Thermo

One part one, thermochemistry. Well, what is thermochemistry? Thermochemistry is a portion of thermodynamics. And thermodynamics is the science of the relationship between heat and other forms of energy. The areas of thermodynamics that exist are enthalpy, which is delta H, entropy, which is delta S, and Gibbs, free energy, which is delta G. Thermochemistry, which is heat energy is associated with enthalpy, or delta H. And thermochemistry, it's the study of heat absorbed and released during a chemical reaction. We are talking about energy. Well, what is energy? Energy is the potential or capacity to do work. There are different forms of energy. There's heat. Is the potential or capacity to do work? There are different forms of energy. There's heat energy, light energy, electrical energy and chemical energy. An example of chemical energy is a battery. A battery contains a chemical substance, when reacted, will produce an electrical current. Well, if energy has the potential or capacity to do work, then what is work? Work is a force acting through a distance. There are types of energy that are associated with work, which are mechanical, sound and electrical energies. With the various forms of energy, they can be converted from one form to another. There are four main types of energy. There is kinetic energy, which is our energy of motion, there is thermal energy, which is our heat energy that is associated with the temperature of a substance. Thermal energy is a kind of kinetic energy because of the association with the temperature and or the molecular motion we have potential energy, which is our stored energy that is associated with the position of an object. For example, the stored energy associated with the ball at the top of a hill, it has a high amount of stored energy that will be released as the ball rolls down the hill. A compressed spring also has stored energy that will be released to push an object or move an object through a distance. Chemical energy, the energy associated with electrons relative to the nuclei, is the stored energy in a chemical bond. This stored energy, or chemical energy, relates to the type of chemical bond and the elements involved in the chemical bonds. Again, these types of energy can be converted between the different forms. We will now look at the conservation of energy. The conservation of energy states that energy may be converted from one form to another, but the total quantity of energy must remain constant. It's neither created or destroyed, it's converted from one form to another and remains constant. Now this energy being converted from one form to another, it can also be exchanged between what we call the system and the surroundings. The system is what is undergoing the change the surroundings is everything around the system. So the system is what is undergoing the change the surroundings is everything around the system. So with every process that occurs, the system and the surroundings need to be identified. So if we are looking at two aqueous solutions that are mixed and a solid is formed, what is the system? What is the surroundings? The system for a chemical process is the reactants and the products. In this scenario, the reactants and the products are the reactant solutes in solution and the solid product form. Those make up the system. The surroundings is what is not the reactants in the products. For our aqueous solutions that are mixed, the solutes are dissolved in the solvent. So the solvent is not a reactant, therefore it is part of the surroundings. The container that the aqueous solutions are in are the surroundings. The bench that you are working on. The air above the solutions, everything else is your surroundings. This will bring us to the first law of thermodynamics, the first law of thermodynamics states the total energy of the universe is a constant. That means the system plus the surroundings equals our universe. For every system, there's a certain amount of energy called our internal energy, or E, that is the sum of the kinetic and potential energies of the particles that make up a system. The internal energy is a state function. Well, what is a state function? The state function is a value that only depends on the state of the system, not how you got there. So basically, what that means is it doesn't matter how you got from point A to point B. What matters is the difference between the two points. State functions reference the state of the system. The state of the system must be identified in terms of the temperature, the pressure, concentration and the physical state. A change in a state function is the difference between the two points, the initial and final values. That is what we call our change in internal energy. And the change in internal energy is the final internal energy minus the initial internal energy. That equals the change in internal energy. So the change in internal energy, or delta E, is equal to the final energy minus the initial internal energy for a chemical reaction, the change in internal energy is the internal energy of the products minus the internal energy of the reactants. Products represent our final scenario, and the reactants are our initial now, the value of the change in internal energy depends on whether the internal energy for the products is greater than or less