Chem103.M3ProblemSet

CHEM Module 3 Review Question 1 Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7 o C is added to a 36.8 gram sample of water at 26.3 o C in a coffee cup calorimeter. c (water) = 4.184 J/g o C heat temp change = q temp change = m x c x Δ t m hot water = 40.5 g t hot water = 85.7 o C m room temp = 36.8 g t room temp = 26.3 o C (m hot water ) x (4.184 J/g o C) x (T mix - t hot water ) = (m room temp ) x (4.184 J/g o C) x (T mix - t room temp ) -[(40.5 g) x (4.184 J/g o C) x (Tmix - 85.7 o C)] = (36.8 g) x (4.184 J/g o C) x (Tmix - 26.3 o C) -[169.452 J/ o C x (Tmix - 85.7 o C)] = 153.9712 J/ o C x (Tmix - 26.3 o C) -169.452 J/ o C(T mix) + 14522.0364 = 153.9712 J/ o C(Tmix) - 4049.44256 323.4232 J/ o C(Tmix) = 18571.47896 Tmix = 18571.47896/323.4232 Tmix = 57.4 o C - (m warn H2O x c warn H2O x ∆ t warn H2O ) = (m cool H2O x c cool H2O x ∆ t cool H2O ) - [40.5 g x 4.184 J/g o C x (T mix - 85.7 o C)] = [(36.8 g x 4.184 J/g o C x (T mix - 26.3 o C)] - [169.452 J/ o C x (T mix - 85.7 o C)] = [(153.9712 J/ o C x (T mix - 26.3 o C)] T mix = 57.4 o C Question 2 Show the calculation of the energy involved in melting 120 grams of ice at 0 o C if the Heat of Fusion for water is 0.334 kJ/g. q s<-->l = mass x Heat of Fusion = m x Δ H fusion m = 120 g Δ H fusion (water) = 0.334 kJ/g

q = 120 g x 0.334 kJ/g q = 40.08 kJ q l ↔ s = m x ∆ H fusion = 120 g x 0.334 kJ/g = 40.08 kJ (since heat is added) = + 40.08 kJ Question 3 Thermochemical Equation Problems Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation: 2 S + 3 O 2 → 2 SO 3 ΔH = - 792 kJ If 42.8 g of S is reacted with excess O 2 , what will be the amount of heat given off? 42.8/32.07 = 1.3346 mol S q = ΔHrx x (new moles/original moles) q = -792 kJ x (1.3346/2) q = -528.5 kJ ΔH rx is for 2 mole of S reaction uses 42.8 g S = 42.8/32.07 = 1.335 mole S q = -792 kJ x 1.335 mole S / 2 mole S = -528.7 kJ Question 4 Thermochemical Equation Problems Hydrosulfuric acid (H 2 S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation: 2 H 2 S + 3 O 2 → 2 SO 2 + 2 H 2 O What is the ΔH of the reaction if 26.2 g of H 2 S reacts with excess O 2 to yield 431.8 kJ? MW of H2S = 34.086 26.2/34.086 = 0.7686 mol H2S q = -431.8 kJ ΔHrx = 1431.8 kJ/(0.7686/2) ΔHrx = -1123.6 kJ/mole

MW of C6H6 = 78.108 7.05/78.108 = 0.09026 mol of C6H6 ΔHrx = -354.1567 kJ / (0.09026mol/2mol) ΔHrx = -7847.478 kJ/mole (2) Isolated system (bomb calorimeter) (3) Exothermic process (heat is given off) 1) q water = s (specific heat of water) x mass x Δt = 4.18 J / g / o K x 500 g x (53.13 o C - 25.00 o C) = - 58,791 J q calorimeter = heat capacity x Δt = (10.5 kJ/ o C) x (53.13 o C - 25.00 o C) = - 295.365 kJ x 1000 J/1 kJ= - 295,365 J q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = - 354.16 kJ (new) moles C 6 H 6 = 7.05 g / 78 = 0.0904 mole C 6 H 6 ΔH = q / (new moles / original moles) ΔH = - 354.16 / (0.09048 / 2) = - 7828 kJ / mole (2) Isolated system (bomb calorimeter) (3) Exothermic (temperature of water rises due to heat given off by combustion reaction ) Question 6 Measuring Heat of Reaction Problems A sample of 14.5 g of sodium bicarbonate (NaHCO 3 ) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction NaHCO 3 (s) → Na + (aq) + HCO 3 - (aq)

