Transportation in Canada: Theme 2 PDF

Theme 2 Transportation 1  Canada's transportation system includes  > 1,400,000 kilometres of roads  10 major international airports  300 smaller airports  72,093 km of functioning railway track  > 300 commercial ports and harbours that provide access to the Pacific, Atl...

Theme 2 Transportation 1  Canada's transportation system includes  > 1,400,000 kilometres of roads  10 major international airports  300 smaller airports  72,093 km of functioning railway track  > 300 commercial ports and harbours that provide access to the Pacific, Atlantic and Arctic oceans as well as the Great Lakes and the St. Lawrence Seaway. In 2005, the transportation sector made up 4.2% of Canada's GDP, compared to 3.7% for Canada's mining and oil and gas extraction industries. 1. "Transportation in Canada". Statistics Canada. Archived 2. An Analysis of the Transportation Sector in 2005". from the original on April 10, 2008. Retrieved March Statistics Canada. Archived from the original on April 9, 26, 2008. 2008. Retrieved March 27, 2008. 2 In 2009, there were 20,706,616 road vehicles registered in Canada These vehicles travelled a total of 333.29 billion kilometres. Canada's vehicles consumed a total of 31.4 million cubic metres of gasoline and 9.91 million cubic metres of diesel. "Canadian Vehicle Survey: Annual" (PDF). Statistics Canada. 2009. Archived from the original on August 20, 2012. Retrieved January 13, 2011. 3  Carbon dioxide (CO₂) emissions from the transportation sector in Canada from 1970 to 2022 In 2022, transportation in Canada released nearly 180 million metric tons of carbon dioxide (MtCO₂) into the atmosphere. This was an increase of 7.3 percent in comparison to 2021 levels. 4 https://www.statista.com/statistics/1290621/transportation-emissions-in-canada/#:~:text=Transportation%20sector%20CO%E2%82%82%20emissions%20in%20 Canada %201970%2D2022&text=In%202022%2C%20transportation%20in%20Canada,in%20comparison%20to%202021%20levels. Carbon dioxide emissions from fossil fuel combustion in Canada from 1960 to 2022, by type(in million metric tons) Oil and gas are the main contributors these days https://www.statista.com/statistics/1198353/ carbon-dioxide-emissions-from-fossil-fuels-in- 5 So where does all that CO2 come from? And how can we predict what happens 5, 10 or 50 years from now? We are going to look at transportation from two very different angles, and very different forms of Chemistry Internal Electric combustion vehicles 6  At the heart of the internal combustion engine is the reaction of a hydrocarbon with oxygen. Internal combustion Gasoline is a mixture of hydrocarbons. One of the main components is 2,2,4-trimethylpentane, or simply isooctane. (CH3)3CCH2CH(CH3)2 7  C8H18 (CH3)3CCH2CH(CH3)2 H HH H HH H HH H C C C CC H CH3 C C C C C C HH33C C3C H C C C CC C C C C C C CH3 H33C H C3C H C H C H H H H H H C HH H H H C C C C C H HH C HH C H H H H H 8  The combustion of isooctane C8H18 + O2 → ? C8H18 + O2 → CO2 + H2O Does this look right? No. The reaction is not BALANCED 9  Balancing Chemical Equations Law of conservation of mass: mass is neither created nor destroyed. The same number and kind of atoms must be present before and after a chemical reaction. C8H18 + O2 CO2 + H2O Before reaction After reaction 8 mole of C 1 mole of C 2 moles of O 3 moles of O 18 moles of H 2 moles of H Overall: 1 mole O created, 7 mole C and 16 moles H destroyed! VIOLATES law of conservation of mass! EQUATION IS NOT BALANCED 10 Balancing Chemical Equations C8H18 + 12.5 O2 8 CO2 + 9 H2O Before reaction After reaction 8 mole of C 8 mole of C 25 moles of O 25 moles of O 18 moles of H 18 moles of H BALANCED! Practice balancing chemical equations! 11 The Combustion of a Carbon-Hydrogen-Oxygen Compound. Liquid triethylene glycol, C6H14O4, is used as a solvent and plasticizer for vinyl and polyurethane plastics. Write a balanced chemical equation for its complete combustion. C6H14O4 + 15 2 O2 → 6CO2 + H 7 2O 1. Balance C. 2. Balance H. 3. Balance O. 4. If you want, multiply by two, but the solution above is perfectly ok. 2 C6H14O4 + 15 O2 → 12 CO2 + 14 H2O and check all elements. 12  STOICHIOMETRY The study of the quantitative aspects of chemical reactions The quantity of product formed from a reaction depends upon the amount of reactant available. 13 Example – how much O2 is needed to combust 1 mole C8H18? - how much CO2 made for 1 mol of C8H18 combusted? C8H18 +12.5 O2 8 → CO 92 + H2O Thus, we need 12.5 moles of O2 to combust 1 mole of C8H18 We make 8 moles of CO2 for every mole of C8H18 combusted. 14  Limiting Reagents Often there is not enough of one reagent to completely consume the other reagent The consumed reagent is the limiting reagent. The leftover reagent is called excess. WARNING: You can not determine the limiting reagent just by looking at the stoichiometric coefficients!! Limiting Reagent Problem: What mass of H2O is produced when 1 g C8H18 is combined with 5 g O2? 15  Molar Mass (Molecular Weight) Molar Mass: mass in grams of 1 mole (6.022 x 1023 atoms or molecules) of a substance molar mass of 12C = 12.0000 g exactly (defined) In nature, elements occur as a mixture of their isotopes. The molar mass of an element is the mass of 1 mole of the naturally occurring mixture of isotopes. 1 mole of O = 15.9994 g (all isotopes) 1 mole of H = 1.0079 g 1 mole of C = 12.011 g Molar mass of an element = Mass Number for the Element (including all naturally occurring (as shown on the periodic table) isotopes) 16 Conversion Between Mass, Moles, and Atoms How many moles of Al are in a 10.0 g sample? con. factor 10.0 g Al x 1 mole Al = 0.371 mol Al atoms 26.98 g Al or simply: 10.0 g Al = 0.371 mol Al atoms 26.98 g/mol How many atoms of Al are in this sample? con. factor 0.371 mol Al atoms x 6.022 x 10 atoms = 2.23 x 1023 atoms 23 1 mole of atoms 17  Molecular Mass Molecular Mass: the average mass (in amu) of one molecule. molecular sum of atomic mass of each mass = atom in the molecule (in amu) Determine the molecular mass of H2O: 2 atoms H x 1.008 amu = 2.016 amu 1 H atom 1 atom O x 15.999 amu = 15.999 amu 1 O atom molecular mass 1 molecule H2O = 2.016 amu + 15.999 amu = 18.015 amu 18 Limiting Reagent Problem: What mass of H2O is produced when 1 g C8H18 is combined with 5 g O2? Step 1: Write the balanced chemical equation C8H18 + 12.5 O2 → 8 CO2 + 9 H2O Step 2: Determine the limiting reagent Step 2a. Determine how many moles of each reagent are available. 