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Portage Learning Chemistry 103 Practice Exams Updated 2023- 2024/ Chemistry 103 Final Exam, Exams of Chemistry

Portage Learning Chemistry 103 Practice Exams Updated 2023- 2024/ Chemistry 103 Final Exam Latest Version

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2022/2023

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Portage Learning Chemistry 103 Practice Exams Updated 2023-

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  1. 3.1: THERMOCHEMISTRY: Thermodynamics is the study of the relationship between heat and other forms of energy, particularly mechanical work. Thermo- chemistry 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.
  2. System: the object (or substance) being studied
  3. Open system: a system that permits the transfer of mass and energy with the surroundings
  4. Closed system: a system that permits the transfer of energy but not mass with the surroundings
  5. Isolated system: a system that does not permit the transfer of energy or mass with the surroundings
  6. Surroundings: the rest of the universe interacting with the system
  7. Energy: the potential or capacity to move matter: the ability to do work (unit is J = joule)
  8. Work: the amount of energy transferred by a force acting through a distance

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

  1. Kinetic energy: the energy possessed by an object by virtue of its motion (unit is J = joule)
  2. Potential energy: the energy possessed by an object by virtue of its position (unit is J = joule)
  3. Heat (q): the thermal energy transferred between system and surroundings due to a difference in temperature between them (unit is J = joule)
  4. Enthalpy: the total energy of a system
  5. Heat of reaction: (”H) the amount of heat (q) gained or lost during a chemical reaction
  6. Exothermic: a reaction with a - ”H
  7. Endothermic: a reaction with a + ”H(absorb heat)
  8. 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.
  9. Temperature change calorimetry: measures the thermal energy change oc- curring 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 sys- tem. This is demonstrated below by adding a 15.6 gram piece of aluminum (heated

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

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. 6496.44 = 204.8144 Tmix Tmix = 6496.44 / 204.8144 = 31.7oC
  1. Phase change calorimetry: measures the energy change occurring as a sub- stance 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: qsi = mass x Heat of Fusion = m x Hfusion Phase Changes of liquid to gas or gas to liquid: qlg = mass x Heat of Vaporization = m x Hvapor
  2. 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. qlg = m x Hvapor = 200 g x 2.26 kJ/g = 452 kJ (since heat is added) = + 452 kJ

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

  1. 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?: qls = m x Hfusion = 200 g x 0.334 kJ/g = 66.8 kJ (since heat is removed/given off) = - 66.8 kJ
  2. 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(lH) 2O(s)) Hvapor = 2.26 kJ/g (H2O(lH) 2O(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) 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 qsl = m x Hfusion = 20 g x 0.334 kJ/g = + 6.68 kJ 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 water at 100oC ’ steam at 100oC qgl = 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. kJ

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

  1. 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.: qlg = m x Hvapor = 150 g x 2.26 kJ/g = 339 kJ (since heat is removed) = - 339 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.: qls = m x Hfusion = 120 g x 0.334 kJ/g = 40.08 kJ (since heat is added) = + 40.08 kJ
  3. 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 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 qlg = 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 qls = 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. kJ
  4. 3.2: THERMOCHEMICAL EQUATIONS & MEASURING HEATS OF REAC- TIONS: 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 NH gas and 91.8 kJ (kilojoules) of heat is given off. N2 (g) + 3 H2 (g) ’ 2 NH3 (g)”Hrx = - 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 ”His reversed

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

(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 H” moles/original moles). times the factor (new moles refers to substance for which mass data is given in the problem q = ”Hrx x new moles (moles given in problem) / original moles (moles given in thermochemical equation)

  1. The heat involved in the reaction of 51 grams NH3 to produce N2 and H is calculated below (note that this is the reverse of the original reaction):: N (g) + 3 H2 (g) ’ 2 NH3 (g)”Hrx = - 91.8 kJ 2 NH3 (g) ’ 3 H2 (g) + N2 (g) ”Hrx is for 2 mole of NH reaction uses 51 g NH3 = 51/17 mole = 3 mole NH q = + 91.8 kJ x 3 mole NH3 / 2 mole NH3 = + 137.7 kJ
  2. 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.
  1. The heat (given off by the reaction) and absorbed by the water in the reac- tion 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 ”t= 4.18 J / g / oK x 400 g x 5.0oK =
  • 8360 J q calorimeter = heat capacity x ”t= 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

