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Energy Energy is the capacity to supply heat or do work. ∆Energy = heat + work ∆E = q + w Chapter 6 Thermochemistry Dr. Peter Warburton [email protected] http://www.chem.mun.ca/zcourses/1010.php There are two types of energy: kinetic or potential. All media copyright of their respective owners Kinetic energy Potential energy Potential energy (Ep) is stored energy, resulting from condition, position or composition. Kinetic energy (Ek) is the energy of motion. Ek = ½ (mass)(velocity)2 Ek = ½ mv2 All media copyright of their respective owners 2 3 All media copyright of their respective owners 4 1 Units of energy Units of energy Energy can be expressed in units other than the joule. But they all represent energy, and can be converted from one unit form to another. Ek = ½ mv2 gives units of kg m2 s-2 = joule (J) Always pay attention to the units of the variables you are working with! You might need to convert! One Tic Tac contains about 8000 J, or 8 kJ (kilojoules) of stored (potential!) chemical energy. All media copyright of their respective owners 5 Calories and calories All media copyright of their respective owners 6 Units of energy Calories are associated with the energy of food. A Calorie should actually be referred to as a kilocalorie (kcal) because 1 Calorie (Cal) = 1000 calories (cal) = 1 kcal By definition 1 cal = 4.184 J (exactly) All media copyright of their respective owners 7 All media copyright of their respective owners 8 2 Forms of energy System and surroundings The system is the part of the universe that undergoes a fundamental change of interest. The surroundings is every other part of the universe. Thermal energy is the kinetic energy of randomly moving molecules. The average thermal energy of the molecules is what we measure as the temperature. All media copyright of their respective owners 9 System and surroundings All media copyright of their respective owners 10 System and surroundings We can further divide a system into one of three different types: 1. Open system – BOTH matter and energy can enter (or exit) the system from (or to) the surroundings. 2. Closed system – ONLY energy can enter or exit the system 3. Isolated system – NEITHER matter nor energy can enter or exit the system. All media copyright of their respective owners 11 All media copyright of their respective owners 12 3 Energy changes and conservation Energy changes and conservation As a person climbs (kinetic energy) Signal Hill, they store an equivalent amount of potential energy due to the work they do changing their height (distance) against the force of gravity. If the person parachutes back down to the harbour this potential energy becomes an equivalent amount of kinetic energy as the person falls. All media copyright of their respective owners 13 Internal energy of a system All media copyright of their respective owners 14 Internal energy of a system However, measuring the total internal energy of a system is next to impossible. It’s somewhat like the amount of water between the bottom of a boat and the bottom of the ocean. The internal energy U is the sum of the kinetic energy Ek of all the molecular motions in the system and the potential energy Ep stored in any chemical bonds, intermolecular attractions and the attractions of electrons to nuclei U = Ek + Ep All media copyright of their respective owners 15 All media copyright of their respective owners 16 4 Internal energy of a system U is a state function Usually we’re more interested in the change in the height of the ocean (how high are the waves), and analogously we are often more interested in the change in internal energy ∆U of our system. Internal energy is a state function just like the height of a hill is a state function. No matter how you move from the top of a hill to the bottom (path A or path B) your change in height is always the same. ∆rU = Ufinal - Uinitial All media copyright of their respective owners 17 First Law of Thermodynamics All media copyright of their respective owners 18 Another consequence of the First Law If the system is isolated from the surroundings, then the internal energy of the system can not change, regardless of the changes that take place in the system. ∆Uisolated system = 0 First Law of Thermodynamics – The internal energy of an isolated system is constant. All media copyright of their respective owners C (s) + O2 (g) → CO2 (g) 19 Imagine the universe