AP Chemistry Unit 7 Homework Problems Equilibrium and K sp


 Amos Bates
 10 months ago
 Views:
Transcription
1 AP Chemistry Unit 7 Homework Problems Equilibrium and K sp Nature of the Equilibrium State 1. Draw on this graph where equilibrium has been reached. [X] equilibrium 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 K c. K c = [C] c [D] d [A] a [B] b Developing K eq 1. For each of the equations below, write the expression for K c : a. 2 H 2 S (g) 2 H 2 (g) + S 2 (g) K c = [H 2 ] 2 [S 2 ] [H 2 S] 2 b. HCN (aq) + H 2 O (l) H 3 O + (aq) + CN 1 (aq) K c = [H 3 O + ] [CN 1 ] [HCN] c. PbCl 2 (s) Pb +2 (aq) + 2 Cl 1 (aq) K c = [Pb +2 ] [Cl 1 ] 2
2 2. For each of the equations below, write the expression for K p : a. SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K p = P SO2 P Cl2 P SO2Cl2 b. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) K p = P CO2 P H2 P CO P H2O c. C 6 H 12 O 6 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O (g) K p = P 6 CO2 P 6 H2O P 6 O2 3. Put the following K values in order of increasing productfavored ability. a. K = 4x105 b. K = 2x109 c. K = 7x105 d. K = 3x103 b < a < c < d Equilibrium Mathematics 1. The equation: C (s) + 2 H 2 O (g) CO (g) + H 2 (g) has a value of K c = 2.5 x106 What is the value of K c for: CO (g) + H 2 (g) C (s) + 2 H 2 O (g)? K c = (2.5x106 ) 1 = 4x10 5 What is the value of K c for: 2 C (s) + 4 H 2 O (g) 2 CO (g) + 2 H 2 (g) K c = (2.5x106 ) 2 = 6.25x The equation: H 2 O (g) H 2 (g) + ½ O 2 (g) has a value of K p = 4.9 x103 What is the value of K p for: 2 H 2 O (g) 2 H 2 (g) + O 2 (g)? K p = (4.9x103 ) 2 = 2.4x105 What is the value of K p for: H 2 (g) + ½ O 2 (g) H 2 O (g)? K p = (4.9x103 ) 1 = The equation: 2 NH 3 (g) N 2 (g) + 3 H 2 (g) has a value of K c = 2.7 x104 At STP, What is the value for K p? K p = K c (RT) n = (2.7 x104 )[(0.0821)(273 K)] 2 = At STP, What is the value of K c for: K c = (2.7 x104 ) 1/2 = 60.9 ½ N 2 (g) + 3/2 H 2 (g) NH 3 (g)? 4. The equation: N 2 O 4 (g) 2 NO 2 (g) has a value of K p = 3.2x104 What is the value for K c at 300 K?? K p = K c (RT) n 3.2x104 = K c [(0.0821)(300 K)] 1 = 1.30x105 What is the value for K c for: NO 2 (g) ½ N 2 O 4 (g) K c = (1.30x105 ) ½ = 277
3 5. Given the following equations: H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Kc = 4.8 FeO (s) + CO (g) Fe (s) + CO 2 (g) K c = 0.48 Calculate the K c value for: Fe (s) + H 2 O (g) FeO (s) + H 2 (g) K c =??? H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Kc = 4.8 Fe (s) + CO 2 (g) FeO (s) + CO (g) K c = (0.48) 1 = 2.08 Fe (s) + H 2 O (g) FeO (s) + H 2 (g) K c = (4.8)(2.08) = Given the following equations: S (s) + O 2 (g) SO 2 (g) K p = SO 3 (g) 2 SO 2 (g) + O 2 (g) K p = Calculate the K p value for: S (s) + 3/2 O 2 (g) SO 3 (g) K p =??? S (s) + O 2 (g) SO 2 (g) K p = 48.2 SO 2 (g) + ½ O 2 (g) SO 3 (g) K p = (0.075) 1/2 = 3.65 S (s) + 3/2 O 2 (g) SO 3 (g) K p = Which of the following equations has K c = K p a. PCl 5 (g) PCl 3 (g) + Cl 2 (g) No; gas = +1 b. 2 NOCl (g) 2 NO (g) + Cl 2 (g) No; gas = +1 c. CaCO 3 (s) CaO (s) + CO 2 (g) No; gas = +1 d. H 2 O (g) + CO (g) H 2 (g) + CO 2 (g) Yes; gas = 0 e. 2 NO (g) N 2 (g) + O 2 (g) Yes; gas = 0 K c and K p Calculations At Equilibrium 1. For the reaction: 2 NO 2 (g) N 2 O 4 (g) At equilibrium [N 2 O 4 ] = 0.25 M & [NO 2 ] = M. Calculate K c K c = [N 2 O 4 ] K c = [0.25] K c = 8.16 [NO 2 ] 2 [0.175] 2 2. For the reaction: 2 NH 3 (g) N 2 (g) + 3 H 2 (g) K p = 32 At equilibrium P NH3 = 0.64 atm & P N2 = 1.18 atm. Calculate P H2 K p = P N2 P 3 H2 P 2 NH3 32 = (1.18)P 3 H2 (0.64) 2 P H2 = 2.23
4 Dissociation 3. For the reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) If the initial pressure of PCl 5 is 2 atm and at equilibrium it is 15% dissociated, what is K p? PCl 5 (g) PCl 3 (g) + Cl 2 (g) P o P eq For the reaction: 2 NO (g) N 2 (g) + O 2 (g) If the initial [NO] = 0.50 M and at equilibrium it is 5% dissociated, what is K c? 2 NO (g) N 2 (g) + O 2 (g) [] o [] eq K p = P PCl3 P Cl2 P PCl5 K p = [0.30] 2 K p = K c = [N 2 ] [O 2 ] [NO] 2 K c = [0.0125] 2 K c = 6.9x104 [0.475] 2 5. For the equation: NH 4 I (s) NH 3 (g) + HI (g) The total pressure at equilibrium is 4.2 atm. What is K p? If P tot = 4.2 atm 2x = 4.2 x = 2.1 atm K p = P NH3 P HI = (2.1) 2 = For the equation: (NH 4 )(H 2 NCO 2 ) (s) 2 NH 3 (g) + CO 2 (g) The total pressure at equilibrium is 0.33 atm. What is K p? If P tot = 0.33 atm 3x = 0.33 x = 0.11 K p = P 2 NH3P CO2 = (0.22) 2 (0.11) = 5.32x For the equation: N 2 (g) + O 2 (g) 2 NO (g), you start with 2 M of each of the reactants. They react away to an extent of 27% to reach equilibrium. Calculate the value of K c. N 2 (g) + O 2 (g) 2 NO (g) [] o [] eq K c = [NO] 2 [N 2 ] [O 2 ] K c = [1.08] 2 K c = [1.46] [1.46] 8. For the equation: 2 NOBr (g) 2 NO (g) + Br 2 (g), you start with 0.75 M of the NOBr. At equilibrium, the NOBr has reacted away by 89%. Calculate the value of K c. 2 NOBr (g) 2 NO (g) + Br 2 (g) [] o [] eq K c = [NO] 2 [Br 2 ] [NOBr] 2 K c = [0.668] 2 [0.334] K c = 22.2 [0.082] 2
