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Welcome to Unit 6, which will be all about thermodynamics. This is technically a branch of physics, but many concepts in thermodynamics will be very important for us to understand. Specifically we will cover endothermic and exothermic processes, more on energy diagrams, heat transfer and thermal equilibrium, heat capacity and calorimetry, the energy of phase changes, enthalpy of reaction, bond enthalpies, enthalpy of formation, and Hess’s law.
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Let’s start out by learning some basic terminology that will allow us to talk about energy exchange in the context of chemical reactions. The property that chemists use to describe the thermodynamics of chemical and physical processes is called enthalpy. This is symbolized with a capital H, and when a system is under constant pressure, we can typically think of change in enthalpy as the change in energy for the system, which will just refer to heat absorbed or
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released. This can apply to a chemical reaction, but also to physical processes like phase changes, or even the dissolution of an ionic solid in water. In such a situation there will be an energy associated with disassembling the lattice, which is equal to the lattice energy, and an energy associated with solvating, or surrounding each individual ion with water molecules, making ion-dipole interactions that release energy, called the hydration energy. Depending on
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which of these parameters is greater, dissolution could be an exothermic or endothermic process. We will use thermochemical equations to list enthalpy data for a given chemical reaction. These are just chemical equations with a delta H value listed. If delta H is negative, the reaction is said to be exothermic, and that quantity of heat energy is released to the surroundings. If delta H is positive, the reaction is said to be endothermic, and that
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quantity of heat energy must be absorbed from the surroundings in order for the reaction to occur. Delta H is a stoichiometric value, which means it is specific to the molar quantities listed in the balanced chemical equation. So for this one here, the implication is that 286 kilojoules of energy will be released to the surroundings when one mole of hydrogen gas and a half a mole of oxygen gas react to produce one mole of liquid water. If these molar quantities are doubled,
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the delta H must also be doubled, as we can see here, since twice as many molecules will require twice the amount of heat exchange. And if the direction of the reaction is reversed, the sign of the accompanying delta H must also be reversed, since a reaction that is exothermic must have an endothermic reverse reaction, so positive becomes negative and negative becomes positive. The enthalpy change of a reaction depends on the physical states of the reactants and products,
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whether solid, liquid, gas, or aqueous, so these must be shown, because phase changes have their own enthalpies associated with them. We can therefore use thermochemical data to predict the heat exchange associated with the reaction of any particular mass of a substance simply by doing stoichiometric calculations. And of course we can recall from examining energy diagrams that the change in enthalpy for a reaction is represented by the distance from reactants to
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products. If a reaction is exothermic, products sit lower and energy is released, if endothermic products sit higher and energy is absorbed. Speaking of heat being released and absorbed, let’s talk about heat transfer. When we talk about heat transfer, we are talking about a transfer of kinetic energy. Hotter objects have particles with a greater average kinetic energy, or moving faster, and colder objects have particles with a lower average kinetic energy,
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or moving slower. When a hot object comes into contact with a cold object, kinetic energy is imparted from the faster-moving objects to the slower-moving objects through collisions, until thermal equilibrium is reached, and the whole system has the same average kinetic energy, which means everything is at the same temperature. With the concept of heat transfer understood, we are ready to learn about heat capacity and calorimetry. Let’s define these terms
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and see how they can be applied. The heat capacity of an object refers to the energy required to raise the temperature of the object by one degree, or the energy released when its temperature is reduced by one degree. Heat capacity will vary according to the composition of the object, because different substances absorb and release heat at different rates due to differences in structure and composition, a fact that we make use
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of in designing different materials. Heat capacity will also depend on the amount of the substance present, because the more of it there is, the more heat it will have to absorb to raise its temperature uniformly. A glass of water will heat up much more quickly than a swimming pool because there is so much less matter for the heat energy to be distributed within. The heat capacity for an object can be calculated by the following equation, where q equals heat and delta T equals
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the change in temperature of the object. This will give us a value in joules per degree Celsius, or the energy required per degree of temperature change. For example, let’s compare two cast iron frying pans. They are made of identical material, but one is five times larger. Therefore the larger one will have a much larger heat capacity, because there is so much more matter that needs to receive sufficient heat energy so as to raise its temperature. Let’s say it takes 18,150
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joules of energy to raise the temperature of the small pan by fifty degrees, and it takes 90,700 joules of energy to raise the temperature of the large pan by fifty degrees. What will be the heat capacity of each pan? Plugging in these values, we should get 363 joules per degree Celsius for the small one, and 1,814 joules per degree Celsius for the large one, which, as we would expect, is right around five times the other one. Whereas the heat capacity of an object depends
