In this video, we're going to focus on thermochemistry and some of the equations that you need for this chapter. So, let's begin. Let's start with internal energy.
The change in the internal energy of a system is equal to Q plus W. Now, Q represents the amount of heat energy that enters or leaves the system. So, here's a question for you.
Let's say if inside the box represents the system, and let's say the outside, anything outside the box is the surroundings. Now, let's say the temperature of the system is 100 degrees Celsius, but the surroundings is only 50 degrees Celsius. Well, actually, let's say 25. Now, will Q be positive or negative? Heat flows from hot to cold, so heat is going to flow out of the system into the surroundings. So Q relative to the system is going to be negative.
Whenever heat is released from the system, it's considered to be an exothermic process. Now for the surroundings, the surroundings absorb heat energy, so Q is positive for the surroundings. So it's endothermic for the surroundings.
Anytime heat is absorbed, Q is positive. Typically, Q is measured in joules. Now you need to know that 1 kilojoule is equal to 1000 joules and a calorie, a lowercase calorie, that's 4.184 joules. A capital calorie is 1000 lowercase calories.
So those are some other conversion factors you need to know. W represents work and work is equal to negative P delta V. When work is done on a system, W is positive. Whenever work is done on a system.
Now, if work is done by the system, W is negative. Now, for this equation, W equals negative P delta V. When the change in volume is positive, whenever gas expands, the work is negative due to this negative sign. And so whenever gas expands, it can do work. It can exert a force to move a piston or something.
Now, to compress the gas, you need to apply a force. So, whenever gas is compressed, the change in volume is negative, so the work overall is positive. So as you apply a force to compress the gas, you're performing work on the system.
So the internal energy of the system goes up whenever you do work on it. But whenever the system does work, or whenever work is done by the system, the internal energy of the system goes down. It loses potential energy because it's using up energy to do work.
So let's see if we can apply this information in a practice problem. But before we do, another thing you need to know is that 101.3 joules is equal to 1 liter times 1 atm. So now, let's say if 300 joules of heat energy was absorbed by the system, and also, let's say the system, which is a gas, let's say the gas expands from 2 liters to 3 liters at a pressure of 5 atm.
Calculate the change in the internal energy of the system. So the first thing we need to do is use this equation, W is equal to negative P delta V. The pressure is 5 atm, the change in volume, final minus initial, it's 3 minus 2. So delta, well, P delta V is going to be negative 5, and the unit is going to be liters times atm. Now, in order to get delta E, We need to convert liters times atm into joules. And as we mentioned earlier, 1 liter times 1 atm is equal to 101.3 joules. So the units, the liters, cancel, and atm cancel as well.
So what's 5 times 101.3? 5 times 100 is 500. 5 times 1 is 5. And 5 times 0.3. If 5 times 3 is 15, 5 times 0.3 is 1.5. So this is going to be negative 506.5.
If you add 500 plus 5 plus 1.5. So that's how many joules of work that has been done by the system. So that's why W is negative. Because the gas expanded and the gas performed work. So the internal energy of the system went down.
So now we can calculate everything. Let's plug in what we have into the delta equation. So since heat energy is absorbed, it's endothermic, Q is positive 300, and W is negative 506.5. So 300 minus 506.5, that's going to be equal to negative 206.5 joules.
So that's how much energy the system lost overall. So now there's some other equations that you need to know. Q can be calculated using the equation mc delta t. m is the mass, c is the specific heat capacity, and in the case of water, it's 4.184 joules per gram per Celsius. So what this means is that if you have a 1 gram sample of water, it takes 4.18 joules to heat up that 1 gram sample of water by 1 degree Celsius.
Delta T is the change in temperature, which could be in Kelvin or Celsius. So let's say if you want to calculate the amount of energy that's required to heat up, let's say 50 grams of water from 25 degrees Celsius to 75 degrees Celsius. How much energy is required? So the mass is 50. The heat capacity of water is 4.184.
And a change in temperature, final minus initial, 75 minus 25, that's 50. So if you have a calculator, 50 times 50 is 2500, and if you multiply that by 4.184, you should get 10,460 joules. So that's how much heat energy is required to heat up 50 grams of water from 25 degrees Celsius to 75 degrees Celsius. And if you divide this by 1000...
