Since the rates of reactions in which acids and bases appear as catalysts depend strongly

Transcription

1 218 Chem 141 Lectures Lecture 29 I stated at the beginning of our study of acids and bases that one reason that acids and bases are important is that they are the most common form of homogeneous catalyst. Since the rates of reactions in which acids and bases appear as catalysts depend strongly upon the concentrations of hydronium ion or hydroxide, it seems apparent that it would be desirable to be able to control the ph of a reacting mixture. Control of ph is also important for studies of biochemical systems, since the activity of many enzymes is extremely dependent on ph. These are just two examples of systems for which ph control is extremely desirable. Why is ph control a problem? Consider a 1 L aqueous biological system at ph 6.9. The concentration of hydronium ion is = 1.26 x 10-7 M. Addition of just 1.00 x 10-3 moles of HCl increases the hydronium ion concentration to 10-3 M = ph 3. This is a hostile ph for biological systems. What we need is a tool that will allow us to add small amounts of acid or base to a system without changing the ph significantly. This is precisely the definition of a buffer. A buffer is a solution that allows us to add small amounts of strong acid or strong base while keeping the ph change small. Notice that buffers resist ph changes when either OH - or H 3 O + is added to them. Therefore they must contain something which can react with OH - and something which can react with H 3 O +. As such they are made up of solutions of a weak acid and its conjugate base or a weak base and its conjugate acid. A common example of the former would be a mixture of acetic acid and sodium acetate, CH 3 COOH and

2 219 Na + CH 3 COO -. An example of the latter would be a mixture of ammonia and ammonium chloride, NH 3 and NH 4 Cl. Buffers control ph most effectively when the conjugate acid/conjugate base ratio [HA]/[A-] = 1:1, but buffers will control ph anytime that these ratios lie between 10:1 and 1:10. Let s look at some specific examples. One example would be a mixture of 0.30 mol of benzoic acid, C 6 H 5 COOH, and 0.15 mol of sodium benzoate, NaC 6 H 5 COO in a 1 L aqueous solution. The K a for benzoic acid is 6.5 x One feature of buffers is that each buffer has a characteristic ph. We can calculate these ph's easily. Let s do it for our benzoic acid-benzoate buffer. We begin by writing the equilibrium constant expression for the ionization of benzoic acid, C 6 H 5 COOH + H 2 O C 6 H 5 COO - + H 3 O +, K a - + [C6 H5 COO ][ H3 O ]. [C H COOH] 6 5 If we solve this equation for the [H 3 O + ], we get [ H O ] 3 [C6 H5COOH] [C H COO ] K a Lets make an equilibrium table for this reaction. Initial [ ] (M) Change in [ ] (M) Equil [ ] (M) C 6 H 5 COOH.30 -x.30-x.30 C 6 H 5 COO -.15 x.15 + x.15 H 3 O + 0 x x

3 220 This means that [H 3 O + ] = x 10 5 = 1.3x10 4 M. Therefore the ph of this buffer is Notice that in retrospect we did not need to bother with the concentration table for this example. We could merely have inserted our concentrations for acetate and acetic acid into the constant expression. This will be the case for all buffers in which the absolute concentrations of the conjugate base pairs are substantial. An example of a basic buffer is a solution made up of methylamine, CH 3 NH 2 and methylaminium chloride, CH 3 NH 3 + Cl -, K b = 3.7 x LET S CALCULATE THE PH FOR THIS BUFFER, WHEN THE CONCENTRATION OF CH 3 NH 2 IS 0.25 M AND THE CONCENTRATION OF CH 3 NH 3 + Cl - IS 1.00 M. Now let s see what happens when we add strong acid to a buffer. Let s calculate the result when we add 0.12 mol of HNO 3 to 1.00 L of a buffer which is 1M in both benzoic acid and benzoate, and which starts out with a ph of We need to calculate this in two stages. First we have to realize that when we add 0.12 mol of nitric acid to a buffer solution, the strong acid will react almost completely with the base of the buffer. We can express this in a concentration table. Initial [ ] (M) Change in [ ] (M) Final [ ] (M) C 6 H 5 COOH C 6 H 5 COO HNO These are the concentrations we now put into our buffer equation,