5 Q vs. K 9. For the reaction: 2 NOCl (g) 2 NO (g) + Cl 2 (g) K c = 1.2x103 If the initial [NOCl] o = 0.15 M, [NO] o = 0.75 M, and [Cl 2 ] o = 0.05 M, is the system at equilibrium? If not, which way will the reaction shift, left or right? K c = [NO] 2 [Cl 2 ] [NOCl] 2 Q = [0.75] 2 [0.05] Q = 1.25 > 1.2x103 so the reaction goes left [0.15] For the reaction: NH 3 (aq) + H 2 O (l) NH 4 +1 (aq) + OH 1 (aq) K c = 1.8x105 If the initial [NH 3 ] o = 0.5 M, [NH 4 +1 ] = M, and [OH 1 ] = M, is the system at equilibrium? If not, which way will the reaction shift, left or right? K c = [NH +1 4 ][OH 1 ] [NH 3 ] Q = [0.0025] 2 Q = 1.25x105 < 1.8x105 so the reaction goes right [0.5] 11. For the equation: CS 2 (g) + 3 Cl 2 (g) S 2 Cl 2 (g) + CCl 4 (g), K c = 4.8x102. If you start with [CS 2 ] = M, [Cl 2 ] = M, [S 2 Cl 2 ]= 0.58 M, and [CCl 4 ] = M, is the reaction at equilibrium? If not, which way will the reaction go to reach equilibrium (left or right)? Q Q = = [S [S s Cl 2 Cl 2 2 ][CCl 4 ] 4 ] Q Q = = (0.58)(0.042) = 2.3 CS 2 (g) + 3 Cl 2 (g) S 2 Cl 2 (g) + CCl 4 [CS [CS 2 ][Cl 2 ] ][Cl 2 ] 3 (0.025)(0.175) 3 [] o Q Q = > 182 K >>>> 4.8 x102 c so reaction goes left so rxn goes left Calculating Equilibrium Conditions 12. For the equation: PCl 5 (g) PCl 3 (g) + Cl 2 (g), you start with 0.25 atm of each of the products as well as the reactants. The K p value is Is the reaction at equilibrium? Prove it. What are the equilibrium pressures of all species? PCl 5 (s) PCl 3 (g) + Cl 2 (g) P o x x x P eq x x x Q = P PCl3 P Cl2 Q = (0.25) 2 = 0.25 P PCl5 (0.25) Q > K p 0.25 > so there are too many products so the reaction moves to the left K p = P PCl3 P Cl = (0.25x) 2 x = P PCl5 (0.25+x) P PCl5 = P PCl3 = P Cl2 = For the equation: H 2 O (g) + CO (g) CO 2 (g) + H 2 (g) K c = If 2 moles of each of H 2 O and CO are put into a 10 L container, what is the concentration of all species at equilibrium? H 2 O (g) + CO (g) CO 2 (g) + H 2 (g) P o x x +x +x P eq x 0.2 x x x K c = [CO 2 ][H 2 ] = (x) 2 [H 2 O][CO] (0.2x) 2 x = [CO 2 ] = [H 2 ] = [H 2 O] = [CO] = 0.135
6 14. For the equation: SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) K p = 4.8 If enough SO 2 Cl 2 is put into a container so its pressure is 8 atm, what is the equilibrium pressure of all species. What is the total pressure? SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) P o x +x +x P eq 8x x x K p = P SO2 P Cl2 P SO2Cl2 K p = [x] 2 K p = x P SO2 = P Cl2 = 4.25 P SO2Cl2 = 3.75 P tot = = For the equation: CaCO 3 (s) CaO (s) + CO 2 (g) K p = 2.4 If 200 g CaCO 3 is put into a 20 L container at 500 K, how many grams of it remain at equilibrium? CaCO 3 (s) CaO (s) + CO 2 (g) P o 0 +x P eq x K p = 2.4 = P CO2 P CO2 = 2.4 PV = nrt (2.4 atm)(20 L) = n( Latm/molK)(500 K) n = 1.17moles reacted away 1.17 moles CaCO 3 * (100.1 g/mole) = 117 g reacted 200 g to begin with 117 g reacted = 83 g left 16. For the equation: 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g), you start with some KClO 3 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 K p. 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) P o P eq 0.29 K P = P 3 O2 =
7 17. For the equation: NH 4 Cl (s) NH 3 (g) + HCl (g), you start with some NH 4 Cl 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 K p. NH 4 Cl (s) NH 3 (g) + HCl (g) P o P eq K P = P NH3 P HCl = (0.9)(0.9) = For the equation: COBr 2 (g) CO (g) + Br 2 (g), you start with 4 moles in a 10 L vessel of COBr 2. The reaction has a K c = What are the equilibrium concentrations of all species? COBr 2 (g) CO (g) + Br 2 (g) [] o x +x +x [] eq 0.4x x x 19. For the equation: H 2 (g) + CO 2 (g) H 2 O (g) + CO (g), you start with 2 atm of each of the reactants and none of the products. The K p = 3.4. What are the equilibrium pressures of all species? H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) P o x x +x +x P eq 2x 2x x x K c = [CO][Br 2 ] [COBr 2 ] 0.76 = x 2 x = 0.29 [0.4x] [COBr 2 ] = 0.11 [CO] = [Br 2 ] = 0.29 K p = P H2O P CO P H2 P CO2 3.4 = x 2 x = 1.3 [2x] 2 P CO2 = P H2 = 0.7 P CO = P H2O = For the equation: 2 CH 2 Cl 2 (g) CH 4 (g) + CCl 4 (g), you start with 0.25 M of CH 2 Cl 2 and is has a K c value of What are the equilibrium concentrations of all species? 2 CH 2 Cl 2 (g) CH 4 (g) + CCl 4 (g) K c = [CH 4 ][CCl 4 ] [] o [CH 2 Cl 2 ] 22x +x +x 0.84 = x 2 x = [] eq x x x [0.252x] 2 [CH 2 Cl 2 ] = [CH 4 ] = [CCl 4 ] = For the equation: NH 4 HS (s) NH 3 (g) + H 2 S (g), you start with 100 grams of NH 4 HS (s) in a 2.5 L flask at 500 K. The K p value is How many grams of the solid remain at equilibrium? NH 4 HS (s) NH 3 (g) + H 2 S (g) P o 0 0 +x +x P eq x x K P = P NH3 P H2S x 2 = 1.45 x = 1.20 atm n = PV/RT = (1.2 atm)(2.5 L)/[(0.0821)(500 K)] n = moles NH 3 so there must have been the same moles of NH 4 HS that broke up ( mol)(51.1 g/mol) = g broke up so: 100 g g = g remain
8 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 + CaCO 3 (s) + CO 2 (g) + H 2 O (l) Ca +2 (aq) + 2 HCO 1 3 (aq) What will be the effect of doing each of the following actions on the above equilibrium? a) Adding CaCO 3 (s) Left Right No Change b) Removing Ca +2 (aq) Left Right No Change c) Removing CO 2 (g) Left Right No Change d) Adding NaHCO 3 (s) Left Right No Change e) Adding Ne (g) Left Right No Change f) Adding CO 2 (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 SO 2 (g) + O 2 (g) 2 SO 3 (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 O 2 (g) Left Right No Change c) Decreasing SO 2 (g) Left Right No Change d) Increasing volume Left Right No Change e) Increasing SO 3 (g) Left Right No Change f) Adding N 2 (g) Left Right No Change
9 K sp 1. For each of the substances below, write the solubility equation as well as the K sp equation. a. AgCl AgCl Ag + + Cl 1 K sp = [Ag + ][Cl 1 ] b. PbI 2 PbI 2 Pb I 1 K sp = [Pb +2 ][I 1 ] 2 c. Ag 2 CO 3 Ag 2 CO 3 2 Ag CO 3 d. Ca 3 (PO 4 ) 2 Ca 3 (PO 4 ) 2 3 Ca PO 4 K sp = [Ag + ] 2 [CO 2 3 ] K sp = [Ca +2 ] 3 [PO 3 4 ] 2 Problem Solving with K sp 1. Calculate the K sp of CaCrO 4 if a saturated solution has [Ca +2 ] = 4.5 x105 CaCrO 4 Ca CrO 4 K sp = [Ca +2 ][CrO 2 4 ] = (4.5 x105 ) (4.5 x105 ) = x Calculate the K sp of Fe(OH) 3 if a saturated solution has [Fe +3 ] = 4.2 x106 Fe(OH) 3 Fe OH 1 K sp = [Fe +3 ][OH 1 ] 3 = (4.2 x106 ) [(3)(4.2 x106 )] 3 = 8.4 x Calculate the solubility (moles/l) of PbCO 3 if K sp = 7.4 x1014 PbCO 3 Pb CO 3 K sp = [Pb +2 ][CO 2 3 ] 7.4x1014 = x 2 x = 2.72 x107 mole/l 4. Calculate the solubility (moles/l) of Ag 2 SO 4 if K sp = 1.2 x105 Ag 2 SO 4 2 Ag SO 4 K sp = [Ag +1 ] 2 [SO 2 4 ] 1.2x105 = 4x 3 x = mole/l 5. Calculate the solubility (mg/l) of FePO 4 if K sp = 9.4 x109 