TestFivePacketSolutions(3)

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Test 5 Packet (New Material) Ideal Gas Constant For this table, the units of R are the units of pressure (column) times the units of volume (row) divided by moles time Kelvin, so for the grey square, the value of R = 8.314 and has units of kPa*L/mole-K R = atm Pa kPa Psi mmHg (torr) mL 82.0528 8,314,000 8,314 1205.85 62360.1 L 0.082053 8,314 8.314 1.24058 62.360 Useful values of R: R = 8.314 kPa*L/mole-K R = 8.314 J/mole R = 0.008314 kJ/mole Conversion Factors atm Pa kPa Psi mmHg (torr) 1 atm = 1 101,325 101.325 14.696 760 1 Pa = 9.87 * 10 -6 1 0.001 0.000145 0.00750 1 kPa = 0.00987 1000 1 0.145 7.50 1 psi = 0.0680 6895 6.895 1 51.7 1 mmHg (torr) = 0.00132 133.3 0.133 0.0193 1 1

Short Table of Thermodynamic Values Compound ∆ H f 0 (kJ/mole) Al 2 O 3(s) -1675.7 AlN (s) -317.98 As 2 O 3(s) -654.8 BaO (s) -553.54 BaTiO 3(s) -1659.8 Be 2 C (s) -116.99 BeO (s) -608.35 C 2 H 6(g) -84.684 CaO (s) -635.09 CaTiO 3(s) -1660.6 CCl 4(g) -100.42 CH 2 Cl 2(g) -95.521 CH 3 Cl (g) -86.316 CH 4(g) -74.873 CHCl 3(g) -101.25 CO 2(g) -393.51 Cu 2 O (s) -170.71 CuO (s) -156.06 Fe 3 C (s) 25.104 FeCO 3(s) -740.57 FeO (s) -272.04 GaF 3(s) -921.53 GaI 3(s) -239.33 HCl (g) -92.312 TiO 2(s) -944.75 2

Short Table of Heat Capacities Compound Cp (J/g- ̊ C) Acetone 2.15 Ag (s) 0.24 Air 1.02 Al (s) 0.90 Au (s) 0.84 Cu (s) 0.385 Ethyl Alcohol (Ethanol) 2.46 Fe (s) 0.444 Glycerin 2.43 H 2(g) 14.267 He (g) 5.3 H 2 O (l) 4.181 Methyl Alcohol (Methanol) 2.51 Mg (s) 1.02 Ni (s) 0.44 Pb (s) 0.16 Sn (s) 0.21 Zn (s) 0.39 3

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Electrolytes What is an electrolyte? A compound that improves the conductivity of liquid by creating charge carriers From the list below, sort each compound into the appropriate category: Non-electrolyte Weak Electrolyte Strong Electrolyte Fructose Sucrose C 2 H 4 CH 2 Cl 2 HNO 2 Mg(OH) 2 CaSO 4 HC 2 H 3 O 2 Ca 3 (PO 4 ) 2 H 2 SO 3 Fe 2 (CO 3 ) 3 HF NaOH HCl NaCl Li 2 CO 3 Cs 3 PO 4 Ca(OH) 2 Mg(NO 3 ) 2 Ba(C 2 H 3 O 2 ) 4

Pressure Conversion Problems Convert 101.3 kPa to atmospheres 1.000 atm Convert 13.25 psi to torr 685.2 torr Convert 3.79 atm to mmHg 2,880 mmHg Convert 578 mmHg to kPa 77.1 kPa Mole Fractions What is the mole fraction of O 2 in a mixture that is 0.250 moles O 2 and 0.500 moles N 2 ? 0.333 What is the mole fraction of O 2 in a mixture that is 0.250 moles O 2 , 0.500 moles N 2 , and 0.250 moles H 2 ? 0.250 What is the mole fraction of N 2 that is 0.250 g O 2 and 0.500 grams N 2 ? 0.696 What is the mole fraction of O 2 in a mixture that is 0.250 grams O 2 , 0.500 grams N 2 , and 0.250 grams H 2 ? 0.0519 5

Simple Ideal Gas Law Problems What is the pressure (kPa) of a gas that is 1.00 moles of N 2 , and is in a 22.4 L container at 21.0 ̊ C? 109 kPa What is the pressure (kPa) of a gas that is 1.00 grams of N 2 , and is in a 22.4 L container at 21.0 ̊ C? 3.90 kPa What is the pressure (kPa) of a gas that is 1.00 grams of O 2 and 1.00 grams of N 2 , and is in a 22.4 L container at 21.0 ̊ C? 7.31 kPa What is the volume of a gas that is 1.00 grams of O 2 and 1.00 grams of N 2 , and is at 10,698 Pa and 21.0 ̊ C? 0.0153 L What is the temperature (in Celsius) of a gas that is 0.156 grams of He, and is in 13.4 L container with a pressure of 9.509 kPa? 122 ̊ C 6