4 221 [H 3 O + [C6 H5 COOH] ] = K a = 6.5 x [C H COO ].88 = 8.27 x 10-5 M. 6 5 Therefore the ph = In comparison, note that if we added 0.12 mol of HNO 3 to 1 L of an unbuffered aqueous solution which had an initial ph of 4.9, the effect would be far more dramatic. The initial [H 3 O + ] in this case is 6.5 x 10-5 M. Adding 0.12 mol of HNO 3 would yield a final concentration of 0.12 M, and a final ph of.75, a much larger change. Let s see what happens if we add 0.12 mol of NaOH to the same benzoic acid benzoate buffer solution. Once again, we begin by recognizing that the strong base will effectively react completely with the buffer acid. The concentration table we get for addition of 0.12 Mol of NaOH to our buffer is Initial [ ] (M) Change in [ ] (M) Final [ ] (M) C 6 H 5 COOH C 6 H 5 COO OH At this point we just use our buffer equation to obtain the new [H 3 O + ]. Once again, [H 3 O + [C6 H5 COOH] ] = K a = 6.5 x [C H COO ] 1.12 = 5.11 x 10-5, and the new ph is We can see from this that a buffer resists changes in ph due to addition of strong acid or strong base. How does a buffer work? When we add a strong acid to an unbuffered aqueous solution, every acid molecule increases the hydronium ion concentration. However, when we add the same amount of acid to a buffer, instead of reacting with the water, it

5 222 reacts with the base in the buffer to form the conjugate acid. Since the acids in buffers are weaker acids than hydronium ion, this results in a lower ph change. Similarly, adding a strong base to water would just result in increase of OH-. When we add the base to the buffer however, it reacts instead with the acid to produce the conjugate base. Once again, since the conjugate base is a weak base, it contributes far less OH- to the solution than the strong base would, and the change in ph is smaller. Now let s do a full buffer problem together. To 1L of a nitrous acid buffer consisting of 1 mol HNO 2 and 1 mol NaNO 2, and having an initial ph of 3.3, we add 0.06 L of 1 M HCl. K a of nitrous acid is 4.5 x WHAT IS THE FINAL PH? Our last subject in the chemistry of acids and bases is titration. Titration is an analytic technique in which we measure the amount of a solution of one substance needed to react completely with a solution of another substance. Generally when we do this we know the concentration of one substance, and use the titration to determine the concentration of the other substance. Although we will be discussing titrations in terms of acids and bases, titrations can be done with any two substances that react with each other to produce a third substance, and which can be dissolved. An important concept in titrations is the concept of the equivalent. For acids and bases, an equivalent is the amount of a substance that produces or reacts with 1 mole of H 3 O +. A mole of a monoprotic acid contains 1 equivalent of protons. A mole of a polyprotic acid contains two or more equivalents of protons, depending on how many acidic protons the polyprotic acid contains. HOW MANY EQUIVALENTS ARE THERE IN 1 MOL OF HCL, H 2 SO 4, CA(OH) 2, KOH? 1 L OF 0.06M HNO 3, 1 L OF 0.06M H 3 PO 4? We use equivalents to define a new unit of concentration called normality. The normality

6 223 of a solution is defined as the number of equivalents per mole of solution. For example, a 1 M solution of HCl is a 1N solution, a 1 M H 2 SO 4 solution is 2N, 1 M H 3 PO 4 is 3N. WHAT IS THE NORMALITY OF 0.5M H 2 SO 4? 0.03 M H 3 ASO 4? 0.25 M HCL? When we do a titration, we are trying to measure the amount of a solution it takes to reach the equivalence point. The equivalence point of an acid base titration is the point at which one reactant has been exactly consumed by the addition of the other. This will occur when the number of equivalents of the substance we add is exactly equal to the number of equivalents in the solution we are titrating. Lets look at some examples. If we have 1 L of 0.05 M HCl and titrate it with NaOH, we begin with 0.05 equivalents of HCl and therefore, we reach the equivalence point when we've added 0.05 mol, or.05 equivalents, of NaOH. Note that it is not the volumes of solution that are equal, but the number of equivalents. IF WE TITRATE L OF.025 M CH 3 COOH WITH KOH, HOW MANY EQUIVALENTS OF ACETIC ACID DO WE START OUT WITH? HOW MANY EQUIVALENTS OF KOH DO WE NEED TO ADD TO REACH THE EQUIVALENCE POINT? HOW MANY MOLES IS THIS? Things are a little more complicated when we consider polyprotic acids. With polyprotic acids we have more than one equivalent per mol of acid and therefore have more than one equivalence point. For example for 1 L of 1M H 2 SO 4, addition of 1 mol of NaOH gives the first equivalence point, which results in the conversion to HSO - 4, while addition of the second mol of OH - gives our second equivalence point, which leads to conversion to SO = 4.

