ENTHALPY, INTERNAL ENERGY, AND CHEMICAL REACTIONS: AN OUTLINE FOR CHEM 101A PART 1: KEY TERMS AND SYMBOLS IN THERMOCHEMISTRY System and surroundings When we talk about any kind of change, such as a chemical reaction or a temperature change, we must specify what object or substance is changing. That object or substance is called the system. For example, if we melt an ice cube, the ice cube is the system, and if we burn a piece of paper, the piece of paper and the oxygen that reacted with it are the system. The surroundings are everything that isn t the system. For example, if we melt an ice cube, the surroundings are everything in the universe except the ice cube. P, V, n and T These mean the same thing that they do in the ideal gas law. Note that T must be in kelvins. R = the gas constant In thermochemistry, we must use the value R = 8.314 J/mol K. C = specific heat capacity The specific heat capacity (or just specific heat ) is the amount of energy required to raise the temperature of 1 gram of a substance by 1 C. For example, the specific heat of liquid water is around 4.18 J/g C and the specific heat of solid iron is about 0.45 J/g C. Specific heat depends somewhat on temperature; for example, the specific heat of water is 4.182 J/g C at 20 C (68 F), but 4.217 J/g C at 0 C (32 F). For gases, the specific heat depends strongly on whether the gas is kept at a constant pressure or at a constant volume, and we use a separate symbol for each situation: the value at constant pressure is called C p and the value at constant volume is called C v. For example, here are the specific heats of oxygen: C p = 0.918 J/g C and C v = 0.658 J/g C. E = internal energy In thermochemistry, E stands for the energy that depends on how atoms are bonded to each other. We call this internal energy to distinguish it from other types of energy such as thermal energy (the energy we sense when something is hot). The internal energy of a chemical depends primarily on its chemical formula and its state. It is impossible to measure the internal energy of a chemical, but we can measure changes in internal energy when chemicals react. For example, we can t measure the internal energy of calcium and sulfur, but we can measure the change in internal energy when calcium and sulfur combine to form calcium sulfide (it is about 6.7 kilojoules per gram of CaS). Some textbooks use E for the overall energy of a substance and U for internal energy. We ll use E for internal energy to be consistent with your textbook. Internal energy also depends on temperature, but only weakly, so we ignore this dependence in most cases.
w = work Work is energy that is used to move an object against some kind of resisting force. Work requires motion: if nothing moves, no work was done. We can measure the work that is done only if we know the resisting force at every moment while the object is moving. For example, the work that can be done by 1 mole of a gas if its volume doubles can range anywhere from 0 to 1.7 kilojoules, depending on the resistance to the expansion. The sign of work depends on whether the system is gaining or losing energy: If the system does the work, w is a negative number, because a system loses energy when it does work. If the system has the work done to it (i.e. if the surroundings do the work), w is a positive number, because a system gains energy when the surroundings do work. q = heat Heat is energy that is used for anything other than doing work. The three main situations that involve heat are: Temperature changes Changes of state (melting, boiling) Changes in chemical composition (reactions) Heating something does not necessarily change its temperature. For instance, if you heat water that is already at 100 C, the water will turn to steam but it will not get hotter. The sign of heat, like the sign of work, depends on whether the system is gaining or losing energy: If the system absorbs heat, q is a positive number. If the system gives off heat, q is a negative number. H = enthalpy Enthalpy is defined as follows: H = E + PV. To understand what this means, let s look at the two terms: E is the internal energy of a