CHEM 103 MODULE 3 NOTES.

CHEM 103 MODULE 3 NOTES. 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. In order to adequately discuss thermochemistry, we need to define some common terms. 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 You must be able to use the terms above to describe a thermochemical system. 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. We will use the term calorimetry to refer particularly to measuring energy changes that accompany temperature and physical change (state change or phase change) processes. 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 The specific heat data for most substances is known and can be found in data tables. The specific heat value for liquid water (not steam or ice) is 4.184 J/g oC and for aluminum is 0.899 J/g oC. In the example given, the hot Al transfers heat to the colder water until the two substances reach the same temperature. This can be predicted by setting the heat lost by the Al equal to the heat gained by the water as in the equation below: (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 This final temperature appears to make sense since the value should be between 100oC and 26.7oC, but closer to 26.7oC since a greater amount of water than Al was used. Incidentally, the actual experimental final temperature of the mixture is only 31.4oC since the coffee cup, cover, and thermometer have absorbed some thermal energy. This absorption of thermal energy by the calorimeter can be corrected for by determining the heat capacity of the calorimeter. 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 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. 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? ql↔g = m x ∆Hvapor = 200 g x 2.26 kJ/g = 452 kJ (since heat is added) = + 452 kJ 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? ql↔s = m x ∆Hfusion = 200 g x 0.334 kJ/g = 66.8 kJ (since heat is removed/given off) = - 66.8 kJ Calorimetry Values for Water ∆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 How much heat is involved to convert of 20 grams of ice at -10oC to 20 grams of steam at 120 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 ql↔s = 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 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 3.2 : THERMOCHEMICAL EQUATIONS & MEASURING HEATS OF REACTIONS Thermochemical Equations 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 ΔHrx = -91.8 kJ – number is negative because it’s an exothermic reaction meaning heat is given off Heat of reaction (ΔHrx) is the amount of heat given off (exothermic) or absorbed (endothermic) – in example - when one mole of N2 reacts with 3 moles of H2 to give two moles of 2 NH3 Problems are given either where they give you the ΔHrx like done in this example and asking you to figure out q (the amount of heat that’s given off for a particular instance where you have certain amounts of each substance being reacted) OR you are going to be given the heat that’s given off and asked to calculate the heat of reaction (given data to get q or in problem will be given q and asked to solve for the heat of reaction or you will be given the heat of reaction and asked to solve for q) 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 a. number is negative when it’s an exothermic reaction meaning heat is given off b. number is positive when its an endothermic reaction meaning heat is absorbed (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). a. moles refers to substance for which mass data is given in the problem q (heat) = ΔHrx x new moles (moles given in problem) / original moles (moles given in thermochemical equation – balanced equation) Applying these rules to the equations above gives the following results: The heat (q) involved (did not say given off or absorbed here – involved meaning you need to figure out of the heat if absorbed or given off) in the reaction of 51 grams NH3 (new moles) to produce N2 and H2 is calculated below (note that this is the reverse of the original reaction – reverse sign of the ΔHrx): 2 NH3 (g) → 3 H2 (g) + N2 (g) ΔHrx = -91.8 kJ (given above) so now the heat of reaction will be. ΔHrx = +91.8 kJ ΔHrx is for 2 mole of NH3 reaction uses 51 g NH3 = 51/17 mole = 3 mole NH3 q (heat) = ΔHrx x new moles (moles given in problem) / original moles (moles given in thermochemical equation – balanced equation) - to get the original moles you divide the molecular weight of NH3  so 14.00 for N and 1.008 for H gives you about 17 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). 