Portage Learning Chemistry 103 Practice Exams Updated 2023-

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H” reaction = - 10.06 kJ / (0.200 mol / 1 mole) = - 50.3 kJ / mole 1 HCl + 1 NaOH ’ 1 NaCl + 1 H2O ”H= - 50.3 kJ / mole

29. 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 ’ CO c (water) = 4.184 J/g oC q water = s x mass x ”t= 4.18 J / g / oK x 2000 g x 1.6oK = - 13376 J q calorimeter = heat capacity x ”t= 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. kJ new moles = 1.00 g / 12 = 0.08333 mole ”Hreaction = - 46.976 kJ / (0.08333 / 1) = - 563.7 kJ / mole C + O2 ’ CO ”H= - 563.7 kJ / mole **30. 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)”H = - 1170 kJ If 26.5 g of NH3 is reacted with excess O2, what will be the amount of heat given off?:** ”Hrx is for 4 mole of NH reaction uses 26.5 g NH3 = 26.5/17 = 1.56 mole NH q = - 1170 kJ x 1.56 mole NH3 / 4 mole NH3) = - 456.3 kJ
  2. **Thermochemical Equation Problems
  3. Sulfur undergoes combustion to yield sulfur trioxide by the following reac- tion equation: 2 S + 3 O2 ’ 2 SO3”H = - 792 kJ If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?:** ”Hrx is for 2 mole of S

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2024 / Chemistry 103 Final Exam Latest Version

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

  1. **Thermochemical Equation Problems
  2. Methane (CH4) reacts with Cl2 to yield CCl4 and HCl by the following reaction equation: CH4 + 4 Cl2 ’ CCl4 + 4 HCl What is the ”H of the reaction if 51.3 g of CH4 reacts with excess Cl2 to yield 1387.6 kJ?:** ”Hrx is for 1 mole of CH4; q = - 1387.6kJ reaction uses 51.3 g CH4 = 51.3/16.042 mole = 3.198 mole CH
  • 1387.6 kJ = ”Hrx x 3.198 mole CH4 / 1 mole CH ”Hrx = - 433.9 kJ
  1. **Thermochemical Equation Problems
  2. 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 ”H of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ?:** ”Hrx 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 = ”Hrx x 0.7686 mole H2S / 2 mole H2S ”Hrx = - 1123.6 kJ
  1. **Measuring Heat of Reaction Problems
  2. 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 ”H 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 ”t= (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

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”H = q / (new moles / original moles) ”H= - 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)

  1. **Measuring Heat of Reaction Problems
  2. 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 ”H 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 ”t= (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 C6H ”H= q / (new moles / original moles) ”H= - 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)
  3. **Measuring Heat of Reaction Problems
  4. 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 calorime- ter = 200 J/oK) increases from 25.00 oC to 25.97oC, (1) what is the ”H 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

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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. M = 0.0100 moles q water = s (specific heat of water) x mass x ”t= 4.18 J / g / oK x 200 g x 0.97oC = - 810.92 J q calorimeter = heat capacity x ”t= 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) ”Hreaction = 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

  1. **Measuring Heat of Reaction Problems
  2. 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 calorime- ter = 150 J/oC) decreases from 25.00oC to 19.86oC, (1) what is the ”H 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 q water = s (specific heat of water) x mass x t” = 4.18 J / g / oK x 100 g x 5.14oK = 2148.52 J q calorimeter = heat capacity x ”t= 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. kJ (1) ”Hreaction = 2.92 kJ / 14.5 g NaHCO3 / 84 g NaHCO3 / mol NaHCO3 = + 16. kJ / mol (2) The calorimeter + contents (no lid) = open system, (3) Endothermic process

Portage Learning Chemistry 103 Practice Exams Updated 2023-

2024 / Chemistry 103 Final Exam Latest Version

38. 3.3: HESS'S LAW & USING HEATS OF FORMATION TO DETERMINE HEATS

OF REACTION: Enthalpy change (”Hrxn) for a chemical reaction is independent of the path by which the products are obtained. Therefore, the ”Hfor an overall reaction which can be written as the sum of two or more steps equals the sum of the ”Hs for the individual steps. This is known as Hess's law.