7 224 Usually when we do acid base titrations we determine the equivalence point by analyzing a titration curve. There are two basic types of titration curves that we will consider. The first is that of a strong acid with a strong base. The second is that of a weak acid with a strong base. The opposite types of curves, where the base is titrated by an acid, have the same general features, so we won't go over them. In general, all titration curves are graphs of ph vs. volume of solution added. In these cases, since we are titrating and acid with a base, the volume will be volume of base added. Let s draw the graphs of each of these types of titrations. For a titration of a strong acid with a strong base, we begin at the ph of our initial solution. Addition of base results in an immediate increase of ph, which has the appearance of a straight line until the solution nears the equivalence point. Then the ph rapidly increases until the equivalence point is passed. At this point the ph will go up only a little bit more. The ph at the equivalence point for a titration of strong acid by strong base is 7.

8 225 For the titration of a weak acid with a strong base, the appearance is somewhat different. We again begin at the ph of the initial solution. Addition of base initially leads to a rapid increase of ph, until 1/10 of the acid has been titrated. From this point until 9/10 of the acid has been titrated our solution acts like a buffer, and the ph changes only slowly. After we pass the buffer region, the ph increases rapidly until it passes the equivalence point. After the region of rapidly increasing ph is passed, the increase in ph slows again, but there will be little further increase in ph. There are three important points on this type of titration curve, the initial point, the halfway point of the titration, when we are in the center of the buffer region, and the equivalence point. It is important to note that for the titration of a weak acid with a strong base the ph at the equivalence point is not 7. Lets calculate the ph at these three points for the titration of a strong acid by a strong base, and then for the titration of a weak acid by a strong base. We'll begin with the former. Let s begin with the titration of 50 ml of 0.25 M HCl titrated with 0.1 M NaOH. WHAT IS THE INITIAL PH? HOW MANY MOLES OF HCL DO WE BEGIN WITH? HOW MANY EQUIVALENTS IS THIS? WHAT VOLUME OF 0.1 M NAOH DO WE NEED TO ADD TO TITRATE 1/2 OF THIS? WHAT IS THE [HCL] AT THIS POINT? WHAT IS

9 226 THE PH? WHAT VOLUME OF 0.1 M NAOH DO WE NEED TO REACH THE EQUIVALENCE POINT? WHAT IS THE [HCL] AT THIS POINT? Now let s look at the titration of 75 ml of 0.37 M CH 3 COOH titrated with 0.13 M KOH. The K a of acetic acid is 1.76 x WHAT IS THE INITIAL PH? HOW MANY MOLES OF ACETIC ACID DO WE BEGIN WITH? HOW MANY EQUIVALENTS IS THIS? WHAT VOLUME OF 0.13 M KOH DO WE NEED TO ADD TO TITRATE 1/2 OF THIS? WHAT IS THE [ACETIC ACID] AT THIS POINT? WHAT STEPS DO WE FOLLOW TO CALCULATE THE PH? WHAT VOLUME OF 0.13 M NAOH DO WE NEED TO REACH THE EQUIVALENCE POINT? WHAT IS THE [CH 3 COOH] AT THIS POINT? WHAT ABOUT ACETATE ION? WHAT STEPS DO WE FOLLOW TO CALCULATE THE PH? Our last acid - base topic is indicators. I don't really think we need to go over them in detail. In short, an indicator is a weak acid or weak base that changes color at a certain ph. For example, phenolphthalein, the indicator that you are most familiar with, turns pink when the ph goes higher than 8.7. The only other thing you need to know about indicators is that when you are doing a titration, if you use an indicator to show when you have reached your endpoint, you want to use the one which changes color as close to the ph of your endpoint as possible. So for example, we just calculated the ph at the equivalence point of our titration of acetic acid to be If we look at a table of indicators, such as can be found in the CRC Handbook of Chemistry and Physics, we find that the best indicator for this titration is good old phenolphthalein. Lecture 30 At this point we are going to turn a bit toward the theoretical end of chemistry and have our first taste of a subject called thermodynamics. Thermodynamics deals with the

