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AP Chemistry Unit 7- Homework Problems Equilibrium and Ksp Nature of the Equilibrium State 1. Draw on this graph where equilibrium has been reached.
equilibrium
[X]
time
2. What are three qualities of any equilibrium equation? a. Reversible b. Dynamic c. Occur when there are a stable ratio of products: reactants 3. For a general equation: aA + bB cC + dD , write the equation for Kc. Kc = [C]c[D]d [A]a[B]b
Developing Keq 1. For each of the equations below, write the expression for K c: a. 2 H2S (g) 2 H2 (g) + S2 (g) Kc = [H2]2[S2] [H2S]2 b. HCN (aq) + H2O (l) H3O+ (aq) + CN-1 (aq) Kc = [H3O+] [CN-1] [HCN] c. PbCl2 (s) Pb+2 (aq) + 2 Cl-1 (aq) Kc = [Pb+2] [Cl-1]2
2. For each of the equations below, write the expression for K p: a. SO2Cl2 (g) SO2 (g) + Cl2 (g) Kp = PSO2 PCl2 PSO2Cl2 b. CO (g) + H2O (g) CO2 (g) + H2 (g) Kp = PCO2 PH2 PCO PH2O c. C6H12O6 (s) + 6 O2 (g) 6 CO2 (g) + 6 H2O (g) Kp = P6CO2 P6H2O P6 O2 3. Put the following K values in order of increasing product-favored ability. a. K = 4x10-5 b. K = 2x10-9 c. K = 7x10-5 d. K = 3x10-3 b>>> reaction 4.8 x10 goes left so rxn goes left
Calculating Equilibrium Conditions 12. For the equation: PCl5 (g) PCl3 (g) + Cl2 (g), you start with 0.25 atm of each of the products as well as the reactants. The Kp value is 0.125. Is the reaction at equilibrium? Prove it. What are the equilibrium pressures of all species? Q = PPCl3PCl2 Q = (0.25)2 = 0.25 PCl5(s) PCl3 (g) + Cl2 (g) PPCl5 (0.25) Po 0.25 0.25 0.25 Q > Kp 0.25 > 0.125 so there are too many x -x -x products so the reaction moves to the left Peq 0.25 + x 0.25 -x 0.25 - x Kp = PPCl3PCl2 0.125 = (0.25-x)2 x = 0.055 PPCl5 (0.25+x) PPCl5 = 0.305 PPCl3 = PCl2 = 0.195
13. For the equation: H2O (g) + CO (g) CO2 (g) + H2 (g) Kc = 0.235 If 2 moles of each of H2O and CO are put into a 10 L container, what is the concentration of all species at equilibrium? Kc = [CO2][H2] 0.235 = (x)2 H2O (g) + CO (g) CO2 (g) + H2 (g) [H2O][CO] (0.2-x)2 Po 0.2 0.2 0 0 x Peq 0.2 - x
-x 0.2 -x
+x x
+x x
x = 0.065 [CO2] = [H2] = 0.065 [H2O] = [CO] = 0.135
14. For the equation: SO2Cl2 (g) SO2 (g) + Cl2 (g) Kp = 4.8 If enough SO2Cl2 is put into a container so its pressure is 8 atm, what is the equilibrium pressure of all species. What is the total pressure? SO2Cl2 (g) Po 8 x Peq 8-x
SO2 (g) + 0 +x x
Cl2 (g) 0 +x x
Kp = Kp =
PSO2PCl2 PSO2Cl2 [x]2 8-x
Kp = 4.25
PSO2 = PCl2 = 4.25 PSO2Cl2 = 3.75 Ptot = 3.75+ 4.25 + 4.25 = 12.25
15. For the equation: CaCO3 (s) CaO (s) + CO2 (g) Kp = 2.4 If 200 g CaCO3 is put into a 20 L container at 500 K, how many grams of it remain at equilibrium? CaCO3 (s)
CaO (s) +
Po Peq
CO2 (g) 0 +x x
Kp = 2.4 = PCO2 PCO2 = 2.4 PV = nRT (2.4 atm)(20 L) = n(0.0821 Latm/molK)(500 K) n = 1.17moles reacted away 1.17 moles CaCO3* (100.1 g/mole) = 117 g reacted 200 g to begin with – 117 g reacted = 83 g left
16. For the equation: 2 KClO3 (s) 2 KCl (s) + 3 O2 (g), you start with some KClO3 that decomposes into the products. At equilibrium, there is some solid remaining and the total pressure in the flask is 0.29 atm. Calculate the value of Kp. 2 KClO3 (s) Po Peq