as a “super” isolated system which we describe in two pieces – a “normal” system (closed, open or isolated) and its surroundings (everything else). Since the internal energy of the universe must be constant, this implies ∆Usystem = -∆Usurroundings All media copyright of their respective owners 20 5 Changes in internal energy Heat Heat is the amount of thermal energy transferred between the system and surroundings due to a difference in temperature. The only way the internal energy of a system can change is if there is energy transfer to or from the surroundings. We’ve already seen the ONLY two types of such energy transfer – heat and work ∆Usys = q + w All media copyright of their respective owners 21 Work All media copyright of their respective owners 22 Units of work Work is the energy change of a force applied over a distance w=Fxd So if w = F x d w=mxaxd since F = mass x acceleration Work must have the same units as energy. Is this true? All media copyright of their respective owners The SI unit of force, the newton (N) is a derived unit 1 N = 1 kg m s-2 23 The unit for work then is a N m, Nm = (kg m s-2) m = kg m2 s-2 = joule (J). All media copyright of their respective owners 24 6 Point of view of the system! More specifically… Net energy transfer out of the system means ∆U is negative Net energy transfer into the system means ∆U is positive All media copyright of their respective owners 25 Heat capacity All media copyright of their respective owners 26 Specific and molar heat capacities The specific heat (capacity) Cs is the amount of heat required to raise the temperature of 1 gram of a specific substance by 1 K. Heat capacity (C) is the heat (q) required to change the temperature (T) of a substance by a given amount (usually 1 K or ºC – they’re the same size!). C = q/∆T ∆T = Tfinal - Tinitial (always!) All media copyright of their respective owners 27 q = m x Cs x ∆ T All media copyright of their respective owners 28 7 Specific heats Specific and molar heat capacities These units can also be expressed as J g-1 K-1 OR the temperature change can be evaluated in ºC. This is the one time you can get away without T conversion! All media copyright of their respective owners q = n x Cm x ∆ T 29 All media copyright of their respective owners 30 Problem Relationship between Cs and Cm Say we heat up a glass of water How much heat in kilojoules is required to raise the temperature of 237 g of cold water from 4.0 ºC to 37.0 ºC? q=q n x Cm x ∆T = m x Cs x ∆T n x Cm = m x Cs Cm = m/n x Cs Cm = molar mass x Cs All media copyright of their respective owners The molar heat capacity Cm is the amount of heat required to raise the temperature of 1 mole of a specific substance by 1 K. 31 All media copyright of their respective owners 32 8 Problem answer Thermal energy transfer q = 32.7 kJ When a hot object touches a cold object we all know what happens. The hot object cools down and the cold object heats up. This thermal energy transfer (heat q) will continue until both objects reach thermal equilibrium – the same temperature! All media copyright of their respective owners 33 Thermal energy transfer All media copyright of their respective owners 34 Problem Heat lost by hot object = heat gained by cold object -qhot = qcold -mhot Cs,hot ∆T = mcold Cs,cold ∆T mhot Cs,hot (Teqm – Thot) = mcold Cs,cold (Teqm – Tcold) A 100.0 g copper sample at 100.0 ºC is added to 50.0 g of water at 26.5 ºC. What is the final temperature of the copper/water mixture? Ultimately we will often want to solve for Teqm! All media copyright of their respective owners 35 All media copyright of their respective owners 36 9 Problem answer PV work If we consider the combustion of gasoline with oxygen to form carbon dioxide and water in a car engine, there will certainly be a qrxn to consider for the energy change of the system. Teqm = 37.9 ºC = 311.1 K All media copyright of their respective owners 37 PV work All media copyright of their respective owners 38 PV work But for the gases to expand against the external pressure, it must “push” (an action with a force) the cylinder out of the way. An amount of work is done depends on “how far” (leading to a volume change) the cylinder is pushed. However, the CO2 and H2O gas generated will try to expand as much as it can against the external barometric pressure Pbar until the PCO2 = PH2O = Pbar. The gas must expand (increase it’s volume) until this happens. All media copyright of their