Our system is simply our chemical reaction. Anything outside of that is considered our surroundings.
Concept: Thermodynamics vs. Thermochemistry1m
Welcome back guys. In this new video we're going to take a look at the energy changes involved in our chemical reactions.
First off, we're going to start off by saying that blank is the branch of physical science concerned with heat and it's transformations to and from other forms of energy. We're going to say that this is called thermodynamics. The part of thermodynamics that we're going to be concerned with is the branch of chemistry that deals with the heat involved in our chemical and physical changes. This branch of chemistry is called thermal chemistry.
Remember when we use the term energy or we use the term heat, they mean the same thing. Heat is just a type of energy. It's thermal energy. Energy of heat.
Thermodynamics deals with the conversion of energy from one form to another. Thermochemistry is the branch of chemistry dealing with thermal (heat) energy.
Concept: System vs. Surroundings2m
When we're talking about thermal energy or heat or energy, we're talking about two basic ideas. We're talking about the specific part of the universe that we're focused on. Since we're in a chemistry course, our system is the part of the universe that we're concerned with. When I say system, I mean our balanced chemical reaction. So any balanced chemical reaction that we're paying attention to when doing word problems, when looking at it on the board when our professor is talking about it, that represents our system, our balanced chemical reaction.
We're going to say everything having to do with that balanced chemical reaction represents our system. Everything outside of it. Everything outside of that, you, me, your professor, the board on which they're talking about, all of that stuff are our surroundings. We have our system and we have our surroundings.
They're not isolated from another. Whatever happens to one, the opposite happens to the other. If our system is gaining energy, the only reason it's gaining energy is because the surroundings are giving that energy to the system, so as a result the surroundings would be losing energy. It also works the opposite way too. Our main concerns are which way is energy traveling. Is it going from our surroundings to our system or our system to our surroundings?
Example: Whether our system (the chemical reaction) releases or absorbs heat or energy will determine if it is exothermic or endothermic.4m
Example: Classify each of the following process as either exothermic or endothermic:
a) Fusion of Ice.
b) Sublimation of CO2.
c) Vaporization of aqueous water.
d) Deposition of chlorine gas.
e) Condensation of water vapor.3m
Concept: Understanding the internal energy of the system3m
Welcome back guys. In this new video, we're going to look at the energy flow to or from a system.
We say that the first law of thermodynamics states that energy can't be created nor destroyed. All that happens is it changes forms. What this really means is anytime we have a balanced chemical reaction, remember we say that the number of atoms and the types of atoms have to be the same on both sides. If I start out with ten hydrogens as reactants, I have to end with ten hydrogens as products. That's what the first law of thermodynamics really is saying.
We're going to say we've discussed systems versus surroundings. These are the two types of concepts we're dealing with when it comes to energy changes. We're going to say that the system either gains or loses energy based on the surroundings. But now we're going to pay more attention to the surroundings.
We're going to say we're normally concerned with just the surroundings and we're going to use this equation here, delta E equals q plus w. We're going to say that delta E represents the internal energy of our system. We're going to say when it comes to the variable q, q also equals delta H. Delta H is enthalpy. When we talk about enthalpy, enthalpy just means that a reaction either absorbs or release heat or energy in order for the reaction to occur. That's all enthalpy is concerned with. We're also going to say that w equals negative pressure times the change in volume. Pressure here will be in atmospheres. This delta V means change in volume. That means final volume minus initial volume. The volume here is usually in liters.
The internal energy (ΔE or ΔU) of the system can be calculated from the heat and work of the system.
Concept: Heat vs. Work2m
When we say q, q means heat and w equals work. When it comes to these two terms, they can either be positive or negative based on certain keywords that your professors like to us. We're going to say that q, it can be positive when we say that the system gains, takes in, or absorbs heat or energy from the surroundings. We say that q can be negative when we say the system loses, evolves, gives off or releases heat or energy to the surroundings.
If the system is absorbing energy, why is it doing that? It's because the surroundings are giving it to it. If the system is losing energy, where is it going? It's going to our surroundings.
For work, we're going to say that work can also be positive or negative. We're going to say when work is done on the system by the surroundings, then work will be positive. They keyword we're going to see here is volume compresses. Basically, volume gets smaller. If the volume is getting smaller, that means that the surroundings are doing work on our system.
Work could be negative if we say that when work done by system on the surroundings. Keyword here, volume expands. The volume increases. If you see the volume increasing, that's because our system is doing that. It's doing the work. Therefore, work would be negative.