2 KCl (s) +
3 O2 (g) 0 +0.29 0.29
KP =
P3O2 = 0.0244
17. For the equation: NH4Cl (s) NH3 (g) + HCl (g), you start with some NH4Cl that decomposes into the products. At equilibrium, there is some solid remaining and the total pressure in the flask is 1.8 atm. Calculate the value of Kp. NH4Cl (s) Po Peq
NH3 (g) + 0 +0.9 0.9
HCl (g) 0 +0. 9 0.9
KP =
PNH3PHCl = (0.9)(0.9) = 0.81
18. For the equation: COBr2 (g) CO (g) + Br2 (g), you start with 4 moles in a 10 L vessel of COBr 2. The reaction has a Kc = 0.76. What are the equilibrium concentrations of all species? COBr2 (g) []o 0.4 -x []eq 0.4-x
Kc = CO (g) + 0 +x x
Br2 (g) 0 +x x
[CO][Br2] [COBr2] 0.76 = x2 x = 0.29 [0.4-x] [COBr2] = 0.11 [CO] = [Br2] = 0.29
19. For the equation: H2 (g) + CO2 (g) H2O (g) + CO (g), you start with 2 atm of each of the reactants and none of the products. The Kp = 3.4. What are the equilibrium pressures of all species? H2 (g) + CO2 (g) H2O (g) + CO (g) Kp = PH2OPCO Po 2 2 0 0 PH2PCO2 -x -x +x +x 3.4 = x2 x = 1.3 Peq 2-x 2-x x x [2-x]2 PCO2 = PH2 = 0.7 PCO = PH2O = 1.3
20. For the equation: 2 CH2Cl2 (g) CH4 (g) + CCl4 (g), you start with 0.25 M of CH2Cl2 and is has a Kc value of 0.84. What are the equilibrium concentrations of all species? 2 CH2Cl2 (g) CH4 (g) + CCl4 (g) Kc = [CH4][CCl4] []o 0.25 0 0 [CH2Cl2] 2 0.84 = x2 x = 0.0809 -2x +x +x [0.25-2x]2 []eq 0.25-2x x x [CH2Cl2] = 0.0882 [CH4] = [CCl4] = 0.0809 21. For the equation: NH4HS (s) NH3 (g) + H2S (g), you start with 100 grams of NH4HS (s) in a 2.5 L flask at 500 K. The Kp value is 1.45. How many grams of the solid remain at equilibrium? NH4HS (s) NH3 (g) + Po 0 +x Peq x
H2S (g) 0 +x x
KP =
PNH3PH2S x2 = 1.45
x = 1.20 atm
n = PV/RT = (1.2 atm)(2.5 L)/[(0.0821)(500 K)] n = 0.0731 moles NH3 so there must have been the same moles of NH4HS that broke up (0.0731 mol)(51.1 g/mol) = 3.735 g broke up so: 100 g – 3.735 g = 96.265 g remain
LeChatelier’s Principle 1. State LeChatelier’s Principle.A change in any factor that effects the equilibrium of a system will cause the system to shift in such a way to reduce or counteract that initial change. 2. For the following reaction: Heat + CaCO3 (s) + CO2 (g) + H2O (l) Ca+2 (aq) + 2 HCO3-1 (aq) What will be the effect of doing each of the following actions on the above equilibrium? a) Adding CaCO3 (s) Left Right No Change b) Removing Ca+2 (aq) Left Right No Change c) Removing CO2 (g) Left Right No Change d) Adding NaHCO3 (s) Left Right No Change e) Adding Ne (g) Left Right No Change f) Adding CO2 (g) Left Right No Change g) Increasing temperature Left Right No Change h) Decreasing volume Left Right No Change 3. For the following reaction: 2 SO2 (g) + O2 (g) 2 SO3 (g) + Heat What will be the effect of doing each of the following actions on the above equilibrium? a) Decreasing temperature Left Right No Change b) Increasing O2 (g) Left Right No Change c) Decreasing SO2 (g) Left Right No Change d) Increasing volume Left Right No Change e) Increasing SO3 (g) Left Right No Change f) Adding N2 (g) Left Right No Change
Ksp 1.
