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Heat Defined as transfer of energy RECALL: Kelvin temperature is a measure of kinetic energy The higher the kelvin temperature of a sample of particles, the more they move. Vectors can be used to represent individual and average velocities of particles at two different temperatures. Notice that the sum of the vectors (length/magnitude) representing particles at a higher temperature is greater than the sum of the particles at a lower temperature. HIGH TEMP LOW TEMP Example: a 500 mL sample of water at 25 oC has less energy than the same volume of water at 35 0C. Doubling the Kelvin temperature of a system, also doubles the kinetic energy as represented by the equation below. KE per molecule = 1 mv2 2 This also must mean a change in the velocity of the particles RECALL (from gas laws): that change in velocity however, is NOT directly proportionate. Rather, the change in velocity is dependent on the mass of the particle. More massive particles will have a less significant change in velocity (root mean square velocity) Heat: Energy transfer When two systems that are at different temperatures come into thermal contact (i.e. the particles collide) Energy is transferred from the hotter one to the cooler one, until equal temperatures to each other is achieved. Average kinetic energy of the particles in each system becomes the same A single, intermediate temperature is achieved. Transfer of energy is called HEAT or HEAT TRANSFER. Heat is NOT a substance When the intermediate, same temperature is achieved, there is no further transfer of energy and THERMAL EQUILIBRIUM is achieved. Specific Heat Capacity Defined as the amount of energy required to raise the temperature of a 1 g sample of a substance by 1 oC or 1 K. Different for different substances (intrinsic property) Implications Transfer of an identical amount of energy to the SAME mass of two different substances will NOT result in the same temperature change of the two substances. Ex: when you heat a pot of water on the stove Is it possible that the mass of water is LESS than the mass of the pot? Which gets hotter first-the pot or the water? Why? The amount of energy transferred (q) can be related to the temperature change of a substance (∆T) by the equation below, where m = mass and c = specific heat capacity (unique value, must be given) q = m c ∆T cwater = 4.186 J/g ·oC Practice Given that the specific heat of water is 4.186 J/g ·oC, determine the amount of energy required to raise the temperature of 125 g sample from room temperature to 37.5 oC A 200. g sample of water requires 4186 J energy to raise the temperature by 5 oC. Given the specific heat values below, predict which sample will have the greatest increase in temperature, if each sample is the same mass as the water sample and the same amount of energy is applied to each. 0.386 J/g ·oC Al: 0.900 J/g ·oC Steel: 0.49 J/g ·oC Calculate the temperature change to each vessel and compare to your prediction. Cu: Heating and Cooling Curves It is possible to investigate the changes in a substance as energy is either added or removed at constant pressure. In these cases, heating and cooling curves result. Starting with a solid below its melting point, the following effects can be observed The T (temperature) of the solid increases at a constant rate until it begins to melt (i.e. reaches its m.p at the given pressure) **RECALL: normal melting and boiling points are the temperature at which these changes occur at 1 atm of pressure) When melting begins, the temperature is constant until the solid has all turned to a liquid The T of the liquid increases at a constant rate unit it begins to boil (i.e reaches it b.p) When boiling begins, the temperature is constant until the liquid has all turned to gas The temperature of the gas increases at a constant rate. In summary, energy is either being used to increase temperature, OR change phase, but not both. In regions where the temperature of the solid, liquid or gas is being increased, the amount of energy being added is q = m c ∆T Where q = energy, m = mass, c = specific heat of the substance (in that state) and ∆T = Tfinal – Tinitial, i.e. the temperature change Where the solid is melting, the amount of energy being added is q = (∆Hfusion) (moles) Where ∆Hfusion is the molar enthalpy of fusion and representes the energy absorbed when 1 mole of a solid melts Where the liquid is boiling, the amount of energy being added is q = (∆Hvaporization) (moles) Where ∆Hvaporization is the molar enthalpy of vaporization and represents the energy absorbed when 1 mole of a liquid vaporizes. **Keep in mind/common points of mistake Conventionally, q has units of Joules (J) and ∆H has units of kiloJoules (kJ) (must convert one or another to ensure proper magnitude and units. temperature change of solid or liquid uses mass, ∆H uses moles (again, must convert one or the other to ensure proper magnitude and units. In a heating curve, energy is always positive (INTER bonds are being broken) Energy The capacity to do work or generate heat. The FIRST LAW OF THERMODYNAMICS is also called the Law of Conservation of Energy, and it states that the total energy of the universe is constant Universe is composed of the system and its surroundings. We ALWAYS want to consider energy change from the system’s perspective. Energy can be gained by the system (+) in which case the surrounding must lose energy (-) or vice versa Energy transfer can occur through HEAT or WORK Heat vs. Work Ball A and ball B at initial position have some internal energy (potential). •Thus, there are their two ways to transfer At their final positions, potential energies have energy— changed. •through work (generally movement) Consider the energy transitions (generally chemical between•through the initialheat and final reaction/particle changes) positions Ball A moves and two energy transitions occur Ball A does work on ball B Ball A loses some energy to the hill due to frictional heating. •What if the hill is bumpy and Ball A comes to rest before hitting ball B. •Potential energy of Ball A is still reduced •ALL of the energy transfer is due to frictional heating. •No work is done Therefore, the way energy is transferred is PATHWAY dependent RECALL: whether Ball A hits and Ball B moves or not determines how the energy was transferred, not how much. However, regardless of the pathway, the total energy of the system and surroundings remains constant The total energy gained and lost remains the same Total energy of the system is INDEPENDENT of the pathway. **Energy is a STATE FUNCTION (property) Depends ONLY on the present state of the system, not how it got there (pathway independent) Energy transitions in chemical systems Consider the products of a hydrocarbon combustion reaction CH4 C2H6 + O2 CO2 + H2O + energy C3H8 Carbon dioxide and water are produced in all combustion of hydrocarbon reactions. However, regardless of HOW CO2 and H2O are produced, the internal energy (bond energies) of CO2 and H2O remains the same Actual reactions may require different mechanisms, but net energy change for the formation of CO2 and H2O is the same but, in the three examples above, the actual NET change will be different for the OVERALL reactions Classification of energy transitions Energy transitions (heat transfer) in chemical reactions are always classified with respect to the system, as exothermic (net output): potential energy of products is less than the reactants, OR endothermic (net input): potential energy of the products is greater than the reactants The opposite change must occur to the surrounding The overall energy of the universe will remain constant Concept check Classify each process as exothermic or endothermic. Explain. The system is underlined in each example. Exo a) Endo b) Endo c) Exo d) Endo e) Your hand gets cold when you touch ice. The ice gets warmer when you touch it. Water boils in a kettle being heated on a stove. Water vapor condenses on a cold pipe. Ice cream melts. Work Work: scientific (physical) definition—use of force, F, to move an object over some distance. Specifically in chemistry, work is considered only in terms of expansion or compression (contraction) of gases. Energy transfer can occur through work. Consider a gas inside a vessel with a movable piston As the gas expands, the particles collide with the piston and energy is transferred from the gas to the piston (and the piston moves away) Work is done BY the gas Energy is lost Quantitative analysis of work Work can by quantitatively analyzed by considering the following change in volume of the gas, and External pressure, P, that the gas is working against Via equation w = -P∆V (see attachment for full derivation) Because of law of conservation of energy, any energy lost (- w) by one system(in this case the gas), must be gained (+w) in equal magnitude by another (in this case the piston). In this work scenario, energy flows from one system (gas) to the other (piston) Practice Calculate the work associated with the expansion of a gas from 46.L to 64 L at a constant external pressure of 15atm. For a gas at constant P, w= -P∆V P = 15 atm And ∆V = 46-64 = -18 L W = 15atm x (-18 L) = -270 L·atm Note that…. For an ideal gas, work can occur ONLY when its V changes. Thus, if a gas is heated at constant volume, the pressure increases but no work occurs. First Law From the calculations of heat (q) and work (w), we can calculate change in internal energy of system. E = q + w Where E = change in system’s internal energy Sign is the direction of the flow. (system point of view) q = + endothermic system E increases q = - exothermic system E decreases AND w = + gas compressed w = - gas expands system E increases system E decreases Joules Energy is measured in Joules kg m J 2 s Work gives units of L·atm Conversion to J 1 L·atm = 101.3 J 2 Practice A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the volume of the balloon changes form 4.00 x 106 L to 4.50 x 106 L by the addition of 1.3 x 108 J of energy as heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process (1L atm = 101.3 J) Calorimetry Experimental technique used to measure energy changes in a chemical system. Process involves Chemical reaction (or a phase change) Thermal contact with heat bath (usually water) Heat capacity of the heat bath must be known. Ideally no heat loss or gain with the universe Only heat transfer is between system (reaction/phase change) and surrounding (heat bath) 26 As the chemical reaction (or phase change) takes place Energy is transferred between reaction/phase change and heat bath Since specific heat capacity of heat bath is known, we can calculate amount of energy transferred (gained or lost) by applying q = m c ∆T If temperature of heat bath goes up, the chemical reaction (or phase change) must have released energy (i.e, the reaction was exothermic). If the temperature of heat bath goes down, the chemical reaction (or phase change) must have absorbed energy (i.e. the reaction was endothermic) Applying the Law of Conservation of Energy (and assuming that there are no energy losses), we can conclude Magnitude of the energy lost or gained by the chemical reaction (or phase change, must be equal to the magnitude of the energy gained or lost y the heat bath. 28 Therefore, for exothermic reaction - q (energy lost by system) = + q (energy gained by surroundings -(m c ∆T)system = + (m c ∆T)surroundings OR For an endothermic reaction - q (energy lost by surrounding) = + q (energy gained by system) -(m c ∆T)surroundings = + (m c ∆T)system 29 Practice Propane is a gas that is commonly used in gas grills. A sample of propane with a mass of 44.0 g is completely burned in oxygen and in the process it releases 2002 kJ of energy. This chemical reaction is brought in contact with a water bath, and the transfer of energy from the reaction to the water takes place. Assuming the specific heat capacity of water is 4.184 J/g·K, and that in such an experiment 100% of the energy generated is transferred to the water causing the water temperature to increase, answer the questions below In which direction does the energy flow? Calculate the change in temperature of 20.00 kg of water Is the combustion (burning) of propane an exothermic or endothermic process? Explain your answer. Practice Consider the following data Specific heat capacity of ice 2.05 J/gK Specific heat capacity of water 4.184 J/gK Specific heat capacity of steam 2.08 J/gK Molar heat of fusion for H2O 6.02 kJ/mol Molar heat of vaporization for H2O 40.7 kJ/mol 1. Determine the energy change when 1.00 mol (18.0 g) of water at 20.0 oC is frozen, and then cooled to a temperature of -2.00 oC a. Is the process exothermic or endothermic 2. Considering ONLY the energy change in the phase change above, what would be the temperature change observed in a heat bath made of gold with a mass of 0.502 kg? The specific heat capacity of gold is 0.129 J/gK a. Is the process exothermic or endothermic? Practice Example: If a 35.0 gram block of Aluminum at 375 K is added to 200. g water at 300 K, calculate final temperature, assuming no heat is lost. Specific heat of Al: 0.92 J/gK Enthalpy, H What is Enthalpy? a thermodynamic quantity equivalent to the total heat content of a system. equal to the internal energy of the system plus the product of pressure and volume. Enthalpy = H = E + PV Since it’s rather difficult to measure heat content directly, we are more interested in the change in Enthalpy or ΔH State function 34 Determining Enthalpy Recall: At constant pressure, the only work allowed is PV qP = defined as heat at constant pressure, and is calculated from E = qP + w = qP PV qP = E + PV= H Therefore, H = energy flow as heat (at constant pressure) **Heat of reaction and change in enthalpy are interchangeable** 35 Enthalpy (H) is used to quantify the heat flow into (or out of) a system in a process that occurs at constant pressure. H = H (products) – H (reactants) H = heat given off or absorbed during a reaction at constant pressure Hproducts < Hreactants H < 0 Hproducts > Hreactants exotherm H > 0 endotherm 6.4 Standard Enthalpy of Formation Defined as the CHANGE IN ENTHALPY associated with the FORMATION of ONE MOLE of a