10 227 flow of energy in physical processes. Thermodynamics is an extremely powerful subject. It is powerful because it is predictive. In fact there are two types of predictions which thermodynamics can make which are of intense interest to chemists. First it can tell us which processes are possible and which are impossible. Obviously knowing this can prevent the waste of a tremendous amount of time. Secondly, it can tell us which processes will occur spontaneously, i.e. without doing work on the system, and which will not. As a corollary to this latter type of prediction, when thermodynamics tells us that a process will not occur spontaneously, we can use thermodynamics to determine the conditions, if any, under which a process will occur spontaneously. All of thermodynamics has to do with changes of energy. There are three laws on which thermodynamics is based. In this section we'll be dealing with the consequences of the first law only. The first law of thermodynamics states that energy is neither created nor destroyed, but simply converted from one form to another. This law is also called the law of conservation of energy. The second law states that the entropy of the universe increases for any spontaneous process, again, a process that will occur without us doing work. Finally, the third law of thermodynamics states simply that the entropy of a perfect crystal of a pure substance at the absolute zero of temperature is 0. We'll come back to these latter two laws of thermodynamics next week. Let s look at some consequences of the first law, especially where it is concerned with the transfer of heat when the pressure is constant. Chemists are particularly concerned with what happens when pressure is constant, because many reactions are done in open vessels, like beakers or Erlenmeyer flasks, where the pressure is just the pressure exerted by the atmosphere. Since we often nudge a process along by heating our

11 228 sample, we are also interested in how much heat is transferred. Heating reactions at constant pressure is so common that we give a special name to the amount of heat given off or absorbed under these conditions. We call the amount of heat absorbed under conditions of constant pressure the change in enthalpy, H. In fact, H is often defined by H = q p, where q is the symbol for heat and the subscript p indicates that the pressure is constant. Notice that I've been talking about the transfer of heat or the transfer of energy. If we are talking about transferring energy from one place to another what places are we talking about? It turns out to be convenient to divide the universe (remember that this is the subject of the first law of thermodynamics) into system and surroundings. The system is simply defined as "what we are interested in" while the surroundings are even more simply defined as "everything else". For example, I could call a flask of acid my system, in which case everything else is the surroundings. If I heat this flask, this means that I am taking energy from outside the flask (the surroundings) and putting it into the flask (the system). If on the other hand, I cool the flask, I would be taking heat from inside the flask (the system) and putting it outside the flask (the surroundings). It s very common for heat to be either released or absorbed in the course of a chemical reaction. Let s consider a couple of quick examples. First consider the case in which acid and base are mixed, for example, HCl and NaOH. This reaction results in the neutralization of the acid and the base. Does anyone remember from the reactions lab what happens when you mix HCl and NaOH? [Test tube gets hot] So the reaction H + (aq) + OH - (aq) H 2 O(l) gives off heat. We say that this reaction is exothermic. There are various ways of expressing this concept of exothermicity:

12 229 1) We could write the equation for the reaction as H + (aq) + OH - (aq) H 2 O(l) + heat, which is the same as saying that heat is a product of the reaction. 2) We can say that heat is given off by the reaction 3) We can say that heat is transferred from the system to the surroundings. 4) We can say that the reaction or process is exothermic. 5) We can say that H is negative. 6) We can say that the reactants have higher energy than the products. Why does this make H negative for exothermic reactions? H rxn = H product - H react. If H product < H react then H is negative, and heat is given off. It is important for you to realize that all of these statements mean the same thing, since at one time or another you'll be seeing all of them. Now let s look at another reaction. Suppose I were to mix some ammonium chloride and some water. This was the basis of your cold packs in the second part of the thermo lab. The result of dissolving the NH4Cl in water was to make the water colder. The water bag felt cold because the reaction H 2 O(l) + NH 4 Cl(s) NH + 4 (aq) + Cl - (aq) was absorbing energy in the form of heat from the water. When a reaction like this one, or some other process, like boiling water, requires heat, we call it an endothermic

13 230 process. Let s run through several ways of describing an endothermic process or reaction. 1) We can write the reaction with heat as a reactant, for example, H 2 O(l) + NH 4 Cl(s) + heat NH + 4 (aq) + Cl - (aq). 2) We can say that heat is absorbed. 3) We can say that heat is transferred from the surroundings to the system. 4) We can say the process is endothermic. 5) We can say H is positive. 6) We can say the products have more energy than the reactants. We can see that this last will be the same as saying H is positive. Since H rxn = H product - H react, if H product > H react then H is positive, and heat is absorbed. Again, all of these statements are equivalent. For some reason people seem to have a harder time understanding endothermic processes than exothermic processes. Just remember that they are complementary: An exothermic process gives off heat at constant pressure while an endothermic process absorbs heat at constant pressure. Now consider the acid-base reaction again. WAS IT DONE AT CONSTANT PRESSURE? [Yes] HOW DID WE KNOW THAT HEAT WAS GIVEN OFF? [Felt the flask] WHAT WAS THE SYSTEM FOR THIS EXPERIMENT? [Flask and liquid] WHAT WERE THE SURROUNDINGS? [Hand, lab bench, air, everything else] So heat was transferred from the liquid reagents to your hand. The amount of heat is H, the change in the enthalpy as the