For each of the substances below, write the solubility equation as well as the K sp equation. a. AgCl AgCl Ag+ + Cl-1 Ksp = [Ag+][Cl-1] +2 -1 b. PbI2 PbI2 Pb + 2 I Ksp = [Pb+2][I-1]2 + -2 c. Ag2CO3 Ag2CO3 2 Ag + CO3 Ksp = [Ag+]2 [CO3-2] +2 -3 d. Ca3(PO4)2 Ca3(PO4)2 3 Ca + 2 PO4 Ksp = [Ca+2]3 [PO4-3]2
Problem Solving with Ksp 1. Calculate the Ksp of CaCrO4 if a saturated solution has [Ca+2] = 4.5 x10-5 CaCrO4 Ca+2 + CrO4-2 Ksp = [Ca+2][CrO4-2] = (4.5 x10-5) (4.5 x10-5) = 2.025 x10-9 2. Calculate the Ksp of Fe(OH)3 if a saturated solution has [Fe+3] = 4.2 x10-6 Fe(OH)3 Fe+3 + 3 OH-1 Ksp = [Fe+3][OH-1]3 = (4.2 x10-6) [(3)(4.2 x10-6)]3 = 8.4 x10-21 3. Calculate the solubility (moles/L) of PbCO3 if Ksp = 7.4 x10-14 PbCO3 Pb+2 + CO3-2 Ksp = [Pb+2][CO3-2] 7.4x10-14 = x2 x = 2.72 x10-7 mole/L -5 4. Calculate the solubility (moles/L) of Ag2SO4 if Ksp = 1.2 x10 Ag2SO4 2 Ag+1 + SO4-2 Ksp = [Ag+1]2 [SO4-2] 1.2x10-5 = 4x3 x = 0.0144 mole/L -9 5. Calculate the solubility (mg/L) of FePO4 if Ksp = 9.4 x10 FePO4 Fe+3 + PO4-3 Ksp = [Fe+3][PO4-3] 9.4x10-9 = x2 x = 9.70 x10-5 mole/L -25 6. Calculate the solubility (mg/L) of Al2(CO3)3 if Ksp = 7.2 x10 Al2(CO3)3 2 Al+3 + 3 CO3-2 Ksp = [Al+3]2 [CO3-2]3 7.2x10-25 = 108x5 x = 5.82 x10-6 mole/L *(234 g/mole) *(1000 mg/g) = 1.36 mg/L 7. How many mg of CuCrO4 will dissolve in 50 mL of water (Ksp = 9.4 x10-10) CuCrO4 Cu+2 + CrO4-2 Ksp = [Cu+2] [CrO4-2] 9.4x10-10 = x2 x = 3.07 x10-5 mole/L *(179.5 g/mole) *(1000 mg/g)*(0.05 L) = 0.276 mg 8. If 100 mg of CaCO3 (Ksp = 3.4 x10-9) is put in 500 mL of water, how many mg remain undissolved? CaCO3 Ca+2 + CO3-2 Ksp = [Ca+2] [CO3-2] 3.4x10-9 = x2 -5 x = 5.83 x10 mole/L *(179.5 g/mole) *(1000 mg/g)*(0.5 L) = 2.92 mg 100 mg – 2.92 mg dissolved = 97.1 mg undissolved 9. Put the following substances in order of least soluble to most soluble. NiCO3 (Ksp = 1.4x10-7) AuCl (Ksp = 2x10-13) MnCO3 (Ksp = 2.3x10-11) PbCrO4 (Ksp = 2.8x10-13) Since they are all the same number of pieces, you can compare them directly by K sp values. Thus: AuCl (Ksp = 2x10-13) < PbCrO4 (Ksp = 2.8x10-13) < MnCO3 (Ksp = 2.3x10-11) < NiCO3 (Ksp = 1.4x10-7) 10. Put the following substances in order of least soluble to most soluble. Zn(CN)2 (Ksp = 8x10-12) AgBr (Ksp = 5.x10-13) Pb(OH)2 (Ksp = 1.4x10-15) BaSO4 (Ksp = 1.1x10-10) These substances have different number of ionic pieces so you can’t compare K sp values directly. If you do a solubility ICE table for each, the ultimate results will be as follows where x is the solubility: Zn(CN)2 4x3 = 8x10-12 x = 1.27x10-4 M 2 -13 AgBr x = 5x10 x = 7.1x10-7 M 3 -15 Pb(OH)2 4x = 1.4x10 x = 7.1x10-6 M 2 -10 BaSO4 x = 1.1x10 x = 1.1x10-5 M Thus
AgBr < Pb(OH)2 < BaSO4 < Zn(CN)2
11. Prove which of each of the substances below is the most soluble. a. AgBr (Ksp = 5.4x10-13) vs. AgI (Ksp = 8.5 x10-17) AgBr Ag+1 + Br-1 Ksp = [Ag+1][Br-1] 5.4x10-13 = x2 AgI Ag+1 + I-1 Ksp = [Ag+1][I-1] 8.5x10-17 = x2
