CHM 101 GENERAL CHEMISTRY FALL QUARTER 2008 Section 2 Lecture Notes 11/17/2008 (last revised: 11/18/08, 11:30 AM) 6.3 Hess Law: Hess Law takes advantage of the fact that enthalpy is a state function. Hess Law applies to the change in enthalpy that takes place when some set of reactants is converted to some set of products. Simply stated, Hess Law says: The change in enthalpy (ΔH) is the same whether the process takes place in a single step or in a series of steps. An Example of Hess Law (pp. 242-3): Let us consider the oxidation of N 2 to produce nitrogen dioxide (NO 2 ). We can write it as a single-step reaction in which 2 oxygen molecules combine with 1 nitrogen molecule to form 2 nitrogen dioxide molecules: N 2 (g) + 2O 2 (g) > 2NO 2 (g) ΔH 1 = 68 kj The alternative is to regard the reaction as taking place in two steps. In the first step, 1 molecule of oxygen combines with 1 molecule of nitrogen to form 2 nitrogen monoxide molecules. Then in the second step, a second oxygen molecule combines with the 2 nitrogen monoxide molecules to form the final products, 2 nitrogen dioxide molecules. We write separate equations (and enthalpy changes) for the two steps, then we add them together to obtain the final, equation and enthalpy change for the overall reaction: N 2 (g) + O 2 (g) > 2NO (g) ΔH 2 = 180 kj 2NO (g) + O 2 (g) > 2NO 2 (g) ΔH 3 = 112 kj N 2 (g) + 2O 2 (g) + 2NO (g) > 2NO 2 (g) + 2NO (g) ΔH 2 + ΔH 3 = (180 112) kj N 2 (g) + 2O 2 (g) > 2NO 2 (g) ΔH 1 = 68 kj Note that the net reaction, and its enthalpy change are identical, whether we run the reaction in a single step or in two steps. A comparison of the one step process versus the two step process is shown in Figure 6.7 from your text: Page 1 of 17
While the foregoing example demonstrates Hess Law, it does not prove it. Any proof lies in the consistency between Hess Law and the First Law. Characteristics of Enthalpy Changes: Suppose we know the stoichiometry and the change in enthalpy for some particular reaction. Take, for example: Xe (g) + 2F 2 (g) > XeF 4 (s) ΔH = 251 kj Suppose we needed to use this reaction in a Hess Law problem, but at double the quantities of reactants and products. What happens to the change in enthalpy? Because enthalpy change is an extensive property, it is proportional to the amounts of material. In this case, we know the enthalpy change for the production of 1 mole of XeF 4 (s). So, if we make 2 moles instead of 1, we double the change in enthalpy. The equation for that case is: 2Xe (g) + 4F 2 (g) > 2XeF 4 (s) ΔH = 502 kj Suppose instead, we needed to use the (first) reaction in another Hess Law problem, but in reverse? What then, would be the change in enthalpy? In the (first) reaction the system transfers 251 kj of heat to the surroundings, so one would expect that the Page 2 of 17
reverse reaction would require that the same amount of heat would be transferred from the surroundings into the system. Thus ΔH would have the same magnitude, but the opposite sign: XeF 4 (s) > Xe (g) + 2F 2 (g) ΔH = +251 kj These characteristics apply in general, and can be expressed as rules: o 1) The magnitude of ΔH is directly proportional to the quantities of reagents (reactants and products) involved in the reaction. If the coefficients in the reaction are multiplied by some factor, the enthalpy change must be multiplied by the same factor. o 2) If the direction of the reaction is reversed, the sign of ΔH is also reversed. (Note that you could derive this rule from the first rule by using 1 as the factor by which the coefficients and ΔH are multiplied.) Sample Exercise 6.7 (p. 244): Graphite and diamond are two allotropic forms of carbon shown in the following illustration: Page 3 of 17
Their properties are strikingly different as are their molecular structures. Graphite is a soft, black, slippery substance used for lubrication and as pencil lead; Diamond is a brilliant, crystalclear substance used for gemstones and as an industrial abrasive. The enthalpies of combustion for graphite and for diamond are, respectively, 394 kj/mol and 396 kj/mol. Calculate ΔH for the conversion of graphite to diamond. (This problem illustrates the use of Hess Law to compute an enthalpy change that is most impractical to measure directly.) o We start by writing equations for the conversion reaction and for the two combustion reactions: C (graphite) > C (diamond) ΔH =? C (graphite) + O 2 (g) > CO 2 (g) ΔH = 394 kj C (diamond) + O 2 (g) > CO 2 (g) ΔH = 396 kj o Now we solve the problem by reversing the reaction for the combustion of diamond and adding it to the reaction for the combustion of graphite: C (graphite) + O 2 (g) > CO 2 (g) ΔH = 394 kj C (graphite) + O 2 (g) + CO 2 (g) CO 2 (g) > C (diamond) + O 2 (g) ΔH = > C (diamond) + O 2 (g) + CO 2 (g) Page 4 of 17 ΔH = +396 kj (396 394) kj C (graphite) > C (diamond) ΔH = 2 kj Sample Exercise 6.8 (p. 245): (This exercise illustrates the kind of systematic procedure often necessary for the solution of a complex Hess Law problem.) Calculate ΔH for the synthesis of diborane (B 2 H 6 (g)) from its elements, boron (B (s)) and hydrogen (H 2 (g)): 2B (s) + 3H 2 (g) > B 2 H 6 (g) Use the following data for the calculation: (a) 2B (s) + 3 /2O 2 (g) > B 2 O 3 (s) ΔH = 1273 kj (b) B 2 H 6 (g) + 3O 2 (g) > B 2 O 3 (s) + 3H 2 O(g) ΔH = 2035 kj (c) H 2 (g) + ½O 2 (g) > H 2 O (l) ΔH = 286 kj (d) H 2 O (l) > H 2 O(g) ΔH = +44 kj o Our objective is to combine the four reactions so that we assemble our reactants, B (s) and H 2 (g), on the left hand side and our product, B 2 H 6 (g), on the right, while we make all the intermediate reagents cancel out. The solution in the text begins by combining (a) with the
reverse of (b), and that is perfectly valid. But it is also perfectly valid to begin by combining (a) with 3 x (c), and that is what we will do. (a) 2B (s) + 3 /2O 2 (g) > B 2 O 3 (s) ΔH = 1273 kj 3 x (c) 3H 2 (g) + 3 /2O 2 (g) > 3H 2 O (l) ΔH = 3( 286 kj) (1) 2B (s) + 3H 2 (g) + 3O 2 (g) > B 2 O 3 (s) + 3H 2 O (l) ΔH = = 1273 858 kj 2131 kj This accomplishes the objective of assembling the right amounts of both reactants on the left hand side. o Now we can continue by combining our first intermediate result (1) with the reverse of (b). (1) 2B (s) + 3H 2 (g) + 3O 2 (g) > B 2 O 3 (s) + 3H 2 O (l) ΔH = 2131 kj (b) B 2 O 3 (s) + 3H 2 O(g) > B 2 H 6 (g) + 3O 2 (g) ΔH = + 2035 kj 2B (s) + 3H 2 (g) + 3O 2 (g) + B 2 O 3 (s) + 3H 2 O(g) (2) 2B (s) + 3H 2 (g) + 3H 2 O(g) > B 2 H 6 (g) + 3O 2 (g) + B 2 O 3 (s) + 3H 2 O (l) > B 2 H 6 (g) + 3H 2 O (l) ΔH = ΔH = 96 kj 96 kj This accomplishes the objective of assembling the right amount of product on the right hand side. It also gets rid of most of the intermediate reagents. But we still need to eliminate 3H 2 O(g) from the left hand side and 3H 2 O (l) from the right. o Now we can finish by combining our second intermediate result (2) with 3 x reaction (d). (2) 2B (s) + 3H 2 (g) + 3H 2 O(g) > B 2 H 6 (g) + 3H 2 O (l) ΔH = 96 kj 3 x(d) 3H 2 O (l) > 3H 2 O(g) ΔH = +3x44 kj 2B (s) + 3H 2 (g) + 3H 2 O (l) + 3H 2 O(g) > B 2 H 6 (g) + 3H 2 O (l) + 3H 2 O(g) ΔH = 96 + 132 kj 2B (s) + 3H 2 (g) > B 2 H 6 (g) ΔH = +36 kj o We are done! The synthesis of one mole of diborane from its elements is slightly endothermic. Assuming that all the measured enthalpy changes for reactions (a) (d) were made under standard conditions (1 atm pressure and 25 C), we have thus determined that the Page 5 of 17
standard enthalpy of formation of diborane is +36 kj/mol. We will discuss standard enthalpies of formation in detail in the next section. 6.4 Standard Enthalpies of Formation: We saw in Section 6.2 that constant pressure calorimetry is a method we can apply to make measurements of enthalpy changes (i. e., heats of reaction) for reactions that take place in solution under ambient pressure. Additionally, constant volume calorimetry (into which we did not delve to any great extent) can be used to determine heats of combustion for a wide variety of substances. But what do we do about reactions that do not lend themselves to calorimetry? How do we obtain enthalpy changes for such reactions? (A good example is the reaction for the conversion of graphite into diamond, where we obtained the enthalpy change from the heats of combustion of graphite and of diamond.) Would it help if someone has already tabulated the enthalpy changes for a list of reactions that encompass all the compounds we might ever be interested in? Actually, there is such a list in the back of your text. Standard Enthalpy of Formation, a Definition: The standard enthalpy of formation (ΔH f ) of a compound is defined as the change in enthalpy that accompanies the formation of one mole of the compound from its elements with all substances in their standard states. The