than the internal energy of the reactants. If the internal energy for the products is greater than the internal energy for the reactants, then delta E will be a positive value. That means the energy is absorbed by the system. You're adding in energy to the system, and that energy came from the surroundings. This is the energy difference that is necessary for the chemical reaction. The other scenario is the energy of the products is less than the energy of the reactants. When the energy of the products is less than the energy of the reactants, then delta E will equal a negative value. If delta E is a negative value, that means energy is released by the system and absorbed by the surroundings. This excess energy that is left over from the chemical reaction is released by the system and absorbed by the surroundings. So the key concepts for our gaining and losing energy is the total energy lost by the system must equal the total energy gained by the surroundings. Or the total energy gained by the system must equal the total energy lost by the surroundings. These are opposite relationships, but the take home message out of all of this is the amount of energy lost has to equal the amount of energy gained. So it doesn't matter if the system is losing energy or gaining energy, or the surroundings is losing energy or gaining energy. The key concept here is the amount of energy lost has to equal the amount of energy gained. So we have our system and our surroundings. So if the change in internal energy for the system is a negative value, which means it is losing energy, then the change in internal energy for the surroundings will be a positive value, which means it is gaining energy with the change the total energy lost must equal the total energy gained. So if delta E for the system is a negative 10 kilojoules, then delta E for the surroundings must be a positive 10 kilojoules. The key relationship here is that we are seeing opposite signs, but equal magnitude. So a negative 10 for the system, a positive 10 for the surroundings, the magnitude, they're both 10, just opposite signs. So the magnitude is the same. The signs are opposite. We can have the opposite scenario, where we have the change in energy for the system being a positive value, and the change in energy for the surroundings being a negative value. Again, positive or negative, it doesn't matter. What matters is the amount of energy lost has to equal the amount of energy gained. So they will be of equal magnitude, but opposite signs. Putting these concepts together, that we have equal magnitudes but opposite signs, then we can deduce that the change in the energy for the system will equal a negative change in the internal energy for the surroundings. If we put them both on the same side of the equation, then the change in internal energy for the system, plus the change in internal energy for the surroundings, will equal zero. We will now look at our units. There are different units for energy, the first being a joule. A joule is a very small unit. It is equivalent to one kilogram, meter squared per second squared. We have calorie with a lowercase c calorie, which is equal to 4.184 joules, exactly so. A calorie is a larger unit than a joule. It is a commonly used unit by chemists, but it is not considered to be an SI unit. The calorie originated by being defined as the amount of energy needed to raise the temperature of one gram of water one degree Celsius. We also have a food calorie. A food calorie is a much larger unit. It is a capital C calorie, which is different than our little c calorie, and that is equivalent to one kilocalorie, or 1000 little c calories. Our final unit is a kilowatt hour, which is 3.6 times 10 to the sixth joules. So it is a very large unit. And with any of our units, you should be able to convert from one unit of measure to another unit of measure, given the relationship between the units, there are two primary forms of energy exchange, those are heat and work. Heat is abbreviated with Q. W represents work. Heat and work are dependent on the conditions. Therefore, the change in internal energy is the sum of the heat transferred and work done by the system. So the change in internal energy for your system is equal to the heat for your system and the work for the system. Now there are relationships that exist with the positive and negative values for heat, work and internal energy. If the heat or Q is a negative value, then that means the system is losing heat. If the heat or Q is a positive value, then that means the system is gaining heat. For work, what is the relationship for positive or negative value? When work is a negative value, that means it is losing energy. That means that it is using energy to do the work. So w will equal a negative value. When work is a positive value, that means the system is absorbing energy. It is absorbing energy because the surroundings is doing work on the system. So therefore work will be a positive value when the surroundings is doing work on the system, and then the relationship that exists with the change in internal energy and a positive and negative value with our change in internal energy. When delta E is a negative value, then the system is losing energy. And when delta E is a positive value, then the system is gaining energy. We will now look at how