chemical (the chemical energy that depends on how atoms are bonded to each other). PV is the kinetic energy that allows a gas to exert pressure and to take up space. (If you want a gas to expand or to exert more pressure, you must add energy to it.) It turns out that H is a better measure of a chemical system s ability to do work or to produce heat than E is. A liter of natural gas (CH 4 ) can do work by burning (in which its chemical energy changes into thermal energy), but it can also do work by simply expanding. Therefore, chemists normally use H when they describe chemical reactions and the energy they produce. PART 2: IMPORTANT CONCEPTS AND RELATIONSHIPS IN THERMOCHEMISTRY 1) How E and H are related to chemical reactions All chemicals contain internal energy (E). The amount of internal energy they contain depends on which atoms are bonded to each other. In a chemical reaction, atoms exchange
partners, so all reactions involve a change in the internal energy of the chemicals. We can make analogous statements about enthalpy (H), so all reactions also involve a change in the enthalpy of the chemicals. We use the symbols ΔE and ΔH for the changes in internal energy and enthalpy, respectively. The difference between E and H is PV, which is generally rather small, so ΔE and ΔH usually have similar magnitudes. For instance, if we make a mole of NaCl from the elements sodium and chlorine, ΔE is -410.0 kj and ΔH is -411.2 kj. The sign of ΔE and ΔH is an important piece of information, because it tells us whether the reaction will produce or absorb heat. If ΔE (ΔH) is positive, the reaction absorbs heat, and the chemicals and their surroundings become colder. We call this type of reaction endothermic. If ΔE (ΔH) is negative, the reaction gives off heat, and the chemicals and their surroundings become hotter. We call this type of reaction exothermic. This may seem confusing: how can chemicals become colder when they absorb heat? To understand, recall that when ΔE or ΔH is positive, the internal energy of the chemicals is increasing. This energy must come from somewhere, though. The chemicals increase their internal energy by converting some of their own thermal energy into internal energy. As a result, the temperature of the chemicals drops. Since the chemicals are now colder than their surroundings, heat flows from the surroundings into the chemicals. The overall result is that the chemicals absorb heat from the surroundings and convert it into internal energy. You can use analogous reasoning to explain how chemicals become hotter when a reaction gives off heat. Example: When a mole of water evaporates, ΔH is 44 kj. Does the evaporation of water absorb heat, or does it give off heat? Does the water become warmer, or does it become colder? ΔH is positive, so water absorbs heat as it evaporates, and the water becomes colder. Most of the heat that the water absorbs becomes internal energy; the rest increases the PV term (since water vapor takes up far more space than liquid water). 2) Interpreting the numerical value of ΔH or ΔE for a chemical reaction The amount of energy we get from a reaction depends on the amount of chemical that reacts; the larger the amount of chemical, the more energy we get. Therefore, just saying something like burning hydrogen produces 565 kj of energy is too vague. How much hydrogen did we use? To clarify this, chemists write a balanced equation next to the corresponding ΔE (or ΔH) value: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔE = -565 kj The ΔE (or ΔH) value you see always corresponds with the numbers of moles in the balanced chemical equation. In this case, the information we are given tells us that the internal energy decreases by 565 kj if we react 2 moles of H 2 with 1 mole of O 2. If we use different amounts of hydrogen and oxygen, the energy change will be different as well. Example: For the reaction 2 NH 3 (g) N 2 (g) + 3 H 2 (g), ΔE = 87.0 kj. What is the relationship between the energy change and the moles of reactants and products? The reaction will absorb 87.0 kj of heat if we break down 2 moles of NH 3, forming 1 mole of N 2 and 3 moles of H 2.