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on the amount of matter in the object, we want a way to describe heat capacity that is independent of the amount of matter, so that we can describe a particular substance in a general way. This will be called the specific heat capacity, or simply specific heat, and it refers to the amount of energy required to raise the temperature of one gram of one substance by one degree Celsius. This means that the two cast iron frying pans of different sizes that had different heat capacities
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will actually have identical specific heats, because they are made of the same material. This means that whereas heat capacity is an extensive property, depending on the amount of substance present, specific heat is an intensive property, and does not depend on the amount of substance present. Iron will heat at a particular rate per gram, regardless of how much there is. The specific heat of an object can be calculated similarly to heat capacity, but this time with an
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additional term for mass. We will need to divide the heat exchanged not just by the temperature change but also by the number of grams of the substance present to get a value that represents joules per gram degree Celsius. Here are the specific heats of various substances. Notice that liquid water has a very high specific heat. Most metals on the other hand, have rather low specific heats, along with a few gases. If we know the specific heat of a material, we can calculate
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the amount of heat exchanged during a process by measuring the mass and temperature change during the process. This requires a simple rearrangement of the original equation to give heat exchanged equaling the specific heat times mass times the change in temperature, or the final temperature minus the initial temperature. This is the version of the equation we will most frequently be using.
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If the temperature goes up, it will have absorbed heat, and q should be positive. If the temperature goes down, it will have released heat, and q should be negative. Let’s see if we can calculate the specific heat of a substance. A piece of unknown metal weighs 348 grams. After absorbing 6.64 kilojoules of heat, its temperature went from 22.4 to 43.6 degrees Celsius. What is the specific heat, and what substance is it likely to be? Let’s use our equation for specific heat
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and plug in all the known values. We should get an answer of 0.9 joules per gram degree Celsius. Looking back at our table of specific heats, it seems likely that this sample is a piece of aluminum. So specific heat data is very useful for identifying unknown substances. Now let’s examine an important application of this concept called calorimetry. A calorimeter is a calibrated device that can measure the amount of heat involved in a physical or chemical process,
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by combining temperature data with a known heat capacity for the device. This can be very useful data if we also know some information about the masses of substances involved, and other aspects of either the system, which refers to the substance or substances undergoing the chemical or physical change, or its surroundings, which refers to the other components of the measuring apparatus that serve to either provide heat to the system or absorb heat from the system, depending on
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whether it’s an exothermic or endothermic process. Many times calorimeters are just well-insulated containers that can directly measure a temperature change due to a physical or chemical process, without letting heat escape to the environment surrounding the calorimeter. This would mean that all of the heat released by a reaction would go directly towards heating up the calorimeter, or the heat absorbed by a reaction would come directly from the calorimeter, thus enabling a
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direct measurement of this temperature change with minimal error. This is often reproduced in a rough way with a coffee cup calorimeter, since styrofoam is a pretty good insulator. Something that can be done with such a calorimeter would be to measure the heat capacity of an unknown object. A piece of metal could be heated to a known temperature, perhaps in boiling water, and then transferred to a known volume of room temperature water in a calorimeter, which
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will then heat up to an intermediate temperature somewhere in between the two initial temperatures. By measuring the temperature change of the water as the hot metal transfers heat into the water, we can measure the amount of heat transferred into the water, since we also know the mass of the water given its volume, as well as the specific heat of water. Specific heat, mass, and change in
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temperature is everything we need to solve for q in this equation for the water. We also know from conservation of energy that there can have been no loss of heat during this transfer, in other words q water, the heat absorbed by the water, plus q metal, the heat released by the metal, equals zero. If we subtract q metal from both sides, we get q water equals negative q metal, which represents the reasonable assumption that all of the heat involved in the process went
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directly from the metal to the water losing only a negligible amount to the environment. Therefore, in calculating the heat absorbed by the water in the calorimeter, we will also learn the amount of heat that was released from the metal, which is simply the negative version of q for the water. With this information, we can plug values into a new equation, but this time using values for the metal. The q value will be the q metal we just got from our calculations for water,
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but this time the mass will be the mass of the metal and the change in temperature will go from the initial temperature of the hot metal to the final temperature in the calorimeter. The delta T will therefore be negative this time, which makes sense as it will cancel out the negative in the q value. The only unknown is the specific heat, so we can then solve for it. Once we have the specific heat, we can hopefully identify the composition of the unknown metal.