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 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 3.3 : HESS’S LAW & USING HEATS OF FORMATION TO DETERMINE HEATS OF REACTION Hess's Law 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) ΔH 0 (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 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: ΔHrxn = Σ n ΔHf0 (products) - Σ m ΔH 0 (reactants) 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 = ΔH 0 (CO ) + 2 ΔH 0 (H O) - ΔH 0 (CH ) - ΔH 0 (O ) ΔHrxn = -393.5 + 2 (-241.8) - (-74.6) + 2 (0) = - 802.5 kJ/mole Standard Enthalpies of Formation (ΔHf) (at 25oC) Formula ΔH0 (kJ/mol) f Formula ΔH0 (kJ/mol) f AlCl3 (s) - 704.2 PbCl2 (s) - 359.4 Al2O3 (s) - 1675.7 PbS (s) - 100.4 PbSO4 (s) - 920.0 BaCl2 (s) - 855.0 BaCO3 (s) - 1213.0 LiF (s) - 616.0 BaSO4 (s) - 1473.2 LiCl (s) - 408.6 Br- (g) - 218.9 MgF2 (s) - 1124.2 Br- (aq) - 120.9 MgCl2 (s) - 641.3 Br2 (g) + 30.9 MgO (s) - 601.6 CaCO3 (s) - 1207.6 MnO2 (s) - 520.0 CaCl2 (s) - 795.4 CaO (s) - 634.9 NH3 (g) - 45.9 CaSO4 (s) - 1434.5 N2H4 (l) + 50.6 NH4Cl (s) - 314.4 CH4 (g) - 74.6 NO (g) + 91.3 C2H2 (g) + 227.4 NO2 (g) + 33.2 C2H4 (g) + 52.4 N2O (g) + 81.6 C2H6 (g) - 84.0 N2O4 (g) + 11.1 C3H8 (g) - 103.8 C6H6 (l) + 49.1 O3 (g) + 142.7 C2H5OH (l) - 277.6 CH3CHO (g) - 166 KBr (s) - 393.8 CO (g) - 110.5 KCl (s) - 436.5 CO2 (g) - 393.5 KClO3 (s) - 397.7 CCl4 (g) - 96 KF (s) - 567.3 KMnO4 (s) - 837.2 HBr (g) - 36.3 KOH (s) - 424.6 HCl (g) - 92.3 K2SO4 (s) - 1437.8 HF (g) - 273.3 HI (g) + 26.5 AgBr (s) - 100.4 H2O (l) - 285.8 AgCl (s) - 127.0 H2O (g) - 241.8 AgI (s) - 61.8 H2S (g) - 20.6 AgNO3 (s) - 124 H2SO4 (l) - 814.0 HNO3 (aq) - 173.2 NaBr (s) - 361.1 NaCl (s) - 411.2 FeCl2 (s) - 341.8 Na2CO3 (s) - 1130.7 FeCl3 (s) - 399.5 NaHCO3 (s) - 950.8 Fe2O3 (s) - 824.2 NaOH (s) - 425.6 Fe3O4 (s) - 1118.4 Na2SO4 (s) - 13.87.1 SO2 (g) - 296.8 SnCl4 (l) - 511.3 SO3 (g) - 395.7 ZnCl2 (s) - 415.1 ZnO (s) - 350.5 3.5: GAS LAWS Gases are often formed as a product of or are used as a reactant in a chemical reaction. These materials (gases) have some special properties and are subject to some special calculations in chemistry. Gases (think of air as an example) are very light, easily expandable, easily compressible, and completely fill a container into which they are released. These properties are due to the fact that the individual molecules of the gas are very far apart from one another, not attracted to one another, and in constant, very rapid motion. This description of gases is known as the kinetic-molecular theory of gases. Due to these properties, the volume of a sample of gas will expand (increase) if the temperature of the gas sample is increased and the volume of a sample of gas will contract (decrease) if the temperature of the gas sample is decreased. The temperature scale used with gases is the Kelvin (oK) scale, which is oC + 273. The volume scale used with gases is the liter scale, which is ml/1000. Also, the volume of a sample of gas will expand (increase) if the pressure on the gas sample is decreased and the volume of a sample of gas will contract (decrease) if the pressure on the gas sample is increased. The pressure unit we will use with gases is atmosphere (atm) scale, which is mm/760. Also, the pressure of a sample of gas will increase if the temperature of the gas sample is increased and the pressure of a sample of gas will decrease if the temperature of the gas sample is decreased. Combined Gas Law These facts can be expressed as a mathematical equation called the combined gas law, which has the following form: P i × V i T i = P f × V f T f This equation states that if the initial pressure (atm) of a gas is multiplied by its initial volume (liters), divided by its initial temperature (oK), and a change is made in the pressure or the temperature causing a corresponding change in the volume, then final pressure (atm) multiplied by the final volume (liters) and divided by the final temperature (oK) will still equal the same numerical value. This equation makes it possible to calculate one of these values if the other