CHEM103 / CHEM 103 Module 3 Exam Questions & Answers | Latest 2026–2027 Update | General Chemistry I with Lab | Portage Learning | Verified Solutions | Grade A

CHEM103 / CHEM 103 Module 3 Exam Questions & Answers | Latest 2026–2027 Update | General Chemistry I with Lab | Portage Learning | Verified Solutions | Grade A 2026 / 2027 Academic Year Q: Heat temp change = qtemp change = m x c x ∆t Answer -(40.5 x 4.184 x (∆t - 85.7)) = 36.8 x 4.184 x (∆t - 26.3) -(169.452 x (∆t - 85.7)) = 153.9712 x (∆t - 26.3) -(169.452∆t - 14522.0364) = 153.9712∆t - 4049..452∆t + 14522.0364 = 153.9712∆t - 4049.44256 18571.479 = 323.423∆t 57.4 C = ∆t 1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter. c (water) = 4.184 J/g C Q: qs↔l = mass x Heat of Fusion = m x ∆Hfusion Answer 120 x 0.334 = 40.08 kJ 2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat of Fusion for water is 0.334 kJ/g. Q: moles = grams/molecular weight Answer moles (S) = 42.8/32.07 = 1.335 mols S = ΔHrx x new moles / original moles q = -792 x ( 1.335/2 )= -526.7 kJ 3. Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation: 2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off? Q: moles = grams/molecular weight Answer moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S ΔHrx = q / (new moles / original moles) q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = - 354.16 kJ -431.8 / ( 0.7686/2) = -1123.6 kJ 4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation: 2 H2S + 3 O2 → 2 SO2 + 2 H2O What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ? 1) q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C - 25.00 C) = - 58,791 J q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365 kJ x 1000 J/1 kJ= - 295,365 J (new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6 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) 5. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) Q: The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 C. The heat capacity of water = 4.18 J/g/C and the heat capacity of the calorimeter = 10.5 kJ/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. Answer q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 100 g x 5.14 K = 2148.52 J q calorimeter = heat capacity x Δt = 150 x 5.14 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 NaHCO3 / 84 g NaHCO3 / mol NaHCO3 = + 16.9 kJ / mol (2) The calorimeter + contents (no lid) = open system, (3) Endothermic process 6. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction Answer NaHCO3 (s) → Na+ (aq) + HCO3- (aq) Q: If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/C) decreases from 25.00oC to 19.86C, (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. Answer 2 (2 NH3 (g) → 3 H2 (g) + 2 N2 (g) ΔH = + 91.8 kJ) 3 (2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ) 2 (N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ) 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔHrxn = - 906.3 kJ ΔHrxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6) = - 906.3 kJ 7. The combustion of ammonia by the following reaction yields nitric oxide and water Answer 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) Determine the heat of reaction (ΔHrxn) for this reaction by using the following thermochemical data: N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = - 91.8 kJ 2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ Q: C2H5OH (l) → 2 C (graphite) + 3 H2 (g) + 1/2 O2 (g) ΔHf0 (C2H5OH) = - (- 277.6 kJ/mole) Answer 2 (C (graphite) + O2 → CO2 (g) 2 ΔHf0 (CO2) = 2 (-393.5 kJ/mole) 3 (H2 (g) + 1/2 O2 (g) → H2O (g) 3 ΔHf0 (H2O) = 3 (- 241.8 J/mole) C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) ΔHrxn = - (- 277.6) + 2 (- 393.5) + 3 (- 241.8) ΔHrxn = - 1234.8 kJ/mole OR more simply and preferably: ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔHf0 (reactants) ΔHrxn = 2 ΔHf0 (CO2) + 3 ΔHf0 (H2O) - ΔHf0 (C2H5OH) – 3 ΔHf0 (O2) ΔHrxn = 2 (- 393.5) + 3 (- 241.8) - (- 277.6) - 3 (0) = - 1234.8 kJ/mole 8. Determine the heat of reaction (ΔHrxn) for the combustion of ethanol (C2H5OH) by using heat of formation data: C2H5OH (l) + 3 O2 (g) → 2 CO2 + 3 H2O (g) Q: Pi x Vi = Pf x Vf Answer Ti Tf 680 ml/1000 = 0.680 liters = Vi 720 mm/760 = 0.947 atm = Pi 820 mm/760 = 1.08 atm = Pf 28 C + 273 = 301 K = Ti 55 C + 273 = 328 K = Tf Put in the data: (0.947) x (0.680) = (1.08) x Vf (301) (328) Q: Solve for Vf: Answer 0. = 0. x Vf Vf = 0.650 literrs Q: 9. A gas sample has an original volume of 680 ml when collected at 720 mm and 28 C. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 55 C? Answer P x V = n x R x T 0.546 mole = n R = 0.0821 700 mm/760 = 0.921 atm = P 25 C + 273 = 298 K = T (0.921) x V = (0.546) x (0.0821) x (298) (0.921) x V = 13.358 V = 13.358 = 14.5 liters 0.921 Q: 10. A gas sample containing 0.546 mole collected at 700 mm and 25 C. would occupy what volume? Answer 46.07 32 44.01 18.016 C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)11.195 g. 23.3 g 21.345 g 13.116 g 0.243 mol 0.728 mol 0.485 mol 0.728 mol 25C 1.25atm 0.728 x ( 1/3) = 0.243 moles C2H5OH 0.728 x (2/3) = 0.485 moles CO2 0.728 x (3/3) = 0.728 moles H2O 0.243 x 46.07 = 11.195 g C2H5OH 0.485 x 44.01 = 21.345 g CO2 0.728 x 18.016 = 13.116 g H2O P x V = n x R x T 1.25 x V = 0.485 x 0.0821 x 298 V = 9.49 L CO2 Q: 11. The combustion of ethanol (C2H5OH) takes place by the following reaction equation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25 C and 1.25 atm? Answer XHe = 4.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2857 Mole%He = 100(XHe) = (100) 0.286 = 28.57% XH2 = 2.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.1429 Mole%H2 = 100(XH2) = (100) 0.143 = 14.29% XCO2 = 3.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2143 Mole%CO2 = 100(XCO2) = (100) 0.214 = 21.43% XAr = 5.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.3571 Mole%Ar = 100(XAr) = (100) 0.357 = 35.71% PHe = XHe (2900 mm) = 0.2857 (2900 mm) = 828.53 mm PH2 = XH2 (2900 mm) = 0.1429 (2900 mm) = 414.41 mm PCO2 = XCO2 (2900 mm) = 0.2143 (2900 mm) = 621.47 mm PAr = XAr (2900 mm) = 0.3571 (2900 mm) = 1035.59 mm from Ideal Gas Law: n H2 = PV / RT = (0.898 atm) (72.0 ml x 1 liter / 1000 ml) / (0.0821) (308oK) Q: 12. A mixture of gases consists of 4.00 moles of He, 2.00 moles of H2, 3.00 moles of CO2 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. Answer PH2 = 725 - P H2O (from table) = 725 - 42.2 = 682.8 x 1 atm / 760 mm = 0.898 atm H2 n H2 = 0.002557 moles grams H2 = moles x MW = 0.002557 moles x 2.016 grams / 1 mole = 0.00515 grams 13. A sample of hydrogen (H2) gas is collected over water at 35 C and 725 mm. The volume of the gas collected is 72.0 ml. How many moles of H2 gas has been collected? How many grams of H2 gas has been collected? r1 /r2 = √ (MW2 /MW1) 1.253/1 = √(MW2 /28.02) MW2 = 43.99 mol 14. The rate of effusion of nitrogen gas (N2) is 1.253 times faster than that of an unknown gas. What is the molecular weight of the unknown gas? 1. Heat absorbed = Endothermic process 2. The air above the cup = Surroundings NH4Cl is being dissolved in water and this process is being studied in a styrofoam coffee cup with a lid and the heat absorbed is measured by means of a thermometer immersed in the reaction mixture. Enter the correct thermochemistry term to describe the item listed. 