FePO 4 Fe PO 4 K sp = [Fe +3 ][PO 3 4 ] 9.4x109 = x 2 x = 9.70 x105 mole/l 6. Calculate the solubility (mg/l) of Al 2 (CO 3 ) 3 if K sp = 7.2 x1025 Al 2 (CO 3 ) 3 2 Al CO 3 K sp = [Al +3 ] 2 [CO 2 3 ] 3 7.2x1025 = 108x 5 x = 5.82 x106 mole/l *(234 g/mole) *(1000 mg/g) = 1.36 mg/l 7. How many mg of CuCrO 4 will dissolve in 50 ml of water (K sp = 9.4 x1010 ) CuCrO 4 Cu CrO 4 K sp = [Cu +2 ] [CrO 2 4 ] 9.4x1010 = x 2 x = 3.07 x105 mole/l *(179.5 g/mole) *(1000 mg/g)*(0.05 L) = mg 8. If 100 mg of CaCO 3 (K sp = 3.4 x109 ) is put in 500 ml of water, how many mg remain undissolved? CaCO 3 Ca CO 3 K sp = [Ca +2 ] [CO 2 3 ] 3.4x109 = x 2 x = 5.83 x105 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. NiCO 3 (K sp = 1.4x107 ) AuCl (K sp = 2x1013 ) MnCO 3 (K sp = 2.3x1011 ) PbCrO 4 (K sp = 2.8x1013 ) Since they are all the same number of pieces, you can compare them directly by K sp values. Thus: AuCl (K sp = 2x1013 ) < PbCrO 4 (K sp = 2.8x1013 ) < MnCO 3 (K sp = 2.3x1011 ) < NiCO 3 (K sp = 1.4x107 ) 10. Put the following substances in order of least soluble to most soluble. Zn(CN) 2 (K sp = 8x1012 ) AgBr (K sp = 5.x1013 ) Pb(OH) 2 (K sp = 1.4x1015 ) BaSO 4 (K sp = 1.1x1010 ) 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 4x 3 = 8x1012 x = 1.27x104 M AgBr x 2 = 5x1013 x = 7.1x107 M Pb(OH) 2 4x 3 = 1.4x1015 x = 7.1x106 M BaSO 4 x 2 = 1.1x1010 x = 1.1x105 M Thus AgBr < Pb(OH) 2 < BaSO 4 < Zn(CN) 2
10 11. Prove which of each of the substances below is the most soluble. a. AgBr (K sp = 5.4x1013 ) vs. AgI (K sp = 8.5 x1017 ) AgBr Ag +1 + Br 1 K sp = [Ag +1 ][Br 1 ] 5.4x1013 = x 2 x = 7.3 x107 mole/l AgI Ag +1 + I 1 K sp = [Ag +1 ][I 1 ] 8.5x1017 = x 2 x = 9.2 x109 mole/l b. PbCl 2 (K sp = 1.7 x105 ) vs. PbBr 2 (K sp = 6.6 x106 ) PbCl 2 Pb Cl 1 K sp = [Pb +2 ][Cl 1 ] 2 1.7x105 = 4x 3 x = mole/l PbBr 2 Pb Br 1 K sp = [Pb +2 ][Br 1 ] 2 6.6x106 = 4x 3 x = mole/l c. AgCl (K sp = 1.8 x1010 ) vs. Ag 2 CrO 4 (K sp = 1.1 x1012 ) AgCl Ag +1 + Cl 1 K sp = [Ag +1 ][Cl 1 ] 1.8x1010 = x 2 x = 1.3 x105 mole/l Ag 2 CrO 4 2 Ag CrO 4 K sp = [Ag +1 ] 2 [CrO 2 4 ] 1.1x1012 = 4x 3 x = 6.5x105 mole/l d. CaCO 3 (K sp = 3.4 x109 ) vs. Mg(OH) 2 (K sp = 5.6 x1012 ) CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO 2 3 ] 3.4x109 = x 2 x = 5.8 x105 mole/l Mg(OH) 2 Mg OH 1 K sp = [Mg +2 ][OH 1 ] 2 5.6x1012 = 4x 3 x = 1.12 x104 mole/l Common Ion Effect 1. Calculate the solubility (moles/l) of MgS (K sp = 5.2 x1016 ) in: a. Pure water MgS Mg +2 + S 2 K sp = [Mg +2 ][S 2 ] 5.2x1016 = x 2 x = 2.3 x108 mole/l b. A 0.25 M solution of MgCl 2 MgS Mg +2 + S 2 K sp = [Mg +2 ][S 2 ] 5.2x1016 = (0.25)(x) x = 2.1 x1015 mole/l 2. Calculate the solubility (moles/l) of PbCl 2 (K sp = 1.7 x105 ) in: a. Pure water PbCl 2 Pb Cl 1 K sp = [Pb +2 ][Cl 1 ] 2 1.7x105 = 4x 3 x = mole/l b. A 0.55 M solution of NaCl PbCl 2 Pb Cl 1 K sp = [Pb +2 ][Cl 1 ] 2 1.7x105 = (x)(0.55) 2 x = 5.6x105 mole/l 3. Calculate the solubility (moles/l) of PbI 2 (K sp = 9.8 x109 ) in: a. Pure water PbI 2 Pb I 1 K sp = [Pb +2 ][I 1 ] 2 9.8x109 = 4x 3 x = 1.35x103 mole/l b. A M solution of AlI 3 PbI 2 Pb I 1 K sp = [Pb +2 ][I 1 ] 2 9.8x109 = (x)(0.015) 2 x = mole/l 4. Calculate what mass of Hg 2 SO 4 (K sp = 6.5 x107 ) will dissolve per liter in: a. Pure water Hg 2 SO 4 2 Hg SO 4 K sp = [Hg +1 ] 2 [SO 2 4 ] 6.5x107 = 4x 3 x = 5.46x103 mole/l(497.3 g/mol) = 2.71 g/l b. A M solution of Na 2 SO 4 Hg 2 SO 4 2 Hg SO 4 K sp = [Hg +1 ] 2 [SO 2 4 ] 6.5x107 = (4x 2 )(0.0075) x = 4.65x103 mole/l (497.3 g/mol) = 2.31 g/l
11 5. Calculate what mass of CaCO 3 (K sp = 3.4 x109 ) will dissolve per liter in: a. Pure water CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO 2 3 ] 3.4x109 = x 2 x = 5.83 x105 mole/l (100.1 g/mole) = 5.84 x103 g/l b. A 0.45 M solution of CaCl 2 CaCO 3 Ca CO 3 K sp = [Ca +2 ][CO 2 3 ] 3.4x109 = (0.45)(x) x = 7.56 x109 mole/l (100.1 g/mole) = 7.56 x107 g/l 6. Calculate what mass of PbCl 2 (K sp = 1.7 x105 ) will dissolve per liter in: a. Pure water PbCl 2 Pb Cl 1 K sp = [Pb +2 ][Cl 1 ] 2 1.7x105 = 4x 3 x = mole/l(278.2 g/mole) = 4.5 g/l b. A M solution of AlCl 3 PbCl 2 Pb Cl 1 K sp = [Pb +2 ][Cl 1 ] 2 1.7x105 = (x)(0.201) 2 x = 4.2x104 mole/l(278.2 g/mole) = g/l 7. AgBr will be the least soluble in 0.10 M: a. NaBr b. CaBr 2 c. AlBr 3 d. Ag 2 SO 4 e. AgNO 3 AlBr 3 will have the most concentrated [Br 1 ] or [Ag +1 ] since there are 3 Br 1 for every 1 AlBr 3. Thus, AgBr will be least soluble in it. 8. MgF 2 will be least soluble in 0.25 M: a. KF b. Mg(NO 3 ) 2 c. MgC 2 O 4 d. BaF 2 e. LiF BaF 2 will have the most concentrated [F 1 ] or [Mg +2 ] since there are 2 F 1 for every 1 BaF 2. Thus, MgF 2 will be least soluble in it. Precipitation 1. Will a ppt of CaCO 3 (K sp = 3.4 x109 ) form if [Ca +2 ] = 4 x106 M and [CO 32 ] = 4 x103? Q = [4 x106 ][4x103 ] = 1.6x108 >> 3.4x109 so yes, ppt 2. Will a ppt of Ag 2 CrO 4 (K sp = 1.1 x1012 ) form if [Ag + ] = 3x104 and [CrO 42 ] = 2x104? Q = [3 x104 ] 2 [2x104 ] = 1.8x1011 >> 1.1x1012 so yes, ppt 3. Will a ppt of BaCO 3 (K sp = 2.6 x109 ) form if 50 ml of 4x105 M Ba +2 is mixed with 50 ml of 8x105 M CO 32? Q = [(50/100)4 x105 ][(50/100)8x105 ] = 8x1010 << 2.6x109 so no ppt 4. Will a ppt of PbBr 2 (K sp = 6.6 x106 ) form if 150 ml of 2x102 M Pb(NO 3 ) 2 is mixed with 50 ml of 3x102 M AlBr 3? Q = [(150/200)2 x102 ][(50/200)(3) 3x102 ] = 7.6x106 >> 6.6x106 so yes, ppt 5. What concentration of [OH 1 ] will cause a ppt of Fe(OH) 2 (K sp = 4.9x1017 ) of a 2x105 M Fe +2 solution? 4.9x1017 = (2x105 )(OH 1 ) 2 [OH 1 ] = 1.56x106 M 6. What concentration of [C 2 O 42 ] will cause a ppt of Ag 2 C 2 O 4 (K sp = 5.4x1012 ) of a 4x104 M Ag +1 solution? 5.4x1012 = (4x104 ) 2 (C 2 O 42 ) [C 2 O 42 ] = 3.4x105 M 7. What mass of Ni(NO 3 ) 2 *7 H 2 O (s) will cause a ppt of NiCO 3 (K sp = 1.4x107 ) in a 2 L solution of 3.5x104 M CO 32? 1.4x107 = [Ni +2 ](3.5x104 ) [Ni +2 ] = 4x104 M Molar mass of Ni(NO 3 ) 2 *7H 2 O = g/mole [Ni +2 ] = 4x104 moles/l * 2 L * g/mole = grams Ni(NO 3 ) 2 *7H 2 O 8. What mass of Pb(NO 3 ) 2 will cause a ppt of PbCl 2 (K sp = 1.7x105 ) in 100 ml of a 2.4x102 M solution of AlCl 3? Since there are 3 Cl 1 ions for every AlCl 3 then [Cl 1 ] = 7.2x102 M so: 1.7x105 = [Pb +2 ](0.072) 2 [Pb +2 ] = 3.28x103 M Molar mass of Pb(NO 3 ) 2 is g/mole Thus: 3.28x103 moles/l * 1 L/1000 ml * 100 ml * g/mole = grams Pb(NO 3 ) 2