3) The relationship between energy and moles: ΔE standard n standard = ΔE nonstandard n nonstandard The amount of energy a reaction produces is proportional to the number of moles of reactant we use. For instance, suppose we burn 0.300 moles of H 2. How much energy will we get? On the previous page, you saw the reaction 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔE = -565 kj This told us that we get 565 kj of energy (in the form of heat) if we burn 2 moles of H 2. These numbers are called the standard values, because they match the balanced equation. We are using a nonstandard number of moles of H 2 (0.300 moles), so we write: -565 kj ΔE = nonstandard 2 mol H 2 0.300 mol H 2 Solving this equation gives ΔE nonstandard = -84.8 kj. Therefore, burning 0.300 moles of H 2 will produce 84.8 kilojoules of energy. Example: If you break down enough NH 3 to make 0.472 moles of H 2, how much energy will be absorbed? The reaction is: 2 NH 3 (g) N 2 (g) + 3 H 2 (g) ΔE = 87.0 kj For this reaction, n standard is 3 moles for H 2, and ΔE standard is 87.0 kj. We are actually making 0.472 moles of H 2, so 87.0 kj ΔE = nonstandard 3 mol H 2 0.472 mol H 2 Solving this equation gives ΔE nonstandard = 13.7 kj. The reaction will absorb 13.7 kj of energy. Example: If you mix 0.500 moles of H 2 with 0.500 moles of O 2 and allow them to react, how much energy will be given off? The reaction is: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔE = -565 kj You must use the limiting reactant to calculate the nonstandard value of ΔE. (This is analogous to stoichiometry problems that ask you to calculate the amount of product.) In this case, the limiting reactant is H 2 (review topic A if you have forgotten how to work this out), so we use H 2 in our calculation: -565 kj ΔE = nonstandard 2 mol H 2 0.500 mol H 2 Solving this equation gives ΔE nonstandard = -141 kj. The reaction will give off 141 kj of energy. Example: When 0.250 moles of Cl 2 reacts with excess P 4 to form PCl 3, 51.1 kj of energy is given off. Calculate ΔE for the reaction P 4 (s) + 6 Cl 2 (g) 4 PCl 3 (g) In this case, we are given nonstandard values for the number of moles of Cl 2 and the energy change, and we must calculate the standard value of ΔE. We can use the same relationship to do this. Note that we are told that the energy is given off, so ΔE nonstandard is a negative number (-51.1 kj). ΔE standard -51.1 kj = 6 mol Cl 2 0.250 mol Cl 2
Solving this equation gives ΔE standard = -1226 kj. (Strictly, this is only valid to three significant figures and should be written -1.23 x 10 3 kj.) 4) The First Law of Thermodynamics: ΔE = q + w This simple equation tells us that there are only two ways in which the energy of a system can change: it can absorb (or give off) heat, and it can do work (or have work done on it). The energy used to do work plus the energy used as heat must add up to the overall energy change. The only tricky part about this equation is keeping track of the signs of q and w. Example: Suppose a reaction absorbs 500 J of heat and does 200 J of work. What is the change in the internal energy of the chemicals? In this situation, q is positive (because our system is absorbing the heat), but w is negative (because our system is doing the work, not having the work done to it). Therefore: ΔE = 500 J + (-200 J) = 300 J 5) The relationships between ΔE, ΔH, and heat When we carry out a chemical reaction, we do so in one of two ways: Option 1: constant volume. This requires that the reaction happen in a sealed, rigid container. As you can imagine, this can be dangerous, so we rarely do this in practice. Option 2: constant external pressure. This requires only that the container either be completely elastic (like a balloon) or that it be open to the surroundings (like an open beaker). We nearly always carry out reactions this way. The heat that is given off or absorbed by a reaction is related to ΔE or ΔH as follows: For a reaction that is carried out at constant volume (option 1): q = ΔE For a reaction that is carried out at constant pressure (option 2): q = ΔH These relationships assume that we are not harnessing the reaction to do work for us. Of course, in many cases the whole point of carrying out a reaction is to do work, like moving a motor vehicle or pushing electrons through a cell phone. However, we will not consider such situations in Chem 101A. Example: Consider the reaction below: CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(l) ΔE = -886 kj ΔH = -891 kj How much heat is given off when one mole of CH 4 reacts with excess oxygen in an open container? In an open container, the external pressure is constant, so the heat will equal ΔH. The reaction will give off 891 kj of heat. Note that ΔH and ΔE do not change if we switch from constant volume to constant pressure. In general, we can treat the ΔH and ΔE values for a reaction as if they were constants.
