Solutions to Chem 03 TT1 Booklet Chem03 TT1 Booklet Solutions to Gases Practice Problems Problem 1. Answer: C Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure. Problem. Answer: 775 mmhg 1atm 760 mmhg ; Patm 775mmHg P atm 1 atm 1.0 atm 760 mmhg 73 310 583 T C P RT dp Mr P Mr P d P RT P 1 1 L atm mol 0.0806 583 MrP.64 g / L 13.8 g / mol 1.0 atm MrP 13.8 g / mol 3.99 4 P4 Mr 31.0 g / mol P Problem 3. Answer: 735 mmhg 1atm 760 torr; P atm 735torr P atm 1 atm 0.967 atm 760 mmhg 73 110 383 T C 4.04 4 1 1 L atm mol 4 8 RT 0.0806 383 M m 0.970 g 56.5 g / mol CH CH PV 0.967 atm 0.559 L Mr Mr CH CH 56.5 g / mol 14.0 g / mol C H Solutions will be posted at www.prep101.com/solutions 1 of 9
Chem03 TT1 Booklet Problem 4. Answer: C H O 6O 6CO 6H O 6 1 6 RT PV nrt VO n O P no 6 n 1 C H O 6 1 6 no 6 n 6 C6H1O 6 6 1 6 6 1 6 g 1 1 atm L mol m 1.50 0.0806 73 C H O RT VO Mr P 180.0 g / mol 1.00 atm 6 1 6 6 6 1.1 C H O 6 1 6 m Mr C H O C H O L Problem 5. Answer: T 73 30.0C 303.0 M X m X RT PV 0.5 g Problem 6. Answer: C 8.3145 J mol1 1 303.0 101.3 kpa 70.4 10 3 L 760mmHg 0.5atmO = 190 mmhg O 1atm Total pressure = 190mmHg O + 440mmHg N = 630mmHg 79.5g/mol Problem 7. Answer: B PV = nrt (1.00atm)(.4L) = n(0.081atml/mol)(303) n = 0.90 moles helium gas Problem 8. Answer: B 10g no 0.315 3 nrt 0.315 8.314 373 P 3.3kPa V 30 Solutions will be posted at www.prep101.com/solutions of 9
Problem 9. Answer: A To determine which gas has the highest density, consider the equation g molar mass ( Density = mol ) pressure(kpa) R(gas constant) temperature() Chem03 TT1 Booklet If measured at the same pressure and temperature, the gas that will have the highest density will have the largest molar mass. The molar masses of the gases are as follows: F, 38g/mol; CH6, 30.08g/mol; HS, 34.086g/mol; NO, 30.01g/mol and SiH4, 3.13g/mol. Therefore, F will have the highest density because it has the largest molar mass. Problem 10. PV = nrt Answer: D PV T PV T 1 1 1 (1.00 atm)(600 ml) (.00 atm)(100 ml) 80 T T = 847C Problem 11. We need to determine what mass is in the gas phase, and the rest must be in the liquid phase. We make the assumption that the liquid takes up a negligible volume of the container. The pressure of the water vapour will be 33.7mmHg P=1 atm / 760mmHg * 33.7 mmhg = 0.3075 atm Now calculate the moles of gas: mass of gaseous water is: n = PV RT = 0.3075 atm 0.5 L = 0.0055 mole 0.0806 L atm mol 1 1 (73.15 + 70) So the mass of liquid water is 0.5-0.098 = 0.40 g. 18 g 0.0055 mol = 0.098 g mol Solutions will be posted at www.prep101.com/solutions 3 of 9
Solutions to Thermodynamics Problems Chem03 TT1 Booklet Problem 1. Moles of AgNO3 = 0.050 L 0.100 M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = 0.050 L 0.100 M = 0.05 moles HCl = 0.05 moles Cl (aq) Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced (you can prove this to yourself using an ICE table) Heat lost by reaction = heat gained by solution Heat gain = m s T = 100.0 g H O 4.18 J 1 (3.40 C.60 C) g C = 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl Therefore, the heat produced per mol of AgCl = 330 J 1kJ 6.60 kj mol 0.050 mol AgCl 1000 J 1 Problem 13. C H 5 OH (l) + 3 O (g) CO (g) + 3 H O (l) Answer: -1370 kj mol-1 Heat capacity of calorimeter = C q=(360 s) (4000 J/s) = 1440 kj q = C T = C (50.3-5.0) rearrange and solve C = 56.9 kjc-1 Q rn 1 (56.9kJ C )(7.4 5.0C) 1370 kj / mol (0.100mol) Problem 14. -14. J Absorbing heat is an endothermic process (positive) Work done by the system on the surroundings is negative by convention, therefore E = q + w = 1.0 J - 15. J= -14. J Solutions will be posted at www.prep101.com/solutions 4 of 9