What you need to remember is what's the equation for the internal energy of the system and will my work and heat be positive or negative based on certain keywords that the professor will ask.
The signs of heat (q) and work (w) of the system can be either negative or positive depending on the key words stated.
Example: Which of the following signs on q and w represent a system that is doing work on the surroundings, as well as losing heat to the surroundings?
q = - , w = - q = +, w = +
q = -, w = + q = +, w = -
Work is one key variable to find the internal energy of the system. It’s equals to – PΔV.
Concept: Calculating work4m
Welcome back guys. In this new video, we're going to work out calculations on the amount of energy that flows to and from systems.
In this first question, it says an unknown gas expands in a container, increasing the volume from 4.3 liters to 8.2 liters at a constant pressure of 931 millimeters of mercury. For part A, it says calculate the work done in kilojoules by the gas as it expands.
Because of the way that the formula for work is set up, we'll have an answer in liters times atmospheres. What we should realize is there's a conversion factor to go from liters times atmospheres to joules. This is the conversion factor you need to know. We're going to say one liter times atmospheres equals 101.3 joules.
We're looking for work and we said that the formula for work is work equals negative pressure in atmospheres times the change in volume, which is just final volume minus initial volume. The units for volume will be in liters.
I give you 931 millimeters of mercury, so the first thing we need to do is convert that to atmospheres. We have 931 millimeters of mercury. Hopefully, you guys remember that for every one atmosphere, we have 760 milliliters of mercury. This and this cancel out. Just remember that conversion. We're going to say one atmosphere equals 760 milliliters of mercury and also one atmosphere equals 760 torrs. These are all just units of pressure. These are the conversion factors you need to remember.
When we do that we get 1.225 atmospheres. There goes my pressure and atmosphere, so I can plug that into my formula. So negative 1.225 atmospheres times the change in volume, which is final volume minus initial volume, so 8.2 liters, which is our final, minus 4.3 liters.
Now what we do there is it's going give me to negative 4.7775 liters times atmospheres. Those are our units. Remember I want the answer in kilojoules, so first we have to convert liters times atmospheres to joules. We want to get rid of liters times atmospheres, so we put that on the bottom. Remember for every one liter times atmospheres we have 101.3 joules.
We want to change joules to kilojoules. Remember we want to get rid of joules, so it goes on the bottom, kilojoules go on the top. Kilo- is a metric prefix. We said that one is associated with the metric prefix. One kilo is 10 to the 3 joules.
Multiply everything on the top. Divide by everything on the bottom. Remember 10 to the 3 is also equivalent to 1000. You could also plug that in. When we do that we get negative 0.483961 kilojoules. Here we'll just round it to three sig fig's. Here I'm not asking you for the number of sig figs directly, so we can just keep it at three sig figs. So it will be negative 0.484 kilojoules. That's the amount of work that the system is doing on the surroundings.
Remember we know that the system is doing work because we say that any time we use the word expand, the volume is increasing. And what's causing that increase? Our system.
Once we’ve calculated work we can calculate the internal energy of the system once we also calculate the heat released or absorbed.
Concept: Calculating heat and the internal energy of the system1m
Using part A calculate the internal energy of the system if the system absorbs 2.3 kilojoules of energy. Remember internal energy of the system is delta E. Delta E equals q plus w. Here we're going to say the system is absorbing. That's one of the keywords we used. It's absorbing that much energy. When we say absorbs that means q, which is energy, which is heat, is positive. We have a positive 2.3 kilojoules plus a negative 0.483961 kilojoules. Work that out and it gives us 1.82 kilojoules at the end. That would be the internal energy of our system. The energy of our system increases by that much.
Under certain conditions either q (heat) or w (work) can be equal to zero. This makes it easier to calculate the internal energy of the system.
Concept: Calculating the internal energy of the system in a vacuum2m
So guys, for Part C it says, using part being calculate the internal energy of the system and the system does work against a vacuum. So, remember internal energy is Delta e equals q, which is heat, plus W, which is work. Now, remember from the Part B they told that, they told us that our system absorbs 2.3 kilojoules of energy. So, that means that our energy here is positive because we're taking it in, we're absorbing but now they're telling us we're doing work against a vacuum. Remember, work equals negative pressure times the change in volume, just realize this, when we're talking about a vacuum where do you imagine vacuums exist? vacuums exist in space and in space you can just think of there being 0 pressure. So, as a result here, our pressure is 0, anything times 0 is equal to 0. So, work done against the vacuum means that no work is being done. So, W here equals 0, therefore the internal energy of our system has to be 2.3 kilojoules, all the energy is coming from the heat that's absorbed from Part B. Now, that we've seen this continue on to the practice question and you the information you learn so far about heat and work to help you figure out the internal energy of your system.