However, q and w do depend on the conditions. If you switch from constant volume to constant pressure, ΔE and ΔH don t change, but q and w do. The fact that q = ΔH at constant pressure (i.e. in almost all real-world situations) is why chemists invented H in the first place, and why all tables of chemical reactions give ΔH instead of ΔE. Many reactions make or consume gases, and these gases take up a lot of space. When a reaction makes a gas, the gas must push the atmosphere out of the way, doing work in the process. Therefore, the reaction produces less heat than we expect from the ΔE value. Likewise, when a reaction consumes a gas, the surrounding atmosphere compresses the remaining chemicals (which take up less space now), doing work in the process. That work becomes part of the energy of the chemicals, which in turn release it back to the surrounding as heat. Therefore, the reaction produces more heat than we expect from the ΔE value. Knowing ΔH means we don t have to account for this expansion/contraction work (called PV work) if we just want to know how much heat a reaction produces. 6) The relationship between ΔH and ΔE: ΔH = ΔE + RTΔn gases You will often need to calculate ΔH from ΔE (or vice versa), and this equation is the practical way to convert one into the other. In the equation, R = 8.314 J/mol K, T is the temperature in kelvins, and Δn gases is the change in the number of moles of gases during the reaction. Example: Consider the following reaction: 2 H 2 (g) + O 2 (g) 2 H 2 O(l) ΔE = -565 kj Calculate ΔH for this reaction, assuming that the temperature is 25 C. We need values for ΔE, R, T, and Δn gases. ΔE is given (-565 kj), R is 8.314 J/mol K, and T is 25 + 273.15 = 298.15 K. As for Δn gases, the balanced equation tells us that we start with 3 moles of gases (2 moles of H 2 and 1 mole of O 2 ), and we end up with 0 moles of gases (note that the water is a liquid!). Therefore, Δn gases = 0 mol 3 mol = -3 mol. Now we substitute into our equation: ΔH = -565 kj + (8.314 J/mol K)(298.15 K)(-3 mol) = -565 kj + (-7436 J) = -565 kj + (-7.436 kj) = -572 kj Watch out for the units when you use this equation. ΔH and ΔE are always given in kj, but the RTΔn gases term will be in joules. Example: Consider the following reaction: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) ΔH = -56 kj Calculate ΔE for this reaction, assuming that the temperature is 25 C. For this reaction, there are no gases involved, so Δn gases = 0. Therefore, the term RTΔn gases is also zero, which means that ΔH equals ΔE. ΔE = -56 kj. It s worth remembering that ΔH equals ΔE for all reactions in which there are no gases involved, such as acid-base reactions and precipitations. ΔH also equals ΔE when the reaction
involves gases but the numbers of moles of gases does not change; an example is the reaction C(s) + O 2 (g) CO 2 (g). 7) Interpreting the difference between ΔH and ΔE: PV work For all reactions in which the number of moles of gases changes during the reaction, ΔH and ΔE are different. What is the significance of this difference? Suppose that a reaction produces a gas, but the temperature does not change. According to the gas laws, either P or V must increase as the number of moles of gas increases. If we keep the volume constant, the pressure will go up; the gases are essentially storing energy, just like a spring. However, if we keep the external pressure constant, the gases will expand, doing work in the process; the excess energy is transferred to the surroundings instead of being stored. The difference between ΔH and ΔE is the amount of work that the reaction mixture will do if its pressure remains constant. We call this work PV work and give it the symbol w PV What if the reaction consumes a gas? In this case, the external pressure will force the reaction mixture to shrink. Again, PV work is done, but this time it is done by the surroundings, so it increases the overall energy of the reaction mixture. In this case, the difference between ΔH and ΔE is the amount of work that the surroundings do if the pressure remains constant. In a gas expansion at constant pressure, the PV work equals P ΔV. The sign is required because the gas loses energy if its volume increases, so w must be negative if ΔV is positive (and vice versa). We can show that P ΔV must equal R T Δn gases if the temperature is constant, using the ideal gas law. Therefore, we can write w PV = RTΔn gases This equation only applies when the pressure is constant. If the volume is constant, the reaction does not do work; instead, it stores the energy by increasing the pressure. We can now expand our statements about ΔH and ΔE to include work: At constant volume, q = ΔE and w = 0 At constant pressure, q = ΔH and w = RTΔn gases Example: For the reaction 2 KClO 3 (s) 2 KCl(s) + 3 O 2 (g), ΔH = -90 kj. What are q and w when 2 moles of solid KClO 3 reacts at constant pressure and 25 C? What is ΔE under these conditions? The reaction is carried out at constant pressure, so q = ΔH. The problem states that we are using 2 moles of KClO 3, which matches the coefficient in the balanced equation, so we can use the standard value of ΔH: q = -90 kj. To calculate w, we use the formula w = RTΔn gases : w = -(8.314 J/mol K)(298.15 K)(3 mol 0 mol) = -7436 J Converting to kilojoules (to be consistent with q) gives w = -7.436 kj. To calculate ΔE, we can use our familiar relationship between ΔE and ΔH. However, it is easier to use the First Law of Thermodynamics: ΔE = q + w. Substituting in the values of q and w we just calculated gives ΔE = -97 kj. This reaction produces 97 kj of energy, of which 90 kj is released as heat and 7 kj is used to do PV work.