1. The type of thermochemical process 2. The air above the cup 1. - (mCu x cCu x ∆tCu) = (mH2O x cH2O x ∆tH2O) - [18.5 g x 0.384 J/g oC x (Tmix - 100oC)] = [(45.2 g x 4.184 J/g oC x (Tmix - 23.2oC)] - [7.104 J/oC x (Tmix - 100oC)] = [(189.1168 J/oC x (Tmix – 23.2oC)] - 7.104 J/oC (Tmix) + 710.4 J = 189.1168 J/oC (Tmix) – 4387.50976 J 5097.90976 J = 196.2208 J/oC (Tmix) Tmix = 25.9oC 2. ql↔g = m x ∆Hvapor = 165 g x 2.26 kJ/g = 372.9 kJ (since heat is removed) = - 372.9 kJ 1. Show the calculation of the final temperature for a 18.5 gram piece of copper heated to 100 C which has been added to a 45.2 gram sample of water at 23.2 C in a coffee cup calorimeter. c (water) = 4.184 J/g C; c (Cu) = 0.384 J/g C 2. Show the calculation of the energy involved in condensation of 165 grams of steam at 100oC if the Heat of Vaporization for water is 2.26 kJ/g. 32.07 g 2 S (s) + 3 O2 (g) → 2 SO3 (g) 20.7 g 0.645 mols q = 255.6kJ ΔHrx = q / (new moles / original moles) ΔHrx = 255.6kJ / (0.6455/2) = -792.9 kJ Show the calculation of the ΔH of the reaction if 20.7 g of S is reacted with excess O2 to yield 255.6 kJ and sulfur trioxide by the following reaction equation. Report your answer to 4 significant figures. 2 S (s) + 3 O2 (g) → 2 SO3 (g) 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ 2 (C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ) 2 (CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ) 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) ΔHrxn = - 2413.84 kJ ΔHrxn = -1270.18 + 2(-393.51) + 2(-178.32) = - 2413.84 Show the calculation of the heat of reaction (ΔHrxn) for the reaction: 2 Ca (s) + 2 C (graphite) + 3 O2 (g) → 2 CaCO3 (s) by using the following thermochemical data: 2 Ca (s) + O2 (g) → 2 CaO (s) ΔH = - 1270.18 kJ C (graphite) + O2 (g) → CO2 (g) ΔH = - 393.51 kJ CaO (s) + CO2 (g) → CaCO3(s) ΔH = - 178.32 kJ CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO2 (g) = -393.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole ΔHrxn = (+74.6) + 2(0) + (-393.5) + 2(-285.8) = - 890.5 kJ/mole Show the calculation of the heat of reaction (ΔHrxn) for the reaction: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) by using the following thermochemical data: ΔHf0 CH4 (g) = -74.6 kJ/mole, ΔHf0 CO2 (g) = -393.5 kJ/mole, ΔHf0 H2O (l) = -285.8 kJ/mole (Pi x Vi ) / Ti = (Pf x Vf ) / Tf 560 ml/1000 = 0.560 liters = Vi 1.05 atm = Pi 1.35 liters = Pf 32oC + 273 = 305oK = Ti 50oC + 273 = 323oK = Tf (1.05) x (0.560) / 305 = Pf x (1.35) / 323 Pf = 0.461 atm Show the calculation of the new pressure of a gas sample which has an original volume of 560 ml when collected at 1.05 atm and 32 C when the volume becomes 1.35 liters at 50 C. P x V = n x R x T SO2 molecular weight = 64 1.32 grams/64 = 0.020625 mole = n R = 0.0821 1.15 liters = V 30 C + 273 = 303 K = T P x (1.15) = (0.020625) x (0.0821) x (303) P = 0.446 atm Show the calculation of the pressure of a 1.15 liter sample of SO2 gas which weighs 1.32 grams at 30 C. 78.108g 32 44.01 18.016 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) 25.5g 0.326 mol 2.445 mol 1.956 mol 0.978 mol 293K 1.25atm moles C6H6 = 25.5/78.108 = 0.326 mol C6H6 moles O2 = 0.326 x (15/2) = 2.445 mol O2 moles CO2 = 0.326 x (12/2) = 1.956 mol CO2 moles H2O = 0.326 x (6/2) = 0.978 mol H2O PV=nRT 1.25 x V = 1.956 x 0.0821 x 293 V = 37.64 L Show the calculation of the volume of CO2 gas formed by the combustion of 25.5 grams of C6H6 at 20oC and 1.25 atm. The combustion of benzene (C6H6) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) XHe = 3.00 / (3.00 + 4.50 + 6.00) = 0.2222 XCO2 = 4.50 / (3.00 + 4.50 + 6.00) = 0.3333 XAr = 6.00 / (3.00 + 4.50 + 6.00) = 0.4444 PHe = XHe (700 mm) = 0.2222 (700 mm) = 155.5 mm PCO2 = XCO2 (700 mm) = 0.3333 (700 mm) = 233.3 mm PAr = XAr (700 mm) = 0.4444 (700 mm) = 311.1 mm Show the calculation of the mole fraction and partial pressure of each gas in a mixture which consists of 3.00 moles of H2, 4.50 moles of CO2 and 6.00 moles of Ar if the total pressure of the mixture is 700 mm. Express your answer in mm. (rO2 /runknown)2 = MWunknown / MWO2 (1.562/1)2 = MWunknown / 32.00 MWunknown = (1.562)2 x 32.00 = 78.01 Show the calculation of the molecular weight of an unknown gas if the rate of effusion of oxygen gas (O2) is 1.562 times faster than that of an unknown gas. Core Electrons (stay at home Netflix) -Electrons in filled shells -Have limited influence on atomic properties -Generally unaltered in chemical reactions Valence Electrons (Free spirit) -Electron in shell with largest N -Determine how an atom interacts with other atoms -Gained/Lost/Shared -Dictate how atoms react with other atoms electron configuration whole sequence (1s2,2s2, etc.) Core electron example (Ne) 1s2, 2s2, 2p6 Valence Electron example (S) [Ne] 3s2, 3p4 Non-Metals: gain valence electron(s) General Principle: Neutral atom (s or p block) will gain or lose enough electrons to become isoelectronic with nearest noble gas Isoelectronic: identical electron count, configuration Metals: lose valence electron(s) d-block metals: form stable cations with different charges that are difficult to predict P3(-) = 18e(-) : 1s2, 2s2, 2p6, 3s2, 3p6 write complete electron configuration of phosphorus and the phosphide ion Ba[Xe]: 6s2 Ba2(+) : [Xe] write condensed electron configuration of barium and the barium ion Pt[Xe] : 6s2, 4f14, 5d8 Pt4(+) : [Xe] 4f11, 5d6 write condensed electron configuration of platinum and the platinum (IV) ion Controlling factor: trends in effective nuclear charge (Z eff) equation for Z (eff) Z(eff) ~ Z - number of core electrons Attractive Force: Valence electrons with experience greater attractions to the nucleus as (Z eff) increases o=shielding constant: "insulation" from nuclear charge provided by core e- General trends in Z (eff) - Z (eff) is constant moving down a group - Z (eff) increases moving Left to Right across a period Z (eff) example.... K Z=19 [1s2, 2s2, 2p6, 3s2, 3p6] 4s 19-18= +1 - answer a physical property: atomic radius atomic radius ______ moving L to R across period decreases atomic radius ______ moving R to L across period increases Atomic radius ______ moving down a group increase When Z (eff) increases moving L to R, _________ atrraction of valence electrons to the neucleus increase Atomic radius ______ moving up a group decreases Higher attraction pulls volume electrons closer to the nucleus, _________ atomic radius decreasing Valence electrons occupy orbitals with ________ moving down a group increasing orbitals extend farther forth nucleus, _________ atomic radius increasing Chemical property: Ionization energy (IE) Ionization energy (IE): Energy required to remove one electron from a neutral Gas-phase atom Why is the IE of oxygen lower than expected? electron-electron repulsion (two angry cats) Another chemical property: electron affinity (EA) Electron affinity (EA) energy change when one electron is added to a neutral, gas phase atom (always negative) _______ distance from the nucleus results In reduced attractive forces, and _____ energy release uopn e- capture Greater, less Orbital overlap region higher probability of finding both electrons Lewis structures (structural formula): illustrates the covalent bonds and