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Let’s get a little more mathematical with this. For example, let’s say a 59.7 gram piece of metal submerged in boiling water is transferred into 60 milliliters of water initially at 22 degrees Celsius. The temperature of the water is raised to 28.5 degrees as a result of the heat transfer. What is the specific heat of the metal? Just as we discussed, let’s first use the given data to figure out the heat absorbed by the water, using this equation. Since water is one gram
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per milliliter, it’s easy to get the mass of the water. The specific heat of water is a known value that is always the same, and we know the initial and final temperature of the water, so let’s plug those values into the equation. We should get 1,632 joules, which is the heat absorbed by the water. Let’s also recall that the heat absorbed by the water must be the heat released by the metal,
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so q metal must be negative 1,632 joules, since a positive q means heat absorbed and a negative q means heat released. So let’s write a new equation using the q metal we just calculated. This time we have the mass of the unknown metal, and the temperature change experienced by the metal, and the only unknown will be the specific heat of the metal. Doing the math, we should get 0.382 J/g oC as the specific heat of the metal. If we take a look at our list of specific heats,
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we might notice that the tabulated specific heat for copper is extremely close to the value we just calculated for our unknown metal, making it very likely that this was a piece of copper. With calorimetry understood, let’s shift gears and look at the change in energy that is associated with phase changes. When we looked at melting and boiling points, we learned about how the types of intermolecular forces that a substance is capable of will
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determine the phase that a substance will be in at a particular temperature. We also discussed what phase changes fundamentally are on the molecular level. Now that we are examining thermochemistry, we are ready to revisit this concept in a quantitative way, so that we can discuss the specific quantities of heat exchanged during any phase change. To begin, let’s think of the temperature at which a liquid changes between a liquid and a gas. If
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going from liquid to gas, this process is called vaporization. If going from gas to liquid, this process is called condensation. The key thing to recall is that in the liquid phase, the molecules are in constant contact and are therefore participating in whatever intermolecular forces that substance is capable of. In the gas phase, the molecules are far apart, and are not making such interactions. So a change from one phase to the other involves either breaking or forming
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these intermolecular attractions. If we look at a liquid in a closed container, we must understand that some molecules at the surface of the liquid may spontaneously vaporize, entering the gas phase. This will happen even far below the boiling point of the liquid. And if a molecule in the gas phase collides with the surface of the liquid, it might condense, and re-enter the liquid phase. If
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these two things are occurring at the same rate, we have what is called a dynamic equilibrium between the liquid and gas phases. The pressure exerted by such a dynamic equilibrium between liquid and gas is called the vapor pressure of the liquid. Vapor pressure will increase with increasing temperature, as more kinetic energy gives molecules a greater likelihood of escaping the attractions, and this means more molecules will vaporize, thus increasing the vapor pressure.
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If the temperature is raised to the point where the vapor pressure is equal to the surrounding pressure, typically atmospheric pressure if the liquid is in an open container, that is said to be the boiling point of the liquid. At the boiling point, all of the liquid will enter the gas phase. All of this means that vaporization will always be an endothermic process, or one that absorbs heat.
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This is because to go from a liquid state where the system is at a lower potential energy due to favorable electrostatic interactions, to a gaseous state where the system is at a higher potential energy because such interactions are not being made, the system must absorb energy. In fact, it must absorb precisely the amount of energy that is associated with those intermolecular contacts.
