadsheet, you can vary the amount of reaction which occurs by adjusting Kc using the Slider; you will see that the amount of CH3OH never exceeds 1.0 mol.
14.24 Equilibrium has been reached when the concentration of reactants and products is constant. The equilibrium region on the graph is where the lines flatten out indicating the concentrations of reactants and products are not changing. At equilibrium, the concentration of A is approaching 2.0 M, and the concentration of B is approaching 1.0 M. Our I-C-E table has the form: 2 A(g) B(g) Initial 4.0 0 moles Change -2.0 +1.0 moles Equilibrium 2.0 1.0 moles
The figure shows that the Changes are: for A: -2 units; for B: +1 unit. Thus the changes are in the ratio of the stoichiometric coefficients, as they must be.
Substituting the equilibrium values into the equilibrium constant expression gives Kc = [B] [A]
2
=
1.0 (2.0)2
= 0.25
14.33 Use the table approach, and give the starting, change, and equilibrium number of moles of each. Amt. (mol) Starting Change Equilibrium 2SO2(g) 0.0400 -2x 0.0400 - 2x + O2(g)
R R
2SO3(g) 0 +2x 2x (= 0.0296)
0.0200 -x 0.0200 - x
From the information for SO3, x = 0.0296 mol/2 = 0.0148 mol and the amounts of the other substances at equilibrium are:
SO2: 0.0400 2 x 0.0148 = 0.0104 mol O2: 0.0200 0.0148 = 0.0052 mol and
14.34 Use the table approach, and give the starting, change, and equilibrium number of moles of each. Note that we are told that the equilibrium amount of CO is 0.1187 mol.
Amt. (mol) Starting Change Equilibrium
CO(g) 0.1500 -x 0.1500 - x (= 0.1187)
+
2H2(g)
R R
CH3OH(g) 0 +x x
0.3000 -2x 0.3000 - 2x
Because 0.1500 - x = 0.1187, x = 0.0313. Therefore, the amounts of substances at equilibrium are 0.1187 mol CO, 0.2374 mol H2, and 0.0313 mol CH3OH. 14.41 Because Kc = 1.84 for 2HI O H2 + I2, the value of Kc for H2 + I2 O 2HI must O O be the reciprocal of Kc for the first reaction. Mathematically, this can be shown as follows: For the
Forward reaction: K c =
[ H 2 ][ I 2 ] = [ 1.84 HI ]2 [ HI ]2 1 1 = = [ H 2 ][ I 2 ] [ H 2 ][ I 2 ] K c (f ) 2 [ HI ]
For the
Reverse reaction: K c =
Thus, for the reverse reaction, Kc is calculated as follows: Kc = 1 1.84 = 5.434 x 10-1 = 0.543 *** The reciprocal of 1 x 10-12 is 1 x 1012; the reciprocal of 2 x 10-12 is 5 x 1011.
14.42
Because Kc = 27.8 for CS2 + 4H2 O CH4 + 2H2S, the value of Kc for O 1/2CS2 + 2H2 O 1/2CH4 + H2S, the second (2) reaction, must be the square root O of Kc for the first (1) reaction. Mathematically, this can be shown as follows: [CH4 ][H2S]2 [CS2 ][H2 ]4 [CH4 ]1/2 [H2S] [CS2 ]
1/2 1/ 2
Kc(1) =
Kc(2) =
[H2 ]
2
[CH4 ][H2S]2 = [CS2 ][H2 ]4
= [Kc(1)]1/2
Thus, for the second (2) reaction, Kc is calculated as follows: Kc = 27.8 = 5.272 = 5.27
14.46
In problem 14.34, we constructed an ICE table. We do not need to repeat this; we just use the Equilibrium values, convert to molarities, and substitute into the Kc equation. [CH3OH] = 0.0313 mol 1.50 L = 0.02086 M [CO] = 0.1187 mol 1.50 L = 0.07913 M [H2] = 0.2374 mol 1.50 L = 0.1582
The Kc equation is: Kc =
(0.02086 ) (0.07913 )(0.1582) 2
[CH3OH] [CO] [H 2 ]2
= 10.53 = 10.5
Kc =
14.50 a. Kp =
PNO2 2 PN2O 4 PSO3 2 PSO2 2 PO2
b. Kp =
PNOBr 2 PNO 2 PBr2 PNO 4 PH2O 6 PNH3 4 PO2 5 [CO 2 ] [N 2 ] 2 [N 2 O] 2 1 [Fe
3+
c. Kp =
d. Kp =
14.56 a. Kc = [NH3] [HCl]
b. K c =
c. Kc = [H2O] [CO2]
d. K c =
] [OH - ] 3
Concept: Solids and liquids do not appear in the Kc or Kp equation.
020. a0. Kc is very small (1031), indicating very little reaction. At equilibrium, we expect relatively large amounts of N2 and O2 and little amount of NO (this is the behavior in our atmosphere!). b0. Kc is very large (1021), indicating nearly complete reaction. We expect that the limiting reagent (either C2H4 or H2) will be almost completely consumed and the stoichiometric amount of C2H6 will be formed.