14 231 reaction proceeds from reactants to products. Why do we care what H is? It tells us whether we need to heat a reaction to make it go, or cool it to keep it from going out of control. If we find that a reaction is endothermic, then we know we have to add heat for it to occur. If it is very endothermic, then we know we have to add a lot of heat. If a reaction is exothermic then we don't need to add heat. If it is very exothermic, then we have to be careful to avoid fires or explosions. Lets look at a couple of examples. For the combustion of methane, the reaction can be written CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat (802 kj) H rxn = -802kJ. Q: IS THIS REACTION EXOTHERMIC OR ENDOTHERMIC? [exothermic] The superscript means that all of the reactants and products were in the standard state, a pressure of one atmosphere. Just about every enthalpy that you look up in a book will be for this one atmosphere standard state. Note that the units we used for H can be joules or calories. A Joule is an SI unit of energy, and the calorie is the old English unit. The Joule is 1 kg m 2 s -2, and is the amount of energy that a 2kg block has if it is moving 1 m/s. A calorie is defined as the amount of energy it takes to heat 1 g of water by 1 C. 1 cal is 4.184J. There is another unit called the Calorie (capitalized). The Calorie is 1 kcal. Guess which one the energy content of your food is measured in! Heat is not the only means of changing the energy of a system. For systems in which the amount of the chemical components doesn t change, the other main means of changing the energy of a system is work. Work is defined as force times distance, where

15 232 the force is any force opposing the motion. What this means is that neither motion nor an opposing force is sufficient for work to take place both must be present. The quantitative definition of work is for motion along the x-axis is: w F * x. If the force is in units of Newtons (N), and the displacement is in meters, then the units of work will be Joules. x For a system in which the chemical composition remains unchanged, only work and heat can change the internal energy of the system, and this is expressed as the equation U q w, where q is the symbol for heat, w for work and U is short for the change in internal energy. Lecture 31 H for a process depends on how much material is involved. This should make sense to you. If I burn 5 grams of charcoal and heat is given off, you'd expect that burning 10 g of charcoal would give off twice as much heat. So now suppose that combustion of 1 g of sugar (MW 180) causes 5J to be given off, i.e. H = J. How much heat would be given off for 1 mol of sugar? Let s use dimensional analysis to get the answer. 1 mol sugar x 180 g sugar 1 mol sugar x 5J 1 = 900J mol 1 g sugar For another example, when 0.32 g of C reacts with S 8 to give carbon disulfide, 2.38 kj is absorbed. Calculate the heat in kj when 1 mol of carbon disulfide is formed

16 233 from carbon and sulfur. We start by writing down a balanced reaction - 4 C(gr) + S 8 (s) 4 CS 2 (l) At this point we know how much energy is absorbed when 0.32 g of C reacts. We want to use dimensional analysis to find out how much heat is required to form 1 mol of CS 2. Our starting point is 2.38 kj /.32 g C and our ending point is kj/ mol CS kj 0.32 g C x 12 g C 1 mol C x 1 mol C 1 mol CS 2 = 89.25kJ / mol CS 2 Let s do one final example, this time for the process of vaporization. When an inch of rain falls on New York City, it results in a rainfall of 1.98 x L. Using the following data determine how much energy is released to the atmosphere when this much rain is formed from water vapor. For H 2 O(l) H 2 O(g) H vap = 44 kj/mol. Since our process is the opposite of vaporization, H = -44kJ/mol. Now we can write 1.98 x 10 L x 1000ml x 1g 1L 1ml x 1mol H O 18 g H O x -44kJ mol = -4.84x 10 The enthalpy, H, is one of a special class of functions called state functions, 2 13 kj which also includes energy, entropy, and temperature. The special feature of state functions is that the change of a state function is dependent only on its initial and final values and not the way you change it. To illustrate this idea, imagine that you are climbing a mountain. There are many different routes that you can take to get to the top, and you will have to walk a different distance for each. Therefore the distance you walk when you go from the bottom to the top depends on how you choose to go and is not a state function. This means that if you want to know the distance you walk you must specify every step of the