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A liquid like water will absorb this energy from its surroundings in the form of heat energy. The energy associated with the vaporization process is called the enthalpy of vaporization. Here is the enthalpy of vaporization for water at room temperature. We can see that it will take about 40 kilojoules of energy to vaporize one mole of liquid water. If vaporization is always endothermic, that means condensation is always exothermic, and when one mole of water vapor
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condenses, it will release this same amount of energy. This makes sense, because if the water molecules suddenly begin participating in hydrogen bonds, the system will go to a lower energy, and that difference in energy will be released. Now let’s examine the phase changes between the liquid and solid phase. These would be melting and freezing. The particles in a solid have very little kinetic energy. Apart from vibrating a little bit in place, the particles don’t move
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with respect to one another. They are fixed in a position that maximizes their interactions with one another. If sufficient heat energy is provided, they begin to move and wiggle enough to break free from the fixed positions and enter the liquid phase, where they continue to interact, but while moving fluidly around one another. As heat is added to a solid, its temperature increases, and once we reach the melting point, more heat will not go towards raising
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the temperature of the solid, it will go towards overcoming the interactions between the particles, and the temperature will remain constant through the process of melting. Once completely melted, additional heat will once again go back to raising the temperature of the substance. If we were to isolate a substance at its melting point, there would be an equilibrium between the liquid and
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solid phases. This means that freezing and melting would be occurring at equal rates. This specific temperature, where the two processes are balanced, is called the melting point of the solid, or the freezing point of the liquid. These refer to the same temperature, and we would use one over the other to indicate a specific direction for the transition. Just like the enthalpy of vaporization, we can quantify the energy needed to melt a solid. This is called the enthalpy
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of fusion. The precise value for the enthalpy of fusion will depend on how strongly the particles are interacting with one another in a crystalline solid. If these interactions are weak, it will result in a low melting point. If they are strong, it will have a high melting point. For water, the enthalpy of fusion is about 6 kilojoules per mole. That means it will take 6 kilojoules of energy to melt one mole of ice, making this an endothermic process, and 6 kilojoules of energy
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will be released when one mole of liquid water freezes, making this an exothermic process. Beyond these, there are two more phase changes that are not as commonly seen, and they involve transitions between the solid and the gas phase. It’s not completely intuitive that substances can skip over the liquid phase, but nevertheless, certain substances even at atmospheric pressure are capable of sublimation, which is when a solid becomes a gas, or deposition, when a
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gas becomes a solid. Sublimation and deposition also have energies associated with them. Sublimation will be endothermic, and deposition will be exothermic. Here is the enthalpy of sublimation for carbon dioxide, and naturally the energy associated with deposition will be of the same magnitude but negative, as it simply represents the reverse process of sublimation. We can now combine our understanding of specific heat and the energy of phase changes to interpret
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things called heating and cooling curves. These show the temperature of an object as heat is exchanged between the object and its surroundings. Such a heating or cooling curve will have sections that rise or fall, but it will also have flat sections, so we want to understand what all of these sections mean, both quantitatively and qualitatively. To do this, let’s take a look at the equation for specific heat that we have already learned. Of course,
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anytime matter changes from one phase to another, there is an amount of energy associated with that change which is specific to the phase and to the material. So if a phase change occurs on a heating and cooling curve, we will need to know the enthalpy change associated with that specific phase change for that substance. If a substance undergoes sufficient heating or cooling so as to go through one or more phase changes, we will have to do this several times. Let’s look at
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this heating curve for water. We can see here that solid water, or ice, will heat at a certain rate, because ice has a particular specific heat. Then, once the melting point is reached, additional heat will go towards melting the sample until it is completely in the liquid phase. Notice that the temperature does not rise during this portion, because the heat energy is going towards overcoming the lattice energy of ice, rather than heating the sample. The heat associated with
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this specific phase change from solid to liquid depends on the enthalpy of fusion for water. Then, once the sample is entirely in the liquid phase, the liquid water will continue to heat up as more energy is absorbed at a rate that depends on the specific heat of liquid water, which is different from the specific heat of solid water. Once the boiling point is reached, additional heat will go towards vaporizing the sample, in a way that is described by the enthalpy of vaporization
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for water. Again, the temperature will not change during this section because the heat energy is going towards disrupting the hydrogen bonding occurring between water molecules, rather than heating the sample. And once completely vaporized, if heat continues to be absorbed, the steam will have its temperature raised at a rate that is determined by the specific heat of steam, which will once again be different than the specific heat for solid or liquid water.