39. Hess's law can be used to obtain H” for chemical reactions for which ”H cannot be directly measured because of some limitation. For instance, the Hrxn 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 ”Hrxn for the synthesis of tungsten carbide as shown below:: W (s) + C (graphite) ’ WC (s)”H 2 W (s) + 3 O2 (g) ’ 2 WO3 (s)”H = - 1680.6 kJ C (graphite) + O2 (g) ’ CO2 (g)”H = - 393.5 kJ 2 WC (s) + 5 O2 (g) ’ 2 WO3 (s) + 2 CO2 (g)”H = - 2391.6 kJ rxn =? 40. Now by combining these three equations as follows, we can obtain the H” for the synthesis of tungsten carbide. To calculate a ”Hrxn 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 H”

  • 1680.6 kJ) of that reaction.: 1/2 (2 W (s) + 3 O2 (g) ’ 2 WO3 (s)”H = C (graphite) + O2 (g) ’ CO2 (g)”H = - 393.5 kJ 1/2 (2 WO3 (s) + 2 CO2 (g) ’ 2 WC (s) + 5 O2 (g)”H = +2391.6 kJ) W (s) + C (graphite) ’ WC (s)”Hrxn = - 38.0 kJ H” rxn = 1/2 (-1680.6 kJ) + (-393.5 kJ) + 1/2 (+2391.6 kJ) = - 38.0 kJ
  1. Using Heats of Formation to Determine Heats of Reaction: Heats of forma- tion 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)”H =? Using the data from the Standard Enthalpy of Formation Table: CH4 (g) ’ C (graphite) + 2 H2 (g) - ”Hf0 (CH4) = - (- 74.6 kJ/mole) C (graphite) + O2 ’ CO2 (g)”Hf0 (CO2) = (-393.5 kJ/mole) 2 H2 (g) + O2 (g) ’ 2 H2O (g) 2”Hf0 (H2O) = 2 x (- 241.8 J/mole)

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge CH4 + 2 O2 ’ CO2 + 2 H2OHrxn = 74.6 + (- 393.5) + 2 (- 241.8) ”Hrxn = - 802.5 kJ/mole

  1. Hrxn = £n H” f0 (products) - £m ”Hf0 (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:
  2. 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 £indicates that the enthalpies of formation should be summed. Notice that the H” 0 f 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: ”Hrxn = ”Hf0 (CO2) + 2 H” f0 (H2O) - ”Hf0 (CH4) - ”Hf0 (O2) Hrxn = - 393.5 + 2 (-241.8) - (-74.6) + 2 (0) = - 802.5 kJ/mole **44. Hess' Law Problems
  3. 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 (”Hrxn) for this reaction by using the following thermochemical data: N2 (g) + O2 (g) ’ 2 NO (g)”H = 180.6 kJ N2 (g) + 3 H2 (g) ’ 2 NH3 (g)H” = - 91.8 kJ 2 H2 (g) + O2 (g) ’ 2 H2O (g)”H = - 483.7 kJ:** 1) N2 (g) + O2 (g) ’ 2 NO (g)H” kJ 2N2 (g) + 2O2 (g) ’ 4NO (g)”H = 180.6 kJ x2 = 361.2 kJ

= 180.

  1. N2 (g) + 3 H2 (g) ’2 NH3 (g)H” = - 91.8 kJ 4 NH3 (g) ’2 N2 (g)+ 6H2 (g)”H = - 91.8 kJ x 2x (- 1 - Reverse reaction) = 183.6 kJ
  2. 2 H2 (g)+O2 (g)’2 H2O (g)”H = - 483.7 kJ 3X (2 H2 (g)+O2 (g))’ 3x2 H2O (g)H” = - 483.7 kJ x 3 = - 1451.10 kJ 6 H2 (g)+ 3O2 (g)’ 6 H2O (g)

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 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) ”Hrxn = 361.2 kJ + 183.6 kJ - 1451.10 kJ = - 906.3 kJ