lone pairs for all atoms in a molecule or polyatomic ion covalent bonds shared pairs of electrons lone pairs electron pairs that are not shared 3.1: THERMOCHEMISTRY Thermodynamics is the study of the relationship between heat and other forms of energy, particularly mechanical work. Thermochemistry is the part of thermodynamics that deals with the quantity of heat given off or absorbed during a chemical reaction. The quantity of heat given off or absorbed during a physical change or temperature change can also be studied, and we will refer to this process as calorimetry. System the object (or substance) being studied Open system a system that permits the transfer of mass and energy with the surroundings Closed system a system that permits the transfer of energy but not mass with the surroundings Isolated system a system that does not permit the transfer of energy or mass with the surroundings Surroundings the rest of the universe interacting with the system Energy the potential or capacity to move matter: the ability to do work (unit is J = joule) Work the amount of energy transferred by a force acting through a distance Kinetic energy the energy possessed by an object by virtue of its motion (unit is J = joule) Potential energy the energy possessed by an object by virtue of its position (unit is J = joule) Heat (q) the thermal energy transferred between system and surroundings due to a difference in temperature between them (unit is J = joule) Enthalpy the total energy of a system Heat of reaction (ΔH) the amount of heat (q) gained or lost during a chemical reaction Exothermic a reaction with a - ΔH Endothermic a reaction with a + ΔH (absorb heat) Calorimetry The energy change that accompanies a physical, temperature, or chemical change is determined by carrying out the process in a device known as a calorimeter. The calorimeter is able to measure the amount of heat absorbed or evolved as a process takes place. A styrofoam coffee cup calorimeter can be used to measure an energy change that takes place at constant pressure. An enclosed bomb calorimeter is used to measure an energy change that takes place at constant volume with a change in pressure. Temperature change calorimetry measures the thermal energy change occurring as a system at higher temperature transfers kinetic energy to a system at lower temperature, which is reflected by a change in temperature for the overall system. This is demonstrated below by adding a 15.6 gram piece of aluminum (heated to 100oC) to a 45.6 gram sample of water at 26.7oC in a coffee cup calorimeter. The final temperature of this system can be predicted using the equations below and several facts about the materials (Al and H2O). Heat temp change = qtemp change = mass x specific heat (heat capacity) x temp change = m x c x ∆t (mAl x cAl x ∆tAl) = (mH2O x cH2O x ∆tH2O) However, since the Al is losing heat, we'll use a negative sign in front of the heat loss equation. - (mAl x cAl x ∆tAl) = (mH2O x cH2O x ∆tH2O) We know ∆t = Tempmixture - Tempinitial, so we can substitute the data to get: - [15.6 g x 0.899 J/g oC x (Tmix - 100oC)] = [(45.6 g x 4.184 J/g oC x (Tmix - 26.70oC)] Now, solve: - [14.0244 J/oC x (Tmix - 100oC)] = [(190.7904 J/oC x (Tmix - 26.7oC)] - 14.0244 Tmix + 1402.44 = 190.79 Tmix - 5094.1 6496.44 = 204.8144 Tmix Tmix = 6496.44 / 204.8144 = 31.7oC Phase change calorimetry measures the energy change occurring as a substance changes from one phase (state) to another, such as water melting or boiling, or ice freezing or steam condensing. In this case, no temperature change occurs, but the energy change causes the particles of the substance to form or break intermolecular bonds and change from one state to another. The equations used to do phase change calorimetry calculations are shown below: Phase changes of solid to liquid or liquid to solid: qs↔i = mass x Heat of Fusion = m x ∆Hfusion Phase Changes of liquid to gas or gas to liquid: ql↔g = mass x Heat of Vaporization = m x ∆Hvapor What is the energy involved in freezing 200 grams of water at 0oC if the heat of fusion for water is 0.334 kJ/g? What is the energy involved in vaporization of 200 grams of water at 100oC if the heat of vaporization for water is 2.26 kJ/g? When energy is given off in any of these phase changes, a negative sign is placed in front of the value, and when energy is added, a positive sign is placed in front of the value. ql↔g = m x ∆Hvapor = 200 g x 2.26 kJ/g = 452 kJ (since heat is added) = + 452 kJ ql↔s = m x ∆Hfusion = 200 g x 0.334 kJ/g = 66.8 kJ (since heat is removed/given off) = - 66.8 kJ How much heat is involved to convert of 20 grams of ice at -10oC to 20 grams of steam at 120 oC? ∆Hfusion = 0.334 kJ/g (H2O(l) ↔ H2O(s)) ∆Hvapor = 2.26 kJ/g (H2O(l) ↔ H2O(g)) c (water) = 4.184 J/g oC c (ice) = 2.09 J/g oC c (steam) = 1.86 J/g oC To do so, ice at -10oC must be converted to ice at 0oC (temp change) then ice at 0oC must be converted to water at 0oC (phase change) water at 0oC → water at 100oC qtemp change = m x c x ∆t = 20 g x 4.184 J/g oC x 100oC = + 8368 J then water at 0oC must be converted to water at 100oC (temp change) then water at 100oC must be converted to steam at 100oC (phase change) and finally steam at 100oC must be converted to steam at 120oC (temp change) as follows: ice at -10oC → ice at 0oC qtemp change = m x c x ∆t = 20 g x 2.09 J/g oC x 10oC = + 418 J ice at 0oC → water at 0oC ql↔s = m x ∆Hfusion = 20 g x 0.334 kJ/g = + 6.68 kJ water at 100oC → steam at 100oC ql↔g = m x ∆Hvapor = 20 g x 2.26 kJ/g = + 45.2 kJ steam at 100oC → steam at 120oC qtemp change = m x c x ∆t = 20 g x 1.86 J/g oC x 20oC = + 744 J converting J to kJ: J / 1000 = kJ Total Heat Required = + 0.418 kJ + 6.68 kJ + 8.368 kJ + 45.2 kJ + 0.744 kJ = + 61.41 kJ Show the calculation of the energy involved in condensation of 150 grams of steam at 100oC if the Heat of Vaporization for water is 2.26 kJ/g. ql↔g = m x ∆Hvapor = 150 g x 2.26 kJ/g = 339 kJ (since heat is removed) = - 339 kJ Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat of Fusion for water is 0.334 kJ/g. ql↔s = m x ∆Hfusion = 120 g x 0.334 kJ/g = 40.08 kJ (since heat is added) = + 40.08 kJ Show the calculation of the heat involved in converting 30 grams of steam at 110oC to 30 grams of ice at -20 oC. ∆Hfusion = 0.334 kJ/g; ∆Hvapor = 2.26 kJ/g; c (water) = 4.184 J/g oC; c (ice) = 2.09 J/g oC; c (steam) = 1.86 J/g oC steam at 110oC → steam at 100oC qtemp change = m x c x ∆t = 30 g x 1.86 J/g oC x 10oC = - 558 J steam at 100oC → water at 100oC ql↔g = m x ∆Hvapor = 30 g x 2.26 kJ/g = - 67.8 kJ water at 100oC → water at 0oC qtemp change = m x c x ∆t = 30 g x 4.184 J/g oC x 100oC = - 12552 J water at 0oC → ice at 0oC ql↔s = m x ∆Hfusion = 30 g x 0.334 kJ/g = - 10.02 kJ ice at 0oC → ice at -10oC qtemp change = m x c x ∆t = 30 g x 2.09 J/g oC x 20oC = - 1254 J converting J to kJ: J / 1000 = kJ Total Heat Required = - 0.558 kJ - 67.8 kJ - 12.552 kJ - 10.02 kJ - 1.254 kJ = -- 92.184 kJ 3.2: THERMOCHEMICAL EQUATIONS & MEASURING HEATS OF REACTIONS The equation written below is a thermochemical equation because it shows that three moles of H2 gas react with 1 mole of N2 gas to form two moles of NH3 gas and 91.8 kJ (kilojoules) of heat is given