You could convert your answer into kilojoules if you want to. So whenever you want to calculate the amount of heat energy absorbed or released whenever there's a temperature change, use the equation Q equals MCAT. Now, whenever you want to calculate the amount of heat energy that's absorbed or released whenever there's a phase change, you can use the equation Q equals M times delta H.
This could be enthalpy of fusion or vaporization. Or you can use Q, which is equal to n times delta H. It really depends on what the units of delta H is. And personally, I think it's easier if you just convert it.
So let me give you an example. Let's say if you have 54 grams of ice at 0 degrees Celsius. And you want to find out how much heat energy is required to melt that ice into liquid water at the same temperature.
Whenever you melt ice, while ice is melting into liquid water, the temperature... It remains constant. So this is a phase change. We want to go from a solid into a liquid. So it's easier if you simply convert it.
Now you need to know the heat of fusion for water. The heat of fusion for water is about 6 kilojoules per mole. So what that means is that if you have 1 mole of water, you require 6 kilojoules to melt ice. It requires 6 kJ of heat energy to melt 1 mol of ice into liquid water.
So what we need to do is convert grams to moles and then moles to kJ. So let's begin. So we have 54 grams of H2O in the solid form.
And if you look at the periodic table, oxygen has an atomic mass of 16 and hydrogen has a mass of 1 times 2. So the molar mass for water is 18 grams per 1 mole. And so the units, grams, cancel. And now we can convert moles to kilojoules. We know that 6 kilojoules of heat is required per 1 mole of ice.
So let's see what this is equal to. 54 divided by 18 is 3. And 3 times 6, well that's going to be 18. So there's 18. kilojoules of heat energy that's required to melt 54 grams of ice into liquid water. Now sometimes you might get what is known as a thermochemical equation.
So consider the combustion reaction of propane with oxygen gas. It's going to produce CO2 and water. Go ahead and balance this reaction.
And let's say this reaction releases 1200 kilojoules of heat energy. Notice that there's no delta H symbol. If you see the kilojoules of heat energy on the right side, it's exothermic. If you see it on the left side, it's endothermic. Now, whenever you want to balance a combustion reaction, balance the carbon atoms first.
So let's put a 3 in front of CO2. Now, there's 8 hydrogen atoms on the left, so we need a 4 in front of water. And so we have 6 oxygens from the 3 CO2 molecules, 4 from H2O, 6 plus 4 is 10. If you divide it by this number 2, you're going to get 5. So we need to put a 5 in front of O2.
So now this reaction is balanced. So how many kilojoules of heat energy will be released if 64 grams of oxygen reacts? How would you answer that question? How would you find the answer?
So just like before, we're going to convert grams into moles, and then moles into kilojoules. So let's start with 64 grams of oxygen. Let's convert it to moles.
Anytime you want to go from grams to moles, you need to use the molar mass found in the periodic table. The atomic mass for oxygen is 16. So for O2, 16 times 2 is 32. So there's 32 grams of O2 per 1 mole of O2. And so as we can see, the units grams of O2 cancel. Now you need to understand what this number means. So if 1 mole of propane reacts, 1200 kJ of heat energy will be released.
Now if 5 moles of O2 reacts, 1200 kJ of heat energy will be released. If 3 moles of CO2 is produced, well that's going to correspond to 1200 kJ. So what we need is the... kilojoule per mole ratio.
So since we're dealing with oxygen, the 5 is important to us. So for every 5 moles of O2 that reacts, 1200 kilojoules of heat energy will be released. And so we can eliminate moles of O2 so now we can do the math so let's see if we can do this without a calculator so 64 divided by 32 is 2 So we have 2 times 1200 and a 5 on the bottom so 1200 is basically 12 times 100 and 100 is 20 times 5 so we can cancel the 5s So, what's 20 times 2? 20 times 2 is 40. And now what's 40 times 12?
Well, 4 times 12 is 120, so 40, I mean, I take that back, 4 times 12 is 48. 8 so 40 times 12 is 480 you can see it this way 40 is 4 times 10 so if you multiply 10 and 12 you get 120 120 times 4 is 480 So let's go back to this example. Now let's say if 3600 kilojoules of heat energy was released in this reaction, how many grams of CO2 was produced? So we want to go backwards, we want to convert kilojoules moles and moles to CO2. So let's start with 3600 kilojoules.