to apply the relationship between internal energy, heat and work. When the air in a balloon is warmed, it absorbs 38.3 kilojoules of heat. The balloon expands 1.5 liters and does 0.463 kilojoules of work. What is the change in internal energy with any of these problems that are involving thermochemistry, we need to identify the system and the surroundings. Now, the system is the balloon,the surroundings is everything else. So now we will look at identifying our change in internal energy is equal to q or heat plus W, or work. The change in internal energy is our unknown. Q is positive or negative and a value so it says it was warmed and absorbed 38.3 kilojoules of heat. It if it absorbed heat, then it has to be a positive value for the work the balloon expanded. When the balloon expands, that means that the balloon is putting pressure or doing work on the surroundings. So the balloon is using energy to do work. So therefore, if it expanded, that means that it's doing work. And if it's doing work, it's using energy. And if it's using energy, then W must be a negative value. So it will be the negative 0.463 kilojoules of work. So solving for our change in internal energy, we have delta E is equal to a positive 37.8 kilojoules of energy. Now everything has a certain amount of heat, or thermal energy, and that thermal energy is measured by the difference in temperature or delta t, which is the amount of heat absorbed or released. So delta T is T final minus T initial. So if delta T is a positive value, then that means that the final temperature has to be greater than the initial temperature. For the final temperature to be greater than the initial temperature, that means the heat was absorbed, and that's associated with an increase in temperature. When delta T is a negative value, then that means that T final is less than the T initial, and that means that heat was released. And if heat is released from the system, there is a decrease in the temperature. So heat or Q is that change in thermal energy. And with any change in thermal energy, we know that energy will be flowing in or out of the system because of the difference in the temperature between the system and the surroundings. Now with heat, there is heat flow, which is a fundamental property of heat. For heat flow to happen, the two objects must be in contact. The direction of flow is always hot to cold. The hot object will get cooler or a decrease in temperature, and the cooler object will get warmer from absorbing the heat, and it will have an increase in temperature. This will continue until thermal equilibrium is reached. Thermal equilibrium is the fancy word for both the system and the surroundings, or the two objects have become the same temperature. Now we can classify a reaction as exothermic or endothermic, depending upon whether it absorbs or releases heat. An exothermic process is associated with a chemical reaction or physical change in which heat is evolved. What that means is, if it is exothermic and heat is evolved, q will be a negative value, delta t will be a negative value. And for the system, if heat is evolved or lost, then there will be a decrease in the temperature. For an endothermic process, a chemical reaction or physical change in which heat is absorbed, then q will be positive. The change in temperature will be positive, which is associated with an increase in temperature. And that heat that is lost or gained during a physical change or a chemical reaction will be proportional to the change in temperature. So with heat exchange, absorbing heat is an increase in internal energy, or releasing heat is a decrease in internal energy. And so if Q is a positive value and the system is absorbing heat, there will be an increase in the temperature for the system and a decrease in the temperature for the surroundings. When heat is released by the system to the surroundings, that is an exothermic process. An exothermic process, the system loses heat or Q is negative. With an exothermic process, the temperature will decrease for the system because it has lost heat energy, which means the surroundings has gained that heat energy, which will have a increase in the temperature for the surroundings. So understanding the relationship between a positive and negative value for Q, a positive negative value for work and a positive and negative value for change in internal energy and the connection with temperature are essential for success in this unit.

heat is proportional to the change in temperature. Because it is proportional, there is a constant of proportionality, or what we call our heat capacity, the equation that represents the relationship between heat and change in temperature is q, which is your heat equals C, which is your proportionality constant times your change in temperature. That heat capacity, or C, is the quantity of heat necessary to raise the temperature of an object or a sample of a substance one degree Celsius, the heat capacity is dependent on the identity of the substance. It is also dependent on the amount of the substance, which makes it an extensive property. Again, an extensive property is dependent on the quantity or the amount of a substance. Common units for our heat capacity are joules per degree Celsius or calorie per degree Celsius. So when you plug that into the equation, you