x = 7.3 x10-7 mole/L x = 9.2 x10-9 mole/L
b. PbCl2 (Ksp = 1.7 x10-5) vs. PbBr2 (Ksp = 6.6 x10-6) PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3 PbBr2 Pb+2 + 2 Br-1 Ksp = [Pb+2][Br-1]2 6.6x10-6 = 4x3
x = 0.0162 mole/L x = 0.0118 mole/L
c. AgCl (Ksp = 1.8 x10-10) vs. Ag2CrO4 (Ksp = 1.1 x10-12) AgCl Ag+1 + Cl-1 Ksp = [Ag+1][Cl-1] 1.8x10-10 = x2 +1 -1 Ag2CrO4 2 Ag + CrO4 Ksp = [Ag+1]2 [CrO4-2] 1.1x10-12 = 4x3
x = 1.3 x10-5 mole/L x = 6.5x10-5 mole/L
d. CaCO3 (Ksp = 3.4 x10-9) vs. Mg(OH)2(Ksp = 5.6 x10-12) CaCO3 Ca+2 + CO3-2 Ksp = [Ca+2][CO3-2] 3.4x10-9 = x2 +2 -1 Mg(OH)2 Mg + 2 OH Ksp = [Mg+2][OH-1]2 5.6x10-12 = 4x3
x = 5.8 x10-5 mole/L x = 1.12 x10-4 mole/L
Common Ion Effect 1. Calculate the solubility (moles/L) of MgS (Ksp = 5.2 x10-16) in: a. Pure water MgS Mg+2 + S-2 Ksp = [Mg+2][S-2] 5.2x10-16 = x2 b. A 0.25 M solution of MgCl2 MgS Mg+2 + S-2 Ksp = [Mg+2][S-2] 5.2x10-16 = (0.25)(x)
x = 2.3 x10-8 mole/L
2. Calculate the solubility (moles/L) of PbCl2 (Ksp = 1.7 x10-5) in: a. Pure water PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3 b. A 0.55 M solution of NaCl PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.7x10-5 = (x)(0.55)2
x = 0.0162 mole/L
3. Calculate the solubility (moles/L) of PbI2 (Ksp = 9.8 x10-9) in: a. Pure water PbI2 Pb+2 + 2 I-1 Ksp = [Pb+2][I-1]2 9.8x10-9 = 4x3 b. A 0.005 M solution of AlI3 PbI2 Pb+2 + 2 I-1 Ksp = [Pb+2][I-1]2 9.8x10-9 = (x)(0.015)2
x = 1.35x10-3 mole/L
x = 2.1 x10-15 mole/L
x = 5.6x10-5 mole/L
x = 0.435 mole/L
4. Calculate what mass of Hg2SO4 (Ksp = 6.5 x10-7) will dissolve per liter in: a. Pure water Hg2SO4 2 Hg+1 + 2 SO4-2 Ksp = [Hg+1]2 [SO4-2] 6.5x10-7 = 4x3 x = 5.46x10-3 mole/L(497.3 g/mol) = 2.71 g/L b. A 0.0075 M solution of Na 2SO4 Hg2SO4 2 Hg+1 + 2 SO4-2 Ksp = [Hg+1]2 [SO4-2] 6.5x10-7 = (4x2)(0.0075) x = 4.65x10-3 mole/L (497.3 g/mol) = 2.31 g/L
5. Calculate what mass of CaCO3 (Ksp = 3.4 x10-9) will dissolve per liter in: a. Pure water CaCO3 Ca+2 + CO3-2 Ksp = [Ca+2][CO3-2] 3.4x10-9 = x2 x = 5.83 x10-5 mole/L (100.1 g/mole) = 5.84 x10-3 g/L b. A 0.45 M solution of CaCl2 CaCO3 Ca+2 + CO3-2 Ksp = [Ca+2][CO3-2] 3.4x10-9 = (0.45)(x) x = 7.56 x10-9 mole/L (100.1 g/mole) = 7.56 x10-7 g/L 6. Calculate what mass of PbCl2 (Ksp = 1.7 x10-5) will dissolve per liter in: a. Pure water PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.7x10-5 = 4x3 x = 0.0162 mole/L(278.2 g/mole) = 4.5 g/L b. A 0.067 M solution of AlCl3 PbCl2 Pb+2 + 2 Cl-1 Ksp = [Pb+2][Cl-1]2 1.7x10-5 = (x)(0.201)2 x = 4.2x10-4 mole/L(278.2 g/mole) = 0.117 g/L 7. AgBr will be the least soluble in 0.10 M: a. NaBr b. CaBr2 c. AlBr3 d. Ag2SO4 e. AgNO3 AlBr3 will have the most concentrated [Br-1] or [Ag+1] since there are 3 Br-1 for every 1 AlBr3. Thus, AgBr will be least soluble in it. 8. MgF2 will be least soluble in 0.25 M: a. KF b. Mg(NO3)2 c. MgC2O4 d. BaF2 e. LiF -1 +2 -1 BaF2 will have the most concentrated [F ] or [Mg ] since there are 2 F for every 1 BaF2. Thus, MgF2 will be least soluble in it.
