Section 7.2: Equilibrium Law and the Equilibrium Constant Tutorial 1 Practice, page (a) 2 CO 2 (g) #!!"
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1 Section 7.: Equilibrium Law and the Equilibrium Constant Tutorial Practice, page 4. (a) CO (g) #!!"! CO(g) + O (g) Products: CO(g); O (g) Reactant: CO (g) [CO [O Equilibrium law equation: [CO (b) Cl (g) + H O(g) #!!"! 4 HCl(g) + O (g) Products: 4 HCl(g); O (g) Reactants: Cl (g); H O(g) 4 [HCl [O Equilibrium law equation: [Cl [H O (c) O (g) #!!"! O (g) Product: O (g) Reactant: O (g) Equilibrium law equation: [O [O (d) 4 NH (g) + O (g) #!!"! N (g) + 6 H O(g) Products: N (g); 6 H O(g) Reactants: 4 NH (g); O (g) [N [HO Equilibrium law equation: [NH [O 6 4. Given: CO(g) + H (g) #!!"! CH OH(g); [CO equilibrium mol/l; [H equilibrium 0.58 mol/l; [CH OH equilibrium 0.0 mol/l Solution: Write the equilibrium law equation using the balanced chemical equation. Then, substitute the equilibrium concentrations into the equilibrium law equation and solve for K. [CH OH [CO[H (0.0) (0.079)(0.58) Statement: The equilibrium constant, K, for the reaction is. Copyright 0 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7.-
2 . Given: [NO [Br [NOBr Required: the balanced chemical equation for the reaction Analysis: Use the general equilibrium law equation to determine the balanced reaction equation. Solution: Products: NO(g); Br (g) Reactant: NOBr(g) Balanced chemical equation: NOBr(g) #!!"! NO(g) + Br (g) Statement: The balanced chemical equation is NOBr(g) #!!"! NO(g) + Br (g). Tutorial Practice, page 44. (a) NH 4 Cl(s) #!!"! NH (g) + HCl(g) Products: NH (g); HCl(g) Reactant: NH 4 Cl(s) Equilibrium law equation: [NH [HCl (b) Unbalanced chemical equation: NaHCO (s) #!!"! Na CO (s) + CO (g) + H O(g) Balanced chemical equation: NaHCO (s) #!!"! Na CO (s) + CO (g) + H O(g) Products: Na CO (s); CO (g); H O(g) Reactant: NaHCO (s) Equilibrium law equation: [CO [H O (c) Chemical equation: Zn(s) + Cu + (aq) + Cl (aq) #!!"! Zn + (aq) + Cu(s) + Cl (aq) Net ionic equation: Zn(s) + Cu + (aq) #!!"! Zn + (aq) + Cu(s) Products: Zn + (aq); Cu(s) Reactants: Zn(s); Cu + (aq) Equilibrium law equation: [Zn+ (aq)] [Cu + (aq)] Section 7. Questions, page 46. (a) SiH (g) + Cl (g) #!!"! SiCl 4 (g) + H (g) Products: SiCl 4 (g); H (g) Reactants: SiH (g); Cl (g) Equilibrium law equation: [SiCl [H 4 [SiH [Cl (b) PBr (g) + Cl (g) #!!"! PCl (g) + Br (g) Products: PCl (g); Br (g) Reactants: PBr (g); Cl (g) [PCl [Br Equilibrium law equation: [PBr [Cl Copyright 0 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7.-
3 (c) H O(l) #!!"! H O (g) Product: H O(g) Reactant: H O(l) Equilibrium law equation: [H O (d) NaHCO (s) #!!"! Na CO (s) + CO (g) + H O(g) Products: Na CO (s); CO (g); H O(g) Reactant: NaHCO (s) Equilibrium law equation: [CO [H O. Given: NOCl(g) #!!"! NO(g) + Cl (g); volume.0 L; quantities at equilibrium: Cl (g).4 mol; NOCl(g).0 mol; NO(g) mol Solution: Step. Write the equilibrium law equation using the balanced chemical equation. [NO [Cl [NOCl Step. Calculate the equilibrium concentrations of the reactant and products..4 mol [Cl equilibrium.0 L [Cl equilibrium 0.8 mol/l.0 mol [NOCl equilibrium.0 L [NOCl equilibrium 0. mol/l [NO equilibrium 4.5! 0" mol.0 L [NO equilibrium.5! 0 " mol/l Step. Substitute the equilibrium concentrations into the equilibrium law equation and solve for K. [NO [Cl [NOCl (.5! 0" ) (0.80) (0.).6! 0 "5 Statement: The equilibrium constant, K, for the reaction is.6! 0 "5. Copyright 0 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7.-
