Thermochemistry. Chapter Energy Storage and Reference. 4.2 Chemical Rxns. 4.3 Hess s law Overview

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1 Chapter 4 Thermochemistry 4.1 Energy Storage and Reference Thermochemistry is the study of the energy released when the atomic-level associations are changed. A chemical reaction (rxn) is one where these associations change. We are free to choose our energy reference, or zero of energy, as we please. We could choose the reference to be separated atoms. However, as you will see, a smarter choice for our standard state is the most stable elemental form at a convenient T and P. Once the zero is chosen, our job is: to follow the energy. That means, if the atom-atom associations change to a more energetically favorable arrangement, there must be an exactly compensating release of energy which we can measure with the aid of calorimeters. 4.2 Chemical Rxns Consider the formation of a molecule of 2 in the gas phase. (By in the gas phase we mean that there are no interactions between 2 units. (This is an idealization achieved only at low.) ( )+ 2 ( ) 2 ( ) REFERENCE state for the energy state function. Thatis,the = 0 2. The combustion of diamond (to 2 ) yields almost 2 kj/mole more energy than does graphite at STP. This means that at modest pressures diamond has a higher enthalpy than graphite, i.e. diamond is metastable wrt graphite at STP. However, the barrier is so large that diamonds will easily outlast human-kind and therefore we can consider diamonds a stationary state. 3. A reaction with a -ve [ (P) (R)] means that the enthalpy is lower for the products, i.e. the reaction is energetically down-hill. This type of reaction is called exothermic, and as energy must be conserved, some energy will come out. If your holding the beaker (a diabatic vessel), your hand will feel the energy released (from the molecular associations sector into molecular kinetic energy), as the reaction proceeds. 4.3 Hess s law Overview 1. Hess s law is no more than a reminder that enthalpy is a state function. H =0= P. 1. We take the most stable elemental form at 1 bar and the specified temperature (usually K) as the 2. For a specified reaction, + + ; = P. The stoichiometric coefficients being +ve for products and -ve for reactants. This sign convention achieves the P R Think about it this way: The energy (heat) released and enthalpy are on opposite sides of an equation that balances energy as well atoms. Therefore energy balanced reaction expressions would read or 31

2 + + + Energy Q h Name out +ve -ve exothermic in -ve +ve endothermic 3. The value of only makes sense in the context of a reaction with specified coefficients. (The subscript is telling you, the reader, to look for the balanced equation that the writer is referring to.) In other words, if an alternate reaction 0 is specified as then 0 =2 Definition 1 The subscript refers to one unit of the reaction with the stoichiometric coefficients as written. This means, any quantity such as has an implied reaction with specified coefficients. The change ( ) refers to taking the stoichiometric coefficient number of moles of R and converting them into P. 4. Hess s law is most often applied to get the standard (P = 1 bar, T = K) enthalpy of reaction, which would read, for a rxn statement with specified stoichiometric coefficients, = P. Definition 2 The standard enthalpic state ( = =0) is taken (for convenience) to be that of the most stable elemental form at STP. Definition 3 If the reaction produces a single mole of product ( =1) from the elements in their standard states then = 5. Consider this example. Note that there are 4 measurable enthalpies, 3 of which would be standard enthalpies (if P=1 bar, T= K) of formation: [ 2 ] = 286 [ [ ]= 635 [ and [ ( ) 2]= 985 [ Therefore, for the rxn 2 +, wehave = [ ( ) 2] ( [ 2 ]+ [ ]) = 64 [ ] Bond Enthalpies (energies) One can get enthalpies corresponding to average bond energies. These bond enthalpies are extremely valuable in building up (in a ticker-toy fashion) estimates of the enthalpies of molecules 1. These energies are useful for systems with no delocalized bonding. Values for organics are far more accurate than those for inorganics. Using tables like those given in the text on pg. 69 (also found on the WWW) are the starting point. To help understand what bond enthalpies are, what they are not and how to use them, consider the following examples. 1. The enthalpy of 4 is 1664 kj/mole less than that of the separated atoms. (a) Therefore, = 416 (b) The energy required to dissociate one H from 4 is 427. = Review example 4.1 in the text, = kj/mole. 3. Do problem 4.30, pg Finally, consider the enthalpy of formation 2 2 ( ), i.e. 2 ( )+ 2 ( ) 2 2 ( ) ; Enthalpy diagram for 2 2 All values in 1 The utility of bond enthalpies is that much (but certainly not all) of chemistry is local. 32