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So if we have all three of these specific heats, as well as the enthalpies of fusion and vaporization, we can calculate the amount of energy required to convert, say, 135 grams of ice at -15 degrees into steam at 120 degrees. To do this, we will need to break up the calculation into five parts. First, the energy needed to heat the ice up to the melting point, and then the energy needed to melt the ice. Next, the energy needed to heat the water up to
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the boiling point, and then the energy needed to boil it. Lastly, the energy needed to heat up the steam to the final temperature. Let’s find these five values one at a time. For the first one, we’ve got the mass, times the specific heat of ice, times the change in temperature, which is the fifteen degrees to get to the melting point. Next, to melt the ice, we need the moles of ice, since our enthalpy of fusion is in kilojoules per mole, and this is what we will get. For the
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next step, the specific heat of liquid water, times the mass and the change in temperature from melting point to boiling point, which is 100. Then we use the enthalpy of vaporization, which again will require the moles of water. And for the last step, we use the mass, times the specific heat of steam, times the final delta T, which will be 20, to get us up to 120. We just add all of these up to get the total, making sure to express everything in the same unit, kilojoules,
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and there is our final value for the amount of energy required to make this process occur.
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So we can see that using this kind of analysis, we can determine the energy change associated with any sample going from one temperature to another, including any phase changes associated with the process. We simply need to know the specific heat of that substance in the particular phase it will be in, as well as the enthalpy of fusion or vaporization for that substance, if the relevant phase change occurs. With physical processes covered, let’s now move
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on to chemical reactions. There are different ways to calculate the enthalpy change associated with a chemical reaction, so let’s learn about them. First we can look at bond enthalpies. Breaking covalent bonds requires energy, and the formation of covalent bonds releases energy, and the amount of energy will depend on the type of covalent bond. We can look at tabulated data regarding bond enthalpies and use this information to calculate the change in enthalpy for a reaction.
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For example, the electrolysis of water to form hydrogen and oxygen gas. The bond enthalpy for an O-H bond is 460 kilojoules per mole. For H-H it’s 436.4, and for an O=O double bond it’s 498.7. We can therefore take the energy required to break all the O-H bonds, and subtract from that the energy released when the H-H and O=O bonds form, and if the result is positive, the reaction must be endothermic, and if negative, it must be exothermic. So let’s take the value for
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O-H and multiply by 4, and then subtract from that the H-H value times 2 plus the O=O value. We will end up with 1840 minus 1371.5, which means this reaction has a change in enthalpy of positive 468.5 kilojoules per mole, which means it is an endothermic process. Now let’s learn a second method for calculating enthalpy changes. Enthalpy of formation, also called heat of formation, is a value that tells us the energy associated with the formation
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of one mole of a compound from its respective elements in their most stable states and under standard conditions. It will be symbolized as delta H with an f subscript for formation, and the degree symbol superscript signifies standard conditions. Enthalpy of formation data is very useful information, because it is kind of like knowing the amount of chemical energy associated with a compound. For example, carbon dioxide is made of carbon and oxygen. In order to form one
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mole of carbon dioxide, we would need one mole of carbon atoms and one mole of oxygen molecules. The carbon is assumed to be graphite rather than coal or diamond, and the oxygen is O2 rather than O3, as these are the most stable states of these elements, and we assume standard conditions, which means room temperature and atmospheric pressure. If this reaction were to occur with the quantities and conditions specified, the change in enthalpy associated with
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it would be negative 393.5 kJ. This is exothermic, meaning the formation of carbon dioxide from its elements is energetically favorable. Looking at another example, in order to form one mole of nitrogen dioxide, we would need a half a mole of nitrogen molecules, and one mole of oxygen molecules. We need to use the one half for nitrogen because we need the correct stoichiometric amounts to specifically make only one mole of the compound in question,
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in this case nitrogen dioxide, even if that results in fractional coefficients. The change in enthalpy associated with this reaction would be 33 kJ, so formation of NO2 from its elements is not energetically favorable. If we want to write a heat of formation reaction equation for a compound, like ethanol, we just have to break it down into each element, in this case carbon, hydrogen, and oxygen, and use the correct coefficients to result in one mole of the
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compound. For ethanol, that would be two moles of carbon graphite to give the two carbon atoms, three moles of hydrogen to give the six hydrogen atoms, and a half a mole of oxygen to give the one oxygen atom. We have tabulated enthalpies of formation for a wide variety of substances, and we can do simple calculations using this tabulated data to discern the change in enthalpy for a particular reaction. All we have to do is use this equation, whereby