degree sign in the symbol, ΔH f, indicates that this is the case. Standard States: The definition for standard enthalpy of formation stipulates that all substances be in their standard states. So we need to define the conditions that make a state standard: o All Substances: A temperature of exactly 25 C is required. o Any Gaseous Compound: The pressure must be exactly 1 atmosphere. (This may soon change. The IUPAC has adopted a new standard pressure of exactly 1 bar, or 100,000 Pa. By comparison, 1 atmosphere = 101,325 Pa.) o Any Pure Substance in a Condensed State (liquid or solid): The standard state is the pure liquid or solid. o Any Substance in Solution: The standard state is a concentration of exactly one mole per liter (1 M). Page 6 of 17
o An Element: The standard state of an element is the form in which the element exists under the conditions of 1 atmosphere pressure and 25 C. (The standard state for oxygen is O 2 (g) at 1 atm.; the standard state for carbon is graphite; the standard state for mercury is Hg (l); the standard state for sodium is Na (s), etc.) Examples of Standard Enthalpies of Formation: We follow the discussion on p. 247 of your text: o Nitrogen Dioxide (NO 2 (g)): We write the reaction for the formation of one mole of product: ½ N 2 (g) + O 2 (g) > NO 2 (g) ΔH f = 34 kj/mol The standard states are all gases at 25 C. The picture shows a sample of NO 2 (g) in a Florence flask. o Methanol (CH 3 OH (l)): We write the reaction for the formation of one mole of product: C (s) + 2H 2 (g) + ½ O 2 (g) > CH 3 OH (l) ΔH f = 239 kj/mol The standard state of carbon is graphite; the standard states of hydrogen and oxygen are both gases; and the Page 7 of 17
standard state of methanol is the liquid, with all reagents at a temperature of 25 C. o Oxygen (O 2 (g)): We write this reaction as: O 2 (g) > O 2 (g) ΔH f = 0 kj/mol Notice that the enthalpy of formation of oxygen, or any other element, in its standard state is necessarily zero, by virtue of no real reaction having taken place. o Lists of Values of Standard Enthalpies of Formation: A short list is given in Table 6.2 in the text: A more complete list can be found in Appendix 4, on pp. A19-A22. The list includes two other properties, ΔG f, and S. We will discuss these properties later in the year. Using Standard Enthalpies to Calculate Standard Enthalpy Changes (Heats of Reaction): We will follow the example from the text, the combustion of methane. We wish to calculate the standard enthalpy of combustion: o We start by writing the reaction. (It should be an old friend by now.) CH 4 (g) + 2O 2 (g) > CO 2 (g) + 2H 2 O (l) Note that the water is in its liquid state. This follows from our requirement that the reaction begin and end at 25 C. This does not restrict it from taking place (as we know it does) at a much higher temperature. We simply invoke the fact that enthalpy is a state function, Page 8 of 17
and we can pick any path, as long as it starts with reactants at standard conditions and ends with products at standard conditions. o We can set the path so as to allow us to break the reaction up into a series of reactions whose enthalpy changes can be found in our tables. A particularly useful series of reactions is the one where we assume that our reactants un-form themselves back to their elements (in standard states) then re-form as the products. We see this diagrammed in Figure 6.8 from your text o This analysis gives us the following set of reactions: (a) CH 4 (g) > C (s) + 2H 2 (g) ΔH a = 75 kj (b) 2O 2 (g) > 2O 2 (g) ΔH b = 0 kj (c) C (s) + O 2 (g) > CO 2 (g) ΔH c = 394 kj (d) 2H 2 (g) + O 2 (g) > 2H 2 O (l) ΔH d = 572 kj Reaction (a) is the reverse of the reaction to form 1 mole of methane from its elements. According to Appendix A4, p. A20, methane has a heat of formation Page 9 of 17