Precipitation 1. Will a ppt of CaCO3 (Ksp= 3.4 x10-9) form if [Ca+2] = 4 x10-6 M and [CO3-2] = 4 x10-3? Q = [4 x10-6][4x10-3] = 1.6x10-8 >> 3.4x10-9 so yes, ppt 2. Will a ppt of Ag2CrO4 (Ksp = 1.1 x10-12 ) form if [Ag+] = 3x10-4 and [CrO4-2] = 2x10-4? Q = [3 x10-4]2 [2x10-4] = 1.8x10-11 >> 1.1x10-12 so yes, ppt 3. Will a ppt of BaCO3 (Ksp = 2.6 x10-9) form if 50 mL of 4x10-5 M Ba+2 is mixed with 50 mL of 8x10-5 M CO3-2? Q = [(50/100)4 x10-5][(50/100)8x10-5] = 8x10-10 > 6.6x10-6 so yes, ppt 5. What concentration of [OH-1] will cause a ppt of Fe(OH)2 (Ksp = 4.9x10-17) of a 2x10-5 M Fe+2 solution ? 4.9x10-17 = (2x10-5)(OH-1)2 [OH-1] = 1.56x10-6 M -2 6. What concentration of [C2O4 ] will cause a ppt of Ag2C2O4 (Ksp = 5.4x10-12) of a 4x10-4 M Ag+1 solution? 5.4x10-12 = (4x10-4)2(C2O4-2) [C2O4-2] = 3.4x10-5 M 7. What mass of Ni(NO3)2*7 H2O (s) will cause a ppt of NiCO3 (Ksp = 1.4x10-7) in a 2 L solution of 3.5x10-4 M CO3-2? 1.4x10-7 = [Ni+2](3.5x10-4) [Ni+2] = 4x10-4 M Molar mass of Ni(NO3)2*7H2O = 308.7 g/mole [Ni+2] = 4x10-4 moles/L * 2 L * 308.7 g/mole = 0.247 grams Ni(NO 3)2*7H2O 8. What mass of Pb(NO3)2 will cause a ppt of PbCl2 (Ksp = 1.7x10-5) in 100 mL of a 2.4x10-2 M solution of AlCl3? Since there are 3 Cl-1 ions for every AlCl3 then [Cl-1] = 7.2x10-2 M so: 1.7x10-5 = [Pb+2](0.072)2 [Pb+2] = 3.28x10-3 M Molar mass of Pb(NO3)2 is 331.2 g/mole Thus: 3.28x10-3 moles/L * 1 L/1000 mL * 100 mL * 331.2 g/mole = 0.109 grams Pb(NO 3)2
Separations by Ksp 1. A solution is made so that [Ca+2] = 0.00050 and [Mg+2] = 0.00050 as well. If Cr2O7-2 is added, answer the following questions. Know that Ksp CaCr2O7 = 4.8 x10-7 and Ksp MgCr2O7 = 7.6 x10-8 a. Which will ppt 1st? At what [Cr2O7-2] will it begin to ppt? CaCr2O7 Ca+2 + Cr2O7-2 Ksp = [Ca+2][ Cr2O7-2] 4.8x10-7 = x(0.0005) x = 9.6 x10-4 mole/L +2 -2 +2 -2 -8 MgCr2O7 Mg + Cr2O7 Ksp = [Mg ][ Cr2O7 ] 7.6x10 = x(0.0005) x = 1.5 x10-4 mole/L MgCr2O7 ppts 1st b. What is the maximum [Cr2O7-2] that can be made to ppt almost all of one and none of the other? 9.6x10-4 moles/L c. What is the concentration of the less soluble ion under the conditions specified in “b”? MgCr2O7 Mg+2 + Cr2O7-2 Ksp = [Mg+2][ Cr2O7-2] 7.6x10-8 = (9.6x10-4)x x = 7.92 x10-5 mole/L d. What % of the less soluble ion remains in solution under the conditions specified in “b”? (7.92x10-5/0.0005)*100% = 15.8% 2. A solution is made so that [Pb+2] = 0.00250 and [Hg+2] = 0.0075. If SO4-2 is added, answer the following questions. Know that Ksp PbSO4 = 2.5 x10-8 and Ksp HgSO4 = 5.4 x10-7 a. Which will ppt 1st? At what [SO4-2] will it begin to ppt? PbSO4 Pb+2 + SO4-2 Ksp = [Pb+2][ SO4-2] 2.5x10-8 = x(0.0025) x = 1 x10-5 mole/L +2 -2 +2 -2 -7 HgSO4 Hg + SO4 Ksp = [Hg ][ SO4 ] 5.4x10 = x(0.0075) x = 7.2 x10-5 mole/L st PbSO4 ppts 1 b. What is the maximum [SO4-2] that can be