will be left with units of joules or calories, which is a unit of energy. Qualitatively, understanding what heat capacity means is important a large proportionality constant, or heat capacity versus a small heat capacity, with a large value of c for heat capacity, that means it will take a large amount of heat to raise the temperature one degree Celsius, with a small heat capacity. That means it does not take a lot of heat, or very little heat is needed to raise the temperature one degree qualitatively, what does that mean in terms of something heating up quickly or slowly? Well, a large value of c is associated with heating up slowly, because at a constant input rate for heat energy, it will take longer to absorb that large amount of heat, and with a small value of c, which means it takes very little heat to raise the temperature one degree, then it doesn't take long to absorb that amount of energy. So with a small value of c, that means it heats up faster. So applying this idea, conceptually to a more real situation, will it take longer to heat up five kilograms of water versus five kilograms of an aluminum pan? This is something that we can connect with if we were actually in the kitchen, if you put the pan on the stove and turn on the heat, versus putting that water on the stove and turning on the heat, which one will heat up faster? Well, looking at the heat capacity values, aluminum has a heat capacity value of 0.897 joules per gram degree Celsius, and water is 4.184 joules per gram degree Celsius, water has a larger heat capacity compared to aluminum, which means that the aluminum pan will heat up faster or the water will take longer. Now, the relationship mathematically to calculate the quantity of heat is dependent on the amount of the substance. When we include the identity of the substance and the amount of the substance that is including what we call our specific heat. The specific heat is the amount of heat needed to raise the temperature of one gram of the substance, one degree Celsius. So your heat capacity is very general. It's the amount of heat needed to raise your substance, or whatever it is, one degree Celsius. Your specific heat is specific because it is dependent on the identity of this substance and the amount so therefore it is the energy needed to raise the temperature of one gram of substance one degree Celsius. The units include the quantity of grams, and the equation includes the quantity. So q is equal to m, which is generally the mass of your substance times the heat capacity for the substance, times the change in temperature mass includes the amount of the substance, and the heat capacity is specific for the identity of the substance. Different substances have different specific heat values. We can also describe the amount of heat in what we call molar heat capacity. Molar heat capacity is very similar to specific heat. It just has different units. The unit is in moles. So the molar heat capacity is the amount of heat necessary to raise the temperature of one mole of the substance, one degree Celsius at a constant pressure. So we have the units of moles, not grams, for the heat capacity. We will now look at applying specific heat and heat calculations. If an aluminum tray weighing 250 grams went from 72 degrees Fahrenheit to 350 degrees Fahrenheit, how much heat was absorbed in joules. The specific heat for aluminum is 0.897 joules per gram, degrees Celsius. So what information is given? We are given a weight of 250 grams. The temperature went from 72 degrees Fahrenheit to 350 degrees Fahrenheit. And we are asked for how much heat in joules, and we have our heat capacity of 0.897 joules per gram, degree Celsius. Well, the information we have, we have a mass which is 250.0 grams. We have our temperature. We actually have two temperatures. We have a T final, and we have a T initial, which is the final, which is the initial. It went from 72 to 350 our final is 350 degrees Fahrenheit, and the temperature of the initial is 72 degrees Fahrenheit. So now we also have our heat capacity, and the heat capacity for aluminum is 0.897 joules per gram degree Celsius. Now the equation for heat or Q is equal to the mass times the heat capacity times the change in temperature. Well, the units for our heat capacity is in Celsius, and our unit for temperature is in Fahrenheit. So we need to convert from Fahrenheit to Celsius and then solve for delta t. So 72 degrees Fahrenheit is 22 degrees Celsius, 350 degrees Fahrenheit is 177 degrees Celsius. Our change in temperature, which is final minus initial, is 155 degrees Celsius. So now, plugging in our information into the equation, Q for the aluminum tray, which is our system, will equal the mass, which is 250.0 grams times the heat capacity, which is 0.897 joules per gram, degrees Celsius, times the change in temperature, 155 and that is degrees Celsius. So our Celsius cancel, our grams cancel, we are left with joules, and our heat is equal to 34,600 34,684 joules. And based on significant figures, our temperature would actually have two significant figures. So using our two significant figures, the amount of heat in joules would be 35,000 joules. We can apply this using also the concept that the amount of heat lost by the system equals the amount of heat gained by the surrounding. What if we actually are now