12 Separations by K sp 1. A solution is made so that [Ca +2 ] = and [Mg +2 ] = as well. If Cr 2 O 72 is added, answer the following questions. Know that K sp CaCr 2 O 7 = 4.8 x107 and K sp MgCr 2 O 7 = 7.6 x108 a. Which will ppt 1 st? At what [Cr 2 O 2 7 ] will it begin to ppt? CaCr 2 O 7 Ca Cr 2 O 7 K sp = [Ca +2 ][ Cr 2 O 2 7 ] 4.8x107 = x(0.0005) x = 9.6 x104 mole/l MgCr 2 O 7 Mg Cr 2 O 7 K sp = [Mg +2 ][ Cr 2 O 2 7 ] 7.6x108 = x(0.0005) x = 1.5 x104 mole/l MgCr 2 O 7 ppts 1 st b. What is the maximum [Cr 2 O 2 7 ] that can be made to ppt almost all of one and none of the other? 9.6x104 moles/l c. What is the concentration of the less soluble ion under the conditions specified in b? MgCr 2 O 7 Mg Cr 2 O 7 K sp = [Mg +2 ][ Cr 2 O 2 7 ] 7.6x108 = (9.6x104 )x x = 7.92 x105 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (7.92x105 /0.0005)*100% = 15.8% 2. A solution is made so that [Pb +2 ] = and [Hg +2 ] = If SO 2 4 is added, answer the following questions. Know that K sp PbSO 4 = 2.5 x108 and K sp HgSO 4 = 5.4 x107 a. Which will ppt 1 st? At what [SO 2 4 ] will it begin to ppt? PbSO 4 Pb SO 4 K sp = [Pb +2 ][ SO 2 4 ] 2.5x108 = x(0.0025) x = 1 x105 mole/l HgSO 4 Hg SO 4 K sp = [Hg +2 ][ SO 2 4 ] 5.4x107 = x(0.0075) x = 7.2 x105 mole/l PbSO 4 ppts 1 st b. What is the maximum [SO 2 4 ] that can be made to ppt almost all of one and none of the other? HgSO 4 Hg SO 4 K sp = [Hg +2 ][ SO 2 4 ] 5.4x107 = x(0.0075) x = 7.2 x105 mole/l c. What is the concentration of the less soluble ion under the conditions specified in b? PbSO 4 Pb SO 4 K sp = [Pb +2 ][ SO 2 4 ] 2.5x108 = x(7.2x105 ) x = 3.5 x104 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (3.5x104 /0.0025)*100% = 13.9% 3. A solution is made so that [Zn +2 ] = and [Ag +1 ] = If F 1 is added, answer the following questions. Know that K sp ZnF 2 = 4.8 x107 and K sp AgF = 7.6 x108 a. Which will ppt 1 st? At what [F 1 ] will it begin to ppt? ZnF 2 Zn F 1 K sp = [Zn +2 ][ F 1 ] 2 4.8x107 = x 2 (0.0025) x = 1.39 x102 mole/l AgF Ag +1 + F 1 K sp = [Ag +1 ][ F 1 ] 7.6x108 = x(0.057) x = 1.3 x106 mole/l AgF ppts 1 st b. What is the maximum [F 1 ] that can be made to ppt almost all of one and none of the other? 1.39x102 mole/l c. What is the concentration of the less soluble ion under the conditions specified in b? AgF Ag +1 + F 1 K sp = [Ag +1 ][ F 1 ] 7.6x108 = 1.39x102 (x) x = 5.47 x106 mole/l d. What % of the less soluble ion remains in solution under the conditions specified in b? (5.47x106 /0.057)*100% = %
13 Combination Problems 1. For the reaction at 2000 K H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) For an experiment, the equilibrium values of each substance are as follows: [H 2 ] = 0.20 M [CO 2 ] = 0.30 M [H 2 O] = [CO] = 0.55 M a. What is the mole fraction of CO in the equilibrium mixture? X CO = (0.55)/( ) = b. Calculate the value of K c, the equilibrium constant for the reaction above. K c = (0.55) 2 /(0.20)(0.30) = c. Determine K p in terms of K c for this system. K p = K c because n gas = 0 d. When the system is cooled from 2000 K to a lower temperature, 30% of the CO is converted back to CO 2. Calculate the value of K c at this lower temperature. Since 30% of CO is converted: 0.55*0.30 = changes for each H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) [] [] eq K c = (0.385) 2 /(0.365)(0.465) = e. In a different experiment, 0.50 mole of H 2 is mixed with 0.50 mole of CO 2 in a 3.0 L reaction vessel at 2000 K. Calculate the equilibrium concentration, in M, of CO at this temperature. H 2 (g) + CO 2 (g) H 2 O (g) + CO (g) [] x x +x +x [] eq x x x x = x 2 /(0.167x) = x/(0.167x) x = [CO] =
14 2. For the reaction: PCl 5 (g) PCl 3 (g) + Cl 2 (g) It is observed that greater amounts of PCl 3 and Cl 2 are produced as the temperature is increased. a) What is the sign of S o 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 G o 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 T S. 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 Cl 2? 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 Cl 2 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 Cl 2 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 Cl 2
15 3. For the reaction: C(s) + CO 2 (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) (a) Write the expression for the equilibrium constant, Kp, for the reaction. K p = P 2 CO/P CO2 (b) Calculate the number of moles of CO 2 (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) = moles (c) For the reaction mixture at equilibrium at 1,160 K, the partial pressure of the CO 2 (g) is 1.63 atm. Calculate: (i) the partial pressure of CO(g), and CO 2 : 5 atm 1.63 atm = 3.37 atm CO 2 reacted *(2 mole CO/1 mole CO 2 )= 6.74 atm CO (ii) the value of the equilibrium constant, Kp. K p = P 2 CO/P CO2 = (6.74) 2 /1.63 = (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 CO 2 (g), each at a partial pressure of 2.00 atm at 1,160 K. Predict whether the partial pressure of CO 2 (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 <<< K p so reaction will go to the right This will cause the partial pressure of CO 2 to decrease