made to ppt almost all of one and none of the other? HgSO4 Hg+2 + SO4-2 Ksp = [Hg+2][ SO4-2] 5.4x10-7 = x(0.0075) x = 7.2 x10-5 mole/L c. What is the concentration of the less soluble ion under the conditions specified in “b”? PbSO4 Pb+2 + SO4-2 Ksp = [Pb+2][ SO4-2] 2.5x10-8 = x(7.2x10-5) x = 3.5 x10-4 mole/L d. What % of the less soluble ion remains in solution under the conditions specified in “b”? (3.5x10-4/0.0025)*100% = 13.9% 3. A solution is made so that [Zn+2] = 0.00250 and [Ag+1] = 0.057. If F-1 is added, answer the following questions. Know that Ksp ZnF2 = 4.8 x10-7 and Ksp AgF = 7.6 x10-8 a. Which will ppt 1st? At what [F-1] will it begin to ppt? ZnF2 Zn+2 + 2 F-1 Ksp = [Zn+2][ F-1]2 4.8x10-7 = x2(0.0025) x = 1.39 x10-2 mole/L +1 -1 +1 -1 -8 AgF Ag + F Ksp = [Ag ][ F ] 7.6x10 = x(0.057) x = 1.3 x10-6 mole/L st AgF ppts 1 b. What is the maximum [F-1] that can be made to ppt almost all of one and none of the other? 1.39x10-2 mole/L c. What is the concentration of the less soluble ion under the conditions specified in “b”? AgF Ag+1 + F-1 Ksp = [Ag+1][ F-1] 7.6x10-8 = 1.39x10-2(x) x = 5.47 x10-6 mole/L d. What % of the less soluble ion remains in solution under the conditions specified in “b”? (5.47x10-6/0.057)*100% = 0.0096%
Combination Problems 1. For the reaction at 2000 K H2 (g) + CO2 (g) H2O (g) + CO (g) For an experiment, the equilibrium values of each substance are as follows: [H2] = 0.20 M [CO2] = 0.30 M [H2O] = [CO] = 0.55 M a. What is the mole fraction of CO in the equilibrium mixture? XCO = (0.55)/(0.20+0.30+0.55+0.55) = 0.355 b. Calculate the value of Kc, the equilibrium constant for the reaction above. Kc = (0.55)2/(0.20)(0.30) = 5.042 c. Determine Kp in terms of Kc for this system. Kp = Kc because ngas = 0 d. When the system is cooled from 2000 K to a lower temperature, 30% of the CO is converted back to CO2. Calculate the value of Kc at this lower temperature. Since 30% of CO is converted: 0.55*0.30 = 0.165 changes for each H2 (g) + CO2 (g) H2O (g) + CO (g) []0 0.20 0.30 0.55 0.55 []eq 0.365 0.465 0.385 0.385 Kc = (0.385)2/(0.365)(0.465) = 0.873 e. In a different experiment, 0.50 mole of H2 is mixed with 0.50 mole of CO2 in a 3.0 L reaction vessel at 2000 K. Calculate the equilibrium concentration, in M, of CO at this temperature. H2 (g) + CO2 (g) H2O (g) + CO (g) []0 0.167 0.167 0 0 -x -x +x +x []eq 0.167-x 0.167-x x x 5.042 = x2 /(0.167-x)2
2.245 = x/(0.167-x)
x = [CO] = 0.1156
2. For the reaction: PCl5 (g) PCl3 (g) + Cl2 (g) It is observed that greater amounts of PCl3 and Cl2 are produced as the temperature is increased. a) What is the sign of So for the reaction? Explain. S should be positive. You are going from one mole of gas on the left to 2 moles on the rights. This shows an increase in disorder and a positive entropy change b) What change, if any, will occur in Go for the reaction as the temperature is increased? Explain. Since