taking and identifying a change in temperature associated with the system and the surroundings. So let's say we took that 250 gram aluminum block, and it started at 150.3 degrees Celsius, and was placed in a beaker with 100 milliliters of water at 22.1 degrees Celsius. What is the temperature of the water and the aluminum block at thermal equilibrium. Again, thermal equilibrium is that fancy way of staying at the same temperature. The specific heat for aluminum is 0.897 joules per gram degree Celsius. And the specific heat for water is 4.184 joules per gram, degree Celsius. So the key concept to understanding how to solve this problem is that the amount of heat lost by the system has to equal the amount of heat gained by the surroundings. So q for the system has to equal a negative q for the surroundings, so the amount of heat lost by one has to equal the amount of heat gained by the other. We observe with our internal energy that the change in internal energy for the system equal the negative change in internal energy for the surroundings. It is the same idea where they will be equal magnitudes but opposite charges. So now, what is the calculation for the system for Q? What is the calculation for the surroundings for Q. Q is the mass times the specific heat, times the change in temperature. So we have the mass of our aluminum, which is our 250 grams of aluminum. We have the mass of the water. It's 100 mils of water. We're going to assume the density to be equal to one for water. The initial temperature for aluminum is 150.3 degrees Celsius. The initial temperature for water is 22.1 degrees Celsius. Our units are joules per gram degrees Celsius. So we have grams of water, grams of aluminum. We have temperature in Celsius, so we don't need to do any additional conversions. But the relationship for Q for the system will be equal to the mass of the aluminum times the heat capacity for the aluminum times the change in temperature for the aluminum, that will equal the same magnitude, but opposite charge. So it will be a negative times the mass, times the heat capacity for the water, times the change in temperature for the water, but it will be opposite charge. So plugging in our values, the mass is 250 grams, times the heat capacity for aluminum, which is 0.897 joules per gram degree Celsius, times the change in temperature. The change in temperature is T final minus T initial, which is 150.3 degrees Celsius, that will equal a negative times your mass, which is 100 grams, times the heat capacity, which is 4.184 joules per gram, degrees Celsius, times the change in temperature. And again, final minus the initial, which is 22.1 degrees Celsius. And now we need to solve for our final temperature. Now, multiplying through you will end up with 224.25 and that is joules per degree Celsius, times T final minus 33,705, and that is joules. And that will equal now we don't forget about the negative. So we have a negative 418.4 and that's joules per degree Celsius times t final. And then we have a negative times a negative. So it becomes plus 9247 joules. Now we can simplify. We end up with 642.65 joules per degree Celsius times t final. And that will equal a positive 42,952 joules divide by 642.65 on both sides, and that gives us a final temperature of 66.8 degrees Celsius. So the final temperature, temperature at thermal equilibrium is 66.8 degrees Celsius. Now a key concept to remember is the relationship between that heat capacity for the specific heat and the change in temperature. When there is a large specific heat, then the change in temperature will be smaller than for a substance that has a small specific heat, there will be a larger change in temperature. So the value of 66.8 degrees Celsius is not equally in the middle, but it is closer to the temperature of the water than it was to the temperature of the aluminum.

We will now look at pressure volume work. Pressure volume work occurs when there's a volume change against an external pressure. Remember, work is a force acting through a distance. So that means work equals force times distance. Well, we can recall that pressure is the amount of force per unit area. If we rearrange the equation, we end up with force equals pressure times area. We can take the pressure times area and substitute it in for force. When we combine the two equations, we end up with, work is equal to pressure times area times distance. Area times distance, or a third length, will give us volume. So that means that we will have work is equal to the pressure times volume, because area times distance will equal volume. Now distance is a change in height. If distance is a change in height, then volume will be a change in volume. This allows us to incorporate that change in volume into our work because gasses will produce a change in volume. So the equation for work is, work is equal to pressure times a change in volume. Now the actual relationship is, work equals a negative p delta V, or a negative pressure times the change in volume. Why is there a negative placed in the equation that has to do with the relationship between work and whether a system is doing work or the system is having work done on it. And so if the system is doing work, then w should be negative because it is using energy to do the work. When the system is doing work, the volume is