3 The value of ( 2 2 ( )) can be estimated from the following data. (a) The energy released to form 2 2 from atomic elements. This energy is a sum of bond enthalpies 2. 2 ( )+2 ( ) 2 2 ( ); [ +2 ]= [ (261)] = 695[ (b) The value of for 2, 2 ( ) 2 ( ) ; = 243 4[ (c) The energy required to take atoms of from the reference state to the gas phase. 2 ( ) 2 ( ) ; = [ (d) Adding the above (a+b+c) we estimate, ( 2 2 ( )) = 2 5 [ ] long indeed for the firstonesinthelist)andtheserulers will be very hard (expensive) to make accurately and easy to use. Remark 3 We are free to choose the zero of energy variables as we please! However we must always be sure ourzeroiscommunicatedtoallpotentialusers. Thisis why we all agree to use the same smart choices of the reference states. 4.4 ( ) To determine how the enthalpy for a reaction changes with temperature ( ), all we need to appreciate is that in changing the temperature all the enthalpies (those of the reactants R and products P) change in accordance with their heat capacities. The individual enthalpies increase with T (C is a strictly positive quantity) but ( ) can increase or decrease with T. a result that relies on estimates of the bond enthalpies. Remark 1 Note this is inaccurate as it involves the difference between large numbers with a resulting small value. This is a situation always to be avoided if at all possible. This points to WHY we choose the reference states to be the stable forms at a convenient T and P rather than say the separated atoms. Let me elaborate on this point. Consider the molecules O 2 and N 2. The enthalpies associated with these bonds are 494 and 942 kj/mole, respectively. (The binding energy of N 2,the triple bond, is one of the strongest molecular bonds.) If we chose our reference state to be the separated atoms, the molecular enthalpies would be H = (O 2 ) = -494 and H = (N 2 ) = -942 kj/mole below the separated atoms. In general, the bonded forms are hundreds of kj (100 kj/mole = 1.03 ev/unit), in association energy, below the separated atoms. As we are almost always interested in the enthalpy difference between bonded forms, it is convenient to take our reference as a bonded form. This largely avoids taking the difference between large numbers. Remark 2 An analogy: Say we are interested in the gravitational potential energy of objects in this room. We need a reference from which to measure the height of the object. We could take this reference as the center of the earth, the sidewalk outside, the floor level of the lab, or the level of the bench top we are doing the experiment on. If we choose the later we can carry the ruler in our hands and it can be made cheaply with precision markings. If we choose any of the others, we have to have a long (very 2 Some of these values have large uncertainties. For example, another source for the Se-Cl gives 243 rather than 261 kj/mole. Consider my favorite reaction: + + ( 2 )= ( 1 )+ R 2 1 [ ( )+ ( ) ( ) ( )] It is convenient (for me) to define, [ ] =C (P) (R) = P P P R Then we just have ( 2 )= ( 1 )+ R 2 1 C ( ) Note that if, as is generally the case, the individual heat capacities are represented as polynomials, C ( ) can be represented as a polynomial, derived by just summingtermsoflikeorder,with+ signs for products and for reactants. (Don t forget the stoichiometric coefficients!) Remark 4 You should now understand why heat capacities are fit to readily integrated standard functional forms (such as polynomials.) Exercise In the figure above, two of four cases are drawn. Draw the other two. For which cases can the sign of the enthalpy change with increasing temperature? 33