off. N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ Two important rules that apply to thermochemical equations are: (1) When the reverse of the thermochemical equation occurs, the sign of the ΔH is reversed (2) When the number of moles of reactants used is changed, the quantity of heat absorbed or evolved is equal to the original value of ΔH times the factor (new moles/original moles). moles refers to substance for which mass data is given in the problem q = ΔHrx x new moles (moles given in problem) / original moles (moles given in thermochemical equation) The heat involved in the reaction of 51 grams NH3 to produce N2 and H2 is calculated below (note that this is the reverse of the original reaction): N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔHrx = -91.8 kJ 2 NH3 (g) → 3 H2 (g) + N2 (g) ΔHrx is for 2 mole of NH3 reaction uses 51 g NH3 = 51/17 mole = 3 mole NH3 q = + 91.8 kJ x 3 mole NH3 / 2 mole NH3 = + 137.7 kJ Measuring Heats of Reactions To be able to determine how much heat is given off (or absorbed) in these two reactions (C + O2 → CO2 and HCl + NaOH → NaCl + H2O), we must be able to determine the amount of heat absorbed by the water as its temperature increases. This quantity of heat is determined by multiplying three pieces of information about the water: specific heat, temp change, and mass. The heat (given off by the reaction) and absorbed by the water in the reaction of 200 ml of 1.0 M NaOH and 200 ml of 1.0 M HCl causes the temperature of the water and the calorimeter (heat capacity of calorimeter = 340 J/oK) to increase by 5.0oK (note total mass of water is 400 grams since the water in the two solutions (HCl and NaOH) is 400 ml = 400 grams). HCl + NaOH → NaCl + H2O c (water) = 4.184 J/g oC q water = s (specific heat of water) x mass x Δt = 4.18 J / g / oK x 400 g x 5.0oK = -8360 J q calorimeter = heat capacity x Δt = 340 J/oK x 5.0oK = -1700 J (signs are negative since heat is given off, exothermic) q reaction = -1700 J + (-8360 J) = -10060 J = -10060 J x 1 kJ / 1000 J = -10.06 kJ new moles of HCl or NaOH = (1.00 M/L) x (0.200 L) = 0.200 mole ΔH reaction = -10.06 kJ / (0.200 mol / 1 mole) = -50.3 kJ / mole 1 HCl + 1 NaOH → 1 NaCl + 1 H2O ΔH = -50.3 kJ / mole The heat (given off) by the combustion of graphite and absorbed by 2000 g of water (in the reaction of 1.00 g of C (graphite) and excess O2 causes the temperature of the water and the calorimeter (heat capacity of calorimeter = 21.0 kJ/oK) to increase by 1.6oK graphite; a gray crystalline allotropic form of carbon C + O2 → CO2 c (water) = 4.184 J/g oC q water = s x mass x Δt = 4.18 J / g / oK x 2000 g x 1.6oK = -13376 J q calorimeter = heat capacity x Δt = 21.0 kJ / oK x 1.6oK = -33.6 kJ x 1000 J/1 kJ= - 33600 J q reaction = -33600 J + (-13376 J) = -46976 J = -46976 J x 1 kJ / 1000 J = -46.976 kJ new moles = 1.00 g / 12 = 0.08333 mole ΔH reaction = -46.976 kJ / (0.08333 / 1) = -563.7 kJ / mole C + O2 → CO2 ΔH = -563.7 kJ / mole Thermochemical Equation Problems 1. Ammonia undergoes combustion to yield nitric oxide and water by the following reaction equation: 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) ΔH = - 1170 kJ If 26.5 g of NH3 is reacted with excess O2, what will be the amount of heat given off? ΔHrx is for 4 mole of NH3 reaction uses 26.5 g NH3 = 26.5/17 = 1.56 mole NH3 q = -1170 kJ x 1.56 mole NH3 / 4 mole NH3) = - 456.3 kJ Thermochemical Equation Problems 2. Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation: 2 S + 3 O2 → 2 SO3 ΔH = - 792 kJ If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off? ΔHrx is for 2 mole of S reaction uses 42.8 g S = 42.8/32.06 = 1.335 mole S q = -792 kJ x 1.335 mole S / 2 mole S = -528.7 kJ Thermochemical Equation Problems 3. Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following reaction equation: CH4 + 4 Cl2 → CCl4 + 4 HCl What is the ΔH of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield 1387.6 kJ? ΔHrx is for 1 mole of CH4; q = -1387.6kJ reaction uses 51.3 g CH4 = 51.3/16.042 mole = 3.198 mole CH4 -1387.6 kJ = ΔHrx x 3.198 mole CH4 / 1 mole CH4 ΔHrx = -433.9 kJ Thermochemical Equation Problems 4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the following reaction equation: 2 H2S + 3 O2 → 2 SO2 + 2 H2O What is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ? ΔHrx is for 2 mole of H2S; q = -431.8 kJ reaction uses 26.2 g H2S = 26.2/34.086 mole = 0.7686 mole H2S -431.8 kJ = ΔHrx x 0.7686 mole H2S / 2 mole H2S ΔHrx = -1123.6 kJ Measuring Heat of Reaction Problems 5. A sample of ethanol (C2H5OH), weighing 6.83 g underwent combustion in a bomb calorimeter by the following reaction: C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (l) If the heat capacity of the calorimeter and contents was 18.1 kJ / oC and the temperature of the calorimeter rose from 25.50 to 36.73, (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. (1) Temperature change for calorimeter= 36.73 oC - 25.50 oC = 11.23 oC heat change for calorimeter and contents q = HC x Δt = (18.1 kJ/oC) X (11.23 oC) = 203.263 kJ heat change for reaction (q) = -203.263 kJ (negative since heat is given off) (new) moles C2H5OH = 6.83 g / 46 = 0.1485 mole C2H5OH ΔH = q / (new moles / original moles) ΔH = - 203.263 / (0.1458 / 1) = - 1368.8 kJ / mole (2) Isolated system (bomb calorimeter) (3) Exothermic (temperature of water rises due to heat given off by combustion reaction) Measuring Heat of Reaction Problems 6. A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a bomb calorimeter by the following reaction: 2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l) If the heat capacity of the calorimeter and contents was 10.5 kJ / oC and the temperature of the calorimeter rose from 25.00 to 53.13, (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. (1) q = HC x Δt = (10.5 kJ/oC) x (53.13 oC - 25.00oC) = - 295.365 kJ (given off) (new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6 ΔH = q / (new moles / original moles) ΔH = - 295.365 / (0.09048 / 2) = - 6534.6 kJ / mole (2) Isolated system (bomb calorimeter) (3) Exothermic (temperature of water rises due to heat given off by combustion reaction) Measuring Heat of Reaction Problems 7. A 100 ml sample of 0.100 M NaOH was reacted with 100 ml of 0.100 M HCl in a coffee cup calorimeter with lid by the following reaction: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (aq) If the temperature of the water and the calorimeter (heat capacity of calorimeter = 200 J/oK) increases from 25.00 oC to 25.97oC, (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. Total Mass of Water: 100 mL of NaOH + 100 mL of HCl = 200 ml = 200 grams New Moles: 100 ml of 0.100 M NaOH or 100 ml of 0.100 M HCl = 100 ml of 0.100 M = 0.0100 moles q water = s (specific heat of water) x mass x Δt = 4.18 J / g / oK x 200 g x 0.97oC = - 810.92 J q calorimeter = heat capacity x Δt = 200 x 0.97oK = - 194 J q reaction = - (194 J + 810.92 J) = - 1004.92 J = - 1004.92 J x 1 kJ / 1000 J = - 1.00492 kJ (1) ΔH reaction = 1.00492 kJ / (0.0100 moles/1 mole) (of HCl or NaOH) = - 100.49 kJ / mole (2) The calorimeter + contents (with lid) = closed system, (3) Exothermic process Measuring Heat of Reaction Problems 8. A sample of 14.5 g of sodium bicarbonate (NaHCO3) was dissolved in 100 ml of