1.0 g C8H18 x 1 mol C8H18 = 8.8 x 10-3 mol C8H18 114.1 g C8H18 5.0 g O2 x 1 mole O2 = 1.6 x 10-1 moles O2 32 g O2 19 Step 2b: Use stoichiometric factors from the balanced chemical equation to determine how many moles of one reagent are needed to consume the other: C8H18 + 12.5 O2 → 8 CO2 + 9 H2O 8.8 x 10-3 mol C8H18 x 12.5 mole O2 = 1.1 x 10-1 mol O2 needed 1 mole C8H18 to consume all the octane Step 2c: Compare moles available to moles needed Have 1.6 x 10-1 moles O2 Need 1.1 x 10-1 moles O2 to consume all of the C8H18 There will be excess O2 C8H18 is the limiting reagent 20 Step 3: Convert moles of limiting reagent to moles of product using the appropriate stoichiometric factor (obtained from the balanced chemical equation): 8.8 x 10-3 mol C8H18 x 9 mole H2O = 7.9 x 10-2 mol H2O 1 mole C8H18 Step 4: Convert moles of product to grams of product 7.9 x 10-2 mol H2O x 18.0 g H2O = 1.4 g H O 2 1 mole H2O 21  Limiting Reagent Problem: What mass of O2 is left unreacted? In step 2c above we determined: Have 1.6 x 10-1 moles O2 but only need 1.1 x 10-1 moles O2 to consume all of the C8H18. This means 0.5 x 10-1 moles O2 are left over. 5 x 10-2 moles O2 x 32 g O2 = 1.6 g O2 left 1 mol O2 22 Example – Compare the mass of CO2 generated by combusting 1 gram isooctane to 1 gram methane C8H18 + 12.5 O2 → 8 CO2 + 9 H2O 1.0 g C8H18 x 1 mol C8H18 = 8.8 x 10-3 mol C8H18 114.1 g C8H18 8.8 x 10-3 mol C8H18 x 8 mol CO2 = 7.0 x 10-2 mol CO2 1 mol C8H18 7.0 x 10-2 mol CO2 x 44 g CO2 = 3.1 grams CO 2 1 mol CO2 23 CH4 + 2 O2 → 1 CO2 + 2 H 2O 1.0 g CH4 x 1 mol CH4 = 6.3 x 10-2 mol CH4 16 g CH4 6.3 x 10-2 mol CH4 x 1 mol CO2 = 6.3 x 10-2 mol CO2 1 mol CH4 6.3 x 10-2 mol CO2 x 44 g CO2 = 2.8 grams CO 2 1 mol CO2 24 All of the species involved (isooctane and O2) and produced are gases inside the cylinder head Q. How do you measure the mass (grams) of a gas??? A. You don’t – you measure pressure. 25 Gases Gases have three characteristics we need to be able to accommodate in any theory of gases we use: Uniformly fill any container Mix completely with any other gas Exert pressure on its surroundings 26 Pressure of Gases Pressure is defined as the force applied per unit area: Pressure = force area Common Units of Pressure: Pascals, Pa = Newton/meter2 = N/m2 (1 Newton = 1 kg m/s2) Standard Atmospheric Pressure = 100 kPa = 1 bar 1 atm = 101.325 kPa = 760 mm Hg 27  Manometers For determining the pressure of an enclosed gas If the liquid level on both sides of the manometer is the same, Pgas = Pbar Pgas = Pressure of gas in chamber Pbar = Barometric (or atmospheric) pressure 28  ca 1 atm thermocouple ionization ~ 10-3 atm ~ 10-9 atm 29 Gas Laws The physical behavior of “ideal gases” can be described by a set of laws: Boyle’s Law: V  1/P (T = constant) Charles’s Law: V  T (P = constant) Avogadro’s Hypothesis: Van T and P are constant These laws were discovered independently, and eventually combined into one equation of state: The Ideal Gas Law: PV = nRT 30  Ideal Gas Law PV = nRT R = proportionality constant R = proportionality constant (called the gas constant) = 0.08206 L atm K-1 mol-1 = 8.314 J K-1 mol-1 P = pressure in atm P = pressure in Pa V = volume in Litres V = volume in m3 n = moles n = moles T = temperature in Kelvin T = temperature in Kelvin This function holds closely at P < 1 atm The ideal gas law is an EQUATION OF STATE for a gas. It relates the initial and final states of a system, no matter how the change is carried out. 31  Limiting Reagent Problem: What mass of H2O is produced when 1 kPa C8H18 is combined with 5 kPa O2 in a 1L cylinder head at 500oC. It is still all about getting to moles PV = nRT Moles C8H18 = PV/RT = 1000 Pa (10-3 m3)/(8.314 JK-1mol-1)(773.15 K) T= 500C = 773.15 K = 1.56 x 10-4 moles 1 L = 10-3 m3 1 kPa = 1000 Pa Moles O2 = PV/RT Since I converted to Si units, = 5000 Pa (10-3 m3)/(8.314 JK-1mol-1)(773.15 K) I will use R = 8.314 J K-1 mol- = 7.78 x 10-4 moles 1 32  12.5 mole O2 = 1.95 x 10-3 mol O needed to 1.56 x 10-4 mol C8H18 x 2 1 mole C8H18 consume all the octane ve 7.78 x 10-4 moles O2. Need 1.95 x 10-3 moles O2 to consume all of the C8H18 ere is not enough O2 to consume all C8H18. Thus, O2 is the limiting reagent Since now O2 is limiting, we use its moles to determine the amount of product C8H18 + 12.5 O2 → 8 CO2 + 9 H2O 7.78 x 10-4 mol O2 x 9 mole H2O= 5.6 x 10-4 mol H2O 12.5 mole O2 5.6 x 10-4 mol H2O x 18.0 g H2O = 1.0 x 10-2 g H O 2 1 mole H2O 33  Gas Density Gas density can be calculated if the gas’s molar mass (MM) is known (and vice versa): m (m = mass) PV nRT MM  n (n = moles) PV m n n RT MM m PV  MM RT Which of the m P( MM ) balloons on the  left contains a V RT gas of greater P( MM ) molar mass? d RT 34  Gas Density Example When considering the fuel to use in a rocket engine, the density of the propellant is a key piece of information. Denser fuels = more mass per unit volume. This has two side effects. More mass per unit volume means the fuel tanks weigh more and thus more thrust is needed to lift the rocket from the ground. However, more mass can mean greater Force (remember F=ma) coming out of the rocket nozzle propelling the rocket up. 35  H2 + ½ O2 → H2O Density liquid H2 = 70.85 kg/m3 Density of gaseous H2O = 0.8 kg/m3 C8H18 + 12.5 O2 → 8 CO2 + 9 H2O Density liquid C8H18 = 0.703 g/ml = 703 kg/m3 Density of gaseous H2O = 0.8 kg/m3 Density of gaseous CO2 = 2 kg/m3 36  Mixtures of Gases For a mixture of gases in a container, the overall pressure is the sum of all the partial pressures of the individual components. PTotal = P1 + P2 + P3 +... Dalton’s Law of Partial Pressure Note that this implies all of the partial pressures are lower than the overall pressure. 37  In a 1.50 L flask at 25oC, 1.10 g of H2O2 is completely decomposed to water and oxygen gas according to the following equation: 2 H2O2(l)  2 H2O(g) + O2(g) What are the pressures of each component, and what is the final pressure of the vessel? Step 1: Calculate moles of H2O2 and then moles of O2 and H2O. n H2O2 = 1.1 g H2O2 x 1 mole H2O2 = 0.032 moles H2O2 n = 0.032 moles H 34.01 O x g1 H 2O2 O = 0.016 mol mole O2 2 2 2 O2 2 moles H O nH2O = 0.032 moles H2O2 x 2 moles 2H22O = 0.032 mol H2O 38 2 moles H O Step 2: Calculate partial pressures from n, R, T, and V PV = nRT P = nRT/V PO2 = 0.016 mol (0.0821 Latm/molK)(298.15 K) 1.5 L PO2 = 0.26 atm PH2O = 0.032 mol (0.0821 Latm/molK)(298.15 K) 1.5 L PH2O = 0.52 atm Step 3: Calculate total pressure using Dalton’s Law Ptotal = P(H2O) + P(O2) = 0.52 atm + 0.26 atm = 0.78 atm 39  A gaseous mixture contains 0.38 moles of nitrogen gas and 0.45 moles of oxygen gas. Determine the mole fractions of oxygen and nitrogen. Solution: 0.38 mol nitrogen + 0.45 mole oxygen = 0.83 moles total moles oxygen 0.45 XOxygen = = = 0.54 moles oxygen + moles nitrogen 0.83 moles nitrogen 0.38 = 0.46 XNitrogen = = moles oxygen + moles nitrogen 0.83 Mole Fraction has NO UNITS Double check answer: 0.54 + 0.46 = 1.00 40  Determining Partial Pressure from Mole Fraction, X moles A XA = S (moles of all species in sample) PA = nART Ptotal = ntotalRT V V nART PA = V nA Therefore: = = XA Ptotal ntotalRT ntotal V PA = Ptotal XA 41 Kinetic Molecular Theory The Ideal Gas Law is empirical (based on observations of macroscopic behavior of gases). But WHY do gases behave the way they do? KMT is a theoretical model to explain these macroscopic behaviors based on microscopic properties of atoms and molecules in the gas. 42 Fundamental assumptions of KMT: 1. Gas molecules are in constant, random, straight-line motion 2. Volume of individual particles is approximately zero. 3. Collisions of the particles with container walls cause pressure exerted by gas. 4. Particles exert no forces on each other except during collisions. (they do not attract or repel each other) 5. The average kinetic energy of the gas is proportional to the temperature of the gas in degrees Kelvin. 43 Using KMT to Determine a Pressure-Volume Relationship for a Gas in a Container Pressure exerted by the gas is proportional to: 1) the momentum with which gas particles hit the wall where momentum, p = mu (mass times speed) 2) the rate of collisions with the wall where collision rate a (#molecules)/(container volume) and collision rate a speed Therefore, we can express pressure as: P a p (collision rate) P a (m u) (u N/V) where N = #molecules P a Nmu2 and V = container V volume 44 P a Nmu2 for 3-D P = Nmu2 V 3V Pressure –volume relationship u2 is the mean square speed: the average of the square of speeds of the particles u12 + u22 + u32 + … + un2 u2 = n 45  KMT and Ideal Gas Law KMT Ideal Gas Law P = Nmu2 P = nRT 3V V (we can show from statistical mechanics that u2 = 3RT which makes both expressions the same) MM KE of an individual particle = ½mu2 Average KE for 1 mole of particles = NA½mu2 = (3/2)RT thus: NAmu2= 3RT P = Nmu2 = 3RT/3V = RT/V (for 1 mole) 3V 46  why u2 ? At a given temperature, not all molecules have the same speed. Particles travel faster when they are lighter and at higher temperatures 47 Distribution of Molecular Speeds um = most probable speed uav= average speed urms= root mean square speed 48 Diffusion and Effusion DIFFUSION is the gradual EFFUSION is the escape mixing of molecules of of a gas from its container different gases. through a small hole. The rate of both of these are proportional to the root mean square of the speeds of the particles involved. 49 Graham’s Law Lighter gases diffuse and effuse more rapidly: 3RT 2 Rate A u A MM A   RateB u 2B 3RT MM B Rate A MMB  Rate B MMA Rate H2 = 28 g/mol H2 effuses 3.7 times faster = 3.7 than N2 Rate N2 2 g/mol 50 Deviations From the Ideal Gas Law There are several problems with the assumptions made by KMT, and this results in non-ideal behavior of gases 1. Real molecules have volume. 2. There are intermolecular forces between gas particles (otherwise a gas could not become a liquid). This will change the pressure. Gases will exhibit behavior closest to ideal behavior at high temperatures and low pressures. In other conditions, the ideal gas law must be corrected to account for non-ideal behavior 51  Van der Waal’s Equation Measured P Measured V 2  n a  P  2 V  nb nRT  V  Volume correction Correction for intermolecular J. van der forces Waals 1837- 1923 Every gas has its own a and b terms, these are determined by experimental detail. 52  Non-Ideal Gas Example The empirical correction factors for The empirical correction factors for Cl2 gas are: O2 gas are: a = 6.49, b = 0.0562 (empirical) a = 1.382, b = 0.032 (empirical) For 8.0 mol Cl2 in a 4.0 L tank at For 8.0 mol O2 in a 1.0 L cylinder at 27 oC. 800 oC. P (ideal) = nRT/V = 49.3 atm P (ideal) = nRT/V = 705 atm P (van der Waals) = 29.5 atm P (van der Waals) = 858 atm Big Difference!! Big Difference!! 53  Remember this? CH4 + 2 O2 → 1 CO2 + 2 H 2O 1.0 g CH4 x 1 mol CH4 = 6.3 x 10-2 mol CH x 1 mol CO2 = 6.3 x 10-2 mol CO 4 2 16 g CH4 1 mol CH 4 6.3 x 10-2 mol CO2 x 44 g CO2 = 2.8 grams CO 2 1 mol CO2 C8H18 + 12.5 O2 → 8 CO2 + 9 H2O 1.0 g C8H18 x 1 mol C8H18 = 8.8 x 10-3 mol C8H18 x 8 mol CO2 = 7.0 x 10-2 mol CO 2 114.1 g C8H18 1 mol C8H18 7.0 x 10-2 mol CO2 x 44 g CO2 = 3.1 grams CO 2 1 mol CO2 Is this the whole story? 54  Thermochemistry e energy changes involved in chemical reactions CHEMICAL ENERGY The capacity to do work or to produce heat. Burning peanuts supply sufficient energy to boil a cup of water. 55 Potential and Kinetic Energy Energy can be classified as either kinetic or potential Potential Energy: Energy available to a system due to the position or composition of the system e.g. a stationary ball on top of a hill Kinetic: Energy available due to the motion of an object e.g. a ball rolling down a hill KE = 1/2 mv2 (m = mass, v = velocity) Kinetic and Potential energy can be interconverted 56 First Law of Thermodynamics: The total energy of the universe is constant. What does this mean? It means we can predict energy changes. Both hikers have experienced the same net increase in potential energy independent of their paths of ascent. A state function is a property of a system that can be defined by the current state of the system and is INDEPENDENT of the pathway by which that state was achieved. e.g. energy, pressure, volume, temperature are state functions Changes in State are represented by “D”: DV, DP, DE DE = Efinal - Einitial 57 Universe = System + Surroundings Chemists evaluate energy changes from the perspective of the system: Energy flowing into a system DE >0 Energy flowing out of a system DE 0 qsystem < 0 DH > 0 DH < 0 ENDOTHERMIC EXOTHERMIC Heat flows into the system Heat flows out of the system 67 Reaction Enthalpy or Heat of Reaction 2H2 (g) + O2 (g) 2H2O (l) + 242 kJ/mol EXOTHERMIC DHrxn = - 242 kJ/mole 2 H2O (l) + 242 kJ/mole 2H2 (g) + O2 (g) ENDOTHERMIC DHrxn = + 242 kJ/mole Reversing a chemical equation will reverse the sign of the enthalpy for the reaction. 68  Determining Reaction Enthalpy There are several ways to determine the enthalpy (or heat) of a reaction: 1. Experimentally (Calorimetry) 2. Use Hess’s Law to determine DH for a reaction based on known DH values of a series of reactions. 3. Calculate DH from the enthalpies of formation of molecules and elements in the reaction by using the equation: DH0rxn = S DH0f (products) - S DH0f (reactants) 4. Using Quantum Mechanics We will look at #1-3 in detail. 