4 . Given: 4 Fe(s) + O (g) #!!"! Fe O (s); volume.0 L; quantities at equilibrium: Fe(s).0 mol; O (g).0! 0 " mol; Fe O (s).0 mol Solution: Step. Write the equilibrium law equation using the balanced chemical equation. [O Step. Calculate the equilibrium concentration of oxygen gas. [O equilibrium.0! 0" mol.0 L [O equilibrium 5.0! 0 "4 mol/l Step. Substitute the equilibrium concentration of oxygen gas into the equilibrium law equation and solve for K. [O (5.0! 0 "4 ) 8.0! 0 9 Statement: The equilibrium constant, K, for the reaction is 8.0! Comparison of Homogeneous and Heterogeneous Equilibria reactants and products equilibrium law equation explanation Homogeneous equilibrium all same state of matter (either gases or solutions) Sample chemical equation: H (g) + I (g) #!!"! HI(g) all reactants and products appear Sample equilibrium law equation: [HI [H [I Concentrations of substances in gas state or solution can vary. Heterogeneous equilibrium present in at least two different states (e.g., gases and solids) Sample chemical equation: PCl 5 (s) #!!"! PCl (l) + Cl (g) pure solids and liquids are not included Sample equilibrium law equation: [Cl Concentrations of pure substances and liquids cannot change; they are constants. Copyright 0 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7.-4
5 5. Formation of methanol gas from carbon monoxide gas and hydrogen gas: (a) Unbalanced chemical equation: CO(g) + H (g) #!!"! CH OH(g) Balanced chemical equation: CO(g) + H (g) #!!"! CH OH(g) [CHOH (b) [CO[H If 6. 0, the concentration of methanol would be expected to be much lower than the concentration of carbon monoxide and hydrogen. (c) This equilibrium system would not be useful for generating large quantities of methanol because the equilibrium favours the reactants. (d) The equilibrium constant for the reverse reaction, K, is the reciprocal of K. K! K 6. " 0 # K! Formation of carbon monoxide from coal and carbon dioxide gas: (a) Unbalanced chemical equation: C(s) + CO (g) #!!"! CO(g) Balanced chemical equation: C(s) + CO (g) #!!"! CO(g) (b) Product: CO(g) Reactants: C(s); CO (g) [CO [CO (c) More carbon monoxide gas will be produced at 000 C than at 5 ºC because the equilibrium constant is much higher at 000 C (.7 0 compared to.6 0 ), indicating that equilibrium favours the product more at the higher temperature than at the lower temperature. (d) Answers may vary. Sample answer: Some advantages of a gaseous fuel over a solid fuel are that gaseous fuels can be delivered through pipelines, so it is easier to control their flow into a combustion chamber and they can disperse throughout the volume so they are likely to burn faster. (e) Sample answer. Some safety issues involved in working with carbon monoxide gas at high temperatures are that leaks of high-temperature gas can cause severe burns and that carbon monoxide is toxic. Copyright 0 Nelson Education Ltd. Chapter 7: Chemical Equilibrium 7.-5
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[CO 2 ] 3 [H 2 O] 4 [C 3 H 8 ][O 2 ] 5 [NO] 2 [H 2 O] 2 [N 2 H 4 ][O 2 ] 2 [Cu(NH c. K eq = [Cu 2+ ][NH 3 ] a. K eq = b.
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