water in a coffee-cup calorimeter with no lid by the following reaction NaHCO3 (s) → Na+ (aq) + HCO3- (aq) If the temperature of the water and the calorimeter (heat capacity of calorimeter = 150 J/oC) decreases from 25.00oC to 19.86oC, (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. Temperature change for calorimeter = 25.00 oC - 19.86 oC = 5.14 oC 3.3: HESS'S LAW & USING HEATS OF FORMATION TO DETERMINE HEATS OF REACTION q water = s (specific heat of water) x mass x Δt = 4.18 J / g / oK x 100 g x 5.14oK = 2148.52 J q calorimeter = heat capacity x Δt = 150 x 5.14oK = 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 NaHCO3 / 84 g NaHCO3 / mol NaHCO3 = + 16.9 kJ / mol (2) The calorimeter + contents (no lid) = open system, (3) Endothermic process Enthalpy change (ΔHrxn) for a chemical reaction is independent of the path by which the products are obtained. Therefore, the ΔH for an overall reaction which can be written as the sum of two or more steps equals the sum of the ΔHs for the individual steps. This is known as Hess's law. Hess's law can be used to obtain ΔH for chemical reactions for which ΔH cannot be directly measured because of some limitation. For instance, the ΔHrxn for the synthesis of tungsten carbide (WC) is not easily obtained since the reaction must be carried out at 1400oC. However, the heats of combustion of tungsten, graphite (C), and tungsten carbide are all easily obtained; they can then be used by way of Hess’s law to obtain the ΔHrxn for the synthesis of tungsten carbide as shown below: W (s) + C (graphite) → WC (s) ΔHrxn = ? 2 W (s) + 3 O2 (g) → 2 WO3 (s) ΔH = -1680.6 kJ C (graphite) + O2 (g) → CO2 (g) ΔH = -393.5 kJ 2 WC (s) + 5 O2 (g) → 2 WO3 (s) + 2 CO2 (g) ΔH = -2391.6 kJ Now by combining these three equations as follows, we can obtain the ΔH for the synthesis of tungsten carbide. To calculate a ΔHrxn by Hess's law, note that the available equations are used in such a way as to eliminate all substances not present in the overall target equation by reversing some equations or multiplying some equations by a numerical factor. This change is also applied to the ΔH of that reaction. 1/2 (2 W (s) + 3 O2 (g) → 2 WO3 (s) ΔH = -1680.6 kJ) C (graphite) + O2 (g) → CO2 (g) ΔH = -393.5 kJ 1/2 (2 WO3 (s) + 2 CO2 (g) → 2 WC (s) + 5 O2 (g) ΔH = +2391.6 kJ) W (s) + C (graphite) → WC (s) ΔHrxn = - 38.0 kJ ΔHrxn = 1/2 (-1680.6 kJ) + (-393.5 kJ) + 1/2 (+2391.6 kJ) = - 38.0 kJ Using Heats of Formation to Determine Heats of Reaction Heats of formation of materials along with Hess' law can be used to predict the enthalpies of most chemical reactions. Let's apply this to the following reaction: CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) ΔH = ? Using the data from the Standard Enthalpy of Formation Table: CH4 (g) → C (graphite) + 2 H2 (g) -ΔHf0 (CH4) = - (- 74.6 kJ/mole) C (graphite) + O2 → CO2 (g) ΔHf0 (CO2) = (-393.5 kJ/mole) 2 H2 (g) + O2 (g) → 2 H2O (g) 2Δ Hf0 (H2O) = 2 x (- 241.8 J/mole) CH4 + 2 O2 → CO2 + 2 H2O ΔHrxn = 74.6 + (- 393.5) + 2 (- 241.8) ΔHrxn = - 802.5 kJ/mole ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔHf0 (reactants) We are breaking down the reactants into their elements and reforming these elements to produce the products. While we combine the enthapies of the formation, we must reverse the sign of the reactant enthalpies and use the product enthalpies as they are given. Since the enthalpies of formation of the products are used as given (since the products are being formed) and the enthalpies of the reactants must be reversed (since the reactants are being broken down), all heat of reaction problems determined from enthalpies of formation can be simplified to the following form: where n and m are, respectively, the coefficients of the product molecules and the reactant molecules in the balanced equation since enthalpy values are in kJ / mole and Σ indicates that the enthalpies of formation should be summed. Notice that the ΔH0f of any element (O2) = 0 since no heat is absorbed or evolved when an element is formed from itself. Applying this form of the solution to the reaction above we get: ΔHrxn = ΔHf0 (CO2) + 2 ΔHf0 (H2O) - ΔHf0 (CH4) - ΔHf0 (O2) ΔHrxn = -393.5 + 2 (-241.8) - (-74.6) + 2 (0) = - 802.5 kJ/mole Hess' Law Problems 1. The combustion of ammonia by the following reaction yields nitric oxide and water 4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g) Determine the heat of reaction (ΔHrxn) for this reaction by using the following thermochemical data: N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ N2 (g) + 3 H2 (g) → 2 NH3 (g) ΔH = - 91.8 kJ 2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ 1) N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ 2N2 (g) + 2O2 (g) → 4NO (g) ΔH = 180.6 kJ x2 = 361.2 kJ 2) N2 (g) + 3 H2 (g) →2 NH3 (g) ΔH = - 91.8 kJ 4 NH3 (g) →2 N2 (g)+ 6H2 (g) ΔH = - 91.8 kJ x 2x (-1 - Reverse reaction) = 183.6 kJ 3) 2 H2 (g)+O2 (g)→2 H2O (g) ΔH = -483.7 kJ 3X (2 H2 (g)+O2 (g))→ 3x2 H2O (g) ΔH = - 483.7 kJ x 3 = -1451.10 kJ 6 H2 (g)+ 3O2 (g)→ 6 H2O (g) 4 NH3 (g)+ 5 O2 (g)→ 4 NO (g) +6 H2O (g) 2N2 (g) + 2O2 (g)+ 4 NH3 (g) + 6 H2 (g) + 3 O2 (g) = 4NO (g) + 2 N2 (g)+ 6H2+ 6 H2O (g) ΔHrxn = 361.2 kJ + 183.6 kJ - 1451.10 kJ = -906.3 kJ Heat of Formation Problems 5. Determine the heat of reaction (ΔHrxn) for the combustion of ethanol (C2H5OH) by using heat of formation data: C2H5OH (l) + 3 O2 (g) → 2 CO2 + 3 H2O (g) C2H5OH (l) + 3 O2 (g) → 2 CO2 + 3 H2O (g) ∆Hof [CO2 (g)] = - 393.5 kJ/mol ∆Hof [H2O (g)] = - 241.8 kJ/mol ∆Hof [C2H5OH (l) ] = - 277 kJ/mol ∆Hof [O2(g) ] = 0 ∆Hrxn = ∆Hof of products - ∆Hof of reactants = 2 x ∆Hof [CO2 (g)] +3 x ∆Hof [H2O (g)] - { ∆Hof [C2H5OH (l) ] + 3 x ∆Hof [O2(g) ] } = 2 x -393.5 + 3 x - 241.8 - { - 277 + 3 x (0) } = - 1235.4 kJ/mol ∆Hrxn = - 1235.4 kJ/mol 3.5: GAS LAWS Kelvin (oK) scale, which is oC + 273. The volume scale used with gases is the liter scale, which is ml/1000. The pressure unit we will use with gases is atmosphere (atm) scale, which is mm/760. Combined Gas Law P1V1/T1=P2V2/T2 Ideal Gas Law P(atm) x V(liters) = n(moles) x R x T(oK) P(atm) x V(liters) = (g/MW) x R x T(oK) R is a numerical constant with the value of 0.0821 at all times. A gas sample has an original volume of 530 ml when collected at 750 mm and 25oC. What will be the volume of the gas sample if the pressure increases to 780 mm and the temperature increases to 50oC? 530 ml/1000 = 0.530 liters = V1 750 mm/760 = 0.987 atm = P1 25oC + 273 = 298oK = T1 780 mm/760 = 1.03 atm = P2 50oC + 273 = 323oK = T2 A gas sample containing 0.256 mole collected at 730 mm and 20oC would occupy what volume? Data in problem gives moles, use P x V = n x R x T 0.256 mole = n R = 0.0821 730 mm/760 = 0.961 atm = P 20oC + 273 = 293oK = T (0.961) x V = (0.256) x (0.0821) x (293) (0.961) x V = 6.158 V = 6.41 liters A sample of CO2 gas that weighs 2.62 grams has a volume of 1.35 liters when collected at 30oC. What would be the pressure of the gas sample? Data in problem gives grams, use P x V = n x R x T CO2 molecular weight = 44 2.62 grams/44 = 0.0595 mole = n R = 0.0821 1.35 liters = V 30oC + 273 = 303oK = T P x (1.35) = (0.0595) x (0.0821) x (303) P x (1.35) = 1.480 P = = 1.10 atm A gas sample has an original volume of 680 ml when collected at 720 mm and 28oC. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperature increases to 55oC? Pi x Vi /Ti = Pf x Vf / Tf 680 ml/1000 = 0.680 liters = Vi 720 mm/760 = 0.947 atm = Pi 820 mm/760 = 1.08 atm = Pf 28oC + 273 = 301oK = Ti 55oC + 273 = 328oK = Tf Put in the data: (0.947) x (0.680) / (301) = (1.08) x Vf / (328) Solve for Vf: 0. = 0. x Vf Vf = 0. / 0.= 0.650 liter A sample of CO2 gas which weighs 1.62 grams has a volume of 1.02 liters when collected at 20oC. What would be the pressure of the gas sample? P x V = n x R x T n= 1.62 g of CO2/44 g/mol = 0.0368 mole V=1.02 L T=20 oC +273 = 293 oK R=0.0821 P= (0.0368*0.0821*293)/1.02 P= 0.868 atm 3.6: GAS VOLUME & LAW OF PARTIAL PRESSURES Gas Volume Stoichiometry The volumes of gases produced in a chemical equation are directly proportional to their coefficients in the balanced equation. This can be shown in the following example: 2 H2 (g) + O2 (g) → 2 H2O (g) Gas Volume Stoichiometry 2 H2 (g) + O2 (g) → 2 H2O (g) If the above reaction is carried out on: Therefore, the volumes of the gases H2, O2 and H2O are 1.5 : 0.75 : 1.5, which is directly proportional to their coefficients 2 : 1 : 2. 1.5 liters H2 (25oC, 2 atm) by ideal gas law n = PV / RT = (2) (1.5) / (0.0821) (298oK) = 0.1226 mol H2 mol O2 = 1/2 x 0.1226 = 0.0613 mol by ideal gas law V = nRT / P = (0.0613)(0.0821)(298)/2 = 0.75 liter mol H2O = 2/2 x 0.1226 = 0.1226 mol by ideal gas law V = nRT / P = (0.1226)(0.0821)(298)/2 = 1.5 liter CaCO3 (s) + 2 HCl (l) → CaCl2 (s) + CO2 (g) + H2O (l) If the above reaction is carried out on 28.5 g of CaCO3 at 25oC and 1 atm, what amount of HCl will be required and what amounts of CaCl2, CO2, and H2O will be produced? The answers will be determined in terms of moles, grams, and volumes (for any gases). (MW = 100) (MW = 36.5) (MW = 111) (MW = 44) (MW = 18) CaCO3 (s) + 2 HCl (l) → CaCl2 (s) + CO2 (g) + H2O (l) 28.5 grams 20.81 grams 31.64 grams 12.54 grams 5.13 grams 28.5gCaCO3X1mol CaCO3/100 g/mol CaCO3X2mol HCL/1mol CaCO3x36.5 gHCl/1molHCl 0.285 mol → 2/1 x 0.285 mol → 1/1 x 0.285 mol → 1/1 x 0.285 mol → 1/1 x 0.285 mol ↓ V = nRT / P = (0.285)(0.0821)(298)/1 = 6.97 liter Law of Partial Pressures - Gas Mixtures When two or more gases exist within a common container, each gas fills the entire container and exerts the same pressure it would exert if it were the only gas in the container. This results in a law known as the law of partial pressures,which states that the total pressure of gases in a container is equal to the sum of the partial pressures of the individual gases present in the container expressed by the following equation: PT = P1 + P2 + P3 ......... where PT = the total pressure of gases in the container and P1... equal the partial pressures of each individual gas. Another term called mole fraction is defined as: (Mole fraction of gas 1) = X1 = n1 / nT Since the individual partial pressures are equal by the ideal gas law to: P1 = (n1) (R x T) / V Since the factor RT / V is constant for all gases in the container, the partial pressure is proportional to moles; therefore, the mole fraction equals the partial pressure divided by the total pressure. n1 / nT = P1 / PT also P1 = n1 / nT (PT) also P1 = X1 (PT) Another useful term is mole percent, which expresses the percentage of a component relative to the total: Mole % = (100) X1 = 100 (n1 / nT) A mixture of three gases consists of 5.00 moles of He, 4.00 moles of H2, 3.00 moles of CO2, and 8.00 moles of Ar. The total pressure of the mixture is 1800 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. XHe = 5.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.250 XH2 = 4.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.200 XCO2 = 3.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.150 XAr = 8.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.400 Mole%He = 100(XHe) = (100) 0.250 = 25.0% Mole%H2 = 100(XH2) = (100) 0.200 = 20.00% Mole%CO2 = 100(XCO2) = (100) 0.150 = 15.00% Mole%Ar = 100(XAr) = (100) 0.400 = 40.00% PHe = XHe (1800 mm) = 0.250 (1800 mm) = 450 mm PH2 = XH2 (1800 mm) = 0.200 (1800 mm) = 360 mm PCO2 = XCO2 (1800 mm) = 0.150 (1800 mm) = 270 mm PAr = XAr (1800 mm) = 0.400 (1800 mm) = 720 mm GAS VOLUME STOICHIOMETRY PROBLEMS The combustion of ethanol (C2H5OH) takes place by the following reaction equation. C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g) What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25oC and 1.25 atm? V = CO2? 23.3 grams of O2 gas Mole of O2 = 23.3g/ 32 g/mol = 0.728125 mol O2 3 mol O2 gas produce 2 mol CO2 0.728125 mol O2 X 2 mol CO2 / 3 mol O2 = 0.4854 mol CO2 T = 25 oC + 273 = 298 oK P = 1.25 atm R=0.0821 V=nRT/P V= 0.4854 mol CO2 X 0.0821 X298/1.25 = 9.5 Liter GAS VOLUME STOICHIOMETRY PROBLEMS Acetic acid (CH3COOH) is formed from its elements by the following reaction equation: C (graphite) + 2 H2 (g) + O2 (g) → CH3COOH (g) What is the volume of acetic acid CH3COOH gas produced by the reaction of 18.6 grams of H2 gas at 35oC and 1.05 atm? 18.6 grams of H2 gas at 35oC and 1.05 atm What Volume of CH3COOH ? 18.6 g H2 / 2 g /mol = 9.3 mol H2 2 mol H2 gas produce 1 mol CH3COOH (g) 9.3 mol H2X 1 mol CH3COOH (g) / 2 mol H2 = 4.65 mol CH3COOH V=nRT/P V= 4.65 X0.0821X (35+273) / 1.05 = 111.98 L GAS VOLUME STOICHIOMETRY PROBLEMS The formation of hydrazine (N2H4) from its elements takes place by the following reaction equation. N2 (g) + 2 H2 (g) → N2H4 (g) What are the volumes of N2 gas and H2 gas required to form 28.5 grams of N2H4 at 30oC and 1.50 atm? 28.5 grams of N2H4 at 30oC and 1.50 atm volumes of N2 gas and H2 gas = ?? 28.5 g N2H4 X 1 mol N2H4/ 32.06g N2H4 /mol = 0.88896 mol N2H4 0.88896 mol N2H4 number of moles of N2 required = 0.88896mol N2 number of moles of H2 required = 0.88896mol × 2 = 1.7779 mol H2 V=nRT/P VN2 = 0.88896mol× 0.0821× 303.15K/1.50atm = 14.74L VH2 = 1.7779mol × 0.0821 × 303.15/1.50 atm = 29.48L GAS VOLUME STOICHIOMETRY PROBLEMS The combustion of benzene (C6H6) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (g) What is the volume of CO2 gas formed by the combustion of 18.5 grams of C6H6 at 30oC and 1.50 atm? the volume of CO2 gas? 18.5 grams of C6H6 at 30oC and 1.50 atm 18.5 g C6H6 X 1 mol C6H6 / 78.11g/mol = 0.236845 mol C6H6 2 C6H6 produces 12 CO2 0.236845 mol C6H6 X 12 CO2 / 2 C6H6 = 1.42107 mol CO2 V=nRT/P V = 1.4207 mol CO2 X 0.0821 X (30+273 oK)/1.50 atm = 23.6 Liter of CO2 PARTIAL PRESSURE - MOLE FRACTION PROBLEMS A mixture of gases consists of 4.00 moles of He, 2.00 moles of H2, 3.00 moles of CO2 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. XHe = 4.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2857 Mole%He = 100(XHe) = (100) 0.286 = 28.57% XH2 = 2.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.1429 Mole%H2 = 100(XH2) = (100) 0.143 = 14.29% XCO2 = 3.