CH 111, Week 8 Notes
CH 111, Week 8 Notes CH111
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This 2 page Class Notes was uploaded by Jordyn Meekma on Monday October 17, 2016. The Class Notes belongs to CH111 at Northern Michigan University taught by Dr. Tom Getman in Fall 2016. Since its upload, it has received 5 views. For similar materials see General Chemistry 1 in Chemistry at Northern Michigan University.
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Date Created: 10/17/16
CH 111 Notes: Week 8 If 25.0 g Cu at 90.0ᵒC is placed in 100.0 g of water at 20.0ᵒC, what will the final temperature of the water be? (c p(Cu)= 24.4 J/mol*ᵒC) o Assume T final the same for both Cu and H O 2 o Assume container absorbs negligible heat → q Cu= -q H2O o 25.0 g Cu (1 mol Cu/63.55g) = 0.3934 mol Cu 100.0 g H 2 (1 mol H O/28.02 g) = 5.5494 mol H O 2 o cp(Cu)*∆T = -( c p(H2O)n*∆T) (0.3934 mol)(24.4 J/mol*ᵒC)(T – f0.0ᵒC) = -(5.5494 mol)(75.3 J/mol*ᵒC) (Tf– 20.0ᵒC) = 9.599 J/ᵒC(Tf– 90.0ᵒC) = -417.87 J/ᵒC(T –f20.0ᵒC) = 9.599 J/ᵒC(Tf) – 863.9 J = -417.87 J/ᵒC(T f – 8357 J = 427.5 J/ᵒC(Tf) = 9221 J Tf= 21.6ᵒC A 2.50 g sample of Zn at 22.5ᵒC is added to 100.0 mL of a 2.0 M HCl solution at 22.5ᵒC in a calorimeter at 22.5ᵒC with a total heat capacity of 481 J/ᵒC. The observed temperature of the system increases to 34.7ᵒC. Calculate ∆H Rxnfor the following equation in kJ/mol Zn: Zn (s)+ 2HCl (aq) ZnCl 2(aq)+H 2(g) o qRxn= -q calorimeter qRxn= ∆H Rxn ∆H Rxn= -q calorimeter o ∆H Rxn= -(481 J/ᵒC)(34.7ᵒC – 22.5ᵒC) = -5868 J = -5.868 kJ o 2.50 g Zn (1 mol Zn/65.39 g) = 0.03823 mol Zn ←limiting reactant 2.0 mol/L HCl (0.1000 L) = 0.20 mol HCl/2 = 0.10 mol Zn o -5.868 kJ/0.03823 mol Zn = -153 kJ/ mol Zn When 1.045 g of CaO is added to 50.0 g of water at 25ᵒC in a calorimeter, the temperature of the solution increases to 32.3ᵒC. If cs(solution).18 J/g*ᵒC, calculate ∆H Rxn in kJ/mol CaO for: CaO (s)+ H 2 (l) Ca(OH) 2(aq) o qRxn(∆H Rxn = -q solution qsol= (cs)(total mass of solution)(∆T) o 1.045 g + 50.0 g = 51.045 g o ∆H Rxn= -(4.18 J/g*ᵒC)(51.045 g)(32.3ᵒC – 25ᵒC) = -1558 J = - 1.558 kJ o 1.045 g CaO (1 mol CaO/56.08 g) = 0.018634 mol CaO - 1.558 kJ/0.018634 mol CaO = -84 kJ/mol CaO ∆HᵒRxn ᵒ → under thermodynamic standard state conditions Thermodynamic standard state conditions… o Solids and liquids are pure substances o Aqueous solutions have a 1.0 M concentration o Gases have a pressure of 1 bar (about 1 atm) o At a specified temperature (usually 25ᵒC) Hess’s Law: If 2 or more chemical equations are added together, then the value of ∆H Rxn(or ∆Hᵒ Rxn for the resulting equation is the sum of the ∆H Rxn(or ∆HᵒRxn) values for the separate equations o SO 2(g) S (s)+ O 2(g) ∆Hᵒ Rxn= 296.8 kJ/mol Rxn 2S (s) 3O 2(g) 2SO 3(g) ∆Hᵒ Rxn= -791.4 kJ/mol Rxn Find ∆Hᵒ Rxnfor: 2SO 2(g)+ O 2(g)→ 2SO 3(g) 2(SO 2(g)→ S (s) O 2(g) = 2SO 2(g) 2S (s) 2O 2(g) 2(296.8) + 2S + 3O → 2SO + -791.4 (s) 2(g) 3(g) 2SO 2(g)+ O 2(g)→ 2SO 3(g) -197.8 kJ/mol Standard Heat (Enthalpy) of Formation: (∆Hᵒ ) the heat change f associated with the formation of a substance from the elements (*Values found in Appendix A4.3 in textbook*) o ∆Hᵒ fO 2(g) C (s)+ O 2(g) CO 2(g) ∆Hᵒ Rxn = ∆Hᵒ f∆Hᵒ fof elements is 0) o ∆Hᵒ Rxn = Sum ∆Hᵒ f (Products)um ∆Hᵒ f (Reactants) Hess’s Law Problems Calculate ∆Hᵒ Rxn for 2C 2 2(g) 5O 2(g)→ 4CO 2(g)+ 2H O2 (g) ∆Hᵒ (kJ/mol): 226.7 0 -393.5 -241.8 f o [4(-393.5) + 2(-241.8)] – [2(226.7) + 5(0)] = -2511 kJ/mol (Products) - (Reactants) ClO (g) O 3(g) Cl(g)+ 2O 2(g) ∆Hᵒ Rxn= -29.90 kJ/mol 2O 3(g) 3O 2(g) ∆Hᵒ Rxn= 24.18 kJ/mol Calculate ∆Hᵒ Rxn for: Cl(g)+ O 3(g) ClO (g)+ O 2(g) o Flip [ClO(g)+ O 3(g) Cl (g) 2O 2(g)so the O 3nd O ca2cel out correctly with the other equation *When you flip an equation, change the sign for the ∆Hᵒ Rxn value* o Cl(g)+ 2O 2(g)→ ClO (g) O 3(g) (+)29.90 + 2O 3(g) 3O 2(g) + 24.18 Cl(g)+ O 3(g) ClO (g)+ O 2(g) 54.08 kJ/mol C (s) O 2(g) CO 2(g) ∆Hᵒ Rxn= -393.5 kJ/mol 2CO (g) O 2(g)→ 2CO 2(g) ∆Hᵒ Rxn= -566.0 kJ/mol 2H 2(g) O 2(g) 2H 2 (g) ∆Hᵒ Rxn= -483.6 kJ/mol Calculate ∆Hᵒ Rxn for: C (s)+ H 2 (g) CO (g)+ H 2(g) C (s) O 2(g)→ CO 2(g) -393.5 ½ (2CO (g)+ O 2(g) 2CO 2(g) = CO 2(g) CO (g)+ ½ O 2(g) ½ (+566.0) ½ (2H 2(g)+ O 2(g) 2H O2)(g)+ H O 2 (g)→ H 2(g) ½ O 2(g) + ½ (+483.6) C (s) H O2 (g)→ CO (g)+ H 2(g) 131.3 kJ/mol
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