69 Experimental Determination of Reaction Enthalpy: Calorimetry 1. Constant-Volume Calorimetry 2. Constant-Pressure Calorimetry 70  1) A “bomb” containing the sample is immersed in a water bath. 2) The sample is ignited 3) The heat produced by combustion of the sample is absorbed by the water and the bomb. 4) The heat of the combustion of the sample is determined from the change in water and bomb temperature. 71  Heat of Reaction from Constant- Volume Bomb Calorimetry No heat enters or leaves the system because it is well insulated: qsystem = 0 Therefore, qwater + qbomb + qrxn = 0 Heat of reaction: qrxn = - (qwater + qbomb) At this point we need to know how much the T will increase for a given amount of heat absorbed = Cp 72  Heat vs. Temperature Heat and temperature are different: TEMPERATURE: A property that reflects the motions of particles in a substance. HEAT: The transfer of energy between two objects due to a temperature difference. Heat is NOT a substance contained by an object (though we often speak of heat as if this were true), it IS a way in which energy is exchanged between a system and its surroundings. 73 1 g of water must absorb 4.18 J of energy (in the form of heat) to increase the temperature by 1oC 25oC 26oC 4.18 J Heat 1 g water 1 g water 1 g of ethanol must absorb 2.46 J of energy (in the form of heat) to increase the temperature by 1oC 25oC 26oC 2.46 J Heat 1 g ethanol 1 g ethanol 74  Heat Capacity, C Heat Capacity, C: Heat energy required to raise the temperature of a given quantity of substance by a given temperature. Heat absorbed (J) C= DT (oC or K) Where DT = change in temperature = Tfinal - Tinitial 75 Specific and Molar Heat Capacity The energy required to raise the temperature of a substance depends upon the amount of the substance. Specific Heat Capacity: Heat Capacity per gram Units: (J/oC g or J/K g) Specific Heat Capacity = Heat absorbed (J)o Mass (g) DT (K or C) Molar Heat Capacity: Heat Capacity per mole Units: (J/oC mole or J/K mole) Molar heat capacity = Heat absorbed (J) moles DT (K or oC) 76  The energy needed to increase the temperature of 1 gram of substance by 1oC. Substance Specific Heat (J/g K) H2O (l) 4.184 H2O (s) 2.03 Ethanol (l) 2.46 Aluminum 0.902 Glass 0.84 A large amount of energy is needed to increase the temperature of water! 77 If 25.0 g of Al cools from 310 oC to 37 oC, how many joules of heat energy are lost by the Al? heat loss = q = (C)(mass)(DT) q = (0.902 J/g oC)(25.0 g)(37 - 310)oC q = - 6160 J 78 What is the heat capacity of 200 g Te Tellurium? Use 4 J g-1 oC-1 (100 oC) for the heat capacity of water. 100 g H2O 50 oC  q Te = q H2O 25 oC 1. 0.5 J g-1 oC-1 2. 1 J g-1 oC-1 3. 2 J g-1 oC-1 4. 4 J g-1 oC-1 5. 10 J g-1 oC-1 we heat 200 g Te up to 100C. This Te then transfers enough heat to 100 g water to raise the Temp by 25C. To do this, 25C x 4 J/g/C x 100 g = 10000 J are transfered. Thus, 200 g Te @ 100 C cools to 50 C by releasing 10000 J heat. 10000J/200g/50C = 1 J/gC 79 You eat a bite of pizza containing 1 g of crust and 1 g of cheese, which are in thermal equilibrium at 56oC. Upon entering your mouth, the temperature of the cheese and crust quickly cool to 42oC. How much energy is released from the cheese and from the crust if their specific heat capacities are 3.5 J/g oC and 2.5 J/g oC respectively? qcheese = (3.5 J/g oC)(1.0 g)(42oC - 56oC) = -49 J qcrust = (2.5 J/g oC)(1.0 g)(42oC - 56oC) = -35 J The cheese releases more energy (in the form of heat) than does the crust. 80  Heat and Changes of State Energy is absorbed or released when a substance undergoes a change in physical state, even if the temperature remains the same. Heat (or Enthalpy) of fusion (solid to liquid): H2O (s) (0oC) H2O (l) (0oC) q = 6.01 kJ/mol Heat (or Enthalpy) of vaporization (liquid to vapor): H2O (l) (100oC) H2O (g) (100oC) q = 40.8 kJ/mol Reversing the process results in a reverse in the sign of q. H2O (l) (0oC) H2O (s) (0oC) q = -6.01 kJ/mol H2O (g) (100oC) H2O (l) (100oC) q = -40.8 kJ/mol 81  Calculate the heat change upon vaporizing 3.0 g of water at 100oC. H2O (l) (100oC)  H2O (g) (100oC) qvap = 40.8 kJ/mol qvap = (40.8 kJ/mol)(# moles water) 3.0 g H2O (s) x 1 mole H2O = 0.17 moles H2O 18.0 g H2O qvap = (40.8 kJ/mol)(0.17 moles H2O) = 6.94 kJ 6.94 kJ of heat is consumed when 3.0 g of water is vaporized at 100oC 82  What quantity of heat is required to melt 500 g (27.8 moles) of ice at 0oC and heat the water to steam at 100oC? Specific heat of water = 4.18 J/g K Heat of fusion of ice = 6.01 kJ/mole Heat of vaporization = 40.8 kJ/mole The total heat needed for this process will be the sum of: 1. Heat required to melt the ice 2. Heat required to raise the water temperature from 0oC to 100oC 3. Heat required to evaporate the water at 100oC qtotal = qfusion + q(100-0) + qvaporization 83 1. Melt the ice: q = qfusion x # moles =(6.01 kJ/mole)(27.8 mol) = 167 kJ 2. Raise the water temperature from 0oC to 100oC: q = (C)(mass)(DT) q = (4.18 J/g K) (500 g)(100-0)K = 209 kJ 3. Evaporate the water at 100oC: q = qvaporization x # moles = (40.8 kJ/mol)(27.8 moles) = 1134 kJ 4. Total heat energy = 167 kJ + 209 kJ + 1134 kJ = 1510 kJ 84  Heat of Reaction from Constant- Volume Bomb Calorimetry, No heat enters or leavesrevisited the system because it is well insulated: qsystem = 0 Therefore, qwater + qbomb + qrxn = 0 Heat of reaction: qrxn = - (qwater + qbomb) qwater = (Cwater)(masswater)(DTwater) = (4.184 J/g oC)(masswater)(DT) qbomb = Cbomb (DT) (note, we measure Cbomb as a whole and don’t report it per mass) Because DV = 0, w = 0 so qrxn  DHrxn, but rather DUrxn 85 A 0.1375 g sample of C8H18 is burned in a constant-volume bomb calorimeter that has a heat capacity of 1769 J/oC. The calorimeter contains exactly 300 g of water and the temperature increases by 1.126 oC. Calculate the heat given off by the burning octane in kJ/gram. qrxn = - (qwater + qbomb) Step 1: Calculate heat transferred from reaction to water: qwater = (Cwater)(masswater)(DTwater) = (4.184 J/g oC)(300g)(1.126oC) = 1413 J or 1.413 kJ heat transferred to water 86 Step 2: Calculate heat transferred from reaction to bomb: qbomb = Cbomb (DT) = (1769 J/oC) (1.126 oC) = 1992 J or 1.992 kJ Step 3: Calculate total heat evolved from the reaction: qrxn = - (qwater + qbomb) = - (1.413 kJ + 1.992 kJ) = - 3.405 kJ Step 4: Calculate heat evolved per gram: -3.405 kJ/0.1375 g = -24.8 kJ/g 87  Pressure = Atmospheric Pressure Since pressure is constant, DHrxn = qrxn Used to determine DH of non- combustion reactions. (Reactions in solution phase) Calculations are the same as for constant-volume calorimetry: qsystem = qsolution + qcalorimeter + qreaction = 0 Therefore, qreaction = - (qsolution + qcalorimeter) 88  Using Hess’s Law There are millions of potential chemical reactions out there and we can’t measure all DH values experimentally. Luckily, the 1st Law of Thermodynamics saves us this burden. Since DH is a state function, we can get from reactants to products any way we want. 89 Aspects of Hess’ Law: Energy changes depend on the AMOUNT of substance H2O (s) H2O (l) DH = 6.01 kJ/mol If the coefficients in a balanced reaction are multiplied by an integer, the value of DH is multiplied by the same integer. 