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Conceptual Gas Law Problems If you halve the volume containing a gas, what happens to the pressure? You double the pressure If a tank of gas is 1.00 atm at room temperature (21.0 ̊ C), what is the pressure if the tank is heated to 50.0 ̊ C? 1.10 atm How much do you have to increase the volume of a gas to get its pressure to go from 101.3 kPa to 15.6 kPa? 6.5 times larger How many more moles (relatively speaking) do you have to add to a container of gas to get the pressure to go from 101.3 kPa to 265.1 kPa? Roughly 2.6 times as many moles as there were to begin with 7

Dalton’s Law If the total pressure is 1.00 atm, what is the partial pressure of He in a gas mixture, if He has a mole fraction of 0.234 in the gas phase? 0.234 atm If the total pressure is 1.00 atm, what is the partial pressure of He if the gas mixture is 1.00 g of He and 1.00 g of Ne? 0.834 atm If the total pressure is 12.3 atm, what is the partial pressure of He if is gas mixture is 1.00 g of He, 1.00 g of Ne, and 1.00 g H 2 ? 3.84 atm If the partial pressure of Ne is 0.112 atm, what is the total pressure if the mole fraction of Ne is 0.500? 0.224 atm If the partial pressure of Ne is 0.112 atm, what is the total pressure if the gas is made by mixing 1.00 g of Ne and 1.00 g of He? 0.678 atm 8

Reactive Gas Law Problems A mixture of 0.478 g of H 2 and 4.22 g of O 2 is pumped into a bomb calorimeter with a volume of 10.0 L, which is submerged into a pool of water maintained at 80.0 °C to preserve isothermal operation a. What is the initial total pressure of the reactor to 3 significant figures? 0.478 g − H 2 ∗ 1 mole − H 2 2.0 g − H 2 = 0.239 mole − H 2 4.22 g − O 2 ∗ 1 mole − O 2 32.0 g − O 2 = 0.132 moles − O 2 P ∗ 10.0 L = ( 0.239 + 0.132 ) ∗ 8.314 L ∗ kPa mole − K ∗ 353 K , P = 109 kPa b. The hydrogen and oxygen react to form water, by the following reaction 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) What is the limiting reagent? 0.478 g − H 2 ∗ 1 mole − H 2 2.0 g − H 2 ∗ 1 2 mole − H 2 = 0.120 4.22 g − O 2 ∗ 1 mole − O 2 32.0 g − O 2 ∗ 1 1 mole − O 2 = 0.132 9

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c. After the reaction is run to completion, what is the final total pressure of the reactor to 3 significant figures? An ICE Table has been given for your convenience, though you are not required to use it H 2 O 2 H 2 O I 0.239 0.132 0.00 C -0.239 -0.120 0.239 E 0.00 0.012 0.239 P ∗ 10.0 L = ( 0.239 + 0.012 ) ∗ 8.314 L ∗ kPa mole − K ∗ 353 K , P = 112 kPa 10

A mixture of 14.2 g of N 2 and 25.3 g of O 2 is pumped into a bomb calorimeter with a volume of 10.0 L, which is submerged into a pool of water maintained at 80 °C to preserve isothermal operation a. What is the initial total pressure of the reactor to 3 significant figures? (10 pts) 14.2 g − N 2 ∗ 1 mole − N 2 28.0 g − N 2 = 0.507 mole − H 2 25.3 g − O 2 ∗ 1 mole − O 2 32.0 g − O 2 = 0.761 moles − O 2 P ∗ 10.0 L = ( 0.507 + 0.761 ) ∗ 8.314 L ∗ kPa mole − K ∗ 353 K , P = 381 kPa b. The hydrogen and oxygen react to form water, by the following reaction 2 O 2 ( g ) + N 2 ( g ) → 2 N O 2 ( g ) What is the limiting reagent? You must show work to receive credit. (5 pts) 14.2 g − N 2 ∗ 1 mole − N 2 28.0 g − N 2 ∗ 1 1 mole − N 2 = 0.507 25.3 g − O 2 ∗ 1 mole − O 2 32.0 g − O 2 ∗ 1 2 mole − O 2 = 0.395 11

c. After the reaction is run to completion, what is the final total pressure of the reactor to 3 significant figures? An ICE Table has been given for your convenience, though you are not required to use it (20 pts) N 2 O 2 NO 2 I 0.507 0.791 0.00 C -0.395 -0.791 0.791 E 0.112 0.00 0.791 P ∗ 10.0 L = ( 0.112 + 0.791 ) ∗ 8.314 L ∗ kPa mole − K ∗ 353 K , P = 265 kPa 12