Now the kilojoule per mole ratio is 1200 kilojoules for every 3 moles of CO2. So we're going to use that to go from kilojoules to moles. joules to moles, so for every 1200 kilojoules of heat energy that was released, 3 moles of carbon dioxide was produced. So the units kilojoules cancel. And now the last thing we need to do is convert moles to grams.
So the molar mass for CO2 is going to be 12 for carbon plus 32 for the two oxygen atoms, so that's going to be 44 grams of CO2 per 1 mole. And so these units cancel. So 12 goes into 36 three times. You can cancel the zeros if you want. So we have 3 times 3 times 44. So 3 times 3 is 9. So what's 9 times 44?
9 times 44 is basically 10 minus 1 times 44. So 44 times 10, that's 440. 44 times 1 is negative 44. so 440 minus 44 how much is that? so 440 minus 40 is 400 and if you subtract another 4 from that you should get 396 and so there's going to be 396 grams of CO2 that's produced in this reaction Consider this reaction. Methane reacts with oxygen gas to produce CO2 and water.
Now let's balance it. So we need a 2 in front of H2. We already have one carbon atom on both sides. Now we have four oxygen atoms on the right side, so we need a 2 in front of O2. So now, let's say if you're given the heats of formation for CO2, for H2O, and CH4.
Let's say for CH4, it's negative 785. And this is delta H with an F on the bottom and a circle on top. That's the enthalpy of formation. The enthalpy of formation is the enthalpy of the heat that's absorbed or released, or the enthalpy of the reaction whenever a compound is produced from its elements in its natural standard state, and you have to produce only one mole of that compound.
That's the definition for the enthalpy of formation, if you wanted to know, but we don't have to worry about that for this example. But let's say for CO2 it's about negative 393, and for water it's negative 286. How can you use this information to calculate or estimate the enthalpy of this entire reaction? If you want to find the enthalpy of the reaction using the heat of formation, it's the sum of the enthalpies for the products. minus the sum of the heat of formation for the reactants. That's how you can calculate it.
So on the product side, which is the stuff on the right side, we have CO2 plus 2H2O. And for the reactants, that's the stuff on the left side. We have CH4 plus 2O2.
Now the value for CO2, that's negative 393. And for H2O, it's going to be 2 times the value for H2O, which is negative 286. And then minus that for CH4, which is negative 785. Now O2 is a pure element in its natural state, so it's 0 for any such elements. So now we just got to do the math. So we have negative 393. 2 times 286, that's, I'm going to use a calculator for that, that's a negative 576, and negative negative 785, that's plus 785. So if you add these numbers up, you should get negative 184. So that's how you can estimate the enthalpy of the reaction using the heat deformation.
And typically, these values are going to be in kilojoules per mole. So this is going to be negative 184 kilojoules per mole. Now let's say if we have this equation, 2A plus 2B turns into D plus E. How can you estimate the enthalpy of this reaction if you're given the enthalpies of two other reactions?
Let's say A plus B. turns into C and the enthalpy for that reaction is 200 and let's say we have D plus E turns into 2C and delta H is negative 400 for that reaction So, how can we use those two reactions to find the enthalpy of this reaction, the one on top? So, we've got to use something called Hess's Law, which means we've got to modify equations.
wanted to such that when we add them can get equation 3 and if that happens we can add the as a piece of number one and two to estimate the SP the action for number three So, let's focus on A. Notice that in the net reaction, we need 2A. So, that tells us that we need to double equation 1. So, let's multiply equation 1 by 2. So, it's going to be...
2a plus 2b and that turns into 2c so we got to multiply the entropy of the reaction as well so it's going to be 400 now. Now notice that D and E are on the right side of the equation that means we got to reverse equation 2. So, it's going to be 2C, which turns into D plus E. And so, whenever you reverse it, the reaction has to change signs. So, it's going to go from negative 400 to positive 400. So, now let's add these two equations. So, notice that we have 2C on the left and 2C on the right.
So, these two are going to cancel. So, everything else is going to cancel. drop down.
On the left side we have 2a plus 2b, so that's going to fall down. And on the right side we have d plus e. So notice that when we add these two equations we get equation 3, which means that we can now add the enthalpies of these two equations. So the enthalpy of reaction for equation 3 is going to be 800. And so that's how you can get it using Hess's law. So that is it for this video, thanks for watching and have a great day.