Solutions to Chem 203 TT1 Booklet

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1 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 T C P RT dp Mr P Mr P d P RT P 1 1 L atm mol MrP.64 g / L 13.8 g / mol 1.0 atm MrP 13.8 g / mol P4 Mr 31.0 g / mol P Problem 3. Answer: 735 mmhg 1atm 760 torr; P atm 735torr P atm 1 atm atm 760 mmhg T C L atm mol 4 8 RT M m g 56.5 g / mol CH CH PV atm L Mr Mr CH CH 56.5 g / mol 14.0 g / mol C H Solutions will be posted at 1 of 9

2 Chem03 TT1 Booklet Problem 4. Answer: C H O 6O 6CO 6H O RT PV nrt VO n O P no 6 n 1 C H O no 6 n 6 C6H1O g 1 1 atm L mol m C H O RT VO Mr P g / mol 1.00 atm C H O m Mr C H O C H O L Problem 5. Answer: T C M X m X RT PV 0.5 g Problem 6. Answer: C J mol kpa 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 nrt P 3.3kPa V 30 Solutions will be posted at of 9

3 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, g/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 (1.00 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 = atm Now calculate the moles of gas: mass of gaseous water is: n = PV RT = atm 0.5 L = mole L atm mol 1 1 ( ) So the mass of liquid water is = 0.40 g. 18 g mol = g mol Solutions will be posted at 3 of 9

4 Solutions to Thermodynamics Problems Chem03 TT1 Booklet Problem 1. Moles of AgNO3 = L M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = L 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 = 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 moles of AgCl Therefore, the heat produced per mol of AgCl = 330 J 1kJ 6.60 kj mol mol AgCl 1000 J 1 Problem 13. C H 5 OH (l) + 3 O (g) CO (g) + 3 H O (l) Answer: kj mol-1 Heat capacity of calorimeter = C q=(360 s) (4000 J/s) = 1440 kj q = C T = C ( ) rearrange and solve C = 56.9 kjc-1 Q rn 1 (56.9kJ C )( C) 1370 kj / mol (0.100mol) Problem 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 J= -14. J Solutions will be posted at 4 of 9

5 Problem 15. q = n s mol ΔT = (0.8 J J mol C 1kJ ) 39.1 mol ( ) C 1000J = 30.9 kj Chem03 TT1 Booklet Pressure-Volume work can be calculated by: w = -PΔV = atm ( 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 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 = kj mol Solutions will be posted at 5 of 9

6 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 6 of 9

7 Chem03 TT1 Booklet Solutions to Equilibrium Practice Problems Problem 1. Answer: c 4 [ NO] [ H ] [ NO ] 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 = ( ) = 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 = 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 = ( ) [(0.081 L atm mol ) (98 )] 1 = Solutions will be posted at 7 of 9

8 Problem 6. Chem03 TT1 Booklet N (g) + C H (g) HCN (g) i: c: e: Q=1> so then rn goes to the left (1.00 ) (1.00 ) c (1.00 ) (1.00 ) c 1 c c Problem 7. COCl (g) CO (g) +Cl (g) Initial: Change: Equilibrium: c = c 0.04c = 0 c c (4)(0.04)( c) Solutions will be posted at 8 of 9

9 Problem 8. c [ CO ][ H ] [ CO][ H O] 5.10 Chem03 TT1 Booklet CO + H O CO +H I 0.1M C E (0.1 ) c 5.10 (0.1 ) M Therefore, at equilibrium [H] = = M = 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 9 of 9

Solutions to Thermodynamics Problems

Solutions to Thermodynamics Problems Chem03 Final Booklet Problem 1. Solution: Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05