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Heat Capacity How much heat is needed to increase the temperature of 500.0 g of methanol 5.00 ̊ C? q = m ∗ C p ∗ ∆T ,q = 6,280 J How much heat is needed to increase the temperature of 500.0 g of ethanol 5.00 ̊ C? q = m ∗ C p ∗ ∆T ,q = 7,380 J What is the temperature change of 1.00 g of water if 100.0 J of heat are supplied? q = m ∗ C p ∗ ∆T ,∆T = 23.9 What is the temperature change of 1.00 g of acetone if 100.0 J of heat are supplied? q = m ∗ C p ∗ ∆T ,∆T = 46.6 What is the temperature change of 1.00 g of water if 100.0 J of heat are removed? q = m ∗ C p ∗ ∆T ,∆T = 23.9 13

If ethanol is heated using 250.0 J and the change in temperature is 10.19, what was the mass of ethanol that was heated? q = m ∗ C p ∗ ∆T ,m = 10.00 g If ethanol is heated using 250.0 J and the change in temperature is 10.19, what was the mass of ethanol that was heated? q = m ∗ C p ∗ ∆T ,m = 10.00 g What is the heat capacity of a solid if 100.0 J of heat increases the mass of 10.0 g block 2.13 ̊ C? q = m ∗ C p ∗ ∆T ,C p = 4.69 J g − C What is the heat capacity of a solid if 256.3 J of heat increases the mass of 9.66 g block 11.3 ̊ C? q = m ∗ C p ∗ ∆T ,C p = 2.35 J g − C 14

Heats of Reaction For the following reaction, 14,351 joules of heat were released with 0.1130 moles of Cl 2 were reacted with Ag. What is ΔH rxn in kJ/mole? Was the reaction endothermic or exothermic? 2 A g ( s ) + C l 2 ( g ) → 2 AgC l ( s ) q = n ∗ ∆ H rxn ,∆ H rxn =− 127.0, exothermic For the following reaction, 1,535.1 joules of heat were released with 1.931 moles of Ca were reacted with Cl 2 . What is ΔH rxn in kJ/mole? Was the reaction endothermic or exothermic? C a ( s ) + C l 2 ( g ) →CaCl 2 ( s ) q = n ∗ ∆ H rxn ,∆ H rxn =− 795.0, exothermic For the following reaction, 332,228.9 joules of heat were consumed when 2.931 grams of H 2 were reacted with carbon. What is ΔH rxn in kJ/mole? Was the reaction endothermic or exothermic? 2 C ( s ) + H 2 ( g ) →C 2 H 2 ( g ) q = n ∗ ∆ H rxn ,∆ H rxn = 226.7, endothermic 15

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For the following reaction, 23,647.1 joules of heat were consumed with 12.556 grams of N 2 were reacted with O 2 . What is ΔH rxn in kJ/mole? Was the reaction endothermic or exothermic? N 2 ( g ) + 2 O 2 ( g ) → 2 N O 2 ( g ) q = n ∗ ∆ H rxn ,∆ H rxn = 33.9, endothermic For the following reaction, how many joules of heat were gained/lost (specify) by the reaction if 0.8675 moles of Br 2 were reacted with potassium and the ΔH rxn = -392.2 kJ/mole? 2 K ( s ) + Br 2 ( g ) → 2 KBr ( s ) q = n ∗ ∆ H rxn ,q = 340,200 J lost For the following reaction, how many joules of heat were gained/lost (specify) by the reaction if 3.09 grams of H 2 S were reacted and the ΔH rxn = 20.1 kJ/mole? H 2 S ( g ) → H 2 ( g ) + S ( s ) q = n ∗ ∆ H rxn ,q = 1,820 J gained 16