16 4. Answer the following questions: a. A saturated solution is prepared by adding excess PbI 2 (s) to distilled water to form 1.0 L of solution at 25 o C. The concentration of Pb +2 (aq) in the saturated solution is found to be 1.3x103 M. The chemical equation for the dissolution of PbI 2 (s) in water is shown below: PbI 2 (s) Pb +2 (aq) + 2 I 1 (aq) i) Write the equilibriumconstant expression for the equation. K sp = [Pb +2 ][I 1 ] 2 ii) Calculate the molar concentration of I 1 (aq) in the solution. [Pb +2 ] = 1.3x103 M *(2 moles I 1 /1 mole Pb +2 ) = 2.6x103 M iii) Calculate the value of the equilibrium constant, K sp. K sp = (1.3x103 )(2.6x103 ) 2 = 8.79x109 b. A saturated solution is prepared by PbI 2 (s) to distilled water to form 2.0 L of solution at 25 o C. What are the molar concentrations of Pb +2 (aq) and I 1 (aq) in the solution? Justify your answer. The concentrations should be the same as above; [Pb +2 ] = 1.3x103 and [I 1 ] = 2.6x103. Changing the volume does not change the concentration if the solution is still saturated. c. Soild NaI is added to a saturated solution of PbI 2 at 25 o C. Assuming that the volume of the solution does not change, does the molar concentration of Pb +2 (aq) in the solution increase, decrease, or stay the same? Justify your answer. By adding NaI, you are adding [I 1 ] and thus, by LeChatelier s Principle, the reaction should go to the left and form a precipitate. To do this, you must use up some of the Pb +2 so thus [Pb +2 ] should decrease 5. The value of K sp for the salt BaCrO 4 is 1.2x When a 500. ml sample of 8.2x106 M Ba(NO 3 ) 2 is added to 500. ml of 8.2x106 M Na 2 CrO 4, no precipitate is observed. i) Assuming the volumes are additive, calculate the molar concentrations of Ba +2 (aq) and CrO 42 (aq) in the 1.00 L of solution. [Ba +2 ] = 8.2x106 M Ba(NO 3 ) 2 *(500/1000) = 4.1x106 M [CrO 42 ] = 8.2x106 M Na 2 CrO 4 *(500/1000) = 4.1x106 M ii) Use the molar concentrations of Ba +2 (aq) and CrO 2 4 (aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer. Q = [Ba +2 ][CrO 2 4 ] = (4.1x106 )(4.1x106 ) = 1.681x1011 Q < K sp 1.681x101 < 1.2x1010 so no ppt forms
17 6. Silver chromate dissociates in water according the equation below: Ag 2 CrO 4 (s) 2 Ag +1 (aq) + CrO 42 (aq) K sp = 2.6x1012 at 25 o C a) Write the equilibriumconstant expression for the dissolving of Ag 2 CrO 4 (s) K sp = [Ag +1 ] 2 [CrO 42 ] b) Calculate the concentration, in M, of Ag +1 (aq) in a saturated solution of Ag 2 CrO 4 at 25 o C 2.6x1012 = 4x 3 x = 8.66x105 M Since there are 2 Ag +1 ions present, [Ag +1 ] = 1.732x104 c) Calculate the maximum mass, in grams, of Ag 2 CrO 4 that can dissolve in 100. ml of water at 25 o C. x = 8.6x105 moles/l *(0.10 L) *(331.8 g/mole) = g d) A mol sample of solid AgNO 3 is added to a 1.00 L saturated solution of Ag 2 CrO 4. Assuming no volume change, does [CrO 42 ] increase, decrease, or stay the same? Justify. Since AgNO 3 will dissolve and make more Ag +1 ions, the reaction will go left due to Le Chatelier s principle. This will cause CrO 42 to react and ppt so [CrO 42 ] will decrease. 7. In a saturated solution of Ag 3 PO 4 at 25 o C, the concentration of Ag +1 (aq) is 5.3x105 M. The equilibrium constant expression for the dissolving of Ag 3 PO 4 (s) in water is shown below: K sp = [Ag +1 ] 3 [PO 43 ] a) Write the balanced equation for the dissolving of Ag 3 PO 4 in water. Ag 3 PO 4 3 Ag +1 + PO 43 b) Calculate the value of K sp for Ag 3 PO 4 at 25 o C. [Ag +1 ] = 5.3x105 but there is only 1 PO 43 for every 3 Ag +1 so: 5.3x105 *(1 mole PO 43 /3 mole Ag +1 ) = 1.77 x105 M PO 43 K sp = [Ag +1 ] 3 [PO 43 ] = (5.3x105 ) 3 (1.77x105 ) = 2.635x1018 c) A 1.00 L sample of saturated Ag 3 PO 4 solution is allowed to evaporate at 25 o C to a final volume of 500. ml. What is the [Ag +1 ] in the solution? Justify your answer. The concentration does not change in a saturated solution. As 500 ml evaporates, solid Ag 3 PO 4 will ppt out at the bottom of the beaker. The solution will still be saturated, though, and therefore not change its concentration.
V. POLYPROTIC ACID IONIZATION. NOTICE: K a1 > K a2 > K a3 EQUILIBRIUM PART 2. A. Polyprotic acids are acids with two or more acidic hydrogens.
EQUILIBRIUM PART 2 V. POLYPROTIC ACID IONIZATION A. Polyprotic acids are acids with two or more acidic hydrogens. monoprotic: HC 2 H 3 O 2, HCN, HNO 2, HNO 3 diprotic: H 2 SO 4, H 2 SO 3, H 2 S triprotic:
More informationSteps for balancing a chemical equation
The Chemical Equation: A Chemical Recipe Dr. Gergens  SD Mesa College A. Learn the meaning of these arrows. B. The chemical equation is the shorthand notation for a chemical reaction. A chemical equation
More informationCHM1 Review for Exam 9
Topics 1. Reaction Types a. Combustion b. Synthesis c. Decomposition d. Single replacement i. Metal activity series ii. Nonmetal activity series e. Double replacement i. Precipitates and solubility rules
More informationSolubility Product Constant
Solubility Product Constant Page 1 In general, when ionic compounds dissolve in water, they go into solution as ions. When the solution becomes saturated with ions, that is, unable to hold any more, the
More informationChapter 16 AcidBase Equilibria and Solubility Equilibria
Chapter 16 AcidBase Equilibria and Solubility Equilibria Student: NOTE: A table of ionization constants and K a 's is required to work some of the problems in this chapter. 1. In which one of the following
More informationGeneral Chemistry II Chapter 20
1 General Chemistry II Chapter 0 Ionic Equilibria: Principle There are many compounds that appear to be insoluble in aqueous solution (nonelectrolytes). That is, when we add a certain compound to water
More informationName AP Chemistry / / Chapter 13 Collected AP Exam Free Response Questions 1980 2010 Answers
Name AP Chemistry / / Chapter 13 Collected AP Exam Free Response Questions 1980 2010 Answers 1980  #6 NH 4 Cl(s) NH 3 (g) + HCl(g) ΔH = +42.1 kilocalories Suppose the substances in the reaction above
More informationSolutions CHAPTER Specific answers depend on student choices.