expanding. Delta V is positive. So in order for work to be negative, we must place a negative sign in the equation. When work is positive, that means the surroundings is doing work on the system, then the change in volume is negative. The gas is being compressed, thus a negative times a negative will make it positive. Now, the units of pressure volume work are ATM, liter, the units for energy are joules. So we must convert pressure, volume work into the units of joules. The relationship that exists is there's 101.31 joules is equivalent to one atm, liter. This can be used as a conversion factor to convert from pressure, volume work to energy unit of joules. Let's calculate the amount of work done for a system. Calculate the work done in kilojoules for a gas. If the volume of the gas changed from 25.8 liters to 14.6 liters at a pressure of 1.204 ATM, we are given two volumes. We are given a pressure, and you were asked for the work in the unit of kilojoules. So we know that work is equal to negative p delta V. So we will need to calculate the change in volume. Make sure the units are in liters and pressure. Make sure the pressure is an ATM, and then we will convert it to joules and then to kilojoules. We have work is equal to a negative 1.204 times the change in volume. And the change in volume is 11.2 liters. So plugging in 11.2 liters for the change in volume, we end up with a calculated work value of 13.485 ATM liter. We now will convert it to joules, multiplying by the conversion factor 101.3 joules per ATM liter, that converts to 1,366.1 joules, which is then converted to kilojoules, which is 1.37 kilojoules. We will now look at constant volume calorimetry. Constant volume calorimetry can be used to determine the change in internal energy for a system. We can do this because we know that work is equal to a negative p, delta V. We know that the equation for the change in internal energy is delta E, is equal to Q plus W. Well, work is equal to a negative p, delta V, and at constant volume, delta V is equal to zero. So if you plug delta v of equals zero into the equation for work, then work equals zero. When work equals zero, then what happens to that energy? This forces all the work energy into heat energy, because we know that energy can't be destroyed. It just converted between different forms of energy. So now the work term is canceled out, and all that energy that was originally in work is now pushed into heat. And so therefore the change in heat energy will equal the change in internal energy for a system at a constant volume. Now the theory of calorimetry can be used to determine the change in energy, because with the theory of calorimetry, the amount of energy changed in the reaction system will equal the change in the energy for the calorimeter. In constant volume calorimetry, all energy is converted to heat energy, and when that happens, then the amount of heat lost by the system will equal the amount of heat gained by the calorimeter, or the amount of heat gained by the system will equal the amount of heat lost by the calorimeter. So the heat for your system will equal a negative heat for the calorimeter. With a calorimeter, a calorimeter has a heat capacity that is called your calorimeter constant, C Cal or K Cal. It is a single value for the calorimeter as a whole, because there are multiple items that make up the calorimeter, the value is unique for each calorimeter. So the K Cal has to be calculated for each device. The amount of heat absorbed or released. The amount of heat absorbed or released by the calorimeter is proportional to the change in temperature. So the equation of Q for the calorimeter is equal to the calorimeter, constant times the change in temperature. And if no heat is lost during the chemical reaction outside of the calorimeter, then we can say that the change in the heat for the calorimeter is equal to a negative change in the heat for the reaction. These are equal quantities, but opposite signs. And for constant volume calorimetry, then the heat for the calorimeter, which equals a negative heat for the reaction, will also equal a negative change in internal energy for the reaction. Now remember, the change in internal energy is usually reported on a per mole basis, so you need to divide the heat of reaction by the number of moles of the substance analyzed. Let's determine the change in internal energy for the combustion of C, 4h 802 if a 10 gram sample of C, 4h, 802, was placed in a bomb calorimeter. The measured temperature changed from 20.4 degrees Celsius to 45.2 degrees Celsius. The calorimeter used had a heat capacity of 2.640 kilojoules per degree Celsius. So now we have for our bomb calorimeter. That means that when we're using a bomb calorimeter, the change in volume is zero. With that change in volume being zero, we know that Q for the reaction is equal to the change in internal energy for the reaction. And with the bomb calorimeter, Q for the calorimeter will equal a negative q for the reaction, the equation for the calorimeter is Q. Cal is equal to the calorimeter constant times the change in temperature. The change in temperature is T, final minus T initial. The change in temperature is 45.2 degrees Celsius, minus 20.4 degrees Celsius, which gives us a change in temperature of 24.8 degrees Celsius. Q Cal is equal to a negative