COMPOUND from its COMPONENT ELEMENTS with ALL substances in their STANDARD STATES. Symbol: ΔHof, where the degree o symbol indicates that all substances are in their STANDARD STATES, and subscript f indicates “formation” Examples: 1/2 N2(g) + O2(g) NO2(g) C(s) + 2H2(g) + 1/2 O2(g) CH3OH(l) ΔHof ΔHof = 34 kJ/mol = -239 kJ/mol **Manipulate equation in such as way that EXACTLY, ONLY 1 mol product is formed Standard Enthalpy of Formation of elements is ALWAYS ZERO “0”. WHY? because elements are NOT formed through chemical reactions. Sometimes difficult to determine directly. Example: Formation of DIAMOND from graphite. ΔHof CANNOT be obtained directly. Process is too slow. ΔHof can be obtained from OTHER KNOWN processes, such as ENTHALPY OF COMBUSTION **unlike standard Enthalpy of Formation, Enthalpy of Combustion is given for EXACTLY 1 mol of the substance combusting. WHY? Combustion products are variable. How can you keep track of which one to count/measure. Easier to consider substance actually undergoing combustion. Factors to Consider Conventional Definitions of Standard States For a Compound The standard state of a gaseous substance (ex: NO2) is a pressure of exactly 1 ATMOSPHERE For a pure substance in a condensed state (liquid or solid), the standard state is the PURE LIQUID OR SOLID . For a substance present in solution, the standard state is a CONCENTRATION OF EXACTLY 1M For an Element The standard state of an element is the form in which the element exists under condition of 1 ATM AND 25oC. Ex: Standard state of Oxygen: O2(g) at a pressure of 1 ATM and 25oC Ex: Standard state of Sodium: Na(s) Ex: Standard state of Mercury: Hg(l) Determining Enthalpy of Reaction from Standard Enthalpy Values Enthalpy values can be used for various types of calculations Enthalpy of reaction: energy transitions in a given reaction Standard enthalpy of combustion Hess’s Law Others Rules to Follow to Calculate Enthalpy of Reaction (and generally) When a reaction is reversed, the magnitude of ΔH remains the SAME, but its SIGN CHANGES When the balanced equation for a reaction is multiplied by an integer (or fraction), the value of ΔH for that reaction MUST ALSO BE MULTIPLIED by the SAME FACTOR Fractional equations are allowed The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products ΔHorxn = Σ np ΔHof (products) - Σ nr ΔHof (reactants) where n is # moles Elements in their standards states are NOT INCLUDED in the ΔHoreaction calculations, because ΔHof for an element in its standard state is ZERO. Example 1: Determine the standard enthalpy of reaction for the combustion of methane Pathway for combustion of methane. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Reactants are first taken apart in (a) and (b) then used to assemble the products in reactions (c) and (d). 41 Figure 6.9: A schematic diagram of the energy changes for the reaction CH4(g) + 2O2(g) CO2(g) + 2H2O(l). Reverse of the formation 42 Apply the equation for change of enthalpy given the standard enthalpy of formation of each substance, given CH4(g) + 2O2(g) CO2(g) + 2H2O(l). (a) formation of CH4 (b) Formation of O2 (c) formation of CO2 (d) formation of H2O(l) ΔHorxn= ΔHorxn Σ np ΔHof (products) ΔHof = -75kJ/mol ΔHºf = 0 (b/c already an element) ΔHºf = -394kJ/mol ΔHºf = -286kJ/mol - Σ nr ΔHof (reactants) where n is # moles = (1 mol CO2(-394 kJ/mol) + 2 mol H2O(-286 kJ/mol)) – (1 mol CH4(-75 kJ/mol ) + 2 mol O2(0 kJ/mol)) NOTE: reverse sign for methane, use 2 moles water, note water (g) vs. (l) 43 Example 2: Using a standard enthalpy of formation reference table, determine the standard enthalpy change for the overall reaction that occurs when ammonia is burned in air to form nitrogen dioxide and water. This is the first step in the formation of nitric acid Using enthalpies of formation, determine the standard change in enthalpy for the thermite reacation, which occurs when a mixture of powdered aluminum and iron III oxide is ignited with a magnesium fuse. Methanol (CH3OH) is often used as a fuel in high performance engines in race cards. Use your reference table to compare the standard enthalpy of combustion per gram of methanol with the combustion per gram of gasoline. Gasoline is actually a mixture of compounds, but assume for this problem that gasoline is a pure substance (C8H18) Standard Enthalpy of Combustion Heat of reaction for the combustion of exactly 1 mole of a substance (commonly hydrocarbon) Practice: Determine ∆Hcombustion for each of the following hydrocarbons, using standard enthalpy of formation values. Methane (CH4) Propane (C3H6) Ethanol (C2H5OH) Hess’s Law Enthalpy (state function) Reactants Products The change in enthalpy is the same whether the reaction takes place in one step or a series of steps. 