Endothermic and exothermic processes

- [Educator] Before we get into the terms, endo and exothermic, We need to look at some other thermodynamics terms that are used. For example, system. The system refers to the part of the universe that we are studying. For our example, we're going to consider a monatomic gas. Let's say we have some helium particles in a container. And the helium gas represents our system. The surroundings are everything else in the universe. So that would include this piston here and the cylinder in which the gas is in. And the universe consists of both the system and the surroundings. Next, let's look at the first law of thermodynamics which can be summarized by writing delta E is equal to Q plus W. Delta E is the change in the internal energy of the system. And the internal energy refers to the sum of all the kinetic and potential energies of the components of the system. Since we have a monatomic gas for our system, we only have kinetic energy. So if you could imagine adding up the kinetic energy for each particle, the sum of those kinetic energies for this example would be equal to the internal energy of the system. Q refers to the heat that's transferred from the system. So it's either transferred to or from the system. And W refers to the work done on or by the system. Let's look at the sign conventions for the first law of thermodynamics. When we think about Q; when heat flows into the system from the surroundings, we say that Q is positive. When heat flows out of the system and into the surroundings, we say that Q is negative. For work, when work is done on the system by the surroundings, the work is positive. But if work is done by the system on the surroundings, work is negative. It's very useful to think about internal energy like a bank account. So if Q is positive and work is positive, that's like money coming into your bank account. But if Q is negative or the work is negative, that's like money leaving your bank account. Let's look at an example of the first law of thermodynamics using our sample of helium that's in a cylinder with a movable piston. So let's say that 6,000 joules of energy. So 6,000 joules flows from the surroundings into the system. And that heats up our helium particles, which then expand and push the piston up. So the piston gets pushed up, which is the system doing work on the surroundings. And let's say that, that's 2000 joules of work done by the system on the surroundings. Using our assigned conventions, since heat flowed into the system from the surroundings, that's a positive value for Q. And since the work was done by the system on the surroundings, that's a negative for the work done. So we can go ahead and plug in to the first law of thermodynamics. The heat transferred is positive 6,000 joules and the work done as negative 2000 joules, therefore delta E or the change in the internal energy is equal to positive 4,000 joules. Thinking about internal energy like our bank account, we've gained 4,000 joules. So that could be like gaining $4,000. Since the system has gained 4,000 joules, that must mean the surroundings has lost 4,000 joules. But since energy is conserved, the total energy of the universe remains constant. Let's apply the first law of thermodynamics to the combustion of propane and an open container at constant pressure. For the combustion of propane, the reactants and products for the combustion reaction are considered to be the system, and everything else is the surroundings. So this combustion reaction gives off 2,044 kilojoules of energy. So that's the heat that's transferred from the system to the surroundings. The system also does two kilojoules of work on the surroundings. So by convention, we make that negative. To find the change in the internal energy of the system, we add Q plus W and get negative 2,046 kilojoules. Since this reaction was carried out under constant external pressure, we can write Q sub P here. So Q sub P is the heat that's transferred at constant pressure, and that is equal to the change in the enthalpy, which is symbolized by delta H. So the change in enthalpy is the heat that's transferred at constant pressure. So the change in enthalpy for the combustion of propane is equal to negative 2,044 kilojoules. And notice how the change in enthalpy is almost the same value as the change in the internal energy. So the work done by the system is a very small amount in this case, and that's usually the case. And since most chemical reactions are done under constant pressure, chemists care more about the change in the enthalpy than they do about the change in the internal energy. When delta H is negative, we call that an exothermic process. So the combustion of propane is an exothermic reaction. An endothermic process is where heat is transferred from the surroundings to the systems. So the system has gained heat from the surroundings. The change in enthalpy, delta H is positive for an endothermic process. An example could be melting an ice cube. If heat flows from the system to the surroundings, the system has released heat to the surroundings, and the change in enthalpy, delta H is negative. And we call this an exothermic process. And we already saw an example of that. The combustion of propane is an exothermic reaction. So a phase change could be an endo or an exothermic process. A chemical reaction could also be an endo or an exothermic process. And even the formation of a solution could be classified as an endo or an exothermic process. Let's think about making a solution. So let's say that we have a beaker and we have it full of water, and we take a solid and we dissolve the solid in the water to form a solution. If the dissolution process is an exothermic process, that means the system releases heat to the surroundings. And since the beaker is part of the surroundings, if we were to put our hand on the beaker and the beaker feels hot, we know that the disillusion of this particular solid is an exothermic process. If we do the same thing with another solid; so we dissolve this solid in water to form a solution. And we put our hand on the beaker, and this time the beaker feels cold. The reason the beaker feels cold is because energy was transferred from their surroundings to the system. And so, the surroundings lost energy. And so, therefore we can say the dissolution of this particular solid was an endothermic process.