  1. **Heat of Formation Problems
  2. Determine the heat of reaction (”Hrxn) 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. 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.
  4. Combined Gas Law: P1V1/T1=P2V2/T
  5. 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.
  6. 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 = V 750 mm/760 = 0.987 atm = P 25oC + 273 = 298oK = T 780 mm/760 = 1.03 atm = P 50oC + 273 = 323oK = T

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge

  1. 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. 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. V = 6.41 liters
  2. 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. 1.35 liters = V 30oC + 273 = 303oK = T P x (1.35) = (0.0595) x (0.0821) x (303) P x (1.35) = 1. P = = 1.10 atm
  3. 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:

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 0.002139402 = 0.003292683 x Vf Vf = 0.002139402 / 0.003292683= 0.650 liter

  1. 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. P= (0.03680.0821293)/1. P= 0.868 atm
  2. 3.6: GAS VOLUME & LAW OF PARTIAL PRESSURES: Gas Volume Stoichiom- etry 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) 55. Gas Volume Stoichiometry 2 H2 (g) + O2 (g) ’ 2 H2O (g): If the above reaction is carried out on: 1.5 liters H2 (25oC, 2 atm) by ideal gas law n = PV / RT = (2) (1.5) / (0.0821) (298oK) = 0.1226 mol H 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 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. 56. 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

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 28.5gCaCO3X1mol CaCO3/100 g/mol CaCO3X2mol HCL/1mol CaCO3x36. 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

  1. 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 + P .........
  2. 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
  3. Since the individual partial pressures are equal by the ideal gas law to:: P = (n1) (R x T) / V
  4. 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 compo- nent relative to the total: Mole % = (100) X1 = 100 (n1 / nT) 61. 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. XH2 = 4.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0. XCO2 = 3.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0. XAr = 8.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0. Mole%He = 100(XHe) = (100) 0.250 = 25.0% Mole%H2 = 100(XH2) = (100) 0.200 = 20.00%

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 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

62. 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 O 3 mol O2 gas produce 2 mol CO 0.728125 mol O2 X 2 mol CO2 / 3 mol O2 = 0.4854 mol CO T = 25 oC + 273 = 298 oK P = 1.25 atm R=0. V=nRT/P V= 0.4854 mol CO2 X 0.0821 X298/1.25 = 9.5 Liter 63. 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 H 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 64. GAS VOLUME STOICHIOMETRY PROBLEMS

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 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 N2H 0.88896 mol N2H number of moles of N2 required = 0.88896mol N number of moles of H2 required = 0.88896mol × 2 = 1.7779 mol H 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

65. 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 C6H 2 C6H6 produces 12 CO 0.236845 mol C6H6 X 12 CO2 / 2 C6H6 = 1.42107 mol CO V=nRT/P V = 1.4207 mol CO2 X 0.0821 X (30+273 oK)/1.50 atm = 23.6 Liter of CO 66. 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. = 35.71%

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge 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

67. 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 68. 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 69. 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 = 0.000129935 mol nNe = gNe / (MWNe) = (2.05 mg x 1 g / 1000 mg) / 20.18 = 0.000101586 mol PT = n R T / V = (0.000129935 + 0.000101586) (0.0821) (298oK) / 1.00 = 0. atm PT = 0.005664 atm x 760 mm/1 atm = 4.305 mm XHe = 0.000129935 / (0.000129935 + 0.000101586) = 0. XNe = 0.000101586 / (0.000129935 + 0.000101586) = 0. 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

  1. 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):

Portage Chemistry 103 Module 3 Exam Study Study online at https://quizlet.com/_8d9ge nship know / r2 equals as. Note that the gases. cular weight.

  1. 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?: PCH = 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 CH 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
  2. 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.
  3. Graham's Law: Both of these processes are described by a relatio as Graham's Law, which is described by the following equation where r the relative rates of effusion and MW is the molecular weight of each g the rate of effusion is inversely proportional to the molecular weights of This means that the rate of effusion is faster for a gas with a lower mole
  4. 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 / MWNH (2.45)2 = MWunknown / 17 MWunknown = (2.45)2 x 17 = 102