2H2O (s) 2H2O (l) DH = 2 mol x (6.01 kJ/mol) = 12.0 kJ 90 Changes in physical states result in enthalpy changes, so must be specified. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DHrxn = - 890.4 kJ BUT CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) DHrxn = - 802.4 kJ The second process produces less energy because in the second process some energy (88.0 kJ) is consumed to convert water from liquid to vapor phase. 2H2O (l) 2H2O (g) DHvap = 88.0 kJ 91 The equation for the combustion of butane is, 13 C4 H10 (g) + O 2 (g) 4CO 2 (g) + 5H 2O(g) 2 Which one of the following generates the least heat? 1. Burning one mole of butane in excess oxygen. 2. Reacting one mole of oxygen with excess butane. 3. Producing one mole of carbon dioxide by burning butane. 4. Producing one mole of water by burning butane. 5. Burning 0.25 moles of butane with excess oxygen. Look at the stoiciometry: 13/2 = 6.5 is the largest ratio. You will get 6.5 times less heat by burning one mole of O2. 92 The formation of 2 moles of NO2 from N2 and O2 can occur in one step or in a series of steps. Either way, the reaction enthalpy will be the same. One step: N2 (g) + 2 O2 (g) 2 NO2 (g) DHrxn = 68 kJ/mol Two steps: N2 (g) + O2 (g) 2 NO (g) DH1 = 180 kJ/mol 2NO (g) + O2 (g) 2 NO2 (g) DH2 = -112 kJ/mol Net Reaction: N2 (g) + 2 O2 (g) 2 NO2 (g) DHrxn = DH1 + DH2 = 68 kJ/mol The sum of the individual steps gives the net reaction and the overall DHrxn. 93 Determine the DH for the formation of diborane (B2H6) from its elements using the known reaction enthalpies shown below: 2 B(s) + 3 H2(g)  B2H6(g) DHrxn = ?? 94  2 B(s) + 3 H2(g)  B2H6(g) DHrxn = ?? Step 1: Find a reaction that uses 2 B (s) as a reactant. 2 B (s) + 3/2 O2 (g)  B2O3(s) DH = -1273 kJ Step 2: B2O3 (s) does not appear as a product in the desired reaction, so the next reaction should have B2O3 (s) on the left side of the arrow. Given: B2H6(g) + 3 O2 (g)  B2O3(s) + 3 H2O(g) DH = -2035 kJ Reverse: B2O3(s) + 3 H2O(g)  B2H6(g) + 3 O2 (g) DH = 2035 kJNOTE: reversing reaction reverses sign of DH 95  Step 3: Add together the first 2 equations and cancel out anything that appears on both the left and the right: 2 B (s) + 3/2 O2 (g)  B2O3(s) DH = -1273 kJ B2O3(s) + 3 H2O(g)  B2H6(g) + 3 O2 (g) DH = 2035 kJ 2 B (s) + 3 H2O(g)  B2H6(g) + 3/2 O2 (g) DH = 762 kJ Now manipulate this new equation to achieve the desired equation Step 4: H2O (g) does not appear in the final reaction. Find a reaction that has 3 H2O (g) on the right side. Given: H2O(g)  H2O(l) DH = -44 kJ Reverse and x 3: 3H2O(l)  3H2O(g) DH = 3(44) = 132 kJ NOTE: Multiply DH by the same constant as the reaction 96  2 B(s) + 3 H2(g)  B2H6(g) Step 5: Add together these 2 equations and cancel out anything that appears on both the left and the right: 2 B (s) + 3 H2O(g)  B2H6(g) + 3/2 O2 (g) DH = 762 kJ 3H2O(l)  3H2O(g) DH = 3(44) = 132 kJ 2 B (s) + 3 H2O(l)  B2H6(g) + 3/2 O2 (g) DH = 894 kJ Now manipulate this new equation to achieve the desired equation Step 6: H2O (l) does not appear in the final reaction. Find a reaction that has 3 H2O (l) on the right side. Given: H2(g) + ½ O2 (g)  H2O(l) DH = -286 kJ Multiply by 3: 3 H2(g) + 3/2 O2 (g)  3H2O(l) DH = 97 3(-286) = -858 kJ Step 7: Add together these 2 equations and cancel out anything that appears on both the left and the right: 2 B (s) + 3 H2O(l)  B2H6(g) + 3/2 O2 (g) DH = 894 kJ 3 H2(g) + 3/2 O2 (g)  3H2O(l) DH = -858 kJ 2 B(s) + 3 H2(g)  B2H6(g) DHrxn = 36 kJ 98  2 B(s) + 3 H2(g)  B2H6(g) 2 B (s) + 3/2 O2 (g)  B2O3(s) DH = -1273 kJ B2O3(s) + 3 H2O(g)  B2H6(g) + 3 O2 (g) DH = 2035 kJ 3H2O(l)  3 H2O(g) DH = 3(44) = 132 kJ 3 H2(g) + 3/2 O2 (g)  3H2O(l) DH = -858 kJ 2 B(s) + 3 H2(g)  B2H6(g) DHrxn = 36 kJ 99  Reaction Enthalpies from Enthalpies of Formation When measured under standard conditions DH values are labeled DHo Standard Conditions for Compounds  For a gas, pressure is exactly 1 atmosphere (bar).  For a solution, concentration is exactly 1 molar.  Liquids or solids are the pure liquid or solid.  Temperature is typically 25oC, but can be different.  Standard Conditions for an Element  The form in which it naturally exists at 1 atm and 25°C e.g. [N2(g), K(s)]. 100  Standard Molar Enthalpy of Formation, DfHo (or DHfo) Change in heat content when 1 mole of a compound is formed from its elements in their standard states at 298 K and 101.3 kPa (1 atm). ½ N2 (g) + O2 (g) NO2 (g) DfH0 (NO2) = 34 kJ/mol 1 mole of compound formed Elements in standard states DfHo = 0 By definition, DfHo for elements in their standard states = 0 101  Standard Enthalpy of Reaction, DHoThe rxn enthalpy of a reaction carried out at 1 atm. This can be calculate from the standard enthalpies of formation. DHorxn = S DfHo (products) - S DfHo (reactants) aA + bB  cC + dD DHorxn = [cDfHo(C) + dDfHo(D)] - [aDfHo(A) + bDfHo(B)] 102 Calculate the heat of combustion of one mole of methanol, i.e., DHorxn for: CH3OH(g) + 3/2 O2(g)  CO2(g) + 2 H2O(g) (Combustion: Reaction of a substance with oxygen to produce CO2 + H2O) DHorxn = S DHof (prod) - S DHof (react) 103  CH3OH(g) + 3/2 O2(g)  CO2(g) + 2 H2O(g) DHorxn = [DHof (CO2) + 2 DHof (H2O)] - [3/2 DHof (O2) + DHof (CH3OH)] = (-393.5 kJ/mol) + 2 (-241.8 kJ/mol) - [0 + (-201.5 kJ/mol)] DHorxn = -675.6 kJ per mol of methanol 104  Heat or Enthalpy of Solution Heat of solution – heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsolution = Hsolution - Hcomponents H2O NaCl (s) Na+ (aq) + Cl- (aq) DHsolution = 4 kJ/mol This is easily measured by constant-pressure calorimetry 105 e generated heat. So what? What moves the cylinder is the WORK this system c Chemical Work The most common type of work associated with chemical process is work done by a gas (expansion) or to a gas (compression). e.g. a piston 106  The Work of an Expanding Gas Pressure = force/area work = force x distance P = F/A w = F x Dh (F = P x A) w = P x A x Dh A gas expanding against an external pressure (P). DV P = F/A P = F/A w = PDV (this is the magnitude, Area = A NOT THE SIGN) Dh Dh DV GAS GAS Initial state final state 107  The Work of an Expanding Gas The Sign for Work For an expanding gas: Pressure is the external pressure (positive) DV is positive (volume is increasing) w is negative (work is being done