CHAPTER 15 1. Specific answers depend on student choices.. A heterogeneous mixture does not have a uniform composition: the composition varies in different places within the mixture. Examples of non homogeneous
More information1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) 3 (s) + H 2 (g)
1. When the following equation is balanced, the coefficient of Al is. Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) A) 1 B) 2 C) 4 D) 5 E) Al (s) + H 2 O (l)? Al(OH) (s) + H 2 (g) Al (s) + H 2 O (l)? Al(OH)
More informationCHAPTER 4. AQUEOUS REACTION CHEMISTRY
CAPTER. AQUEOUS REACTION CEMISTRY solution  homogeneous mixture of or more substances; uniform distribution of particles and same properties throughout. A solution is composed of a solute dissolved in
More informationCHEMICAL EQUILIBRIUM
Chemistry 10 Chapter 14 CHEMICAL EQUILIBRIUM Reactions that can go in both directions are called reversible reactions. These reactions seem to stop before they go to completion. When the rate of the forward
More informationBuffer Solutions. Buffer Solutions
Chapter 18 Common Ion Effect Buffers and Titration Curves A/B Titrations Salts and Solubility Product The Common Ion Effect and If a solution is made in which the same ion is produced by two different
More informationChapter 4: Solution Stoichiometry Cont. Aqueous Solutions
Chapter 4: Solution Stoichiometry Cont. 1 Aqueous Solutions Molarity (dilution calculations, solution stoichiometry); Solubility and Solubility Rules Molecular, Ionic and Net Ionic Equations Precipitation
More informationChemistry Ch 15 (Solutions) Study Guide Introduction
Chemistry Ch 15 (Solutions) Study Guide Introduction Name: Note: a word marked (?) is a vocabulary word you should know the meaning of. A homogeneous (?) mixture, or, is a mixture in which the individual
More information2. Write a balanced chemical equation which corresponds to the following equilibrium constant expression. 1/2 3/ 2
Practice Problems for Chem. 1B Exam 1 F2011 These represent the concepts covered for exam 1. There may be some additional net ionic equations from chem. 1A. This is not the exact exam! Sections 16.116.3
More informationAnswers and Solutions to Text Problems
Chapter 7 Answers and Solutions 7 Answers and Solutions to Text Problems 7.1 A mole is the amount of a substance that contains 6.02 x 10 23 items. For example, one mole of water contains 6.02 10 23 molecules
More information1. Balance the following equation. What is the sum of the coefficients of the reactants and products?
1. Balance the following equation. What is the sum of the coefficients of the reactants and products? 1 Fe 2 O 3 (s) + _3 C(s) 2 Fe(s) + _3 CO(g) a) 5 b) 6 c) 7 d) 8 e) 9 2. Which of the following equations
More informationChemistry 12 Review Sheet on Unit 3 Solubility of Ionic Substances
Chemistry 12 Review Sheet on Unit 3 Solubility of Ionic Substances 1. Identify each of the following as ionic or molecular substances: a) NaCl (aq)... b) CH 3 COOH (aq)... c) CCl 4(l)... d) HNO 3(aq)...
More informationSOLUBILITY CURVES WORKSHEET
SOLUBILITY CURVES WORKSHEET 1.) Which compound is least soluble at 20 o C? At 80 o C? 2.) Which substance is the most soluble at 10 o C? At 50 o C? At 90 o C? 3.) The solubility of which substance is most
More informationSolubility Equilibria
Chapter 17: AcidBase Equilibria and Solubility Equilibria Key topics: Common ion effect Buffers Acidbase equilibria Solubility equilibria; complex ion formation The Common Ion Effect If we have two solutes
More informationCh 8 Notes: Chemical Equations and Reactions
Name: Ch 8 Notes: Chemical Equations and Reactions I. Chemical Equations A properly written chemical equation can summarize any chemical change. The following requirements will help you write and read
More informationCHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS. = 2 mol PH 3. a g b = 0.4 g PH. = 12 mol H 2 O.
HEINS09095117v4.qxd 1/30/06 1:58 PM Page 95 CHAPTER 9 CALCULATIONS FROM CHEMICAL EQUATIONS SOLUTIONS TO REVIEW QUESTIONS 1. A e ratio is the ratio between the e amounts of two atoms and/or ecules involved
More informationChemistry 132 NT. Solubility Equilibria. The most difficult thing to understand is the income tax. Solubility and Complexion Equilibria
Chemistry 13 NT The most difficult thing to understand is the income tax. Albert Einstein 1 Chem 13 NT Solubility and Complexion Equilibria Module 1 Solubility Equilibria The Solubility Product Constant
More informationSolutions Review Questions
Name: Thursday, March 06, 2008 Solutions Review Questions 1. Compared to pure water, an aqueous solution of calcium chloride has a 1. higher boiling point and higher freezing point 3. lower boiling point
More informationName Class Date. Chapter: Chemical Equations and Reactions
Assessment Chapter Test A Chapter: Chemical Equations and Reactions In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question. 1. You
More informationK c = [C]c [D] d [A] a [B] b. k f [NO 2 ] = k r [N 2 O 4 ] = K eq = The Concept of Equilibrium. Chapter 15 Chemical Equilibrium
Chapter 15 Chemical Equilibrium Learning goals and key skills: Understand what is meant by chemical equilibrium and how it relates to reaction rates Write the equilibriumconstant expression for any reaction
More informationEquilibrium Notes Ch 14:
Equilibrium Notes Ch 14: Homework: E q u i l i b r i u m P a g e 1 Read Chapter 14 Work out sample/practice exercises in the sections, Bonus Chapter 14: 23, 27, 29, 31, 39, 41, 45, 51, 57, 63, 77, 83,
More informationMoles. Moles. Moles. Moles. Balancing Eqns. Balancing. Balancing Eqns. Symbols Yields or Produces. Like a recipe:
Like a recipe: Balancing Eqns Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) coefficients subscripts Balancing Eqns Balancing Symbols (s) (l) (aq) (g) or Yields or Produces solid liquid (pure liquid)
More informationReaction Type & Balancing Review Questions
Reaction Type & Balancing Review Questions There are 20 questions sets in this review. The corresponding answers can be found in the second half of this document. 1. Match the term with its proper description.
More informationSolute and Solvent 7.1. Solutions. Examples of Solutions. Nature of Solutes in Solutions. Learning Check. Solution. Solutions
Chapter 7 s 7.1 s Solute and Solvent s are homogeneous mixtures of two or more substances. consist of a solvent and one or more solutes. 1 2 Nature of Solutes in s Examples of s Solutes spread evenly throughout
More informationA k 1. At equilibrium there is no net change in [A] or [B], namely d[a] dt
Chapter 15: Chemical Equilibrium Key topics: Equilibrium Constant Calculating Equilibrium Concentrations The Concept of Equilibrium Consider the reaction A k 1 k 1 B At equilibrium there is no net change
More informationDr. Fus CHEM 1220 CHAPTER 17 PRACTICE EXAM
CHEMISTRY 1220 CHAPTER 17 PRACTICE EXAM All questions listed below are problems taken from old Chemistry 123 exams given here at The Ohio State University. Read Chapter 17.4 17.7 and complete the following
More informationChapter 5 Chemical Quantities and Reactions
Chapter 5 Chemical Quantities and Reactions 1 Avogadro's Number Small particles such as atoms, molecules, and ions are counted using the mole. 1 mole = 6.02 x 10 23 items Avogadro s number 602 000 000
More informationTutorial 5 NET IONIC EQUATIONS
T33 Tutorial 5 NET IONIC EQUATIONS The efficiency and extent of a chemical reaction is very much dependent upon the physical state (solid, liquid, gas, or solution) of reactants and products. Not surprisingly,
More informationAP* Chemistry Solubility Equilibrium
AP* Chemistry Solubility Equilibrium SOLUBILITY EQUILIBRIA (The SolubilityProduct Constant, K sp ) We ve got good news and we ve got bad news Which do you want first? The good news: Solubility equilibrium
More informationc. PCl 3 (l) + 3H 2 O(l) H 3 PO 3 (l) + 3HCl(g)
Chapter 9 Chemical Quantities 1. Although we define mass as the amount of matter in a substance, the units in which we measure mass are a human invention. Atoms and molecules react on an individual particlebyparticle
More informationChapter 16 Review Packet
Chapter 16 Review Packet AP Chemistry Chapter 16 Practice Multiple Choice Portion 1. For which process is ΔS negative? Note: ΔS = S final S initial therefore, if ΔS is positive, S final > S initial if
More informationBalancing Equations Notes
. Unit 9 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written
More informationChapter 8 Review and Study Guide
Name: Class: Date: Chapter 8 Review and Study Guide Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Knowledge about what products are produced in a chemical
More information2. The percent yield is the maximum amount of product that can be produced from the given amount of limiting reactant.