respective owners 39 All media copyright of their respective owners 40 10 Amount of PV work PV work is energy transfer Remember from slide 34 of last chapter (blowing up a balloon) pressure multiplied by volume is equivalent to energy units! Here we have a gas in a piston cylinder where the piston can move in or out. 1 Pa m3 = 1 J 1 bar L = 100 J 1 atm L = 101.325 J ݓ௦ = −ܲ ∆ܸ௦ All media copyright of their respective owners 41 w = -P∆V All media copyright of their respective owners 42 Problem Positive ∆V (expansion so V gets bigger) means work is negative. The system does work on the surroundings Negative ∆V (compression so V gets smaller ) means work is positive. The system is worked upon by the surroundings In compressing a gas, 355 J of work is done on the system. At the same time 185 J of heat escapes from the system. What is ∆U for the system? No volume change (∆ ∆V = 0). There is no PV work done. All media copyright of their respective owners 43 All media copyright of their respective owners 44 11 Problem answer Problem ∆U = +170 J If the internal energy of a system decreases by 125 J at the same time 54 J of heat is absorbed by the system, does the system do work or have work done on it? How much? All media copyright of their respective owners 45 Problem answer 46 Chemical energy Chemical energy is potential energy stored in chemical bonds of molecules. As bonds are broken or formed in a chemical reaction, there will be a heat change qrxn in the reaction system. w = -179 J. The system does work. All media copyright of their respective owners All media copyright of their respective owners 47 All media copyright of their respective owners 48 12 Endothermic and exothermic Endothermic and exothermic Endothermic reaction - heat moves from the surroundings into the system. qrxn is positive. Exothermic reaction - system releases heat to the surroundings. qrxn is negative. All media copyright of their respective owners 49 Bomb calorimeter 50 Bomb calorimeter A bomb calorimeter is an isolated system where a heat change of a (often combustion) reaction results in a temperature increase of the calorimeter, which is made up of water, the bomb and any other parts that are not part of the reaction system All media copyright of their respective owners All media copyright of their respective owners Because the bomb is a rigid, tightly sealed container, the chemical system CAN NOT change its volume. There is no PV work! 51 All media copyright of their respective owners 52 13 Bomb calorimeter Problem ∆rU = qrxn,V = -qcalorimeter = -(qwater + qbomb + qother parts) = - Ccalorimeter x ∆Tcalorimeter All media copyright of their respective owners The combustion of 1.013 g of vanillin, C8H8O3 in a bomb calorimeter with a Ccalorimeter of 4.90 kJ K-1 results in a temperature rise from 24.89 to 30.09 ºC. What is the heat of combustion (∆rU) of vanillin in kJ mol-1? Molar mass of vanillin is 152.147 g mol-1 53 Problem answer All media copyright of their respective owners 54 Internal energy and enthalpy ∆rU =qcombustion = qrxn,V = -3.83 x 103 kJ mol-1 ∆U = q + w ∆U = q - P∆ ∆V So heat transferred in a process or reaction is q = ∆U + P∆ ∆V A process at constant volume where ∆V = 0 (like in a bomb calorimeter – slide 52) has qV = ∆U All media copyright of their respective owners 55 All media copyright of their respective owners 56 14 Internal energy and enthalpy Endothermic and exothermic At a constant pressure (like an open beaker), the volume of the system can change, and so P∆ ∆V is not always zero, and therefore the heat is Endothermic reaction – the products are higher in enthalpy than the reactants ∆H is positive. Exothermic reaction - the products are lower in enthalpy than the reactants ∆H is negative. qP = ∆U + P∆ ∆V = ∆H Notice we give the heat at constant pressure a special name and symbol. We call it the change in enthalpy (∆ ∆H) All media copyright of their respective owners 57 Enthalpy is a state function All media copyright of their respective owners 58 Internal energy and enthalpy Enthalpy (H), like internal energy (U), is a state function. Since ONLY one path can be taken “at constant pressure”, qp behaves like a state function! ∆H = Hfinal - Hinitial All media copyright of their respective owners 59 All media copyright of their respective owners 60 15 Enthalpy is a state function A convenient relationship The difference between ∆U and ∆H is generally pretty small. We will almost exclusively