47 Figure 6.7: The principle of Hess's law. The same change in enthalpy occurs when nitrogen and oxygen react to form nitrogen dioxide, regardless of whether the reaction occurs in one (red-right) or two (blue-left) steps. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 48 Oxidation of N2 to produce NO2 N2(g) + 2O2(g) 2NO2(g) ΔH1=68kJ N2(g) + O2(g) 2NO(g) + O2(g) N2(g) + 2O2(g) Or.. 2NO (g) 2NO2(g) 2NO2(g) ΔH2=180kJ ΔH3=-112kJ ΔH=68kJ 49 Calculations via Hess’s Law 1. If a reaction is reversed, H is also reversed. N2(g) + O2(g) 2NO(g) 2NO(g) N2(g) + O2(g) 2. H = 180 kJ H = 180 kJ If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g) 3N2(g) + 3O2(g) H = 540 kJ 50 3. Fractional equations **Coefficients do not have to be whole numbers, as long as ALL values are adjusted appropriately Example: Determine enthalpy change if half mole of nitrogen monoxide gas is produced, given the equation: N2(g) + O2(g) 2NO(g) Hrxn = 180 kJ 0.25 N2(g) + 0.25 O2(g) 0.50 NO(g) Hrxn = 45 kJ Two forms of carbon are graphite, the soft, black, slippery material used in “lead” pencils and as a lubricant for locks, and diamond, the brilliant, hard gemstone. Using the enthalpies of combustion for graphite (-394 kJ/mol) and diamond (-396kJ/mol), calculate ∆H for the conversion of graphite to diamond: Cgraphite(s) Cdiamond(s) Copyright©2000 by Houghton Mifflin Company. All rights reserved. 52 Cgraphite(s) Cdiamond(s) Cgraphite(s) +O2 (g) CO2 (g) Cdiamond(s) + O2 (g) CO2 (g) ∆H=-394 kJ ∆H=-396 kJ Cgraphite(s) +O2 (g) CO2 (g) ∆H=-394 kJ CO2 Cdiamond(s) + O2 (g) ∆H=-(-396 kJ) Cgraphite(s) Cdiamond(s) Copyright©2000 by Houghton Mifflin Company. All rights reserved. 53 Diborane (B2H6) is a highly reactive boron hydride, which was once considered as a possible rocket fuel for the US space program. Calculate ∆H for the synthesis of diborane from its elements, according to the equation 2B(s) + 3H2 (g) B2H6(g) Using the following data: Reactions: 2B(s) + 3/2 O2(g) B2H6(g) + 3O2(g) H2(g) + 1/2O2(g) H2O(l) B2O3(s) B2O3(s) + 3H2O(g) H2O(l) H2O(g) Copyright©2000 by Houghton Mifflin Company. All rights reserved. ΔH ΔH ΔH ΔH -1273 kJ -2035 kJ -286 kJ 44kJ 54 BOND ENERGIES and ENTHALPIES RECALL: Atoms are attracted to each other when the outer electrons of one atom are electrostatically attracted to the nuclei of another atom Attraction between the two atoms makes them increasingly stable, giving lower and lower potential energies However, if atoms continue to approach one another, and get increasingly close, there comes a point at which the two nuclei (and/or the more densely packed core electrons) will start to repel each other As they start to repel, potential energy is raised, and the atoms become less stable Happy medium (i.e. bond formed) at a distance where the force of attraction and repulsion results in the LOWEST potential energy Distance between the nuclei at this point is the bond length Potential energy at this point is the bond strength Plotting potential energy vs. bond length Typical plot of potential energy vs. bond length appears as follows Ex: Hydrogen, H2 bond length: 74 pm bond strength: 436 kJ/mol Since forces of attraction stabilize atoms, the greater the number of electrons involved, the stronger the attraction Result: triple bonds tend to be stronger than double bonds, and double bonds are stronger than single bonds Shorter bonds also tend to be stronger than longer bonds Practice Assume that the potential energy vs. bond length plot represents Oxygen, draw a curve representing Nitrogen. Bond dissociation energy RECALL: Bond breaking is generally an endothermic process (ΔH is +) and bond making is generally an exothermic process (ΔH is -) (exceptions exist) The strength of a bond can be measured by measuring the amount of energy required to break a bond Ex: H2 2H Bond Dissociation Energy (ΔH) = +436 kJ Consider a polyatomic molecule, such as methane, CH4 Composed of 4 identical C-H bonds. However, each bond dissociation energy is actually quite different CH4(g) CH3(g) CH2(g) CH(g) CH3(g) CH2(g) CH (g) C(g) + + + + TOTAL: H(g) H(g) H(g) H(g) 435 kJ/mol 453 kJ/mol 425 kJ/mol 339 kJ/mol 1652 kJ/mol TO overcome complexity of accounting for each individual bond dissociation energy, we define bond energy as the AVERAGE of each of the four, separate C-H bonds. Therefore 1652/4 = 413 kJ/mol Determining ΔH values based on Bond strength The energy change in a reaction can be calculated by Summing up the energy required to break each of the bonds in the reactants (a positive value) Summing up the energy released by making each of the bonds in the products (a negative value) Summing the totals of the two values above If the total energy needed to break the reactant bonds is greater than the total energy released in