by the system) Therefore we define wsystem = - PDV AT CONSTANT EXTERNAL PRESSURE 108  The Work of an Expanding Gas Solid CO2 absorbs thermal energy and vaporizes. As the vaporized CO2 molecules expand in a sealed bag, the bag expands. The gas expansion is used to perform the work of lifting a book. DUCO2 = qsublimation + workexpansion 109 The Work of an Expanding Gas Flexible container: DUCO2 = qsublimation - PDV heat is absorbed work is done by by CO2 to cause CO2 as it sublimation expands (energy transfer (energy transfer from surroundings from system to to system) surroundings) Ridged container: DV = 0 DUCO2 = qsublimation 110  The Work of an Expanding Gas Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm. w = -PDV P = 15 atm DV = 64 L – 46 L = 18 L w = - (15 atm)(18 L) = -270 L atm Note: In expansion, work is done on the surroundings (energy is transferred from system to surroundings), so w is negative 111  Energy Units Units of L atm are not very meaningful when we are interested in knowing about energy. Answers should be converted to units of kJ or kcal. From previous slide: w = -270 L atm Conversion to kJ: (1 L atm = 101.3 J) -270 L atm x 101.3=J - 2.73 x 104 J = - 27.3 kJ 1 L atm -270 L atm x 8.314 J/molK = - 2.73 x 104 J = - 27.3 kJ 0.08206Latm/molK 112 What work can be done by the combustion of 1 mole of C8H18 in a 1 L single-stroke engine at 800oC and a constant external pressure of 1 atm? C8H18 + 12.5 O2 → 8 CO2 + 9 H2O Volume initial will be approximately half the final volume The max work we can do with this engine has almost nothing to do with the heat released by the combustion and is dictated by the size of the cylinder. The heat released makes sure maximum expansion is achieved and that the reaction goes to completion. w = - PDV Convert work to Joules: -0.5L atm x 101.3 J = -51 J = - (1.0 atm)(1 L – 0.5L) 1 L atm = - 0.5 L atm 113 Electric vehicles 114  Redox Chemistry Reactions involving transfer of ELECTRONS (e-) 2 Ca (s) + O2 (g) 2CaO (s) 0 0 +2 -2 Electricity is the flow of electrons. Chemical reactions involving flow of electrons can be used to generate electricity 115 Redox reactions have a broad range of applications ELECTROPLATING PRODUCTION OF METALS BATTERIES Rust 116  Recognizing Redox Reactions How do we know if electron transfer occurs in a reaction?? 2 Ca (s) + O2 (g) 2CaO (s) Step 1: Determine the oxidation number (or oxidation state) of each element in the reactants and products. Step 2: If the oxidation state of a given element is different in the products than it is in the reactants, then electrons were transferred. 117  Oxidation Numbers Oxidation numbers are a mathematical tool for “dividing up” the electrons in a molecule. 2 Ca (s) + O2 (g) 2CaO (s) 0 0 +2 -2 Gaining electrons cause oxidation number to decrease Losing electrons causes oxidation number to increase Element Initial Final Change______ Ca 0 +2 Ca lost 2 electrons O 0 -2 O gained 2 electrons 118  Rules for Assigning Oxidation Numbers (Nox) 1) The oxidation number, Nox, for a free elements (not in a compound) is zero. e.g. Na Nox = 0 Mg Nox = 0 Al Nox = 0 2) The oxidation number of a monatomic ion (ion composed of only 1 atom) is equal to the ion charge: e.g. Na+ Nox = +1 Mg2+ Nox = +2 O2- Nox = -2 Cl- Nox = -1 119 3) The oxidation number of a group 1A element in a molecule is always +1 e.g. NaCl Nox (Na) = +1 KClO3 Nox (K) = +1 4) The oxidation number of a group 2A element in a molecule is always +2 e.g. MgCl2 Nox (Mg) = +2 Ba(OH)2 Nox (Ba) = +2 5) The oxidation number for a halogen in a molecule is usually –1. There are exceptions. e.g NaCl Nox (Cl) = -1 MgF2 Nox (F) = -1 HCl Nox (Cl) = -1 120 6) The oxidation number of each atom in a compound containing only 1 type of element is zero. e.g. H2 each H atom has Nox = 0 O2 each O atom has Nox = 0 P4 each P atom has Nox = 0 7) In a neutral, polyatomic compound, the sum of the oxidation numbers of each atom in the compound is zero: e.g. NaCl [Nox (Na) + Nox (Cl)] = 0 SO2 [Nox (S) + (2 x Nox (O)] = 0 CO [Nox (C) + Nox (O)] = 0 8) For a polyatomic ion, the sum of oxidation numbers of each atom in the compound is equal to the overall ion charge: e.g. CO32- [Nox (C) + (3 x Nox (O))]= -2 HSO4- [Nox (H) + Nox (S) + (4 x Nox (O))]= -1 121 9) The oxidation number of oxygen in most compounds is –2. Exception: In H2O2 and in O22-, each oxygen has Nox = -1 e.g MgO Nox (O) = -2 H2O Nox (O) = -2 CO2 Nox for each O = -2 H2O2 Nox for each O = -1 10)The oxidation number of hydrogen in most compounds is +1. Exception: When H is bound directly to a group 1A or group 2A metal, it has Nox = -1. e.g. H2O Nox for each H = +1 CH4 Nox for each H = +1 HNO2 Nox (H) = +1 NaH Nox (H) = -1 MgH2 Nox for each H = -1 122  Examples Assigning Oxidation Numbers (Nox) Assign oxidation states to all atoms in CO2 From the rules, we know Nox for each O = -2 From the rules, we know [Nox(C) + (2xNox(O))] = 0 Nox (C) + (2 x (-2)) = 0 Nox (C) – 4 = 0 Nox (C) = +4 CO 2 +4 -2 for each oxygen 123  Assign oxidation states to all atoms in NO3- From the rules, we know Nox for each O = -2 From the rules, we know [Nox(N) + (3xNox(O))] = -1 Nox (N) + (3 x (-2)) = -1 NO3- Nox (N) – 6 = -1 Nox (N) = -1 + 6 = + 5 +5 -2 for each oxygen Assign oxidation states to all atoms in SF6 From the rules, we know Nox for each F = -1 From the rules, we know [Nox(S) + (6xNox(F))] = 0 Nox (S) + (6 x (-1)) = 0 SF6 Nox (S) – 6 = 0 Nox (S) = +6 +6 -1 for each fluorine 124  General Catagories of Redox Reactions 1. Combination reactions: Two or more substances combine to form a single product: A+BC 2. Decomposition reactions: A breakdown of a compound into two or more compounds: AB+C 3. Displacement reactions: An ion (or atom) in a compound is replaced by an ion (or atom) of another element: A + BC AC + B 4. Disproportionation Reactions: a single element both gains electrons and loses electrons 125  Combination Redox Reactions A+BC 2 Na (s) + Cl2 (g) 2 NaCl (s) Oxidation states: 0 0 +1 -1 Element Initial Final Change______ Na 0 +1 each Na lost 1 electron Cl 0 -1 each Cl gained 1 electron 126  Decomposition Redox Reactions AB+C +1 -1 2 KClO3 (s) 2 KCl (s) + 3 O2 (g) +1 +5 -2 for each oxygen 0 for each oxygen Element Initial Final Change______ K +1 +1 No change in Nox for K O -2 0 each O lost 2 electrons Cl +5 -1 Cl gained 6 electrons 127  Displacement Redox Reactions A + BC AC + B Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) 0 +2 +1 -1 -1 for each Cl 0 for each hydrogen Element Initial Final Change______ Zn 0 +2 Zn lost 2 electrons H +1 0 Each H gained 1 electron Cl -1 -1 No change in Nox for Cl 128  Disproportionation Reactions In a disproportionation reaction a single element both gains electrons and loses electrons!! 