Problem 15. q = n s mol ΔT = (0.8 J J mol C 1kJ ) 39.1 mol (38.0 0.0) C 1000J = 30.9 kj Chem03 TT1 Booklet Pressure-Volume work can be calculated by: w = -PΔV = -1.00 atm (998 876 L) = -1 L atm 101.3J Latm 1kJ 1000J = -1.4 kj So, the total internal energy, ΔE = q + w = 30.9 kj + ( 1.4 kj) = 18.5 kj Problem 16. NO (g) + O 3(g) NO (g) + O (g) ΔH = 199 kj 3 O (g) O 3(g) ΔH = ½ ( 47 kj) O (g) ½ O (g) ΔH = ½ (+ 495 kj) ------------------------------------------------------------ NO (g) + O (g) NO (g) ΔH = 33 kj Problem 17. -1367 kj C H 5 OH (l) C H 4(g) + H O (l) ΔH = +44 kj C H 4(g) + 3 O (g) CO (g) + H O (l) ΔH = 1411 kj C H 5 OH (l) + 3 O (g) 3 H O (l) + CO (g) ΔH = 1367 kj Problem 18. Answer: ΔH = (33.8) (90.9) 0 = 11.94 kj mol Solutions will be posted at www.prep101.com/solutions 5 of 9
Chem03 TT1 Booklet Problem 19. CH 3 CH OH (g) + 3 O (g) CO (g) + 3 H O (g) The energy lost from breaking bonds and energy gained in bond formation must be determined. Bonds Broken Bonds Formed 5 C-H Bonds (413 kj mol-1) 4 C=O bonds (CO) (799 kj mol-1) 1 C-C Bond (347 kj mol-1) 6 O-H bonds (467 kj mol-1) 1 C-O Bond (358 kj mol-1) 1 O-H Bond (467 kj mol-1) 3 O=O Bonds (498 kj mol-1) Total = 4731 kj Total = 5998 kj H = BE(Broken) BE(Formed) H = 4731 kj 5998 kj = -167 kj Problem 0. Answer: a) Water at 105 o C + b) Na (aq) + Cl (aq) c) HO/MeOH miture d) Reactants e) F (g) Solutions will be posted at www.prep101.com/solutions 6 of 9
Chem03 TT1 Booklet Solutions to Equilibrium Practice Problems Problem 1. Answer: c 4 [ NO] [ H ] [ NO ] 4 4 3 Pure solids and pure liquids and solvents are not included in the epression. Products go over Reactants, and coefficients in the equation are written as superscripts. Problem. c = [CO (g) ] Remember: Solids and liquids are not included in the epression. Problem 3. Equation is equal the double and reverse of equation1 therefore p = p1 = ( 1 0.157 ) = 40.6 Problem 4. Answer: Equilibrium Constant = 1/ 1 = 1 1 The second equation if reversed (1/) and halved (1/). Combining these two gives 1/1/. Problem 5. Answer: So the reaction equations is: CO (g) + O (g) CO (g) c = 3.3 10 91 The relationship between c and p is: p = c (RT) n gas In this case there are 3 moles of gas in the reactants and moles of gas in the products, so n = -1 So solving for p : p = c (RT) n gas = (3.3 10 91 ) [(0.081 L atm mol ) (98 )] 1 = 1.35 10 90 Solutions will be posted at www.prep101.com/solutions 7 of 9
Problem 6. Chem03 TT1 Booklet N (g) + C H (g) HCN (g) i: 1.00 1.00 1.00 c: + + - e: 1.00+ 1.00 + 1.00- Q=1> so then rn goes to the left (1.00 ) (1.00 ) c (1.00 ) (1.00 ) c 1 c c 0.488 Problem 7. COCl (g) CO (g) +Cl (g) Initial: 0.04 0 0 Change: - + + Equilibrium: 0.04- c = 0.04 + c 0.04c = 0 c c (4)(0.04)( c) 5.9 10 3 Solutions will be posted at www.prep101.com/solutions 8 of 9
Problem 8. c [ CO ][ H ] [ CO][ H O] 5.10 Chem03 TT1 Booklet CO + H O CO +H I 0.1M 0.1 0.1 0.1 C - - + + E 0.1-0.1-0.1+ 0.1+ (0.1 ) 0.01 0. c 5.10 (0.1 ) 0.01 0. 0.01 0. 0.0511.0 0 0.041 1. 1. 0.041 0.034M Therefore, at equilibrium [H] = 0.1 + = 0.1 + 0.034M = 0.134M Problem 9. Only C is true An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO at equilibrium. Problem 30. a) shift to the left b) no effect c) shift to the right d) shift to the right Problem 31. The reaction will shift left forming Ni(CO) 4(g) to reach equilibrium Problem 3. Answer: C Q = [NO][Cl]/[NOCl] = (1.)(0.56)/(1.3) = 0.51 =, therefore we are already at equilibrium. Solutions will be posted at www.prep101.com/solutions 9 of 9