making the product bonds, then the reaction (system) is endothermic Energy is absorbed from surrounding Temperature of surrounding decrease If the total energy needed to break the reactant bonds is less than the totoal energy released in making the product bonds then the reaction (system) is exothermic Energy is released to the surrounding Temperature of surrounding rises Common Bond Enthalpy Values Practice Calculate the standard enthalpy change of reaction below CH3CH=CH2 + H2 CH3CH2CH3 2. CH3CH=CH2 + Br2 CH2BrCHBrCH3 3. CH4 + 2Cl2 + 2F2 CF2Cl2 + 2HF + 2HCl 4. CH3CH2OH + CH3COOH CH3CO2CH2CH3 + H2O (solve WITHOUT any calculations) 1. Born Haber Process Describes the thermochemistry of ionic bonds Process of ionic bond formation can be broken down into several stages. Ex: two possible routes for formation of sodium chloride Single step process (Enthalpy change = standard enthalpy of formation of NaCl) Na(s) + ½ Cl2(g) NaCl(s) Recall: standard enthalpy of formation is the enthalpy change when one mole of a substance is formed from its component elements in their standard states. 63 A multi-step process involving five separate changes 1. Atomization of sodium Na(s) Na(g) ∆Ha(M) = standard enthalpy of atomization is the energy required to form one mole of gaseous atoms from the element under standard conditions (endothermic) 2. Ionization of sodium Na(g) Na+1(g) + e∆Hi(1) = First ionization energy is the energy required to remove one mole electrons from one mole of atoms in the gaseous phase (endothermic) 64 3. Dissociation of chlorine molecules ½ Cl2(g) Cl(g) ∆Ha(NM) =Also considered as atomization of Chlorine molecules, equal to 1/2 standard bond dissociation energy of chlorine (since only ½ mole Cl2 is required to form 1 mol NaCl The standard enthalpy of bond dissociation is the energy required to dissociate one mole of molecules into atoms. (endothermic) 65 4. Formation of gaseous choride ions from gaseous chlorine atoms Cl(g) + e Cl-1(g) ∆HEa =The first electron affinity is the enthalpy change when one mole of gaseous atoms gains an electron to form a mole of gaseous ions. (exothermic) 4. Bring together the gaseous ions Na+1(g) Cl-1(g) NaCl(s) ∆HLE =The standard lattice enthalpy is the enthalpy change when one mole of a solid is formed from its constituents gaseous ions (exothermic) 66 Born-Haber Cycle Diagram 67 RECALL: Hess’s Law states that the energy change for a reaction is independent of the route taken, therefore, ΔHf = ΔHa(Me) + ΔHi(1) + ΔHa(NM) + ΔHEa + ΔHLE Standard conventions Arrows pointing upwards represent endothermic changes (ΔH = +) Arrows pointing downwards represent exothermic changes (ΔH = -) 68 Practice 69 Practice Which ions are present in MgO(s)? Using the data given, draw a schematic diagram representing each process in the formation of MgO. Calculate the 2nd electron affinity of Oxygen Is the 2nd electron affinity exothermic or endothermic? Explain. • The actual ΔHf for this reaction is -602 kJ/mol. What does this information indicate about the ionic characteristic of this compound? ΔHa(O) ΔHa(Mg) ΔHi(1)(Mg) ΔHi(2)(Mg) ΔHEa(1)(O) ΔHLE(MgO) ΔHf(MgO) +249 kJ/mol +148 kJ/mol +738 kJ/mol +1451 kJ/mol -141 kJ/mol -3791 kJ/mol -548 kJ/mol 70 Considerations When considering the relative attractions of ions for one another, it can be useful to consider charge density. Small, highly charged ions have HIGH charge densities Tend to attract one another to greater degree Leads to higher melting points and higher lattice energies 71 Discrepencies between calculated and experimental lattice enthalpy Assumption-a compound is essentially 100% ionic Possible to calculate theoretical value of lattice enthalpy (Born Haber Process) Compare to experimental value. If theoretical values agree closely to experimental values, bond is considered to be essentially ionic 72 Compound Theoretical Lattice Formation Enthalpy (kJ/mol) Experimental Lattice Formation Enthalpy (kJ/mol) Agreement? NaCl -766 -781 Good Match ZnS -3427 -3565 Poor Match If theoretical values do not agree closely, the assumption of 100% ionic bond is incorrect. Differences are due to polarization Ionic bond is said to take on a degree of covalent character The greater the discrepancy, the greater the covalent character. 