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.2143 Mole%CO2 = 100(XCO2) = (100) 0.214 = 21.43% XAr = 5.00 / (4.00 + 2.00 + 3.00 + 5.00) = 0.3571 Mole%Ar = 100(XAr) = (100) 0.357 = 35.71% PHe = XHe (2900 mm) = 0.2857 (2900 mm) = 828.53 mm PH2 = XH2 (2900 mm) = 0.1429 (2900 mm) = 414.41 mm PCO2 = XCO2 (2900 mm) = 0.2143 (2900 mm) = 621.47 mm PAr = XAr (2900 mm) = 0.3571 (2900 mm) = 1035.59 mm PARTIAL PRESSURE - MOLE FRACTION PROBLEMS Determine the total pressure of a mixture of 0.400 mole of He and 0.600 mole of Ne in a 2.00 liter container at 25oC. P total = nRT/V = (0.4+0.6) X0.0821X (25+273)/ 2L = 12.23 atm PARTIAL PRESSURE - MOLE FRACTION PROBLEMS Determine the mass of Ar in 1.00 liter of a gas mixture at 25oC which contains 0.300 atm of Ne and has a total pressure of 4.00 atm. P total = 4.00 atm PNe = 0.300 atm PAr = PT - PNe = 4.00 -0.300 = 3.7 atm n (Ar) = PV/RT = 3.7 X 1.00 / 0.0821 X (25+273) = 0.1512 mol Ar Mass or Ar = 0.1512 mol Ar X 39.95 g Ar/ mol = 6.040 g PARTIAL PRESSURE - MOLE FRACTION PROBLEMS A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne in the flask. What is the total pressure? nHe = gHe / (MWHe) = (0.52 mg x 1 g / 1000 mg) / 4.002 = 0. mol nNe = gNe / (MWNe) = (2.05 mg x 1 g / 1000 mg) / 20.18 = 0. mol PT = n R T / V = (0. + 0.) (0.0821) (298oK) / 1.00 = 0.005664 atm PT = 0.005664 atm x 760 mm/1 atm = 4.305 mm XHe = 0. / (0. + 0.) = 0.5612 XNe = 0. / (0. + 0.) = 0.4388 PHe = XHe (4.305 mm) = 0.5612 (4.305 mm) = 2.416 mm PNe = XNe (4.305 mm) = 0.4388 (4.305 mm) = 1.889 mm 3.7: COLLECTION OF GASES OVER WATER & DIFFUSION and EFFUSION OF GASES Collection of Gases over Water Gases are commonly collected by displacement of water from a volume-calibrated water filled container (as shown in the diagram below). When this is done, the gas mixes with water vapor in the container and produces a gas-water vapor mixture that can be studied by the partial pressure/mole fraction principles. The partial pressure of the water vapor (water gas) present in the container can be determined from a readily available table of water vapor pressures (shown below): A sample of methane (CH4) gas is collected over water at 26oC and 745 mm. The volume of the gas collected is 55.5 ml. How many moles of CH4 gas has been collected? How many grams of CH4 gas has been collected? PCH4 = 745 - P H2O (from table) = 745 - 25.2 (water vapor pressure at 26o =25.2 mm) = 719.8 x 1 atm / 760 mm = 0.947 atm CH4 from Ideal Gas Law: n CH4 = PV / RT = (0.947 atm) (55.5 ml x 1 liter / 1000 ml) / (0.0821) (299oK) n CH4 = 0.00214 moles grams CH4 = moles x MW = 0.00214 moles x 16.042 grams / 1 mole = 0.0343 grams Diffusion and Effusion of Gases Effusion is the process by which a gas escapes through a small opening into a vacuum. Diffusion is the process by which a gas spreads out through a space, which may be a vacuum or occupied by another gas to occupy the space uniformly. Graham's Law Both of these processes are described by a relationship know as Graham's Law, which is described by the following equation where r1 / r2 equals the relative rates of effusion and MW is the molecular weight of each gas. Note that the rate of effusion is inversely proportional to the molecular weights of the gases. This means that the rate of effusion is faster for a gas with a lower molecular weight. The rate of effusion of ammonia gas (NH3) is 2.45 times faster than that of an unknown gas. What is the molecular weight of the unknown gas? (rNH3 / runknown)2 = MWunknown / MWNH3 (2.45)2 = MWunknown / 17 MWunknown = (2.45)2 x 17 = 102 Law of Conservation of Matter matter cannot be created or destroyed Reactant Substances undergoing the reaction Products Substances that are generated by the reaction Solvent Component of solution that has a greater concentration Solution Homogenous mixture of two more substances Solute Component of solution that has a lesser concentration What trades do solutions exhibit? The homogenous, the components of a solution are dispersed on a molecular scale, the dissolve salute, and a solution will not settle out or separate from solvent What happens when an ionic compound is mixed with water the different ions of the compound are pulled apart , or dissolved, by the water molecules. ion-dipole attraction electrostatic attraction between an ion and a polar molecule Electrolyte Substances that produce ions in solution when dissolved Non-electrolytes substances that form no ions in water and cannot conduct electricity Strong electrolytes If the physical or chemical process generates is 100% efficient Weak electrolytes Small fraction of dissolve, substance undergoes I am productivity process Precipitation reaction Went to aqua solutions are mixed to form an insoluble product Endothermic (of a chemical reaction or compound) occurring or formed with absorption of heat Are ionic compounds endothermic Yes, Ionic compounds with stronger bonds require greater amount of energy to separate anions and cations (higher melting point) Dissolving Breaking apart, ionic lattice, but also requires saluation Saluation Energy releasing process as solvent molecules, surround and interact with an ions and ions Hydration solution process with water as the solvent Precipitation reaction Dissolve substances, reactive, form, one or more solid product. They are often double replacement reactions. Exothermic reactions Chemical reactions that release a significant amount of heat. Soluation energy Energy of attraction between a cat ion or an ions and water Solubility The maximum concentration of a substance that can be achieved under specific conditions Precipitate Occurs when a solution conditions are such that concentration exceeds solubility Is the compound soluble if it contains an alkali, metal ion, ammonium, nitrate, Perchlorate, chlorate, or acetate Yes is a compound containing chlorine bromine or iodine soluble? A compound containing chlorine bromine and iodine is soluble as long as it does NOT contain Ag^+, Hg2^2-, Pb^2+ Is a compound containing sulfate soluble A compound containing SO4^2- is soluble as long as it does NOT contain Ca^2+, Sr^2+, Ba^2+,Pb^2+ Is a compound containing hydroxide soluble A compound containing hydroxide is soluble as long as it contains an alkali metal ion or NH4+ What would mixed with hydroxide is slightly soluble Sr(OH)2, Ba(OH)2, Ca(OH)2 When a compound contains carbonate and phosphate is its soluble A compound containing PO4^3- or CO3^2- is soluble if it contains an alkali metal ion or NH4^+ If a compound contains chromate is it soluble A compound containing CrO4^2- is is soluble as long as it contains an alkali metal ion, NH4^+, or Mg^2+ If a compound contains sulfur with a negative 2 charge is it soluble If it contains an alkali metal ion or ammonium, it is soluble What when mixed with sulfur is slightly soluble? MgS, CaS, BaS Generally speaking, what do insoluble compounds contain Carbonate, chromate, phosphate, fluoride, sulfide, hydroxide Generally speaking, what do soluble compounds contain Group one metal cations, ammonium ion, halide, ions, acetate, nitrate, chloride, perchlorate, and sulfate Overall equation Bouncy equation that doesn't explicitly represent the ionic species that are present Complete ionic equation Equation that represents all dissolved ion results Spectator ions Ions that do not take part in a chemical reaction and are found in solution both before and after the reaction Not ionic equation Chemical equation in which only these dissolved ionic reactants and products that undergo chemical or physical change are represented Acid base reaction One in which a proton is transferred from one chemical species to another When an acid is dissolved