2H2O2 (aq) 2H2O (l) + O2 (g) -1 each O -2 0 Element Initial Final Change______ O -1 -2 2 oxygens gain 1 electron each O -1 0 2 oxygens lose 1 electron each 129  Oxidation and Reduction The element that loses electrons is said to be oxidized The element that gain electrons is said to be reduced 2 Na (s) + Cl2 (g) 2 NaCl (s) 0 0 +1 -1 Na is oxidized Cl is reduced (loses 1 electron) (gains 1 electron) Zn (s) + 2HCl (aq) ZnCl2 (aq) + H2 (g) 0 +2 +1 -1 -1 for each Cl 0 for each hydrogen Zn is oxidized H is reduced (loses 2 electrons) (gains 1 electron) 130  Half Reactions To visualize electron flow, we can write any redox reaction as if it occurs in two different steps called half-reactions. 2 Ca (s) + O2 (g) 2CaO (s) Step 1: The oxidation half-reaction shows electrons being lost by an element (electrons are a product) 2 Ca 2 Ca2+ + 4 e- Each Ca loses 2 electrons Step 2: The reduction half-reaction shows electrons being gained by an element (electrons are a reactant) O2 + 4 e- 2 O2- 131 The sum of the half reactions gives the overall reaction: Oxidation: 2 Ca 2 Ca2+ + 4 e- Reduction: O2 + 4 e- 2 O2- Overall: 2 Ca + O2 + 4 e- 2 Ca2+ + 2 O2- + 4 e- Cancel electrons that appear on both sides: 2 Ca + O2 2 Ca2+ + 2 O2- Combine Ca2+ with O2- to make CaO 2 Ca + O2 2 CaO 132  # of electrons consumed in the reduction = # of electrons produced by oxidation Oxidation: 2 Ca 2 Ca2+ + 4 e- Reduction: O2 + 4 e- 2 O2- Overall: 2 Ca + O2 + 4 e- 2 Ca2+ + 2 O2- + 4 e- # electrons in reactant = # electrons in products NO ELECTRONS ARE CREATED OR DESTROYED 133  Oxidizing and Reducing Agents Oxidation: 2 Ca 2 Ca2+ + 4 e- Reduction: O2 + 4 e- 2 O2- Overall: 2 Ca + O2 + 4 e- 2 Ca2+ + 2 O2- + 4 e- Reducing Agent: The species that is oxidized Calcium is a reducing agent because it donates electrons to oxygen causing oxygen to be reduced. Oxidizing Agent: The species that is reduced Oxygen is an oxidizing agent because it accepts electrons from calcium cause calcium 134 to oxidized.  Balancing Redox Reactions by the Ion- Can I shorten this? Electron Method Step 1: Determine oxidation states of each atom in the reaction. Step 2: Divide the reaction into the contributing half-reactions. Step 3: Balance each half reaction (We will see how!) Step 4: Equalize the # of electrons in the two half reactions. (# electrons produced by oxidation must equal # electrons consumed by reduction.) Step 5: Add the balanced half reactions together to obtain the overall reaction. 135  Balance the following redox reaction in acidic solution: H+ (aq) + Cr2O72- (aq) + C2H5OH (l) Cr3+ (aq) + CO2 (g) + H2O (l) Step 1: Determine oxidation states of each atom in the reaction. -2 each -2 +3 -2 each -2 each H+ (aq) + Cr2O72- (aq) + C2H5OH (l) Cr3+ (aq) + CO2 (g) + H2O (l) -2 each +4 +1 each +1 +6 each +1 each Each carbon loses 6 electrons (-2 to +4 oxidation state) Each chromium gains 3 electrons (+6 to +3 oxidation state) 136 Step 2: Divide the reaction into the contributing half-reactions. +6 each -2 each +3 +4 H+ (aq) + Cr2O72- (aq) + C2H5OH (l) Cr3+ (aq) + CO2 (g) + H2O (l) Each carbon loses 6 electrons (-2 to +4 oxidation state) Oxidation Half-Reaction: C2H5OH (l)  CO2 (g) (unbalanced) Each chromium gains 3 electrons (+6 to +3 oxidation state): Reduction Half-Reaction: Cr2O72- (aq) Cr3+ (aq) (unbalanced) 137 Step 3: Balance each half reaction Rules for Balancing Half-Reactions in ACIDIC solution Treat each half-reaction separately 1. Balance all elements except hydrogen and oxygen 2. Balance oxygen using H2O 3. Balance hydrogen using H+ 4. Balance charge using electrons charge on left charge on right = side of arrow side of arrow 138  Oxidation Half-Reaction: C2H5OH (l)  CO2 (g) 1. Balance all elements except hydrogen and oxygen Before: C2H5OH (l)  CO2 (g) 2 carbons 1 carbon After: C2H5OH (l)  2 CO2 (g) 2 carbons 2 carbons 2. Balance oxygen using H2O Before: C2H5OH (l)  2 CO2 (g) 1 oxygen 4 oxygens After: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) 4 oxygens 4 oxygens 139 3. Balance hydrogen using H+ Before: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) 0 hydrogens 12 hydrogens After: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) 12 hydrogens 12 hydrogens 4. Balance charge using electrons Before: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) No charge +12 After: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) + 12e- No charge No charge 140  Check your answer: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) + 12e- Left side: Right Side: 12 H 12 H 4O 4O 2C 2C No charge No charge The oxidation half-reaction is balanced! 141  Reduction Half-Reaction: Cr2O72- (aq) Cr3+ (aq) 1. Balance all elements except hydrogen and oxygen Before: Cr2O72- (aq) Cr3+ (aq) 2 Chromium 1 Chromium After: Cr2O72- (aq) 2 Cr3+ (aq) 2 Chromium 2 Chromium 2. Balance oxygen using H2O Before: Cr2O72- (aq) 2 Cr3+ (aq) 7 oxygen 0 oxygen After: Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) 7 oxygen 7 oxygen 142 3. Balance hydrogen using H+ Before: Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) 0 hydrogen 14 hydrogen After: 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) 14 hydrogen 14 hydrogen 4. Balance charge using electrons Before: 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) +12 charge +6 charge After: 6 e- + 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) +6 charge +6 charge 143  Check Your Answer! 6 e- + 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) Left Side: Right Side: 2 Chromium 2 Chromium 7 Oxygen 7 Oxygen 14 Hydrogen 14 Hydrogen Charge = +6 Charge = +6 The reduction half-reaction is balanced! 144  Step 4: Equalize the # of electrons in the two half reactions. (# electrons produced by oxidation must equal # electrons consumed by reduction.) Oxidation: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) + 12e- 12 electrons are produced by the oxidation half-reaction Reduction: 6 e- + 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) 6 electrons are consumed by the reduction half-reaction Number of electrons consumed by reduction MUST BE EQUAL TO number of electrons produced by oxidation! 