Hi there! I’m Jeremy Krug, and welcome to my ten minute review of AP Chemistry Unit 6 – Thermochemistry. If you find this video helpful, I would really appreciate it if you hit that LIKE button and SUBSCRIBE to my channel, so you don’t miss a thing! Thank you SO MUCH for sharing this video with your classmates and getting the word out about my videos. If you’re looking for practice free-response and multiple choice, and a comprehensive practice and review program for EVERY AP unit, then check out Ultimate Review Packet dot com, where I’ve prepared all the resources you need to get that FIVE on your AP Chemistry exam. Now, let’s jump right into Thermochemistry… A process can be exothermic or endothermic. Endothermic processes are when the molecules participating in a process – we call that the system – gain energy from their surroundings. Since energy is conserved, that means when we monitor the temperature, we’ll find that the surroundings get colder in an endothermic process. On the other hand, exothermic processes have molecules that lose energy to the surroundings. So since the molecules in the system lose energy, we find that the temperature of the surroundings gets warmer in an exothermic process. On the AP exam you’re expected to know why the formation of a solution might be exothermic or endothermic. Remember, forming a bond releases energy, and breaking a bond absorbs energy. When some ionic compounds dissolve, they break a very high energy bond, and form a somewhat weaker attraction to water molecules. That would be a net endothermic process. On the other hand, other compounds might break a relatively weak bond and then form a stronger attraction to water molecules. That would be a net exothermic process. Energy diagrams help us see how the energy of a reaction changes, from the reactants, when it attains its activation energy and gets to the transition state or activated complex, then finally to the products. Notice that when a system has a net loss of potential energy to the surroundings, we have an exothermic reaction; that’s why if you put your hand near an exothermic reaction, it feels warm to the touch. On the other hand, when the system has a net gain of potential energy, we have an endothermic reaction. The system absorbs heat from the surroundings, so if you put your hand next to that reaction, it will feel cold to the touch. Realize that average kinetic energy is just a synonym for temperature. The higher the temperature an material has, the more average kinetic energy its molecules have. When a warmer material makes contact with a cooler material, the collisions of molecules are able to transfer that heat, or that thermal energy, and the heat is always transferred from the warmer object to the cooler object. The warmer object will get cooler, and the cooler object will get warmer, until both of their temperatures are equal. We call that thermal equilibrium. We can calculate the amount of heat transferred between two systems with the equation Q equals M C delta T. Q represents the amount of heat transferred in Joules, M is the mass in grams, C is the specific heat capacity of a material, and delta T is the change in temperature. Specific heat capacity, usually given in Joules per gram degree Celsius, is a measure of how well a material resists temperature change. So some materials have a very low value for C, and adding just a small amount of heat causes a huge jump in the temperature. Some materials have a high value for C, and adding heat causes a relatively small jump in temperature. A couple of things to remember about this equation: Since energy is conserved, the heat gained by one system is equal to the heat lost by another. So if we have two systems in contact, we might use this equation twice and say that the heat gained by one system equals the heat lost by another, which can also be written like this. Also, we usually compute specific heat capacity in Joules per gram degree Celsius, but we can also express it in Joules per mole degree Celsius by simply converting grams of the material to moles, like this. One last thing, we use this equation for measuring temperature changes during a heating or cooling process. If we have a phase change or an actual chemical reaction, we’d use another way to calculate heat change. This heating curve represents how temperature changes as we add heat to a substance. Notice how the temperature stays constant during the process of melting, how it also stays constant as it boils. When we add heat to a solid or a liquid or a gas, its temperature goes up. But during any phase change, the temperature stays constant. This tells us that during melting and boiling, energy is being absorbed by the substance, so melting and boiling are endothermic processes. Now the opposite is true, too. During condensation and freezing, the molecules are shedding energy, so condensation and freezing are exothermic processes. Freezing and melting are two processes that undo each other, as are boiling and condensation. So if the heat of vaporization of water is +40.7 kilojoules per mole, that means that the heat of condensation of water would have to be negative 40.7 kilojoules per mole. We often write a chemical reaction accompanied by its enthalpy of reaction, which we abbreviate as Delta H. So this reaction is saying that if we take two moles of sodium and react it with one mole of chlorine gas, we will make two moles of sodium chloride, and we’ll release 822 kilojoules of heat. Now, if we double the recipe and react 4 moles of sodium with 2 moles of chlorine, we’ll get twice the heat, 1644 kilojoules. So we can use these values in reaction stoichiometry. Perhaps we’re forming 10.0 grams of sodium chloride and want to know how much energy is released. We just follow our typical stoichiometry process. Step one is convert to moles, then in our second step we write the mole ratio with kilojoules on top, and moles of sodium chloride on bottom. We use negative 822 kilojoules equivalent to 2 moles of sodium chloride. Then we can solve and get the answer. One way to determine the change in enthalpy of a reaction is to use bond enthalpies. Some chemical bonds require more energy to break than others. So with a list of bond enthalpies, we can determine the Delta H of a reaction by taking the total enthalpies of the bonds broken minus the total enthalpies of the bonds formed. So in this reaction, we take the four C-H bonds and get 1652 and add that to two O-O double bonds and get 990. So the total bonds broken would be 2642. Then we have two C-O double bonds, for 1598, and 4 O-H single bonds, for 1868. So the total energy of bonds formed is 3466 kilojoules. So subtract bonds broken, minus bonds formed, and you get the total change of enthalpy as negative 824 kilojoules. Sometimes, there is a variation of this problem where you’re given the total Delta H, and all of the bond enthalpies except for one, and you have to use algebra to solve for the missing bond enthalpy. Another way to solve for Delta H is to use the standard enthalpies of formation for the substances in a reaction. The change in enthalpy for any reaction is equal to the sum of the enthalpies of formation of all the products, minus the sum of the enthalpies of formation of all the reactants. So in this reaction, we take the enthalpies of formation for each of the substances and multiply them by the moles of each substance present. Then we take the sum of the products and subtract the sum of the reactants, and we get the answer. They will give you these constants on the exam. You don’t have to memorize them; however, you do need to be aware that the enthalpy of formation for any element in its most natural state is 0 kilojoules per mole, just like you see for oxygen here. The last way to solve for Delta H is to use Hess’s Law. The idea here is that if two reactions add up to give you a new reaction, those two Delta H’s will add up to give the Delta H of the new reaction. Sometimes, you need to flip around reactions and multiply reactions to make them add up, like in this example. In this example, we have to flip around reaction number 2, and that changes the sign of its Delta H. Then, we have to double all its coefficients. When we double the coefficients, that also doubles the Delta H value. Finally, we can add the two Delta H values and get the answer for our final reaction. That’s Unit 6! Don’t forget that my full daily videos for Unit 6 are right here on my channel – just click on Playlists, and look for the AP Chemistry Full Course. I’m Jeremy Krug, thanks for watching, and don’t forget to join me for my 10 minute review of Unit 7, over Chemical Equilibrium. I’ll see you soon!