talk about ∆H (open beakers) for reactions unless we are dealing with a bomb calorimeter. There any heat change at constant volume will give us ∆U. All media copyright of their respective owners 61 How “small” is P∆V? All media copyright of their respective owners 62 How “small” is P∆V? The PV work done on the system is +2.5 kJ, which happens to be RT when we lose 1 mole of gas per mole of rxn. This 2.5 kJ represents a value of work that is about 0.44% of the enthalpy (heat) change for this reaction! Consider the reaction 2 CO(g) + O2 (g) 2 CO2 (g) ∆U for this reaction at 298 K and constant V is -566.0 kJ mol-1 ∆H for this reaction at 298 K and constant P is -563.5 kJ mol-1 All media copyright of their respective owners In a reaction involving just gases treated with the ideal gas law, then at a constant temperature and constant pressure PV = nRT w = P∆ ∆V = ∆(nRT) w = P∆ ∆V = ∆nRT The work done at constant P and T can be related to the change in the total number of moles of gas! 63 All media copyright of their respective owners 64 16 Problem Problem answer ∆U = -2.008 x 103 kJ mol-1 and ∆H = -2.012 x 103 kJ mol-1 The heat of combustion of liquid 2-propanol (C3H8O) is determined to be -33.41 kJ g-1 at 298 K in a bomb calorimeter. For the combustion of 1 mole of 2-propanol, what is ∆U and ∆H? All media copyright of their respective owners 65 “Coffee-cup” calorimeter 66 “Coffee-cup” calorimeter A “coffee-cup” calorimeter is treated as an isolated system where a heat change of a reaction results in a temperature change of the calorimeter, which is made up of water, the cup and any other parts that are not part of the reaction system. All media copyright of their respective owners All media copyright of their respective owners ∆rH = qrxn,P = -qcalorimeter = -(qwater + qcoffee cup + qother parts) We OFTEN assume the cup, other parts and any non-water reactants and products have small heat capacities relative to water, so ∆rH = qrxn,P ≈ -qwater qrxn ≈ -mwater x Cs,water x ∆Twater qrxn ≈ -(dwater x Vwater) x Cs,water x ∆Twater where d and V are the density and volume of the water in the cup 67 All media copyright of their respective owners 68 17 Problem Problem answer 100.0 mL of 1.00 M AgNO3 (aq) and 100.0 mL of NaCl (aq), both at 22.4 ºC, are mixed in a coffee-cup calorimeter. If the reaction is Ag+ (aq) + Cl- (aq) AgCl (s) and the temperature rises to 30.2 ºC, then what is qrxn per mole of AgCl formed. Assume you can ignore qcoffee cup (since Styrofoam has a high heat capacity) and the mixed solution has Cs and density equal to that of pure water (1.00g mL-1) All media copyright of their respective owners 69 Enthalpy is extensive All media copyright of their respective owners 70 Enthalpy is extensive Enthalpy is an extensive property – it changes directly with the amount of reaction we consider, whether stoichiometricly or in actual amounts. If ½ N2 (g) + ½ O2 (g) NO (g) ∆rH = 90.25 kJ mol-1 then N2 (g) + O2 (g) 2 NO (g) ∆rH = 2 x 90.25 kJ mol-1 = 180.50 kJ mol-1 All media copyright of their respective owners qrxn = -65 kJ mol-1 ½ N2 (g) + ½ O2 (g) NO (g) ∆rH = 90.25 kJ mol-1 N2 (g) + O2 (g) 2 NO (g) ∆rH = 2 x 90.25 kJ = 180.50 kJ mol-1 Notice that regardless of how we’ve written the balanced equation, in each case we have an enthalpy change of 90.25 kJ mol-1 of NO formed 71 All media copyright of their respective owners 72 18 Enthalpy for reversed process State functions If we reverse a process (switch the initial and final states around), the sign of the enthalpy change will change as well. If ½ N2 (g) + ½ O2 (g) NO (g) Internal energy is a state function. A state function is a value or property that depends only on the current condition of the system. ∆rH = 90.25 kJ mol-1 then NO (g) ½ N2 (g) + ½ O2 (g) ∆rH = -90.25 kJ mol-1 All media copyright of their respective owners 73 State functions revisited A change in a state function only requires us to know the initial state and the final state – not the “path” that got us from one to the other. ∆rH = Hfinal - Hinitial 74 State functions revisited Height above sea level is a state function. Here the enthalpy change for A + 2B → 2D is the same as the sum of the enthalpy changes for the rxns A + 2B → C C → 2D All media copyright of their respective owners All media copyright of their respective owners 75 All media copyright of their respective owners 76 19 Path-dependent functions Distance travelled is a pathdependent function. All media