five are known. This equation should be used to do any gas problem in which two temperatures, volumes, or pressures are given or asked for. If one of the quantities (pressure, volume, or temperature) remains constant or is not given, that quantity is dropped from both sides of the equation. Ideal Gas Law There is another very important gas law called the ideal gas law that relates the pressure, volume, and temperature of a gas to the number of moles of gas present (or grams since moles = grams/molecular weight). The ideal gas law can be written in either of the following forms: 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. This equation makes it possible to calculate the number of moles (or grams) of gas present if the pressure, volume, and temperature are known. These equations should be used to do any gas problem in which grams or moles are given or asked for. EXAMPLES: (A) 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? Two pressures, two temperatures, one volume given, one volume asked for, so we use: P i × V i T i = P f × V f T f 530 ml/1000 = 0.530 liters = Vi 750 mm/760 = 0.987 atm = Pi 780 mm/760 = 1.03 atm = Pf 25oC + 273 = 298oK = Ti 50oC + 273 = 323oK = Tf Put in the data: (0.987) x (0.530) = (1.03) x Vf: (298) (323) Solve for Vf: 0. = 0. x Vf: Vf: = 0. = 0.550 liter 0. (B) A gas sample has an original volume of 850 ml when collected at 1.00 atm and 28oC. If a change is made in the gas pressure that causes the volume of the gas sample to become 1.25 liters at 50oC, what is the new pressure? Two volumes, two temperatures, one pressure given, one pressure asked for, so we use: P i × V i T i = P f × V f T f 850 ml/1000 = 0.850 liters = Vi 1.00 atm = Pi 1.25 liters = Vf 28oC + 273 = 301oK = Ti 50oC + 273 = 323oK = Tf Put in the data: (1.00) x (0.850) = Pf x (1.25) (301) (323) Solve for Pf: 0. = 0. x Pf: Pf: = 0. = 0.73 atm 0. (C) A gas sample has an original volume of 650 ml when collected at 730 mm and 20oC. If a change is made in the gas temperature that causes the volume of the gas sample to become 840 ml at 1.05 atm, what is the new temperature? Two volumes, two pressures, one temperature given, one temperature asked for, so we use: P i × V i T i = P f × V f T f 650 ml/1000 = 0.650 liters = Vi 730 mm/760 = 0.961 atm = Pi 840 ml/1000 = 0.840 liters = Vf 20oC + 273 = 293oK = Ti 1.05 atm = Pf Put in the data: (0.961) x (0.650) = (1.05) x (0.840) (293) Tf Solve for Tf: 0.62465 x Tf = 258.426 Tf: = 258.426 = 414oK (By cross-multiplication) 0.62465 (D) 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 = 6.158 = 6.41 liters 0.961 (E) 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 = 1.480 = 1.10 atm 1.35 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) 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 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 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. Now let's look at how grams, moles, and volumes are related in stoichiometric calculations by use of the following reaction: 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 ↓ ↑ ↑ ↑ ↑ 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 (298)/1 = 6.97 liter ↓ V = nRT / P = (0.285)(0.0821) 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 Mole%He = 100(XHe) = (100) 0.250 = 25.0% XH2 = 4.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.200 Mole%H2 = 100(XH2) = (100) 0.200 = 20.00% XCO2 = 3.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.150 Mole%CO2 = 100(XCO2) = (100) 0.150 = 15.00% XAr = 8.00 / (5.00 + 4.00 + 3.00 + 8.00) = 0.400 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 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): Table of Water Vapor Pressures Temp (oC) Vapor Pressure Temp (oC) Vapor Pressure 0o 18.7 mm 1o 19.8 mm 2o 21.1 mm 3o 22.4 mm 4o 23.8 mm 5o 25.2 mm 6o 26.8 mm 7o 28.4 mm 8o 30.1 mm 9o 31.8 mm 10o 42.2 mm 11o 9.8 mm 40o 55.4 mm 12o 10.5 mm 45o 71.9 mm 13o 11.2 mm 50o 92.6 mm 14o 118.0 mm 12.0 mm 55o 15o 149.5 mm 12.8 mm 60o 16o 187.5 mm 13.6 mm 65o 17o 233.7 mm 14.5 mm 70o 18o 289.1 mm 15.5 mm 75o 19o 355.1 mm 16.5 mm 80o 20o 433.6 mm 17.5 mm 85o 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 = 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. 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. or 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