14.60
Kc is extremely large, indicating nearly complete reaction at room temperature. The problem states that, at equilibrium, [SO2] = [O2]. As we are given all the equilibrium concentrations, there is no need to build an ICE table; rather, this is a Type 1 problem and we just substitute the known values into the Kc equation. So for the reaction of SO2, [SO 3 ] 2 [SO 2 ] 2 [O 2 ] [SO 3 ] 2 [SO 2 ] 3
Kc =
=
, because [O2] = [SO2] here.
Solving for [SO2] [SO 3 ] 2 [SO2] = Kc
1/ 3
(1.0) 2 = 8.0 x 10 35
1/ 3
= 1.07 x 10-12 = 1.1 x 10-12 M
This agrees with what we expect from the magnitude of K c.
14.66
Substitute into the expression for Kc and solve for [NO]: Kc = 0.0025 = [NO]2 [NO]2 = [N2 ][O2 ] (0.031)(0.023)
[NO] = [(0.0025)(0.031)(0.023)]1/2 = 1.33 x 10-3 = 1.3 x 10-3 M
Note that no ICE Table is needed for a Type 1 problem Another way to look at this is that the stoichiometric coefficient in the balanced equation for NO, which is 2, appears only as the power of [NO] in the Kc expression. DO NOT try to multiply the answer by 2 or anything like that!!!
14.68 Observation: We are given only initial amounts and the Kc value. To find the equilibrium amounts, we need to set up an ICE table, including expressions with x for the unknown amounts. Then solve the resulting Kc equation for x. Divide moles of substance by the volume of 8.00 L to obtain concentration. The starting concentrations are 0.10625 M for both [N2] and [O2]. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) Starting Change Equilibrium
Kc = 0.0123 =
N2(g)
+
O2(g)
R R
2NO(g) 0 +2x 2x
0.10625 -x 0.10625 - x
0.10625 -x 0.10625 - x
[NO]2 (2x)2 = [N2 ][O2 ] (0.10625 - x)(0.10625 - x)
Taking the square root of both sides gives 0.1109 = (2x ) (0.10625 - x )
Rearranging and simplifying the right side yields (0.10625 - x ) = (2x ) = (1 8.03 x ) 0.1109
x = 5.583 x 10-3 M Thus [N2] = [O2] = 0.1006 = 0.101 M, and [NO] = 1.116 x 10-2 = 1.12 x 10-2 M. Using Excel, Slider sheet: set the stoichiometric coefficients and initial amounts, then move the Slider until equilibrium is reached. (Solver sheet: set the stoichiometric coefficients and the initial amounts, then select Tools/Solver to solve for the equilibrium amounts.) 14.69 The strategy is the same as for 14.68. The difference is that the resulting equation is quadratic in x. Divide moles of substance by the volume of 1.25 L to obtain concentration. The starting concentration is 1.00 M for [CO2], but the concentration of carbon, a solid, is omitted. Assemble a table of starting, change, and equilibrium concentrations.
Conc. (M) Starting Change Equilibrium
CO2(g) 1.00 -x 1.00 - x
+
C(s)
R R
2CO(g) 0 +2x 2x
Substituting into the equilibrium expression for Kc gives K c = 14.0 = [CO] 2 (2x ) 2 = [CO 2 ] (1.00 - x )
No approximation is possible as Kc is quite large. However, you can use Solver on your calculator.
If you prefer to use algebra, you need the quadratic solution:
Rearranging and solving for x yields 14.0 - 14.0x = 4x2 4x2 + 14.0x - 14.0 = 0 (quadratic equation) Using the solution to the quadratic equation gives (14.0)2 - 4(4)(-14.0) 2(4) x = -4.31 (impossible; reject), or x = 0.8117 = 0.812 M (logical) x= -14.0 Thus, [CO2] = 0.19 M, and [CO] = 1.62 M.
14.103 a. The molar mass of PCl5 is 208.22 g/mol. Thus, the initial concentration of PCl5 is 35.8 g PCl5 x 1 mol PCl5 208.22 g PCl5 5.0 L
= 0.0344 M
Use the table approach, and give the starting, change, and equilibrium concentrations.
Conc. (M) Starting Change Equilibrium
PCl3(g) 0 x x
+
Cl2(g) 0 x x
R R
PCl5(g) 0.0344 -x 0.0344 - x
Substituting into the equilibrium-constant expression gives K c = 4.1 = [ PC l5 ] 0.0344 - x = [ PC l3 ][ C l 2 ] x2
Once again, use the Solver or the quadratic equation method:
Rearranging and solving for x gives a quadratic equation. 4.1x2 + x - 0.0344 = 0 Using the quadratic formula gives -1 (1) 2 - (4)(4.1)(-0.0344) 2(4.1)
x=
= 0.0306 (positive root)
Thus, at equilibrium, [PCl3] = [Cl2] = x = 0.031 M. The concentration of PCl5 is [PCl5] = .0344 - x = 0.0344 - 0.0306 = 0.0038 = 0.004 M