copyright of their respective owners 77 All media copyright of their respective owners 78 All media copyright of their respective owners 79 All media copyright of their respective owners 80 20 Enthalpy is a state function too! ½ N2 (g) + O2 (g) NO2 (g) The overall change in height is -425 ft from Signal Hill to Fort Amherst, no matter what steps we take. It is a state function! We’re interested in the enthalpy change of the overall reaction above. We can use Hess’s Law on the reactions: ½ N2(g) + O2(g) NO(g) + ½ O2(g) ∆rH = 90.25 kJ mol-1 NO(g) + ½ O2(g) NO2(g) ∆rH = -57.07 kJ mol-1 Hess’s Law states the overall enthalpy change for a reaction is equal to the sum of enthalpy changes for all the individual steps that lead to the same overall reaction. All media copyright of their respective owners 81 How do we use Hess’s Law? 82 How do we use Hess’s Law? - number the step reactions - make a list of the chemicals from the overall reaction, including states - in the chemical list put the numbers of all step reactions where the same chemical appears We need a balanced equation for the reaction whose enthalpy change we are interested in. This is the overall reaction. We also need balanced equations with enthalpy change data for the step reactions. All media copyright of their respective owners All media copyright of their respective owners 83 All media copyright of their respective owners 84 21 How do we use Hess’s Law? How do we use Hess’s Law? - identify the chemicals that appear in ONLY ONE step reaction - we want these step reaction chemicals to match the overall reaction both in amounts and as reactants or products All media copyright of their respective owners 85 How do we use Hess’s Law? All media copyright of their respective owners 86 How do we use Hess’s Law? - add the modified step reactions - group together and cancel out similar chemicals based on which side of the arrow they appear - ∆rH for this sum reaction is the added ∆rH values of the modified step reactions All media copyright of their respective owners - to make the match we are allowed to reverse the step reaction and multiply the step reaction amounts - if we reverse the step reaction we must change the sign of ∆rH - if we multiply the step reaction amounts we must multiply ∆rH by the same amount 87 - if this sum reaction matches the overall reaction, we are done and have found ∆H - if not, use other step reactions to cancel out the chemicals that do not appear in the overall reaction All media copyright of their respective owners 88 22 How do we use Hess’s Law? Problem - multiply to match amounts but have step reaction chemicals on opposite side of arrow (by reversing) as in the sum reaction - REMEMBER to change ∆rH of the step reaction - add sum reaction and step reactions to give overall reaction and add ∆rH values All media copyright of their respective owners Methane, the main constituent of natural gas, burns in oxygen to yield carbon dioxide and water: CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (l) Use the following data to calculate ∆rH (in kJ mol-1) CH4 (g) + O2 (g) CH2O (g) + H2O (g) ∆rH = -284 kJ mol-1 CH2O (g) + O2 (g) CO2 (g) + H2O (g) ∆rH = -518 kJ mol-1 H2O (l) H2O (g) ∆rH = 44.0 kJ mol-1 89 Problem answer All media copyright of their respective owners 90 Problem CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l) ∆rH = -8.90 x 102 kJ Use the following data to find ∆rH (in kJ mol-1) 4 C (s) + 6 H2 (g) + O2 (g) 2 CH3CH2OH (l) ∆H° = -555.2 kJ mol-1 2 C (s) + 2 H2 (g) + O2 (g) CH3COOH (l) ∆H° = -484.3 kJ mol-1 2 H2 (g) + O2 (g) 2 H2O (g) ∆H° = -483.6 kJ mol-1 H2O (l) H2O (g) ∆H° = 44.0 kJ mol-1 All media copyright of their respective owners 91 All media copyright of their respective owners 92 23 Problem answer Standard states We must be careful when looking at energy changes. Any differences in the system will mean differences in enthalpy changes as well ∆H° = -492.5 kJ We must pay attention to differences in amounts, chemical states, pressure, temperature, and volume All media copyright of their respective owners 93 All media copyright of their respective owners 94 Why are they different? Common experience tells us that gaseous water has higher internal energy than liquid water (we heat water to boil it). 