UNIT 6 stoichiometry practice test True/False Indicate whether the statement is true or false. moles F 1. The mole ratio is a comparison of how many grams of one substance are required to participate in
More informationAnswers to Practice Problems 6.1, p Answers to Practice Problems 6.2, p. 227
Answers to Practice Problems 6.1, p. 226 1. This is a synthesis reaction (between a metal and a nonmetal). The balanced chemical equation is: 2Li + Cl 2 2LiCl. 2. This is a synthesis reaction (between
More informationName Class Date. Chapter: Chemical Equations and Reactions
Assessment Chapter Test B Chapter: Chemical Equations and Reactions PART I In the space provided, write the letter of the term or phrase that best completes each statement or best answers each question
More informationChemistry 51 Chapter 8 TYPES OF SOLUTIONS. A solution is a homogeneous mixture of two substances: a solute and a solvent.
TYPES OF SOLUTIONS A solution is a homogeneous mixture of two substances: a solute and a solvent. Solute: substance being dissolved; present in lesser amount. Solvent: substance doing the dissolving; present
More informationCHEMICAL REACTIONS. When sodium bicarbonate, e.g. baking soda, is combined with an acid, a gas is produced
CHEMICAL REACTIONS OBJECTIVES 1. To study reactions between ions in aqueous solutions 2. To observe exothermic and endothermic reactions 3. To study oxidationreduction reactions 4. To practice balancing
More informationCopyright 2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
Sample Exercise 17.1 Calculating the ph When a Common Ion is Involved What is the ph of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of
More information3. How many moles of KCl and O 2 are formed from the decomposition of 6 moles of KClO 3?
What coefficients mean: 2 Na + Cl 2 2NaCl 2 Na 1 Cl 2 2NaCl 4 Na 2 Na + Cl 2 4Cl 2 6 moles Na 2NaCl 10 atoms Na ONLY WORKS FOR MOLES, MOLECULES, ATOMS 1. How many moles of H 2 and O 2 must react to form
More informationName AP CHEM / / Collected AP Exam Essay Answers for Chapter 16
Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16 1980  #7 (a) State the physical significance of entropy. Entropy (S) is a measure of randomness or disorder in a system. (b) From each of
More information) g /mol = 3.86!10"3 moles
Chem01, Winter 006 Midterm N1 01/6/06 Name Answer key SID 1. A solution is prepared by dissolving 5.8 grams on magnesium chloride (MgCl ) in water to produce 50.0 ml of solution. Molecular weight of the
More informationAP*Chemistry Solubility Equilibria
AP*Chemistry Solubility Equilibria SOLUBILITY EQUILIBRIA (K sp, THE SOLUBILITY PRODUCT) Saturated solutions of salts are another type of chemical equilibria. Remember those solubility rules? The fine print
More informationExperiment 1 Chemical Reactions and Net Ionic Equations
Experiment 1 Chemical Reactions and Net Ionic Equations I. Objective: To predict the products of some displacement reactions and write net ionic equations. II. Chemical Principles: A. Reaction Types. Chemical
More informationChapter 6 Notes Science 10 Name:
6.1 Types of Chemical Reactions a) Synthesis (A + B AB) Synthesis reactions are also known as reactions. When this occurs two or more reactants (usually elements) join to form a. A + B AB, where A and
More informationChemical reactions. Classifications Reactions in solution Ionic equations
Chemical reactions Classifications Reactions in solution Ionic equations Learning objectives Distinguish between chemical and physical change Write and balance chemical equations Describe concepts of oxidation
More informationTutorial 4 SOLUTION STOICHIOMETRY. Solution stoichiometry calculations involve chemical reactions taking place in solution.
T27 Tutorial 4 SOLUTION STOICHIOMETRY Solution stoichiometry calculations involve chemical reactions taking place in solution. Of the various methods of expressing solution concentration the most convenient
More informationChapter 14. CHEMICAL EQUILIBRIUM
Chapter 14. CHEMICAL EQUILIBRIUM 14.1 THE CONCEPT OF EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT Many chemical reactions do not go to completion but instead attain a state of chemical equilibrium. Chemical
More informationEQUILIBRIUM. Consider the reversible system initially consisting of reactants only.
EQUILIBRIUM When non reversible chemical reactions proceed to completion, the concentration of the reactants gradually decrease, until there is NO limiting reactant remaining. Most chemical reactions,
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
A.P. Chemistry Practice Test  Ch. 13: Equilibrium Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) At equilibrium,. A) the rates of the forward
More informationEasy Examples: Hd 10 + Bu 8 HdBu. Introduction. (l) or (aq)? Introduction. Symbols. Yields or Produces. Balancing. (l) or (aq)?
Chemical Equations Chemical Equations Like a recipe: Introduction Reactants Products 2H 2 (g) + O 2 (g) 2H 2 O(l) Symbols (s) (l) (aq) (g) Introduction Yields or Produces solid liquid aqueous gas (l) or
More informationChemical Reactions and Equations. Chapter 8
Chemical Reactions and Equations Chapter 8 Describing Chemical Reactions A chemical reaction is the process by which one or more substances are changed into different substances Reactants Products When
More informationCHM1045 Practice Test 2 v.1  Answers Name: Fall 2013 (Ch. 3 & 4)
CHM1045 Practice Test v.1  Answers Name: Fall 01 (Ch. & 4) MULTIPLE CHOICE AND FILL IN THE BLANK. Circle the one alternative that best completes the statement or answers the question or fill in the blank.
More informationCHEM 101/105 Aqueous Solution Chemistry Lect06
CHEM 101/105 Aqueous Solution Chemistry Lect06 Ionic solutes are involved in three types of reactions in aqueous solutions: precipitation reactions  sparingly soluble ionic substances produced when two
More information5. Which of the following reactions is a nonredox decomposition reaction? A) 2CuO 2Cu + O2 B) 2KClO3 2KCl + 3 O2 C) CaCO3 CaO + CO2 D) SO2 + H2O H2SO3
CHEM 120 Online Course Chapter 9. Date: 1. Which of the following general reaction types is characterized by there being a single reactant? A) combination B) decomposition C) singlereplacement D) doublereplacement
More informationAP Chem Unit 1 Assignment 3 Chemical Equations
Symbols used in chemical equations: Symbol Meaning + used to separate one reactant or product from another used to separate the reactants from the products  it is pronounced "yields" or "produces" when
More informationSTOICHIOMETRY STOICHIOMETRY. Measurements in Chemical Reactions. MoleMole Relationships. MassMass Problem. MoleMole Relationships
STOICHIOMETRY STOICHIOMETRY The analysis of the quantities of substances in a chemical reaction. Stoichiometric calculations depend on the MOLE MOLE relationships of substances. Measurements in Chemical
More informationProblems you need to KNOW to be successful in the upcoming AP Chemistry exam.
Problems you need to KNOW to be successful in the upcoming AP Chemistry exam. Problem 1 The formula and the molecular weight of an unknown hydrocarbon compound are to be determined by elemental analysis
More informationCHE141 Chapter 10. Chapter 10 Gases
Chapter 0 Gases. A sample of gas (4.g) initially at 4.00 atm was compressed from 8.00 L to.00 L at constant temperature. After the compression, the gas pressure was atm. (a). 4.00 (b)..00 (c)..00 (d).
More informationNet Ionic Equations Making Sense of Chemical Reactions
14 Making Sense of Chemical Reactions OBJECTIVE Students will be able to write net ionic equations from balanced molecular equations. LEVEL Chemistry NATIONAL STANDARDS UCP.1, UCP.2, B.3 T E A C H E R
More informationQUALITATIVE ANALYSIS
QUALITATIVE ANALYSIS Objectives: 1. To perform spot test precipitation reactions. 2. To write and balance precipitation reaction equations. 3. To learn how to balance equations. 4. To learn the solubility
More informationTest 4: Equations and Math of Equations Review
Name: Tuesday, November 27, 2007 Test 4: Equations and Math of Equations Review 1. Given the balanced equation: 2KClO 2KCl + 3O 3 2 Which type of reaction is represented by this equation? 1. synthesis
More informationCHEMICAL REACTIONS. Chemistry 51 Chapter 6
CHEMICAL REACTIONS A chemical reaction is a rearrangement of atoms in which some of the original bonds are broken and new bonds are formed to give different chemical structures. In a chemical reaction,
More informationUnit 6: Reactions and Stoichiometry (Link to Prentice Hall Text: Chapters 7, 8 & 9)
R e a c t i o n s a n d S t o i c h i o m e t r y N o t e s P a g e 1 Unit 6: Reactions and Stoichiometry (Link to Prentice Hall Text: Chapters 7, 8 & 9) Name: Date Due Assignments Page Number: Problem
More informationChapter 5 Chemical Quantities and Reactions. Collection Terms. 5.1 The Mole. A Mole of a Compound. A Mole of Atoms.