Note that the PV work we just calculated is the standard value, corresponding to 2 moles of KClO 3 reacting. The next example shows how to deal with a situation where you have a nonstandard amount of reactant. Example: If you want to get 2.000 kj of PV work from the reaction in the previous example, how much KClO 3 should you use? In the previous example, we found that 7.436 kj of PV work is done when 2 moles of KClO 3 reacts. The PV work is proportional to the amount of reactant (just like ΔE and ΔH), so we can use the usual technique to calculate the amount of KClO 3 here. 7.436 kj 2.000 kj = 2 mol KClO 3 n nonstandard Solving this equation gives n nonstandard = 0.5379 moles of KClO 3. 8) Putting it all together: relating nonstandard values of ΔH and ΔE In many problems, you will need to interconvert ΔH and ΔE and you will have to deal with nonstandard values. In any situation where you must interconvert ΔH and ΔE, you should always use the standard values. In other words, your strategy map should look like this: nonstandard ΔH! standard ΔH! standard ΔE! nonstandard ΔE (heat at constant pressure) (heat at constant volume) You can do this sequence either forward or backward, and you can start and end anywhere on the map. The following examples illustrate some typical problem types. Example: Consider the following reaction: 2 CO(g) + O 2 (g) 2 CO 2 (g) ΔE = -563.5 kj How much heat will be given off if 7.130 moles of CO reacts with excess O 2 in an open container at 25 C? In an open container (constant pressure), the heat equals ΔH, so we are going to need to convert ΔE into ΔH. We were given the standard value of ΔE, so we can start by using the equation ΔH = ΔE + RTΔn gases to calculate the standard value of ΔH. Once we know that, we can work out the nonstandard value of ΔH, which will equal the heat. Using ΔH = ΔE + RTΔn gases gives us: ΔH = -563.5 kj + (8.314 J/mol K)(298.15 K)(2 mol 3 mol) = -563.5 kj + (-2479 J) = -563.5 kj + (-2.479 kj) = -566.0 kj This is ΔH standard for the reaction, corresponding to the reaction of 2 moles of CO with 1 mole of O 2. Now we calculate the nonstandard value of ΔH that corresponds to the reaction of 7.130 moles of CO. -566.0 kj 2 mol CO = ΔH nonstandard 7.130 mol CO Solving this equation gives ΔH nonstandard = -2018 kj, so the reaction will give off 2018 kj of heat.