73 CONGRATULATIONS. YOU’VE JUST FINISHED TOPIC 5C of THERMOCHEMISTRY 74 Review Unit 2: Chemical Bonding Recall: 3 types of INTERmolecular bonds London dispersion forces Present in ALL substances Weakest Increases with increasing polarizability Increase in amount of electrons. Generally can be directly related to increasing molar mass. Ex: noble gases, halogens (reason why Cl2 is liquid and I2 is solid, even though individual molecules are entirely non-polar covalent). Substances that are gases at room temp almost always display LDF as primary forces, alkanes, alkenes, alkynes Dipole-dipole forces Present in polar covalent molecules Stronger Strength increases with increasing polarity, Ex: esters, HCl, HBr Hydrogen bonding H connected to F, O, N AND Attracted to lone pair of electrons on F, O or N in another molecule Ex: HF, H2O, NH3, alcohols, carboxylic acids Effects of intermolecular forces Amount AND relative strengths of intermolecular bonds affect physical properties of covalently bonded compounds such as boiling point and melting point For molecules of SIMILAR molar mass, primary factor affecting b.p./m.p. is relative strength of type of INTER bond For molecules of similar type (ex: comparing ALKANES ONLY) primary factor affecting b.p./m.p. is relative molar mass Although pure alkanes display ONLY LDFs as primary inter bond, starting with 6C (C6H14-hexane) exists as liquid and with greater mass as solid due to INCREASING polarizability (greater number of electrons). Conclusion: greater molar mass = greater polarizability = stronger overall combined LDF When comparing across both type and mass, pay extra careful attention to information given. Ex: alkane X is a solid, and alcohol Y is a liquid. Given this information, you can conclude that the combined LDF forces in alkane X MUST BE stronger than the hydrogen bonds in the alcohol Dissolution in water Hydration (dissociation) of ionic substances involve three steps Separation of ion (input of energy-endothermic) Separation of water molecules (input of energy-endothermic) Ions being surrounded by water (attractive forces-exothermic) Predicting whether a reaction will occur RECALL: spontaneous reaction are those that proceed to the product side without external influence. RECALL: particles with LOWER potential energy are more stable than particles with higher potential energy. By this definition, one may assume that ALL exothermic reaction always proceed since a negative ∆H implies that products have lower potential energy. However, this is not necessarily true. For ex: if a reaction has a very high activation energy, then it will not occur and is described as kinetically stable or under kinetic control. In these circumstances, it is possible that a reaction that one might predict as being highly likely, produces little or no product. Ex: oxidation of ammonia by oxygen to produce nitrogen monoxide and water has an enthalpy change value of -909 kJ/mol. However, the reaction does not proceed due to the high activation energy. Similarly, one may assume that an endothermic reaction will never occur, because by definition the products have higher potential energy than reactants However, many endothermic reactions (ex: dissolution of KCl (+19.2 kJ/mol) or acetic acid and sodium bicarbonate (XXXXXXXXX) do proceed to completion. Since one can assume that endothermic reactions must also have some required activation energy that must be obtained before the reaction will proceed, we can conclude that ∆H and activation energy cannot be the only factors that determine whether a chemical reaction will or will not occur. Further, careful examination of these reactions shows that there is a change from a relatively ordered (organized) to relatively less ordered (disorganized) state. KCl(s) K+(aq) + Cl-(aq) 1 mole 2 mole Solid to aqueous CH3COOH(aq) + NaHCO3(s) NaC2H3O2(aq) + CO2(g) + H2O(l) 2 moles 3 moles Solid/aqueous aqueous/liquid/gas We call this degree of disorder or dispersal of energy ENTROPY Entropy Defined as degree of disorder or disarray in a system. A measure of the randomness of a system. Symbol, S Units, (J / K·mol) RECALL: reactions are either reversible or irreversible. By definition the entropy of the universe Increases for spontaneous (thermodynamically favorable) processes (+ ∆S) Does not change for reversible processes (= ∆S) **A system however can display an increase (+) or decrease () in entropy. Ex: gas liquid, system is more organized, and the system has a decrease in entropy (-) Perfect Crystal An element is considered to be a pure, perfect crystal at 0 K. It is completely organized or ordered. RECALL: By definition, at absolute zero, all motion stops. The absolute entropies of substances in the real world (not at 0 K) are measured relative to a perfect crystal As such all substances in he real world must have entropy values greater than 0. Reasonable, as we recognize that all particles have motion; the greater the motion, the greater the entropy. The state, temperature, number of particles, bonds, molar mass and volume of gases are all important factors in entropy. Comparing Entropies Absolute entropy of Solids < Liquids Small number of particle Gases of smaller volume Solids < < < < Gases large number of particles