17 234 path you choose, which is extremely inconvenient. In contrast, the change in altitude when you walk from the bottom of the mountain to the top of the mountain is always the same, no matter what path you choose. This means that altitude is a state function, and also that all you need to know to calculate a change in altitude is the height you end at and the height you begin at. This is clearly a much simpler calculation. Functions, such as the distance you walk, which depend on the path you choose are known as path functions. Another example of a path function is life itself, since the poet Robert Frost has written: I took the road less traveled by, and that has made all the difference. If life was a state function, taking the road less traveled would not have made any difference. The big advantage of enthalpy being a state function is that it means that we can determine enthalpy changes for processes that we have not measured directly. We do this by finding two or more processes which if done in sequence are the same as the process we are interested in. We then measure the enthalpy for the two processes and add them. The sum of these two enthalpies is the same as the enthalpy for the overall process. Suppose for example we are interested in the reaction A + D B + E, but we can't run the reaction to measure the enthalpy. Now suppose we have two reactions where we've measured the enthalpies and whose sum is the reaction we are interested in: A B + C D + C E H = 10 kj H = 20 kj A + D B + E H= 30kJ Not only can we add them to get the reaction we are interested in, but the sum of their enthalpy changes will be the enthalpy change for the reaction we are interested in. This

18 235 H is the same we would have gotten if we had measured the enthalpy for the reaction A + D B + E directly. It doesn't matter how we got from A and D to B and E, the enthalpy change is the same. For example suppose we want to know H for the reaction H 2 O(g) H 2 (g) + 1/2O 2 (g). It would be hard to measure H for this because of a property of water called hydrogen bonding. However, we can still figure out H for this reaction by measuring H for two other processes. If we know the enthalpies of the following two reactions: H 2 O(g) H 2 O(l) H 2 O(l) H 2 (g) + 1/2O 2 (g) H = -41 kj/mol H 2 O H = 283 kj/mol H 2 O We can add them up to get H 2 O(g) H 2 (g) + 1/2O 2 (g) H = 243 kj/mol H 2 O. We could represent this in a diagram:[ Start with H 2 O(g), then go to H 2 O(l) then up to H 2 + O 2. Net enthalpy is the same as if we went straight up.] Let s do another example. Suppose you want to find out how much heat it takes to make acetylene, C 2 H 2, from its elements. The reaction is 2C(gr) + H 2 (g) C 2 H 2 (g). We are given the following data: C 2 H 2 (g) + 5/2 O 2 2 CO 2 (g) + H 2 O(l) H = kj/mol

19 236 C(gr) + O 2 (g) CO 2 (g) H = kj/mol H 2 (g) + 1/20 2 (g) H 2 O(l) H = kj/mol The goal is to add these three reactions in such a way that their sum will be the reaction for the formation of acetylene. The reaction we want has two carbon atoms on the left side of the equation. We'll use our second reaction to get these carbon atoms, but we have to multiply it by two: 2 x[c(s) + O 2 (g) CO 2 (g)] H = 2x kj/mol. Remember that when we multiply the reaction by two we also have to multiply the H by 2, since H depends on the amount of chemical we react. Now we need a hydrogen molecule on the left side of the equation, and we use the third equation for this: H 2 (g) + 1/20 2 (g) H 2 O(l) H = kj/mol Finally we need an acetylene on the right. We can use our first reaction for this, although we need to flip it around to get the acetylene on the correct side: 2 CO 2 (g) + H 2 O(l) C 2 H 2 (g) + 5/2 O 2 H = 1300 kj/mol Notice that when we write the reaction backwards we have to change the sign of the H from negative to positive. If we add the three reactions, we find that everything we don't want cancels out to give 2C(s) + H 2 (g) C 2 H 2 (g). Now we just add up all of our H's to give H = kj/mol. What we just did in our last example is to use Hess's Law. Hess's law says "If two or more chemical equations are added together, the H for the resulting equation is equal to the sum of the H's for the separate equations."