q reaction. We then have the calorimeter constant times the change in temperature will equal a negative q for the reaction, plugging in our values of 2.640 kilojoules per degree Celsius, times the change in temperature of a positive 24.8 degrees Celsius, that will equal a negative q for the reaction. A negative q for the reaction equals 65.472 kilojoules, and we end up with a negative 65.472 kilojoules is equal to q for the reaction. We now need to convert it to the change in internal energy. The change in internal energy is reported as q divided by the number of moles. The number of moles needs to be calculated. We have 10 grams of our C, 4h, 802 which has a molecular weight of 88.10 so 10 grams times one over 88.10 grams gives us a quantity of 0.11348 moles of c4, 8802, we will then plug in our values. We have our negative 65.472 kilojoules divided by 0.11348 moles. And that comes out to be a negative 577 kilojoules per mole for the change in the internal energy for the reaction. Now, reactions aren't always done in a closed system of a bomb calorimeter. They are also done in open to the atmosphere. When they are open to the atmosphere, they are at what we call constant pressure, because atmosphere is essentially at a constant pressure. So when you have a reaction that is open to the atmosphere, both work and heat energy still exchange. So the change in internal energy is equal to Q plus work for the total internal energy change, we cannot force the work energy into heat energy, because the change in volume in a constant pressure system is not equal to zero, but the change in heat energy is still proportional To the change in temperature. Now, enthalpy is a thermodynamic property of a substance that can be used to obtain the heat absorbed or evolved in chemical reaction at a constant pressure. It is an extensive property, which means it is dependent on the mass and the volume. It is a state function that depends only on its present state. The common units are kilojoules per mole. The enthalpy of the system is dependent on the change in enthalpy of the reactants to the products, because the enthalpy of reactants is different than the enthalpy of products, therefore delta H, or the change in enthalpy is equal to the final enthalpy minus the initial enthalpy. When we combine that into a chemical reaction, then the change in enthalpy for reaction is the enthalpy of the products minus the enthalpy of the reactants. Now, enthalpy is defined as the sum of internal energy and pressure volume work, or H equals e plus p times b, because it is a state function, then we can apply the idea that we have delta H is equal to delta e plus p, delta V, or we can rearrange the equation and have E, the change in internal energy is equal to delta H minus p, delta V. We know that delta E is equal to the heat plus the work, and we know that P delta V is equal to a negative work. Plugging in these two equations into our original equation of delta H equals and delta e plus p delta v, we end up with delta H equals Q plus W for delta E minus work for p delta V, the work will cancel out, because a positive work plus a negative work will equal zero. And so for a reaction at constant pressure, the change in enthalpy will equal Q. And so we have the change in enthalpy for reaction will equal the heat of reaction at a constant pressure. So what is the relationship between the change in internal energy and the change in enthalpy? Well, if the change in volume is very small, or we can consider it to be negligible, or delta V is equal to zero, then that P, delta v term in our equation is gone, and we can say that the change in enthalpy for reaction will equal the change in internal energy for reaction at a constant pressure. Well, what happens if the change in volume is large or large enough to be not negligible, then that pressure volume term is not excluded, and then the change in enthalpy will be slightly different than the change in internal energy for a reaction, and generally, the change in internal energy is greater than the enthalpy if work is done by the system on the surroundings. So let's look at applying the concept of the change in internal energy and the change in enthalpy. The reaction of one mole of sodium and water, yields hydrogen gas and sodium chloride and produces 367.5 kilojoules of heat and does 2.5 kilojoules of work when open to the atmosphere, calculate delta H and delta E for the reaction if the reaction in open to the atmosphere produces 367.5 kilojoules of heat, then Q is equal to the amount of heat. If it produces heat, then it needs to be negative. So q equals a negative 367.5 367.5 kilojoules of heat. Now delta H is in the units of kilojoules per mole, and since it is in the units of kilojoules per mole, then we need to normalize it and divide by the number of moles of reactant. So we would have a negative 367.5 kilojoules, and divide that by the total number of moles. And we reacted one mole of sodium. So delta H for the reaction is equal to a negative 367.5 kilojoules per mole, we now need to calculate the change in internal energy for the reaction. We know that the equation of delta H is equal to delta e plus p delta V. If we rearrange the equation, we end up with delta E is equal to delta H minus p, delta V, a negative p, delta V is equal to work. Well, now we need