gases of larger volume (larger volume has more space to be disordered) aqueous solution Increases with increasing temperature since particles move around more and are more dispersed Practice Predict the sign of ∆So in these reactions MgO(s) + H2O(l) Mg(OH)2(s) Na2CO3(s) Na2O + CO2(g) Which of the following has the greater entropy? A metal at 273 K or the same metal at 400 K? A flexible soft metal like lead, or a rigid solid like diamond? Two samples of the same gas, at the same temperature, but one at a pressure of 1 atm and the other at a pressure of 0.5 atm Calculating Entropy of a Reaction The change in Entropy (∆So) can be calculated by subtracting the sum of the absolute entropies of the reactants from the sum of the absolute entropies of the products This is analogous to calculating change in Enthalpy. ∆Sorxn = Σ moles Soproducts - Σ moles Soreactants Example Calculate the entropy change, Sorxn for the formation of ammonia from its component elements, given the following information N2(g) + 3H2(g) Absolute entropies values N2(g) 192 J/(K mol) H2(g) 131 J/(K mol) NH3(g) 193 J/(K mol) 2 NH3(g) ∆Sorxn=[(2 moles NH3 * 193 J/(K mol)] –[(1 mole N2 * 192 J/(K mol)+(3 mol H2 * 131 J/(K mol)] -199 J/(K molrxn) Is the calculated value reasonable? Yes—negative value implies that the system has become more ordered (entropy has decreased) since four moles of gas molecules have converted to two moles of gas molecules Gibbs Free Energy RECALL: we established that given that exothermic reactions don’t always occur and that many endothermic reactions do occur, enthalpy change and activation energy are not the only factors affecting favorability of a reaction. The second law of thermodynamics (entropy) must also be considered. Enthalpy and entropy are brought together in the Gibbs-Helmholtz free energy equation ∆Go = ∆Ho - T ∆So ∆Go = ∆Ho - T ∆So Gibbs Free Energy allows us to predict the spontaneity or thermodynamic favorability of a reaction occuring. A reaction is said to be thermodyamically favorable, and therefore will occur, when the value of ∆G is negative Considering the relationship above, indicates that a negative ∆Go is favored by a negative ∆Ho and a positive ∆So Other combinations may also yield a negative ∆Go value **Know that relationships below-common topic of AP questions. Additional Considerations In order to apply the Gibbs Free Energy equation, all reactants and products in an equilibrium mixture (Unit 6) must be present in their standard states (standard states of elements, gases at 1.0 atm, solutios at 1.0 M). Temperature should be stated, and is usually (but not always) 298 K Entropy values tend to be given in units that involve Joules, J, whereas enthalpy values are given in kiloJoules, kJ. This means that when performing calculations that involve both entropy and enthalpy, appropriate conversions must be applied. Thermodynamica favorability does NOT indicate any information about the kinetics (rate, speed) of a reaction. A reaction with very negative ∆Go value (spontaneous) may still be very slow. Calculating Gibbs Free Energy of a Reaction Gibbs Free Energy (∆Go) can be calculated by subtracting the sum of the ∆Go values of the reactants from the sum of the ∆Go values of the products This is analogous to calculating change in Enthalpy and Entropy. ∆Gorxn = Σ moles Goproducts - Σ moles Goreactants If ∆Go is positive, then the reaction is NOT thermodynamically favorable. This usually means that the reverse reaction IS thermodynamically favorable. If ∆Go is zero, then the reaction is favored EQUALLY in both the forward and backward directions System is at equilibrium Free Energy Changes At temperature of 25oC, use appendix 4 At temperatures other than 25°C, G° = H TS How does G change with temperature? Free Energy and Temperature How does G change with temperature? There are two parts to the free energy equation: H: the enthalpy term TS : the entropy term The temperature dependence of free energy, then comes from the entropy term. 1. 2. 3. 4. H = TS H < TS H > TS We cannot determine without additional information. 1. 2. 3. 4. It indicates the process is spontaneous under standard conditions. It indicates the process has taken place under standard conditions. It indicates the process has taken place at 273K and 1 barr. It indicates the process has taken place at 1 atm and 0K. 1. 2. 3. 4. Entropy of universe increases and free energy of the system decreases. Entropy of system decreases and free energy of the universe increases. Entropy of system increases and free energy of the universe decreases. Entropy of universe decreases and free energy of the system increases. 1. 2. 3. always increase always decrease sometimes increase and sometimes decrease, depending on the process