145  # electrons in oxidation = # electrons in reduction In this case, for every single oxidation half-reaction that occurs, the reduction half-reaction must occur TWICE! Oxidation: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) + 12e- 12 electrons are produced by the oxidation half-reaction Reduction: 6 e- + 14 H+ (aq) + Cr2O72- (aq) 2 Cr3+ (aq) + 7 H2O (l) X 2 12 e- + 28 H+ (aq) + 2 Cr2O72- (aq) 4 Cr3+ (aq) + 14 H2O (l) NOW 12 electrons are consumed by the reduction 146  Step 5: Add the balanced half reactions together to obtain the overall reaction. Oxidation: 3 H2O (l) + C2H5OH (l)  2 CO2 (g) + 12 H+ (aq) + 12e- Reduction: 12 e- + 28 H+ (aq) + 2 Cr2O72- (aq) 4 Cr3+ (aq) + 14 H2O (l) Overall Reaction: 16 3 H2O (l) + C2H5OH (l) + 12 e- + 28 H+ + 2 Cr2O72- 11 2 CO2 (g) + 12 H+ (aq) + 12 e- + 4 Cr3+ (aq) + 14 H2O (l) C2H5OH (l) + 16 H+ (aq) + 2 Cr2O72- 2 CO2 (g) + 4 Cr3+ (aq) + 11 H2O (l) 147  Check Your Answer! C2H5OH (l) + 16 H+ (aq) + 2 Cr2O72- 2 CO2 (g) + 4 Cr3+ (aq) + 11 H2O (l) Left Side Right Side 2 Carbon 2 Carbon 22 Hydrogen 22 Hydrogen 15 Oxygen 15 Oxygen 4 Chromium 4 Chromium +12 charge + 12 charge REACTION IS BALANCED! 148  Basic Solutions In basic solution, H+ is not available! Thus it cannot appear in our final balanced equation. Rules for Balancing Half-Reactions in BASIC solution Treat each half-reaction separately 1. Balance all elements except hydrogen and oxygen 2. Balance oxygen using H2O 3. Balance hydrogen using H + For every H that is added, add an equal number of + OH- ions to both sides of the equation. Wherever H+ and OH- appear on the same side of the equation, combine them to make water. (No H+ should appear anywhere in the equation!) 4. Balance charge using electrons 149  Balance the following chemical equation in basic solution: MnO4- (aq) + SO32- (aq) MnO42- (aq) + SO42- (aq) Step 1: Determine oxidation states of each atom in the reaction. MnO4- (aq) + SO32- (aq) MnO42- (aq) + SO42- (aq) +7 -2 +4 -2 +6 -2 +6 -2 Each Mn gains 1 electron (+7 to +6) Each S loses 2 electrons (+4 to +6) 150 Step 2: Divide the reaction into the contributing half-reactions. MnO4- (aq) + SO32- (aq) MnO42- (aq) + SO42- (aq) +7 +4 +6 +6 Each S loses 2 electrons (+4 to +6) Oxidation Half-Reaction: SO32- (aq) SO42- (aq) (unbalanced) Each Mn gains 1 electron (+7 to +6) Reduction Half-Reaction: (unbalanced) MnO4- (aq) MnO42- (aq) 151 Step 3: Balance each half reaction Oxidation Half-Reaction: SO32- (aq) SO42- (aq) 1. Balance all elements except hydrogen and oxygen SO32- (aq) SO42- (aq) 1 sulphur 1 sulphur 2. Balance oxygen using H2O Before: SO32- (aq) SO42- (aq) 3 oxygen 4 oxygen After: H2O (l) + SO32- (aq) SO42- (aq) 4 oxygen 4 oxygen 152  3. Balance hydrogen using H+ Before: H2O (l) + SO32- (aq) SO42- (aq) 2 hydrogen 0 hydrogen After: H2O (l) + SO32- (aq) SO42- (aq) + 2 H+ (aq) 2 hydrogen 2 hydrogen For every H+ that is added, add an equal number of OH- ions to both sides of the equation 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H+ (aq) + 2 OH- (aq) Wherever H+ and OH- appear on the same side of the equation, combine them to make water. 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) 153  4. Balance charge using electrons Before: 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) charge = -4 charge = -2 After: 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) + 2 e- charge = -4 charge = -4 154  Check your answer 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) + 2 e- Left Side Right Side 1 sulphur 1 sulphur 6 oxygen 6 oxygen 4 hydrogen 4 hydrogen charge = -4 charge = -4 The oxidation half-reaction is balanced 155  Reduction Half-Reaction: MnO4- (aq) MnO42- (aq) 1. Balance all elements except hydrogen and oxygen MnO4- (aq) MnO42- (aq) 1 Manganese 1 Manganese 2. Balance oxygen using H2O MnO4- (aq) MnO42- (aq) 4 oxygen 4 oxygen 3. Balance hydrogen using H+ MnO4- (aq) MnO42- (aq) no hydrogen no hydrogen 156 4. Balance charge using electrons Before: MnO4- (aq) MnO42- (aq) charge = -1 charge = -2 After: 1 e- + MnO4- (aq) MnO42- (aq) charge = -2 charge = -2 Check your answer 1 e- + MnO4- (aq) MnO42- (aq) 1 Mn 1 Mn 4O 4O charge = -2 charge = -2 Reduction half-reaction is balanced! 157 Step 4: Equalize the # of electrons in the two half reactions. (# electrons produced by oxidation must equal # electrons consumed by reduction.) Oxidation half-reaction: 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) + 2 e- 2 electrons are produced Reduction half-reaction: 1 e- + MnO4- (aq) MnO42- (aq) 1 electron is consumed The reduction half-reaction must occur twice for every time the oxidation half-reaction occurs. 1 e- + MnO4- (aq) MnO42- (aq) x2 158 Step 5: Add the balanced half reactions together to obtain the overall reaction. Oxidation: 2 OH- (aq) + H2O (l) + SO32- (aq) SO42- (aq) + 2 H2O (l) + 2 e- Reduction: 2 e- + 2 MnO4- (aq) 2 MnO42- (aq) Overall Reaction: 2 OH- (aq) + H2O (l) + SO32- (aq) + 2 e- + 2 MnO4- (aq) 1 SO 4 2- (aq) + 2 H2O (l) + 2 e- + 2 MnO42- (aq) 2 OH- (aq) + SO32- (aq) + 2 MnO4- (aq) SO42- (aq) + H2O (l) + 2 MnO42- (aq) 159  Check Your Answer: 2 OH- (aq) + SO32- (aq) + 2 MnO4- (aq) SO42- (aq) + H2O (l) + 2 MnO42- (aq) Left Side Right Side 1 sulfur 1 sulfur 2 manganese 2 manganese 13 oxygen 13 oxygen 2 hydrogen 2 hydrogen charge = -6 charge = -6 Reaction is Balanced (in basic solution)! 160  Balancing Redox Reactions in Acidic vs. Basic Solution NOTE: The overall reaction depends on whether it is in acidic or basic solution! IN ACID: 8 H+ + 2 NO3- + 3 Cu  3 Cu2+ + 2 NO + 4 H2O IN BASE: 4 H2O + 2 NO3- + 3 Cu  3 Cu2+ + 2 NO + 8 OH- 161 Summary of Steps for Balancing Redox Reactions: 1. Determine oxidation numbers of each element: 2. Identify the two half reactions (oxidation and reduction). 3. Balance each half-reaction following this sequence in order: a) Balance elements (except H, O) b) Balance O using H2O c) Balance H using H+ For basic solutions i) For every H+ ion, add on OH- to both sides of the equation ii) When H+ and OH- appear on the same side of the equation, combine them to make H2O. d) Balance 4. Equalize charges count the electron of half-reactions in oxidationusing electrons half-reactions and reduction 5. Add together the oxidation and reduction half-reactions 6. Check that all atoms and overall charges are balanced 162 a car battery for an EV – what kind of battery is used? What is the Redox Chemistry? Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) Zinc atoms are oxidized to Zn2+(aq): Zn(s) → Zn2+(aq) + 2e- (loss of electrons) Copper(II) ions are reduced to Cu(s): Cu2+(aq) + 2e- → Cu(s) (gain of electrons) https://byjus.com/chemistry/electrochemical-cell/ 163 164 165  What work can be done by an Electrochemical Cell? Now, we are no longer limited by temperature, pressure, and volume. This makes electrochemical cells much more efficient than PV-work engines. wmax=nFE cell # electrons moving Cell potential Faraday’s const. EVs convert over 77% of the electrical energy from the grid to power at the wheels. Conventional gasoline vehicles only convert about 12%–30% of the energy stored in gasoline to power at the wheels. 166 167