4 4.5 Experimental and Data from: 1. Calorimetry - The heat released by the reaction is captured in calorimeter which is ideally adiabatically separated from the remainder of the universe. 2. Spectroscopic studies 3. And most importantly, from other and values using Hess s law. Here you have to be confident of the values for the other legs. This has been done for tabulated values by measuring all legs and checking for consistency, (i.e. going in a complete cycle and ending up with no change.) 4. QMical calc (isodesmic) (d) The desired = (e) However if really in an adiabatic enclosure = (f) the object of the experiment is really to get so that the R in (c) can be done. 2. Bomb calorimetry (Constant V, = ) is used when there are volatile or gaseous species in the reaction. As dv=0, = is the measured quantity and must be calculated by adding the change in PV energy. Overview Calorimetry Lets discuss the three most common types of calorimetry. 1. Constant-pressure calorimetry is the simplest type and can be used if there are no gases or volatile species involved. As (in the ideal experiment) no heat is lost to the outside, the heat given up by the reaction is captured by the calorimeter, = However, the process that actually occurs (A) is not the reaction enthalpy change at a fixed temperature (as heating or cooling) can occur during the reaction. Therefore to get the reaction enthalpy, we need to break down the real process into two steps. While the process by which accurate values can be obtained is quite involved, the basic process can be analyzed with the aid of another triangle diagram, this time referring the changes in the internal energies (rather than H) between various thermodynamic states.. (a) The actual process (A) is R( 1 ) P( 2 )isthe sum of: (b) The desired step (B), the conversion of reactants to products R P at a constant (initial) temperature R( 1 ) P( 1 ) (c) The heating of the products to the final temperature (C): P( 1 ) P( 2 ) = R 2 1 [ (P)+ ] Clearly one needs a separate experiment to determine (P)+ (a) The actual process (A) R( 1 ) P( 2 )canbe decomposed into: (b) A heating step (B), = R 2 1 [ (R)+ ] (again requires one or more separate experiments) and 34

5 (c) The desired internal energy change is (C). If the calorimeter is really adiabatic, = = (d) Again the object of the experiment is really to get so that the R in (b) can be done. (e) Corrections must be applies for the energy input from ignition and stirring. To get standard state values requires an additional set of corrections 3. Step-by-step Bare-bones logic. (a) Determine the heat capacity of the calorimeter (C =allpartsincluding the water in inner bath, not a molar value.) Measure the change in temperature when you dump in a known amount of energy. This can be done by i) passing a known current through a known resistor ( = = 2 = 2 ) or ii) using a reference reaction for which the heat released (under constant volume conditions) is already known. C ) = [ ] ) = (b) run reaction and measure temperature increase in the heat bath exp (c) Calculate exp = [ ] exp[k] (d) Calculate the = exp and the molar value by dividing by the number of moles of product = (e) Calculate = + ( ) 0 if only l s ( )= ( ) if g involved This procedure illustrates the first-pass logic but cannot give the standard reference values. To get these, the previously mentioned (Washburn) corrections must be made (see footnote) for which you need to know the pressure of the bomb and the EoS of the gases involved. (a) The experimental and reference sample are each connected to heaters which try to keep the temperatures of the two samples the same. (That is, the temperatures of both samples start at 298 K and heat at a rate of, say, 5 C/min. (b) The power [ ]( = = 2 ) is measured for both samples. There will be a power difference ( = ) due to the fact that the samples will have different heat capacities. (The sample with the larger heat capacity will require more power to heat.) i) If the heat capacities are different but constant will be a constant. ii) If the heat capacity of the sample grows linearly while the heat capacity of the reference is constant, will increase linearly. If they both increase linearly, will increase linearly with the difference in slopes of the heat capacities. (c) Or iii) when a PT occurs in the sample (one must not occur in the reference), much more heat will have to be dumped into the sample than in the reference. The total heat required to execute the PT is the excess energy, which is calculated from the excess power times the time = = R [ with = (d) As DSC is employed for condensed phases Fig. of DSC 3. Differential Scanning Calorimetry (DSC) is a simple and fast method of determining the enthalpies of condensed phase (s-s or s-l) exothermic phase transitions (PTs). The way to view this processinasfollows: 3 Search WEB for Washburn corrections. These procedures correct for: a) Compression of the condensed phase and gas phase species to 1 atm pressure, b) Enthalpy of solution and of dilution of the product gases in the aqueous phase (e.g. HCl, 1:600), c) Vaporization of water because of the heat liberated by the reaction, and d) energy changes associated with non-isothermal reactions. 35

Thermochemistry. Chapter Energy Storage and Reference. 4.2 Chemical Rxns. 4.3 Hess s law Overview

Thermochemistry. Chapter Energy Storage and Reference. 4.2 Chemical Rxns. 4.3 Hess s law Overview Chapter 4 Thermochemistry 4.1 Energy Storage and Reference Thermochemistry is the study of the energy released when the atomic-level associations are changed. A chemical reaction (rxn) is one where these