If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/ o C) decreases from 25.00 o C to 19.86 o C, (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process. qwater = s (specific heat of water) x mass x Δ t s = 4.18 J/g/ o C mass = 100 g Δ t = 19.86 - 25.00 = -5.14 o C qwater = (4.18 J/g/ o C) x (100 g) x (-5.14 o C) qwater = +2148.52 J (positive bc heat is absorbed) qcalorimeter = heat capacity x Δ t heat capacity = 150 J/ o C Δ t = 19.86 - 25.00 = -5.14 o C qcalorimeter = (150 J/ o C) x (-5.14 o C) qcalorimeter = +771 J (positive bc heat is absorbed) (+2148.52) + (+771) = +2919.52 J/1000 = +2.9195 kJ (1) MW of NaHCO3 = 84.008 14.5/84.008 = 0.1726 mol of NaHCO3 ΔHrx = 2.9195 kJ / (0.1726 mol/1mol) ΔHrx = 16.9148 kJ/mol (2) Open system (coffee cup calorimeter with no lid) (3) Endothermic process (heat is absorbed) q water = s (specific heat of water) x mass x Δt = 4.18 J / g / o K x 100 g x 5.14 o K = 2148.52 J q calorimeter = heat capacity x Δt = 150 x 5.14 o K = 771 J q reaction = 771 J + 2148.52 J = + 2919.52 J = + 2919.52 J x 1 kJ / 1000 J = 2.92 kJ (1) ΔH reaction = 2.92 kJ / 14.5 g NaHCO 3 / 84 g NaHCO 3 / mol NaHCO 3 = + 16.9 kJ / mol (2) The calorimeter + contents (no lid) = open system, (3) Endothermic process Question 7 Hess' Law Problems

ΔH rxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6) = - 906.3 kJ Question 8 Heat of Formation Problems Determine the heat of reaction (ΔH rxn ) for the combustion of ethanol (C 2 H 5 OH) by using heat of formation data: C 2 H 5 OH (l) + 3 O 2 (g) → 2 CO 2 + 3 H 2 O (g) Click here for Standard Enthalpies of Formation Table. ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔHf0 (reactants) ΔHrxn = 2 ΔHf0(CO 2 ) + 3 ΔHf0(H 2 O) - ΔHf0(C 2 H 5 OH) - 3 Hf0(O 2 ) ΔHrxn = 2 (-393.5) + 3 (-241.8) - (-277.6) - 3 (0) ΔHrxn = ( -787) + (-725.4) + (277.6) - (0) ΔHrxn = -1234.8 kJ/mole Determine the heat of reaction (ΔH rxn ) for the combustion of ethanol (C 2 H 5 OH) by using heat of formation data: C 2 H 5 OH (l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O (g) C 2 H 5 OH (l) → 2 C (graphite) + 3 H 2 (g) + 1/2 O 2 (g) ΔH f 0 (C 2 H 5 OH) = - (- 277.6 kJ/mole) 2 (C (graphite) + O 2 → CO 2 (g) 2 ΔH f 0 (CO 2 ) = 2 (-393.5 kJ/mole) 3 (H 2 (g) + 1/2 O 2 (g) → H 2 O (g) 3 ΔH f 0 (H 2 O) = 3 (- 241.8 J/mole) C 2 H 5 OH (l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O (g) ΔH rxn = - (- 277.6) + 2 (- 393.5) + 3 (- 241.8) ΔH rxn = - 1234.8 kJ/mole OR more simply and preferably: ΔH rxn = Σ n ΔH f 0 (products) - Σ m ΔH f 0 (reactants) ΔH rxn = 2 ΔH f 0 (CO 2 ) + 3 ΔH f 0 (H 2 O) - ΔH f 0 (C 2 H 5 OH) – 3 ΔH f 0 (O 2 ) ΔH rxn = 2 (- 393.5) + 3 (- 241.8) - (- 277.6) - 3 (0) = - 1234.8 kJ/mole