When work is done against a vacuum the pressure is equal to 0 atm. Since ΔE = q + w, the equation becomes only ΔE = q.
Problem: The reaction of nitrogen with hydrogen to make ammonia has an enthalpy, ?H = - 92.2 kJ: N2 (g) + 3 H2 (g) ----> 2 NH3 (g) What is in the internal energy of the system if the reaction is done at a constant pressure of 20.0 atm and the volume compresses from 10 L to 5 L?3m
At constant pressure, which of these systems do work on the surroundings?
A. 2A(g) + 3B(g) → 4C(g)
B. A(s) + B(s) → C(g)
C. 2A(g) + 2B(g) → 3C(g)
D. 2A(g) +B(g) → 4C(g)
E. More than one of the above
At constant pressure, which of these systems do work on the surroundings?
a. 2A(g) + 3B(g) → 4C(g)
b. A(s) + B(g) → 2C(g)
c. A(g) + B(g) → 3C(g)
d. A(s) + 2B(g) → C(g)
e. More than one of the above
A system absorbs 159 kJ of heat, and performs 84 kJ of work on the surroundings. What is ΔE of the system?
Calculate the work for the expansion of CO2 from 1.0 to 3.0 liters against a pressure of 1.0 atm at constant temperature.
A system that does no work but which receives heat from the surroundings has:
a) q < 0, ΔE > 0
b) q > 0, ΔE < 0
c) q = ΔE
d) q = -ΔE
e) w = ΔE
Which statement is not correct?
a. Internal energy, E, is a state function.
b. Heat and work are state functions.
c. Heat is given off to the surroundings in an exothermic reaction.
d. The enthalpy change is the heat of reaction at constant pressure.
e. Enthalpy is a state function.
A gas is allowed to expand at constant temperature from a volume of 2.0 L to 11.2 L against an external pressure of 0.750 atm. If the gas absorbs 128 J of heat from the surroundings, what are the values of q, w, and ΔE respectively?
a. 128 J, 6.9 J, 135 J
b. 128 J, -6.9 J, 121 J
c. 128 J, 697 J, 825 J
d. 128 J, -697 J, -569 J
e. -128 J, -6.9 J, -135 J
The gas in a piston is warmed and absorbs 655 J of heat. The expansion of the piston performs 344 J of work on the surroundings. Find the change in internal energy of the system.
A system releases 415 kJ of heat and does 125 kJ of work on the surroundings. What is the change in internal energy of the system?
DO NOT FORGET TO WRITE A BALANCED CHEMICAL EQUATION!!
The combustion of one mole of octane (C8H18(l)) to produce carbon dioxide and liquid water has ∆Hr = −5471 kJ · mol−1 at 298K. What is the change in internal energy for this reaction?
1. −5460 kJ · mol−1
2. −5493 kJ · mol−1
3. −5471 kJ · mol−1
4. −5449 kJ · mol−1
5. −5482 kJ · mol−1
1 L.atm =101.325 J
When 2.00 kJ of energy is transferred as heat to nitrogen in a cylinder fitted with a piston at an external pressure of 2.00 atm, the nitrogen gas expands from 2.00 to 5.00 L against this constant pressure. What is ∆U for the process?
2. −0.608 kJ
3. +1.39 kJ
4. +2.61 kJ
5. −2.61 kJ
Consider the process of 2 H(g) → H2(g) where ΔH = −436 kJ/mol
Determine if the sentence below is true or false.
The temperature of the surroundings would increase as this reaction proceeds.
More heat is derived from cooling one gram of steam at 100°C to water at 50°C than from cooling one gram of liquid water at 100°C to 50°C because
a) the steam is hotter than the water.
b) the steam occupies a greater volume than the water.
c) the density of water is greater than that of steam.
d) the heat of condensation is evolved.
Which statement is true of the internal energy of a system and its surroundings during an energy exchange with a positive Esystem?
A refrigerator uses 900 kWh of electrical energy per year. How many joules of electricity are used in two years?
A. 3.24 x 109 J
B. 6.48 x 109 J
C. 2.50 x 10-4 J
D. 5.00 x 10-4 J
When heat is absorbed by the surroundings from the system, the process is said to be ______________ , and the sign of Δq is____________? (circle one)
a.) exothermic, positive
b.) exothermic, negative
c.) endothermic, positive
d.) endothermic, negative