of 75 kj/mol, so we change the sign to get ΔH a = 75 kj. Reaction (c) is the reaction to form 1 mole of CO 2 (g) from its elements and we can find ΔH c = 394 kj in Table 6.2. Reaction (d) is the reaction to form 2 moles of liquid water from its elements, so we double the value of 286 kj/mol we find in Table 6.2 for the standard enthalpy of formation of water and obtain ΔH d = 572 kj. (We have already discussed why ΔH b = 0 kj.) o Now we can recombine the four reactions and add the 4 enthalpy changes to get the final result: (a) CH 4 (g) > C (s) + 2H 2 (g) ΔH a = 75 kj (b) 2O 2 (g) > 2O 2 (g) ΔH b = 0 kj (c) C (s) + O 2 (g) > CO 2 (g) ΔH c = 394 kj (d) 2H 2 (g) + O 2 (g) > 2H 2 O (l) ΔH d = 572 kj CH 4 (g) + 2O 2 (g) > CO 2 (g) + 2H 2 O (l) ΔH = o Let s look at this problem another way: CH 4 (g) + 2O 2 (g) > CO 2 (g) + 2H 2 O (l) ΔH f 75 kj 0 kj 394 kj 286 kj x 1 2 +1 +2 ΔH = +75 kj +0 kj 394 kj 572 kj ΔH = 891 kj 891 kj The (balanced) equation for the reaction is written in the first row. The standard (molar) enthalpies of formation then are written in the second row for each component of the reaction. The third row contains a set of multipliers where the magnitude of each multiplier is the stoichiometric coefficient from the reaction equation and the signs are for each reactant and + for each product. The fourth row contains terms that are obtained by multiplying together the entries in the second and third rows. Page 10 of 17
When the terms in the fourth row are added together, the result (fifth row) is the standard enthalpy change for the reaction, our desired result. Equation 6.1 (p. 249): The procedure we have just used to obtain the standard enthalpy of combustion of methane can be generalized so that it applies to any reaction where the initial and final states correspond to standard conditions. The standard enthalpy change for a given reaction can be calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products. (These enthalpies of formation are calculated for the actual numbers of moles of each component in the reaction.) Mathematically, this statement can be written as Equation 6.1: ΔH reaction = Σn p ΔH f (products) Σn r ΔH f (reactants) (6.1) In Equation 6.1, the symbol, Σ (sigma), means to compute the sum of the terms, the n p s represent the stoichiometric coefficients of the products, and the n r s represent the stoichiometric coefficients of the reactants. The five row procedure that we just used is a systematic and compact way to apply Equation 6.1 to a standard enthalpy change problem. We listed the molar enthalpies of formation (ΔH f ) in the second row for each component in the reaction, and we listed the stoichiometric coefficients (the n p s and the n r s) in the third row (with signs for the n r s). Then we computed the individual terms of the summation in row four (n p ΔH f and n r ΔH f ) and did the final summation in row five to get the desired answer. Key Concepts for Performing Enthalpy Calculations: Your text offers the following summary (found in the middle of p. 249): Page 11 of 17
Sample Exercise 6.9 (pp. 249-51): Compute the standard enthalpy change for the reaction of ammonia burning in air to produce nitrogen dioxide and water. Refer to Table 6.2 for standard enthalpies of formation. The text spreads this calculation over two full pages; we ll do it in five lines: 4NH 3 (g) + 7O 2 (g) > 4NO 2 (g) + 6H 2 O (l) ΔH f 46 kj 0 kj 34 kj 286 kj x 4 7 +4 +6 ΔH = +184 kj +0 kj +136 kj 1716 kj ΔH = 1396 kj Sample Exercise 6.10 (p. 251): Compute the standard enthalpy change for the reaction of aluminum metal with iron Page 12 of 17
(III) oxide to form aluminum oxide and iron metal. (This is commonly called the thermite reaction.) We will again do it in five lines: 2Al (s) + Fe 2 O 3 (s) > Al 2 O 3 (s) + 2Fe (s) ΔH f 0 kj 826 kj 1676 kj 0 kj x 2 1 +1 +2 ΔH = +0 kj +826 kj 1676 kj +0 kj ΔH = 850. kj Sample Exercise 6.11 (p. 252): Compare the standard enthalpy of combustion of one gram of (liquid) methanol (CH 3 OH (l)) with that of one gram of gasoline. Assume that gasoline is pure liquid octane (C 8 H 18 ). We will first compute the standard molar enthalpies of the two fuels, then convert to enthalpies per gram for the comparison. The calculation for (one mole of) methanol is: CH 3 OH (l) + 3/2O 2 (g) > CO 2 (g) + 2H 2 O (l) ΔH f 239 kj 0 kj 394 kj 286 kj x 1 3/2 +1 +2 ΔH = +239 kj +0 kj 394 kj 572 kj ΔH = 727. kj This result is the heat of combustion of methanol in units of kj/mol. To convert this to kj/g, we divide by the molar mass of methanol (32.0 g/mol). Thus the desired result is: ΔH = (727. kj/mol)/(32.0 g/mol) = 22.7 kj/g The calculation for (one mole of) octane (C 8 H 18 ) is: C 8 H 18 (l) + 25/2O 2 (g) > 8CO 2 (g) + 9H 2 O (l) ΔH f 269 kj 0 kj 394 kj 286 kj x 1 25/2 +8 +9 ΔH = +269 kj +0 kj 3152 kj 2574 kj ΔH = 5457. kj This result is the heat of combustion of octane in units of kj/mol. To convert this to kj/g, we divide by the molar mass of octane (114.2 g/mol). Thus the desired result is: ΔH = (5457 kj/mol)/(114.2 g/mol) = 47.8 kj/g Page 13 of 17