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change in enthalpy can be calculated by finding the sum of the enthalpies of formation for all of the products, with respect to their stoichiometric coefficients, and subtracting from that the sum of the enthalpies of formation for all the reactants, with respect to their stoichiometric coefficients. Don’t be scared of this upper case sigma, it just means to find the sum, or to add everything up. So
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we just get this data from a table of given information, and plug it in where it goes, making sure to multiply it by a stoichiometric coefficient if necessary. Take for example the combustion of methane. We can identify methane gas, oxygen gas, carbon dioxide gas, and water vapor. The enthalpies of formation for these substances are as follows. Notice that oxygen does not have a heat of formation, as it is an element in standard state, meaning there
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is nothing to form, so that one will be zero. Now we take the products first, so we plug in our value for CO2, as well as our value for water vapor, making sure to multiply that one by two, since it has a coefficient of two in the balanced equation. Let’s place all of that in brackets. Then let’s list the reactants. We have methane, and then the zero for oxygen, times two, which will still be zero, but let’s write it out just to be thorough. Let’s place all of that in brackets
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as well, and subtract this whole second term from this first term. This part is very important, because simple arithmetic errors and mistakes when inputting these numbers into a calculator is a very common source of error on this type of question. You can either simplify each term and subtract, or type it all in your calculator using parentheses and brackets correctly. Either way, we do the math, and we get negative 887 kilojoules per mole as an answer, a strongly exothermic
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value to match this combustion reaction. Let’s learn just one more way to do this kind of calculation regarding changes in enthalpy for chemical reactions, which is called Hess’s law. A summary of Hess’s law tells us that the change in enthalpy for a chemical reaction is a state function, meaning that it is independent of the pathway. This means that if we can rewrite a chemical equation that has an unknown enthalpy change as the sum of several other chemical
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reactions with known enthalpy changes, then the sum of the enthalpy changes of those reactions will be the enthalpy change for the reaction in question. This concept is best demonstrated by example, so let’s see what this means. Let’s say we want to know the enthalpy of formation for FeCl3, which means the energy associated with the generation of one mole of this compound from its constituent elements, but we only have the following enthalpy data,
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which tells us the enthalpy of forming FeCl2 which then forms FeCl3. We can make use of these equations by adding them together, because if we combine them, FeCl2 will cancel out, one mole plus a half a mole of Cl2 will give us three halves moles of Cl2, and we end up with the reaction in question. Because we added these reactions together, we simply add up their delta H’s, and we get -399.5 kJ, which must therefore be the enthalpy of formation for FeCl3.
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Now it won’t always be as easy as just adding two equations together, sometimes we need to manipulate them first. So let’s recall some rules. First, if we have a chemical equation, and we multiply or divide the stoichiometric coefficients by some number, we must multiply or divide the corresponding delta H by that same number. Look at this case, involving the formation of water. We can double the coefficients but we must therefore double the delta H. We also could
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have multiplied by three, cut in half, whatever we like, as long as we change the delta H in the same way. And second, if we reverse the direction of an equation by swapping reactants and products, such as we do here by making water the reactant and the gases the products, we must reverse the sign of the corresponding delta H, in this case from negative to positive, to show that energy released
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in one direction is energy absorbed in the other. Let’s do one full example. Say we want to find the delta H for this reaction involving the reaction of water and elemental carbon to produce carbon monoxide and hydrogen gas, and we are given the following thermochemical data. Let’s see how we need to manipulate the given information. To start, we see that this first equation has both carbon and carbon monoxide, but it has two of each. The equation in question has only one of
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each. So let’s cut this one in half, so that the coefficients are one, one half, and one, and we can cut the delta H in half as well. Now the second equation has hydrogen gas, but it has two of them, and it has it on the wrong side. We need it as a product. So let’s cut it in half and reverse it, giving us this, and we must do the same to the delta H, which will now be half the original value and positive. And the third gives us liquid water on the left,
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so that one is actually fine just the way it is. We can add these all up and notice that oxygen gas and water vapor are present on both sides and will cancel out, leaving us with exactly what we are looking for. And adding up the respective delta H values, we get positive 175 kilojoules. And with that we know of three ways to calculate the change in enthalpy for a chemical reaction. We can use bond enthalpies, enthalpy of formation data, or Hess’s law. That concludes
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our Unit 6 review, so I’ll see you over in Unit 7, where we will discuss equilibria.
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