in water, what does it generate? H3O^- and anion What is produced when a base is dissolved in water? Hydroxide and cation Strong acids do what when added to water Completely ionizes What are strong acids? Hydrobromic acid, hydrochloric acid, hydroiodic acid, perchloric acid, sulfuric acid, nitric acid Sulfuric Acid H2SO4 (Strong Acid) Perchloric Acid HClO4 (Strong Acid) Nitric Acid HNO3 (Strong Acid) hydrobromic acid HBr hydrochloric acid HCl (Strong Acid) Hydroiodic Acid HI (Strong Acid) carbonic acid H2CO3 What happens when you mix a weak acid with water? They partially ionize and leave a majority of dissolved acid intact, and only generate a small number of hydronium ions What are weak the acids? acetic acid, carbonic acid, phosphoric acid acetic acid CH3COOH Phosphoric Acid H3PO4 What happens when a strong base is added to water It will completely ionize What are the strong bases? LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 LiOH lithium hydroxide (strong base) NaOH sodium hydroxide (strong base) RbOH rubidium hydroxide Ca(OH)2 calcium hydroxide (strong base) Ba(OH)2 barium hydroxide (strong base) Sr(OH)2 strontium hydroxide (strong base) What happens when a weak base is added to water It partially ionizes and leaves a majority of the dissolved base intact and only generates a small number of hydroxide ions What are the weak bases? Ammonia (NH3) neutralization reaction a reaction in which an acid and a base react in an aqueous solution to produce a salt and water Gas forming reaction with metal carbonate When metal carbonates react with acids to produce an aqueous salt, water, and CO2 gas Base a compound that produces hydroxide ions in solution Gas forming reaction with metal sulfites Metal sulfides react with acids to produce an aqueous salt, water, and sulfur dioxide gas Acid compound that forms hydrogen ions (H+) in solution acid-base reaction reaction involving the transfer of a hydrogen ion between reactant species neutralization reaction a reaction in which an acid and a base react in an aqueous solution to produce a salt and water Salt An ionic compound made from the neutralization of an acid with a base. Acid pH less than 7 Base on pH scale above 7 Solvent Present in greatest abundance Solute Substance being dissolved (present in lesser quantities) Electrolyte Any substance that disassociates into ions were dissolved into a suitable medium, and thus forms, a conductor of electricity Solvation/hydration energy Energy of attraction between ions and water. Energy is released when the water interacts with the ions making it exothermic. Lattice energy Energy of attractions between cat ions and anions in salt lattice When something is soluble salt, what is the relation relation to solvation energy, and lattice energy? Solvation energy lattice energy Ion H2O interactions ion ion lattice energy When something is insoluble salt, what is the relation between solvation energy and lattice energy? Solvation energy lattice energy Ion H2O interaction ion ion lattice energy precipitation reaction equation AB(aq) + CD(aq)-- AD(aq) + CB(aq) What is the oxidization number for a pure element? 0 What is the oxidation number of hydrogen when reacting with non-alkali metals +1 What is the oxidization number for monatomic ions? The oxidation number is equal to the charge What is fluorine's oxidization number when it is combined with another element -1 What is the oxidation number of hydrogen when it is reacting with alkaline metals? -1 What is the oxidation number of oxygen except when combined with fluorine and peroxides -2 What is oxygen oxidation number when with fluorine +1 What is oxygen oxidation number one with hydrogen? +1 What is the oxidation number of chlorine bromine and iodine except when combined with oxygen or fluorine? -1 What must the sum of oxidization numbers be equal to? The charge of a neutral compound or polyatomic ion When solid, alkaline metals are added to water what are the products? An aqueous salt, hydrogen gas, and energy When acid is added to solid metal, what are the products? Salt (aq), hydrogen gas, and energy Combustion reaction Vigorous redox reaction producing a significant amount of energy and form of heat, and sometimes light Half reaction Un equation that shows whether each reactant loses or gains electrons in reaction Oxidation Process and wedge in elements oxidation number is increased by loss of electrons Oxidation reduction reaction Also known as a redox reaction, is a reaction involving a change in oxidation number for one or more reactant elements Oxidation number Charge each atom of an element would have in compound if it were ionic Oxidation agent Substance that brings about the oxidation of another substance, and becomes reduced in the process Reduction Process in which an elements oxidation number is decreased by gain of electrons Reducing agent Substance that brings out the reduction of another substance, and becomes oxidized Single displacement reaction A redox reaction involving the oxidation of an elemental substance by an ionic species What do redox reactions involve and yield? Redox reaction involve transfer electrons between reactant species to yield ionic products Which medals are easiest to oxidize, strongest reducing agents, and react with cold water, replacing the hydrogen Lithium, potassium, barium, strontium, calcium, sodium Which metals react with steam, but not cold water replacing the hydrogen Magnesium aluminum, zinc, chromium, iron, and cadmium (2nd best to oxidize) Which elements do not react with water, but react with acids replacing hydrogen Cobalt, nickel, lead, tin (3rd best to oxidize) What will an acid and a carbonate form Salt + H2O(l) + CO2(g) Which metal elements are most difficult to oxidize and are the weakest reducing agents that are unreactive with water or acids Copper, silver, mercury, gold, platinum Which reactions form gas Acid + carbonate Acid + sulfite Acid + sulfide What will an acid and sulfite form Salt + H2O(l) + SO2(g) What will acid and sulfide produce Salt + H2O (l) + H2S(g) Sulfite SO3 2- Sulfide S 2- Carbonate CO3 2- neutralization reaction a reaction in which an acid and a base react in an aqueous solution to produce a salt and water What happens when you add heat to a metal carbonate It forms CO2 gas and metal oxide What happens when you add heat to a metal hydroxide You get water (g) and a metal oxide When you have an active metal and react itbwith an acid what will you always get H2 gas and a metal compound What does carbonic acid break down to in normal atmospheric pressure H2CO3 - H2O + CO2 Na2CO3 + HCl - NaCl + CO2 + H2O What ever is oxidized is the — and whatever is reduced is the — Reducing agent Oxidizing agent reducing agent The electron donor in a redox reaction. oxidizing agent The electron acceptor in a redox reaction. When dealing with oxidation/reduction/single replacement reactions what do you need to include in work Oxidation numbers, say what is oxidized, say what is reduced, what's reducing agent, and what's oxidation agent When dealing with precipitation reactions what do you need to include Net ionic equation Oxidized - electrons Reduced - electrons Lost Gains what does Columbs law protentional energy e