Example: When 0.07563 mol of gaseous C 4 H 10 reacts with oxygen in a sealed, rigid container, 217.6 kj of heat is given off. Calculate ΔH for the following reaction at 25 C: 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O(l) First, let s be sure that we understand what we re trying to calculate; the problem is asking for the standard value of ΔH for this reaction. In other words, we need to calculate the enthalpy change when 2 moles of C 4 H 10 burns. Now, let s understand the numbers we are given. We are given the amount of heat that is produced when 0.7563 moles of C 4 H 10 burns in a rigid container. The heat is given off, so q is a negative number: q = -217.6 kj. Under these conditions, the volume is constant, so the heat equals ΔE (not ΔH). This is a nonstandard value, because we didn t use 2 moles of C 4 H 10, so we can say that ΔE nonstandard = -217.6 kj. To convert ΔE into ΔH, we need the standard value of ΔE. Therefore, our strategy will be: nonstandard ΔE standard ΔE standard ΔH We are now ready to start calculating. First, we convert the nonstandard value of ΔE into the standard value: ΔE standard -217.6 kj = 2 mol C 4 H 10 0.07563 mol C 4 H 10 Solving this equation gives ΔE standard = -5754.33 kj Finally, we convert the standard value of ΔE into the standard value of ΔH: ΔH = ΔE + RTΔn gases = -5754.33 kj + (8.314 J/mol K)(298.15 K)(8 mol 15 mol) = -5754.33 kj + (-17352 J) = -5754.33 kj + (-17.352 kj) = -5772 kj Example: When 1.500 g of MgS reacts with excess oxygen at constant pressure, 9.207 kj of heat is given off. How much heat would be given off if the same amount of MgS reacted with oxygen at constant volume? Assume a temperature of 25 C for both reactions. The chemical reaction is MgS(s) + 2 O 2 (g) MgSO 4 (s) We are given the heat at constant pressure, which equals ΔH. This is a nonstandard value, because we did not use 1 mole of MgS. Our final answer will be a nonstandard value of ΔE, since the heat equals ΔE when the volume is constant. Our strategy will be: nonstandard ΔH standard ΔH standard ΔE nonstandard ΔE To begin, we convert the nonstandard ΔH value into the standard value. The molar mass of MgS is 56.37 g/mol, so 1.500 g equals 0.0266099 mol of MgS. Using our normal setup: ΔH standard 1 mol MgS = -9.207 kj 0.0266099 mol MgS Solving this equation gives ΔH standard = -345.999 kj. Next, we translate our standard ΔH into the standard ΔE: ΔH = ΔE + RTΔn gases -345.999 kj = ΔE + (8.314 J/mol K)(298.15 K)(0 mol 2 mol)
Solving this equation gives ΔE = -341.041 kj. This is ΔE standard. Finally, we calculate ΔE nonstandard for the reaction of 0.0266099 moles of MgS, which will equal our heat. -341.041 kj 1 mol MgS = ΔE nonstandard 0.0266099 mol MgS Solving this equation gives ΔE nonstandard = -9.075 kj. We conclude that the reaction gives off 9.075 kj of heat at constant volume. Appendix: Bomb Calorimetry Chemists measure ΔE and ΔH for a reaction using a technique called calorimetry. In calorimetry, we begin by measuring the amount of heat a reaction produces under specific conditions (either constant pressure or constant volume). Then we relate the heat to ΔE and ΔH. The most common type of calorimetry is called bomb calorimetry. In this technique, the reaction is carried out in a sealed, rigid steel container that is surrounded by water. As the reaction produces heat, the heat flows into the steel container and the surrounding water, raising their temperature. The temperature change can be converted into the corresponding amount of heat using the heat capacity of the calorimeter, which is the number of joules needed to raise the temperature of the steel and the water by 1 C. This heat can in turn be related to ΔE and ΔH for the reaction. Here is an example. Example: A 1.000 g sample of sodium is burned in a bomb calorimeter that has a heat capacity of 3631 J/ C. During the reaction, the temperature of the calorimeter rises from 18.31 C to 23.28 C. Calculate ΔE and ΔH for the reaction 4 Na(s) + O 2 (g) 2 Na 2 O(s) at 25 C. To begin, we calculate the heat that was absorbed by the calorimeter, using the temperature change and the heat capacity of the calorimeter: temperature change = 23.28 C 18.31 C = 4.97 C heat absorbed by calorimeter = 4.97 C 3631 J/ C = 18,046 J Note that this value is positive, because the energy of the calorimeter increased. Next, we recognize that the reaction must have given off 18,046 J of heat, since energy is conserved. When a system gives off heat, the sign of q is negative, so: q reaction = -18,046 J = -18.046 kj A bomb calorimeter keeps the reaction at a constant volume, so q = ΔE here. This is a nonstandard value, corresponding to the reaction of 1.000 g of Na, so we must convert it to a standard value (for 4 moles of Na, according to the balanced equation). 1.000 g of Na equals 0.0434972 mol of Na, so ΔE standard 4 mol Na = -18.046 kj 0.0434972 mol Na Solving this equation gives ΔE standard = -1660 kj. Finally, we relate ΔE standard to ΔH standard using ΔH = ΔE + RTΔn gases. Substituting the values we know (Δn gases = -1 mol here) and solving for ΔH gives ΔH standard = -1662 kj.