Chapter 5 Chemical Quantities and Reactions 5.1 The Mole Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans 1
More informationAP Review: Predicting Reactions (Question 4) Answers
AP Review: Predicting Reactions (Question 4) Answers For each of the following three reactions, in part (i) write a balanced equation for the reaction and in part (ii) answer the question about the reaction.
More informationTypes of Chemical Reactions
Why? Types of Chemical Reactions Do atoms rearrange in predictable patterns during chemical reactions? Recognizing patterns allows us to predict future behavior. Weather experts use patterns to predict
More informationNET IONIC EQUATIONS. A balanced chemical equation can describe all chemical reactions, an example of such an equation is:
NET IONIC EQUATIONS A balanced chemical equation can describe all chemical reactions, an example of such an equation is: NaCl + AgNO 3 AgCl + NaNO 3 In this case, the simple formulas of the various reactants
More informationTypes of Chemical Reactions PreLab Assignment
Types of Chemical Reactions PreLab Assignment Name Period Date The prelab is due at the beginning of the laboratory period 1. Read the entire procedure before beginning this prelab assignment. What
More informationChemistry Final Exam Review
Name: Date: Block: Chemistry Final Exam Review 20122013 Unit 1: Measurement, Numbers, Scientific Notation, Conversions, Dimensional Analysis 1. Write 0.000008732 in scientific notation 8.732x106 2. Write
More informationCHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS
1 CHEM 110: CHAPTER 3: STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS The Chemical Equation A chemical equation concisely shows the initial (reactants) and final (products) results of
More informationPractise Exam Chapter 4
Practise Exam Chapter 4 1. Which one of the following is a nonelectrolyte? A) aqueous barium nitrate solution B) aqueous calcium chloride solution C) aqueous lithium phosphate solution D) aqueous methyl
More informationREVIEW QUESTIONS Chapter 8
Chemistry 51 ANSWER KEY REVIEW QUESTIONS Chapter 8 1. Identify each of the diagrams below as strong electrolyte, weak electrolyte or nonelectrolyte: (a) Nonelectrolyte (no ions present) (b) Weak electrolyte
More informationChemistry B11 Chapter 4 Chemical reactions
Chemistry B11 Chapter 4 Chemical reactions Chemical reactions are classified into five groups: A + B AB Synthesis reactions (Combination) H + O H O AB A + B Decomposition reactions (Analysis) NaCl Na +Cl
More informationChapter 4 Chemical Composition
Chapter 4 Chemical Composition 4.1 (a) mole; (b) Avogadro s number; (c) empirical formula; (d) solute; (e) molarity; (f) concentrated solution 4. (a) molar mass; (b) percent composition by mass; (c) solvent;
More informationExperiment 5. Chemical Reactions A + X AX AX A + X A + BX AX + B AZ + BX AX + BZ
Experiment 5 Chemical Reactions OBJECTIVES 1. To observe the various criteria that are used to indicate that a chemical reaction has occurred. 2. To convert word equations into balanced inorganic chemical
More informationSECTION 14 CHEMICAL EQUILIBRIUM
11 SECTION 1 CHEMICAL EQUILIBRIUM Many chemical reactions do not go to completion. That is to say when the reactants are mixed and the chemical reaction proceeds it only goes to a certain extent, and
More informationChemistry 212 EXAM 2 February 17, 2004 KEY
1 Chemistry 212 EXAM 2 February 17, 2004 100_ (of 100) KEY Name Part 1: Multiple Choice. (1 point 4. The equilibrium constant for the each, circle only one answer, reaction 1. For the equilibrium that
More informationChemistry: Chemical Equations
Chemistry: Chemical Equations Write a balanced chemical equation for each word equation. Include the phase of each substance in the equation. Classify the reaction as synthesis, decomposition, single replacement,
More informationW1 WORKSHOP ON STOICHIOMETRY
INTRODUCTION W1 WORKSHOP ON STOICHIOMETRY These notes and exercises are designed to introduce you to the basic concepts required to understand a chemical formula or equation. Relative atomic masses of
More informationAqueous Solutions. Water is the dissolving medium, or solvent. Some Properties of Water. A Solute. Types of Chemical Reactions.
Aqueous Solutions and Solution Stoichiometry Water is the dissolving medium, or solvent. Some Properties of Water Water is bent or Vshaped. The OH bonds are covalent. Water is a polar molecule. Hydration
More informationTutorial 7 HEATS OF REACTION
T54 Tutorial 7 HEATS OF REACTION In a chemical reaction bonds are broken in the reactants and new bonds formed in the products. Energy is required to break bonds and energy is released when bonds are
More informationMoles. Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations
Moles Balanced chemical equations Molar ratios Mass Composition Empirical and Molecular Mass Predicting Quantities Equations Micro World atoms & molecules Macro World grams Atomic mass is the mass of an
More informationChapter 5 Chemical Reactions and Equations
Chapter 5 Chemical Reactions and Equations 5.1 (a) singledisplacement reaction; (b) anhydrous; (c) molecular equation; (d) decomposition reaction; (e) balanced equation; (f) reactant; (g) spectator ion;
More informationChapter 7: Chemical Reactions
Chapter 7 Page 1 Chapter 7: Chemical Reactions A chemical reaction: a process in which at least one new substance is formed as the result of a chemical change. A + B C + D Reactants Products Evidence that
More informationChemical Equations. Chemical Equations. Chemical reactions describe processes involving chemical change
Chemical Reactions Chemical Equations Chemical reactions describe processes involving chemical change The chemical change involves rearranging matter Converting one or more pure substances into new pure
More information2. Write the chemical formula(s) of the product(s) and balance the following spontaneous reactions.
1. Using the Activity Series on the Useful Information pages of the exam write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products phases as either (g)as,
More informationCHEMISTRY COMPUTING FORMULA MASS WORKSHEET
CHEMISTRY COMPUTING FORMULA MASS WORKSHEET Directions: Find the formula mass of the following compounds. Round atomic masses to the tenth of a decimal place. Place your final answer in the FORMULA MASS
More informationCP Chapter 11 Notes Reactions and Equations
CP Chapter 11 Notes Reactions and Equations Evidence of Chemical Reactions How can you tell a reaction has taken place? Temperature change Color change Gas/bubbles Appearance of a (precipitate) Chemical
More informationstoichiometry = the numerical relationships between chemical amounts in a reaction.
1 REACTIONS AND YIELD ANSWERS stoichiometry = the numerical relationships between chemical amounts in a reaction. 2C 8 H 18 (l) + 25O 2 16CO 2 (g) + 18H 2 O(g) From the equation, 16 moles of CO 2 (a greenhouse
More informationUNIT (4) CALCULATIONS AND CHEMICAL REACTIONS
UNIT (4) CALCULATIONS AND CHEMICAL REACTIONS 4.1 Formula Masses Recall that the decimal number written under the symbol of the element in the periodic table is the atomic mass of the element. 1 7 8 12
More informationName period Unit 9: acid/base equilibrium
Name period Unit 9: acid/base equilibrium 1. What is the difference between the Arrhenius and the BronstedLowry definition of an acid? Arrhenious acids give H + in water BronstedLowry acids are proton
More information7. 1.00 atm = 760 torr = 760 mm Hg = 101.325 kpa = 14.70 psi. = 0.446 atm. = 0.993 atm. = 107 kpa 760 torr 1 atm 760 mm Hg = 790.
CHATER 3. The atmosphere is a homogeneous mixture (a solution) of gases.. Solids and liquids have essentially fixed volumes and are not able to be compressed easily. have volumes that depend on their conditions,
More informationEquilibrium Practice Problems
Equilibrium Practice Problems 1. Write the equilibrium expression for each of the following reactions: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 I 2 (s) + Cl 2 (g) 2 ICl (g) K = [ICl]
More information