(Since our data are precise to 3 significant figures, we express this result to 3 significant figures.) We have just demonstrated that the enthalpy of combustion of gasoline (octane) is slightly more than twice that of methanol. 6.5 Present Sources of Energy But First, a Little History: o Solar Power The Ultimate Source of All Energy: All the energy we have used throughout human history comes from stellar sources. Even our nuclear fuels were forged in stellar interiors. Most of the energy we use today was captured from the sun by photosynthesis in prehistoric times and laid down where it would be transformed into fossil fuels like petroleum, natural gas, and coal. Even wind and water power are driven by solar energy. Wind arises from solar energy heating the atmosphere and setting it into motion. Water power is driven through solar heating that evaporates water (mostly from the sea) which ultimately falls as rain and which can be channeled to flow downhill through electric generators. o The Pre-Industrial Age: Biomass, mostly wood, was the primary source of energy, and it was used mostly for heating. There was also some exploitation of wind power (e. g., windmill-driven pumps in Holland) and water power (e. g., water-driven grist mills). What little light that could be generated after dark came from (whale) oil lamps, tallow candles, and open wood fires. o The Industrial Revolution: The Industrial Revolution began in England in the 18 th Century. This period was marked by advances in coal-mining technology, and the development of coal-fired steam engines to drive machinery for industrial production. Then steam ships and steam-driven railways revolutionized transport in the 19 th century. And the science of thermodynamics was developed as part of the quest for more-and-more efficient steam engines. o Petroleum: The first oil well went into production in 1859 to supply kerosene for lighting when demand for lamp oil outstripped the supply of oils derived from animal fat. Gasoline was an unwanted by-product until the advent of automobiles. Another initially unwanted Page 14 of 17
by-product was natural gas. It was typically flared (burned) at the wellhead for lack of any system of gas pipelines. In those days, gas for street and domestic lighting was made by coal gasification in plants located near the urban areas they supplied. o Sources of Energy in the USA Since 1850: In 1850, we depended primarily on wood, by 1900 coal had supplanted wood; but nowadays, petroleum and natural gas reign supreme, as shown in Figure 6.11: Petroleum and Natural Gas: Petroleum and Natural Gas are fossil fuels consisting largely of hydrocarbons. With the singular exception of methane (CH 4 ), they feature carbon carbon bonds with chain lengths that can be quite long. Table 6.3 contains the names and formulas of the hydrocarbons with between 1 and 8 carbons: Page 15 of 17
o Natural Gas: Natural Gas is largely methane, but it also contains significant quantities of ethane, propane, and butane, all of which are gases at room temperature. It is usually found in combination with petroleum. o Petroleum: Petroleum (literally rock-oil ) is largely a mixture of liquid hydrocarbons ranging from pentane (C 5 H 12 ) to hydrocarbons containing 25 or more carbon atoms. Before petroleum becomes really useful, it needs to be separated into its various fractions, by a process called fractional distillation. These petroleum fractions and their uses are listed in Table 6.4: Page 16 of 17
In the early days of the petroleum industry, fractional distillation was the extent of petroleum refining. The kerosene would be packaged and sold, but a substantial proportion of the other fractions would have no market and would have to be discarded. The advent of the electric light and the automobile changed all that. Demand for kerosene plummeted while demand for gasoline skyrocketed. The process of catalytic cracking was developed to break kerosene and higher fractions down to four-carbon (C 4 ) compounds, and the process of reforming was developed to react pairs of C 4 molecules together to produce desirable branched isomers of octane (C 8 H 18 ). (We will conclude Section 6.5 in class on November 19.) (Link to Nov. 19 Lecture) Page 17 of 17