1 H2O (l) 1 H2O (g) 4 H2O (l) 4 H2O (g) Notice the one difference between these reactions, and the differences in ∆rH The difference between the enthalpy changes is (-2043 kJ mol-1) – (-2219 kJ mol-1) = 174 kJ mol-1 All media copyright of their respective owners 95 ∆H = 44.0 kJ mol-1 ∆H = 176 kJ mol-1 Notice the change in enthalpy of 4 liquid water to 4 gas water is essentially the same as the difference on the last slide. All media copyright of their respective owners 96 24 Thermodynamic standard state Thermodynamic standard state - pure substances - physical state must be explicitly mentioned - gases at one bar of pressure - solution concentrations of one mole per litre temperature is NOT PART of the standard state definition, but must be mentioned (usually 298.15 K) All media copyright of their respective owners 97 Standard enthalpies of formation Property values measured at thermodynamic standard state conditions are emphasized by the special symbol °. So if we measure the enthalpy change for a reaction at the thermodynamic standard state conditions, then we have the standard enthalpy of the reaction ∆rH° All media copyright of their respective owners 98 Standard enthalpies of formation The standard enthalpy (or heat) of formation (∆ ∆fH°) is the enthalpy change for the formation of 1 mole of a substance in its standard state from its constituent elements in their reference forms in their standard states. All media copyright of their respective owners 99 All media copyright of their respective owners 100 25 What is ∆H°f for O2(g)? O2 (g) O2 (g) Reference form? formation reaction Obviously, no change occurs here, and so ∆fH° for oxygen gas is zero. In fact, for any pure element in its reference form and standard state, we define the standard enthalpy of formation as zero. All media copyright of their respective owners 101 Reference form? All media copyright of their respective owners 102 ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) We can find the standard enthalpy change for a chemical reaction by ∆rH° = Σνp∆fH°(prod) - Σνr∆fH° Here ν is the stoichiometric coefficient from the balanced equation! The reference form of an element is almost always the most stable form of the element at thermodynamic standard state conditions. However solid “red” phosphorus is more stable than solid “white” phosphorus, but we choose white phosphorous as the reference form All media copyright of their respective owners The reference form of an element is almost always the most stable form of the element at thermodynamic standard state conditions. Therefore we choose to use C(graphite) instead of C(diamond) and Br2(l) instead of Br2(g) as the reference forms for these elements. 103 All media copyright of their respective owners 104 26 Example ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) Let’s look at the reaction ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) CH3CH2OH (l) + O2 (g) CH3COOH (l) + H2O (l) We add together νp∆fH°(prod) for ALL products, add together νr∆fH° °(react) for ALL reactants, and take the difference of the sums (products minus reactants) For the four chemicals involved we can come up with the formation reactions, and get ∆fH° values from the text. ∆H°f = -277.6 kJ mol-1 2 C (s) + 3 H2 (g) + ½ O2 (g) CH3CH2OH (l) ∆H°f = 0.0 kJ mol-1 O2 (g) O2 (g) ∆H°f = -484.3 kJmol-1 2 C (s) + 2 H2 (g) + O2 (g) CH3COOH (l) ∆H°f = -285.8 kJ mol-1 H2 (g) + ½ O2 (g) H2O (l) All media copyright of their respective owners 105 ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) 106 ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) The result ∆H°r = -492.5 kJ looks familiar because this is the same overall reaction we looked at in slides 92 and 93 where we applied Hess’s Law. ∆rH° = Σνp∆fH°(prod) - Σνr∆fH°(react) is Hess’s Law! Here the very specific set of step reactions we use are the formation reactions for all chemicals involved All media copyright of their respective owners All media copyright of their respective owners 107 The application of the equation represents the hypothetical case where we break our reactants down into the elements in their standard states and then recombine the elements to make the products. All media copyright of their respective owners 108 27 Problem Problem answer Use data from Table 6.5 to calculate the standard enthalpy of combustion of ethanol at 298.15 K. All media copyright of their respective owners 109 ∆rH° = -1367 kJ mol-1 All media copyright of their respective owners 110 Chapter 6-10 Energy Use and the Environment Read and study this part of the textbook for yourself. All media copyright of their respective owners 111 28