Calorimetry Problems A piece of metal weighing 90.0 g and at a temperature of 100.0 ̊ C is dropped into a calorimeter containing 100.0 g of water at a temperature of 21.3 ̊ C. When the metal and water reach thermal equilibrium, the temperature is 28.2 ̊ C. What is the heat capacity of the metal? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− ( T f − T i ) ∗ m metal ∗ C p, metal ( 28.2 − 21.3 ) ∗ 100 ∗ 4.181 =− ( 28.2 − 100 ) ∗ 90 ∗ C p,metal C p, metal = ( 28.2 − 21.3 ) ∗ 100 ∗ 4.181 ( 100 − 28.2 ) ∗ 90 = 0.446 J g − C 10.0 g of AgNO 2 are dissolved in 100.0 g of water, that is initially at 21.1 ̊ C. The final temperature of the water, after the AgNO 2 dissolves, is 15.4 ̊ C. What is the heat of dissolution (ΔH diss ) of AgNO 2 in kJ/mole? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− n ∗ ∆ H diss ∆ H diss = − n ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O = 36.9 kJ mole 17

A piece of metal weighing 108.2 g and at a temperature of 97.3 ̊ C is dropped into a calorimeter containing 100.0 g of water at a temperature of 20.3 ̊ C. When the metal and water reach thermal equilibrium, the temperature is 32.1 ̊ C. What is the heat capacity of the metal? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− ( T f − T i ) ∗ m metal ∗ C p, metal ( 32.1 − 21.3 ) ∗ 100 ∗ 4.181 =− ( 32.1 − 100 ) ∗ 90 ∗ C p,metal C p, metal = ( 32.1 − 21.3 ) ∗ 100 ∗ 4.181 ( 100 − 32.1 ) ∗ 90 = 0.672 J g − C 10.0 g of RbF are dissolved in 100.0 g of water, that is initially at 22.1 ̊ C. The final temperature of the water, after the RbF dissolves, is 28.1 C. What is the heat of dissolution (ΔH diss ) of RbF in kJ/mole? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− n ∗ ∆ H diss ∆ H diss = − n ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− 26.1 kJ mole 18

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10.2 g of CsI are dissolved in 100.0 g of water in a calorimeter. Before dissolving the CsI, the temperature of the water was 23.4 ̊ C. If the ΔH diss is 33.35 kJ/mole, then what is the final temperature of the water after dissolution? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− n ∗ ∆ H diss T f = T i − n ∗ ∆H diss m H 2 O ∗ C p, H 2 O = 19.9 13.8 g of CsF are dissolved in 100.0 g of water in a calorimeter. Before dissolving the CsI, the temperature of the water was 20.4 ̊ C. If the ΔH diss is -36.86 kJ/mole, then what is the final temperature of the water after dissolution? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− n ∗ ∆ H diss T f = T i − n ∗ ∆H diss m H 2 O ∗ C p, H 2 O = 28.4 19

A 100.0 g piece of copper at 90.0 ̊ C is dropped into a calorimeter filled with 100.0 g of water that is initially at 21.0 ̊ C. When the water and copper reach thermal equilibrium what are their final temperatures? ( T f − T i ) ∗ m H 2 O ∗ C p, H 2 O =− ( T f − T i ) ∗ m Cu ∗ C p,Cu ( T f − 21 ) ∗ 100 ∗ 4.181 =− ( T f − 90 ) ∗ 100 ∗ 0.385 T f = 90 ∗ 100 ∗ 0.385 + 21 ∗ 100 ∗ 4.181 100 ∗ 4.181 + 100 ∗ 0.385 = 26.8 20

Hess’s Law Problems Problem 1 C ( s ) + O 2 ( g ) →C O 2 ( g ) ∆ H 1 =− 393.5 kJ mole 3 Ni ( s ) + C ( s ) → ∋ ❑ 3 C ( s ) ∆ H 2 = 67.4 kJ mole 2 N i ( s ) + O 2 ( g ) → 2 ∋ O ( s ) ∆ H 3 =− 293.7 kJ mole 2 N i 3 C ( s ) + 5 O 2 ( g ) → 6 ∋ O ( s ) + 2 CO 2 ( g ) ∆ H = ? ∆ H = 2 ∗ ∆ H 1 − 2 ∗ ∆H 2 + 3 ∗ ∆ H 3 =− 1802.9 kJ mole Problem 2 H 2 ( g ) + F 2 ( g ) → 2 H F ( g ) ∆ H 1 =− 272.5 kJ mole Pb ( s ) + F 2 ( g ) →Pb F 2 ∆ H 2 =− 677.0 kJ mole 2 H F ( g ) + Pb ( s ) → H 2 ( g ) + PbF 2 ( s ) ∆ H = ? ∆ H =− ∆ H 1 + ∆ H 2 =− 404.5 kJ mole 21