S is positive, G should decrease as temperature is increased. This is due to the equation G = H –TS. Since T is in K and therefore positive, you will be subtracting a positive value from G. c) If He gas is added to the original mixture at constant volume and temperature, what will happen to the partial pressure of Cl2? Explain. Dalton’s law of partial pressures says that the total pressure is equal to the sum of the individual pressures. Adding more He gas will increase the total pressure but not change the pressure the Cl2 exerts. d) If the volume of the reaction mixture is decreased at constant temperature to half the original volume, what will happen to the number of moles of Cl2 in the reaction vessel? Explain. By decreasing the volume, you are increasing the pressures of each of the above species. Since there are unequal amounts of gas on the reactant and product sides of the equation above, the system will try to reduce the pressure increase it experienced. It can do this by going to the left side which will reduce the number of moles of Cl2
3. For the reaction: C(s) + CO2(g) 2 CO(g) Solid carbon and carbon dioxide gas at 1,160 K were placed in a rigid 2.00 L container, and the reaction represented above occurred. As the reaction proceeded, the total pressure in the container was monitored. When equilibrium was reached, there was still some C(s) remaining in the container. Results are recorded in the table below. Time (hours) Total Pressure of gases (atm) 0.0 5.00 2.0 6.26 4.0 7.09 6.0 7.75 8.0 8.37 10.0 8.37 (a) Write the expression for the equilibrium constant, Kp , for the reaction. Kp = P2CO/PCO2 (b) Calculate the number of moles of CO2(g) initially placed in the container. (Assume that the volume of the solid carbon is negligible.) n = PV/RT = (5 atm)(2 L)/(0.0821)(1160 K) = 0.105 moles (c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO2(g) is 1.63 atm. Calculate: (i) the partial pressure of CO(g) , and CO2: 5 atm – 1.63 atm = 3.37 atm CO2 reacted *(2 mole CO/1 mole CO2 )= 6.74 atm CO (ii) the value of the equilibrium constant, Kp . Kp = P2CO/PCO2 = (6.74)2/1.63 = 27.87 (d) If a suitable solid catalyst were placed in the reaction vessel, would the final total pressure of the gases at equilibrium be greater than, less than, or equal to the final total pressure of the gases at equilibrium without the catalyst? Justify your answer. (Assume that the volume of the solid catalyst is negligible.) Nothing would change. It would just get to equilibrium faster. (e) In another experiment involving the same reaction, a rigid 2.00 L container initially contains 10.0 g of C(s) , plus CO(g) and CO2(g) , each at a partial pressure of 2.00 atm at 1,160 K. Predict whether the partial pressure of CO2(g) will increase, decrease, or remain the same as this system approaches equilibrium. Justify your prediction with a calculation. Q = (2)2/2 = 2 Q