20 237 One of the consequences of Hess's law, which we've just seen, is that if we have a reaction and we know H, that H backward = - H forward. Similarly we can say H forward = - H backward. The enthalpy of formation is a special and very useful type of heat of reaction. The enthalpy of formation, which used to be called heat of formation, is the amount of heat it takes to make one mole of a compound from its elements in their stablest form under conditions of constant pressure. When you look up heats of formation in a book, the symbol you'll see is H 0 f, which is the symbol for standard heat of formation. Once again the standard condition is a pressure of one atmosphere. Standard enthalpies of formation are most commonly reported at 298K. By definition: H 0 f of any element in its reference state at 298K is zero. H 0 f = amount of heat to form 1 mol of a substance under 1 atm pressure. If we want to form gaseous water from its elements, the reaction would be H 2 (g) + 1/2O 2 (g) H 2 O(g) H 0 f (H 2 O(g)) = -243 kj/mol Notice that the stablest forms of hydrogen and oxygen respectively are H 2 (g) and O 2 (g). If we want to form liquid water from its elements, the reaction would be H 2 (g) + 1/2O 2 (g) H 0 f H 2 O(l) H 0 f (H 2 O(l)) = -283 kj/mol Notice from these examples that the enthalpy of formation depends on the phase of the product. For the case of methane the reaction would be C(gr) + 2H 2 (g) CH 4 (g) H 0 f (CH 4 (g)) = kj/mol When we did our first example of Hess's law, one of the reactions we used was for the

21 238 heat of formation of liquid water. 0 Enthalpies of formation are useful because if we want to find H rxn for some reaction we can just say 0 H rxn = Sum of H f 0 (products) - Sum of H f 0 (reactants). For example, for the combustion of methane, the overall reaction is CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) 0 To get H rxn we subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products: 0 H rxn = H f 0 (CO 2 ) + 2 H f 0 (H 2 O) - H f 0 (CH 4 ) - 2 H f 0 (O 2 ) Now the question is, where do we get the heats of formation to use in this calculation? The answer is from a book. An excellent source of heats of formation is the CRC "Handbook of Chemistry and Physics" of which we have several copies in the library. Your book also has a moderate sized table in Appendix B on pages A-9 to A-13. If we look these up we find H 0 f (CO 2 (g)) = kj/mol H f 0 (H 2 O(l)) = -283 kj/mol H f 0 (CH 4 (g)) = kj/mol We don't find a value for H f 0 (O 2 (g)). This is because, as we've already said it is defined as zero. Remember, the H f 0 of any element in its reference state at 298K is zero. So now 0 H rxn = kj/mol + 2(-283 kj/mol) - ( kj/ mol) - 2 x 0 = -884 kj/mol.

22 239 This is a particularly useful version of Hess's law, because it allows us to calculate H for just about any reaction just by using H f 0. Lecture 32 0 Just a quick additional note about H's. If we simply write H rxn, the H could refer to any kind of reaction. It could be for formation of water or combustion of octane or anything. In order for people to know what we're talking about we usually have to write out the specific reaction we're referring to. So for example, if our reaction involved a physical change like the dissolution of ammonium chloride we'd write out the reaction for the dissolution of ammonium chloride. If it involved a chemical change like the reaction between NaOH and HCl, we write out the neutralization reaction. Sometimes however, a reaction is so common that rather than write out the reaction we identify it by a subscript on H. The one of these common reactions that we have already seen is formation of a compound from its elements, where the fact that a formation reaction is involved is indicated by the subscript f in H f 0. Another common reaction is combustion, indicated by the subscript c, as in H c. In both of these cases, the subscript replaces the full reaction equation. Another special enthalpy is the called the bond enthalpy, the heat required to break a bond. Bonds are held together with fairly large amounts of energy. If we take a water molecule and break all its bonds, the reaction is H 2 O(g) 2 H(g) + O(g) and we have broken two OH bonds. The enthalpy is H = 925 kj/mol. Since we've broken two bonds, to get the enthalpy per bond, symbolized as H BE, we divide the overall enthalpy by two to get H BE = 463 kj/mol.

23 240 Why are bond dissociation enthalpies of interest? There are two primary reasons: First, if we compare H BE for two different bonds we can get an idea about which bonds are strongest. For example, a typical bond between C and H takes 414 kj/mol to break, while a typical SiH bond takes only 293 kj/mol to break. So by comparing the bond enthalpies, we can learn that CH bonds are stronger than SiH bonds. Second, we can use bond dissociation enthalpies to estimate heats of reaction when we don't have more 0 accurate information at hand. We can do this because H rxn = the energy it takes to break all the bonds - the energy released when we combine the atoms into new bonds in the molecules we want. There are a couple of reasons that the answers we get using H BE are only approximate. When you read a bond dissociation enthalpy in a table the number reported is an average number. They get it by measuring the heat when that kind of bond is broken for many different compounds and averaging them. And this works fairly well, because, for example, most C-H bonds will be about the same strength. However, the strength of CH bonds in different molecules is a bit different, so our results won't be completely accurate. The second is that these dissociation enthalpies are typically for gas phase dissociation, and the actual dissociation energy depends on whether the molecule is in the form of a gas or aqueous. Nonetheless using bond dissociation enthalpies this way is a useful tool if enthalpies of formation are not available. For example, let s estimate H 0 f (HF) from the bond dissociation enthalpies of H 2 (g), F 2 (g) and HF(g). The reaction is H 2 (g) + F 2 (g) 2HF(g). Since two HF's are formed, H f 0 0 will be 1/2 H rxn. To use bond dissociation energies, we break this reaction into three steps