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Problem 3 Pb ( s ) + S ( s ) →PbS ( s ) ∆ H 1 =− 96.6 kJ mole PbS ( s ) + 2 O 2 ( g ) → PbSO 4 ( s ) ∆ H 2 =− 923.1 kJ mole S ( s ) + O 2 ( g ) →S O 2 ( g ) ∆H 3 =− 296.8 kJ mole Pb ( s ) + S O 2 ( g ) + O 2 ( g ) →PbSO 4 ( s ) ∆ H = ? ∆ H = ∆ H 1 + ∆ H 2 − ∆H 3 =− 722.9 kJ mole Problem 4 S ( s ) + F 2 ( g ) →S F 2 ( g ) ∆ H 1 =− 296.6 kJ mole Pb ( s ) + F 2 ( g ) →Pb F 2 ∆ H 2 =− 677.0 kJ mole Pb ( s ) + SF 2 ( g ) →Pb F 2 + S ( s ) ∆ H = ? ∆ H =− ∆ H 1 + ∆ H 2 =− 380.4 kJ mole 22

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Problem 5 S ( s ) + 3 F 2 ( g ) →S F 6 ( g ) ∆ H 1 =− 1220.5 kJ mole S ( s ) + O 2 ( g ) →S O 2 ( g ) ∆H 3 =− 296.8 kJ mole S F 6 ( g ) + O 2 →SO 2 ( g ) + 3 F 2 ( g ) ∆ H = ? ∆ H =− ∆ H 1 + ∆ H 2 = 923.7 kJ mole Problem 6 C ( s ) + O 2 ( g ) →C O 2 ( g ) ∆ H 1 =− 393.5 kJ mole 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g ) ∆H 2 =− 483.6 kJ mole C ( s ) + 2 H 2 ( g ) →C H 4 ( g ) ∆ H 3 =− 74.9 kJ mole 2 C 2 H 5 O H ( l ) + 6 O 2 ( g ) → 4 C O 2 ( g ) + 6 H 2 O ∆ H 4 =− 2471.1 kJ mole 2 C H 4 ( g ) + H 2 O ( g ) → 2 H 2 ( g ) + C 2 H 5 O H ( l ) ∆ H = ? ∆ H = 2 ∗ ∆ H 1 + ∆H 2 − 2 ∗ ∆H 3 − 1 2 ∗ ∆ H 4 = 114.6 kJ mole 23

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Practice Problems (Tables) C H 4 ( g ) + C l 2 ( g ) →C H 3 C l ( g ) + HCl ( g ) ΔH = -103.755 kJ/mole 2 F e 3 C ( s ) + 5 O 2 ( g ) → 4 FeO + 2 FeC O 3 ( s ) ΔH = -2619.520 kJ/mole 2 C H 4 ( g ) + Cl 2 ( g ) → 2 C H 3 Cl ( g ) + H 2 ( g ) ΔH = -22.886 kJ/mole 2 A s ( s ) + Al 2 O 3 ( s ) → A s 2 O 3 ( s ) + 2 Al ( s ) ΔH = 1020.896 kJ/mole C H 4 ( g ) + C l 2 ( g ) →CCl 4 ( g ) + H 2 ( g ) ΔH = -25.543 kJ/mole B e 2 C ( s ) + 2 O 2 ( g ) → 2 BeO ( s ) + CO 2 ( g ) ΔH = -1493.228 kJ/mole 2 Ga I 3 ( s ) + 3 F 2 ( g ) → 2 Ga F 3 ( s ) + 3 I 2 ( g ) ΔH = -1364.402 kJ/mole 3 O 2 ( g ) + 4 Al N ( s ) → 2 Al 2 O 3 ( s ) + 2 N 2 ( g ) ΔH = -2079.448 kJ/mole 2 C H 4 ( g ) + 3 C l 2 ( g ) → 2 CHCl 3 ( g ) + 3 H 2 ( g ) ΔH = -52.760 kJ/mole BaO ( s ) + TiO 2 ( s ) →BaTiO 3 ( s ) ΔH = -161.507 kJ/mole C H 4 ( g ) + C l 2 ( g ) →C H 2 C l 2 ( g ) + H 2 ( g ) ΔH = -20.648 kJ/mole 24

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2 Al ( s ) + N 2 ( g ) → 2 Al N ( s ) ΔH = -635.968 kJ/mole CaO ( s ) + TiO 2 ( s ) →CaTiO 3 ( s ) ΔH = -80.760 kJ/mole 2 C u 2 O ( s ) + O 2 ( g ) → 4 CuO ( s ) ΔH = -282.838 kJ/mole 25

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TestFivePacketSolutions(3)

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