to figure out what the work is, so that we can include it in the equation, because we will have delta E is equal to delta H plus the work. Well, as far as work, the system did 2.5 kilojoules of work, the amount of work done was 2.5 kilojoules. Of work for the one mole of sodium. So we end up with a value for work of 2.5 kilojoules per mole. So plugging in our values, our delta H is a negative 367.5 kilojoules per mole plus the work. Now it did work, and since it did the work, it used energy. It is a negative value. So negative 2.5 kilojoules per mole, and that will equal the change in internal energy for a reaction. So the change in internal energy for the reaction will equal a negative 367.5 kilojoules per mole, plus a negative 2.5 kilojoules per mole, which equals a negative 370 kilojoules per mole. We can use constant pressure calorimetry, or our coffee cup calorimetry, to determine the change in enthalpy for aqueous reactions, because the change in volume is negligible for our constant pressure calorimetry, our system is the chemical reaction our surroundings is the water, because it's aqueous reaction and the calorimeter. So we will have, for our surroundings, Q for our solution or water plus q for the calorimeter. Applying the principle that the amount of heat lost has to equal the amount of heat gained, we can end up with two scenarios. If the calorimeter has zero absorption of heat, then Q reaction will equal a negative q for the solution. But calorimeters do absorb heat, and so therefore we have Q for the reaction is equal to a negative times the sum of the Q for the solution plus the Q for the calorimeter. Now the equations for Q for the solution is the mass of the solution times the specific heat for the solution. In most instances, it is that of water for aqueous solutions, and the change in temperature for the solution, the heat for the calorimeter, is equal to the calorimeter constant times the change in the temperature of the calorimeter. Now when we're talking about these aqueous solution reactions, if the reaction is exothermic and it loses heat, then the heat lost from the system will be the heat absorbed by the solution. So there will be an increase in the temperature of the solution. If the heat is absorbed by the system, then where does the heat come from? The heat comes from the solution. And so therefore, if the heat is absorbed from the solution, then the temperature of the solution will decrease for an endothermic process. And again, since q is happening at a constant pressure and there's no change in volume for our coffee cup calorimeter, then we can apply the principle that Q for the reaction will equal Q for our constant pressure calorimeter, which we can then calculate the change in enthalpy for reaction By converting the quantity to on a per mole basis. So delta H will be in the units of kilojoules per mole. Let's look at an example of your coffee cup calorimetry, or constant pressure calorimetry, calcium reacts with hydrochloric acid to produce calcium chloride and hydrogen gas, as shown in the following reaction. When 0.200 grams of calcium is placed in a coffee cup calorimeter with 50 milliliters of aqueous hydrochloric acid solution at a high enough concentration to ensure a complete reaction. The temperature of the solution changed from 22.5 degrees Celsius to 42.7 degrees Celsius. Assume the density of the solution to be 1.00 grams per milliliter. The heat capacity of the solution is the heat capacity of water, which is 4.184 joules per gram, degrees Celsius, and the heat capacity for the calorimeter is 15.46 joules per degree Celsius. What is the heat of reaction? Well, first again, we need to consider that Q for the system is equal to negative q for the surroundings. The surroundings is water and the calorimeter, we need to set up our equation, including Q for the water, which is mass of water times the heat capacity of water times the change in temperature for the water plus q for the calorimeter, which is the heat capacity for the calorimeter times the change in temperature for the calorimeter, the change in temperature for both the calorimeter and the solution is 20.2 degrees Celsius, because they both started at the same temperature and finished at the same temperature. Setting up our equation, we have Q for the reaction equals a negative times 50.0 grams times 4.184 joules times 20.2 degrees Celsius plus the product of 15.46 joules per degree Celsius times 20.2 degrees Celsius equals our Q for the reaction We then have Q for a reaction is equal to a negative times 4,225.84 joules plus 312.292 joules. Adding those together, we have Q for the reaction is equal to a negative 4,538.132 joules. Converting it to kilojoules, the heat for the reaction in kilojoules is 4.54 kilojoules. We now need to convert it to the enthalpy of reaction is on a per mole basis, taking and converting our amount of calcium into moles, we have 0.200 grams times the molecular weight for calcium, which is 40.08 grams equals 4.99 times 10 to the minus three moles for calcium. Setting up our relationship for delta H is equal to q divided by the number of moles, or Q divided by n, we end up with a negative 4.54 kilojoules divided by 4.99 times 10 to the minus three moles, gives us a negative 909 kilojoules per mole for the change in enthalpy for the reaction.