24 241 H 2 (g) 2H(g) H BE = 436 kj F 2 (g) 2F(g) H BE = 159 kj 2F(g) + 2H(g) 2HF(g) H = -2 H BE (HF) = kj. 0 Adding them up gives H rxn 436 kj kj kj = -543 kj. Dividing by two gives us H f kj/mol. If we look up the actual H f 0 (HF) = kj, we see that our approximation is pretty good. You may be wondering how H's are determined. The answer is a technique called calorimetry. Typically what is measured is a heat of combustion. We put a sample of known weight in a device called a bomb calorimeter, add a whole bunch of oxygen, and ignite the substance with a spark. When the sample is finished burning, the heat from the reaction causes the temperature of the calorimeter to rise. Now at this point we know the number of moles of our sample, and the change in temperature, i.e. we know T/mole. We want kj/mol so this means that we need a conversion factor with units kj/t. This conversion factor is called the heat capacity, C p. The heat capacity is the amount of heat necessary to raise the temperature of a sample by 1K. This means that C p = H/ T. The heat capacity depends on how much material we have. For example, to raise the temperature of 1 gram of water takes 4.18J, while to raise the temperature of 100 g of water takes 418 J. We use heat capacity to calculate the heat added to the calorimeter when we know

25 242 a temperature change T. The equation we use is H = C p T. So suppose we know that C p for a sample is 75.2 J/K, and the temperature goes up 2K. H = 2K x 75.2 J/K = J. Again it is important to remember that the size of your sample matters - the bigger the sample, the bigger the heat capacity. There are two different kinds of heat capacity in books. One is C p, the molar heat capacity, = C p per mole of substance. The other is C sp = specific heat = C p per gram of substance. So for example for water C_ p = J/mol K, while C sp = J/mol K x (1 mol H 2 O/ 18.0 g H 2 O) = 4.18 J/g K. Note that you can tell immediately whether you have been given a specific heat or a molar heat capacity simply by looking at the units. It is also important to understand that when using heat capacities to calculate heat, the equation H C T refers neither to the specific heat or the molar heat capacity but the overall heat capacity. If you are given a molar heat capacity it must be converted to heat capacity by p C p nc p before being used to calculate H. and if you are given a specific heat capacity it must be converted to heat capacity by C p mc sp before being used to calculate H. Let s do an example of using heat capacities. When 4 kg of water were used to cool an engine the temperature rose from 298K to 323K. C sp for water is 4.18 J / g K.

26 243 How much heat did the engine give off? Heat = H = C p T. T = 323K - 298K = 25K. C p = 4000 g H 2 O x 4.18 J/g K = 1.67 x 10 4 J/K. Therefore heat = 1.67 x 10 4 J/K x 25K = 4.18 x 10 5 J = 4.18 x 10 2 kj. Now let s do an example of a calorimetry problem. NH 4 NO 3 combusts according to the reaction NH 4 NO 3 2N 2 (g) + 4H 2 O(g) + O 2 (g) g of NH 4 NO 3 is exploded in a bomb calorimeter, with a heat capacity of 4.92 kj/k. The temperature of the calorimeter increases by 3.06 K. What is H c for one mole of NH 4 NO 3. C p x T = H for 2 g = 4.92 kj/k x 3.06 K = kj/2.00 g NH 4 NO 3. Now we just use dimensional analysis to get kj/mol kj 2.00 g NH NO g NH4 NO3 x = kj / mol NH NO 1mol NH NO Now this is really important. This heat we just calculated is the increase in heat of the calorimeter. In order for the calorimeter to get all that heat, it had to be given up by the exploding NH 4 NO 3. So the NH 4 NO 3 lost this energy, and therefore H for the reaction is kj/mol.