Question 9 A gas sample has an original volume of 680 ml when collected at 720 mm and 28 o C. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 55 o C? Vi = 680ml/1000 = 0.68 L Pi = 720 mm/760 = 0.947 atm Ti = 28 o C + 273 = 301 o K Vf = ? Pf = 820mm/760 = 1.079 atm Tf = 55 o C + 273 = 328 o K (Pi x Vi)/Ti = (Pf x Vf)/Tf (0.947 x 0.68)/301 = (1.079 x Vf)/328 (0.947) x (0.68) x (328) = (1.079) x (Vf) x (301) 211.21888 = Vf x 324.799 211.21888/324.799 = Vf x (324.799/324.799) Vf = 0.65 L P i x V i = P f x V f T i T f 680 ml/1000 = 0.680 liters = V i 720 mm/760 = 0.947 atm = P i 820 mm/760 = 1.08 atm = P f 28 o C + 273 = 301 o K = T i 55 o C + 273 = 328 o K = T f Put in the data: (0.947) x (0.680) = (1.08) x V f : (301) (328) Solve for V f : 0.002139402 = 0.003292683 x V f : V f : = 0.002139402 = 0.650 liter

23.3/32 = 0.728 mol of O2 0.728 x (2/3) = 0.4853 mol of CO2 P = 1.25 atm V = ? n = 0.4853 R = 0.0821 T = 25 o C + 273 = 298 o K P x V = n x R x T V = nRT/P Vco2 = (0.4853 x 0.0821 x 298)/1.25 Vco2 = 9.5 L (MW = 46) (MW = 32) (MW = 44) (MW = 18) C 2 H 5 OH (l) + 3 O 2 (g) → 2 CO 2 (g) + 3 H 2 O (g) 23.3 grams 9.5 liters ↓ ↑ by V = nRT / P = (0.4854)(0.0821)(298) /1.25 0.7281 mol → 2/3 x 0.7281 mol Question 12 PARTIAL PRESSURE - MOLE FRACTION PROBLEMS A mixture of gases consists of 4.00 moles of He, 2.00 moles of H 2 , 3.00 moles of CO 2 and 5.00 moles of Ar. The total pressure of the mixture is 2900 mm. Determine the mole fraction of each gas in the mixture. Determine the mole percent of each gas in the mixture. Determine the partial pressure of each gas in the mixture. P T = 2900 mm n1 = 4.00 moles of He n2 = 2.00 moles of H 2 n3 = 3.00 moles of CO 2 n4 = 5.00 moles of Ar

nT = 4.00 + 2.00 + 3.00 + 5.00 = 14.00 moles x1 = n1/nT xHe = 4.00/14.00 = 0.2857 x 100 = 28.57% He xH2 = 2.00/14.00 = 0.1429 x 100 = 14.29% H2 xCO2 = 3.00/14.00 = 0.2143 x 100 = 21.43% CO2 xAr = 5.00/14.00 = 0.3571 x 100 = 35.71% Ar P 1 = X 1 x P T PHe = 0.2857 x 2900mm = 828.53mm PH2 = 0.1429 x 2900mm = 414.41mm PCO2 = 0.2143 x 2900mm = 621.47mm PAr = 0.3571 x 2900mm = 1035.59mm mole fraction in purple mole percent in green partial pressure in blue X He = 4.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2857 Mole% He = 100(X He ) = (100) 0.286 = 28.57% X H2 = 2.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.1429 Mole% H2 = 100(X H2 ) = (100) 0.143 = 14.29% X CO2 = 3.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2143 Mole% CO2 = 100(X CO2 ) = (100) 0.214 = 21.43% X Ar = 5.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.3571 Mole% Ar = 100(X Ar ) = (100) 0.357 = 35.71% P He = X He (2900 mm) = 0.2857 (2900 mm) = 828.53 mm P H2 = X H2 (2900 mm) = 0.1429 (2900 mm) = 414.41 mm P CO2 = X CO2 (2900 mm) = 0.2143 (2900 mm) = 621.47 mm P Ar = X Ar (2900 mm) = 0.3571 (2900 mm) = 1035.59 mm

EFFUSION AND DIFFUSION PROBLEMS The rate of effusion of nitrogen gas (N 2 ) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the unknown gas? MW of N2 = 28.02 rn2 = 1.253 (r N2 /r unknown ) 2 = MW unknown /MW N2 (1.253/1) 2 = MW unknown /28.02 cross multiply MW unknown /1 = (1.253) 2 x 28.02 MW unknown = 43.992 (r N2 / r unknown ) 2 = MW unknown / MW N2 (1.253/1) 2 = MW unknown / 28.02 MW unknown = (1.253) 2 x 28.02 = 43.99