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
What is the change in internal energy (in kJ) when a gas does 135 J of work on the surroundings and at the same time absorbs 156 J of heat?
a. 0.021 kJ
b. 21 kJ
c. -0.291 kJ
d. 0.291 kJ
e. 0.458 kJ
Automobile airbags contain solid sodium azide, NaN3, that reacts to produce nitrogen gas when heated, thus inflating the bag.
Calculate the value of work, w, for the following system if 14.2g of NaN3 reacts completely at 1.00 atm and 22℃.
Enter your answer in the rpovided box.
In a gas expansion, 71 J of heat is absorbed by the system, and the energy of the system decreases by 124 J. Calculate the work done.
Calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a constant pressure of 16.7 atm.
One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 8.41 liters against a constant external pressure of 1.00 atm. How much work (in joules) is performed on the surroundings? Ignore significant figures for this problem. (T= 300 K: 1 L•atm = 101.3 J)
A. 375 J
B. 751 J
C. 225 times 103 J
D. 852 J
E. none of these
A piston has an external pressure of 5.00 atm. How much work has been done in joules if the cylinder goes from a volume of 0.180 liters to 0.530 liters?
Express your answer with the appropriate units.
An expanding gas does 173 J of work on its surroundings at a constant pressure of 1.01 atm. If the gas initially occupied 68.0mL, what is the final volume of the gas?
Use the data from this table of thermodynamic properties to calculate the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane, CH3CH3(g) at 25 "C and standard conditions.
A Carnot engine receives 280 kW of heat from a heat-source reservoir at 500ºC and rejects heat to a heat-sink reservoir at 25ºC. What are the power developed and the heat rejected?
A gas expands in volume from 26.7 mL to 89.3 mL at constant temperature. Calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of 1.5 atm, and (c)against a constant pressure of 2.8 atm.
Consider the reaction
If 3 moles of H2 react with 3 moles of Cl2 to form HCl, calculate the work done (in joules) against a pressure of 1.0 atm at 25°C. What is ΔU for this reaction? Assume the reaction goes to completion.
Consider the reaction
If 2.0 moles of H2O(g) are converted to H2(g) and O2(g) against a pressure of 1.0 atm at 125°C, what is ΔU for this reaction?
Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and 100°C. Assume that the volume of liquid water is negligible compared with that of steam at 100°C, and ideal gas behavior.
Calculate the work done when 50.0 g of tin dissolves in excess acid at 1.00 atm and 25°C:
Sn(s) + 2H+(aq) → Sn2+(aq)+ H2(g)
Assume ideal gas behavior.
The work done to compress a gas is 74 J. As a result, 26 J of heat is given off to the surroundings. Calculate the change in energy of the gas.
A gas expands and does P-V work on the surroundings equal to 325 J. At the same time, it absorbs 127 J of heat from the surroundings. Calculate the change in energy of the gas.
A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. Calculate the work done in joules if the gas expands (a) against a vacuum, (b) against a constant pressure of 0.80 atm, and (c) against a constant pressure of 3.7 atm.
What is heat? How does heat differ from thermal energy? Under what condition is heat transferred from one system to another?
Define these terms: system, surroundings, open system, closed system, isolated system, thermal energy, chemical energy, potential energy, kinetic energy, law of conservation of energy.
A certain liquid has Δvap Ho = 26.0 kJmol -1. Calculate q, w, ΔU, and ΔH when 0.50 mol is vaporized at 250K and 750 Torr.
All of the following statements are true EXCEPT:
a. energy is a state property.
b. the magnitude of delta H is proportional to the limiting reagent.
c. the magnitude of delta H for a reaction is equivalent to that of the reverse reaction.
d. the magnitude of delta H is proportional to the amount of product produced.
e. the magnitude of delta H is dependent upon the intermediate steps in the chemical reaction.
An ideal gaseous reaction (which is a hypothetical gaseous reaction that conforms to the laws governing gas behavior) occurs at constant pressure of 30.0 atm and releases 67.9 kJ of heat. Before the reaction, the volume of the system was 8.00 L. After the reaction, the volume of the system was 2.80 L.
Calculate the total internal energy change, ΔE, in kilo joules.
Classify the following by the sign of delta U for the system: the following can be classified into these three categories (negative,positive, not enough data)
1. The system contracts and the surroundings get cooler.
2. The system expands and the surroundings get cooler.
3. The system contracts and the surroundings get hotter.
4. The system expands and the surroundings get hotter.
A sample of gas occupies a volume of 51.1 mL. As it expands, it does 137.7 J of work on its surroundings at a constant pressure of 783 torr.
What is the final volume of the gas?
A balloon is inflated from 0.0100 L to 0.400 L against an external pressure of 10.00 atm. How much work is done in joules? (101.3 J = 1 L. atm)
a. -395 J
b. -0.395 J
c. -39.5 J
d. 39.5 J
e. 0.395 J
The oxidation of copper(I) oxide, Cu 2O(s), to copper(II) oxide, CuO(s), is an exothermic process.
2CuO2(s) + O2(g) → 4CuO(s)
The change in enthalpy upon reaction of 67.68 g of CuO (s) is -69.06 kJ
Calculate the work, w, and the energy change, ΔU rxn, when 67.68 g of Cu 2O (s) is oxidized at constant pressure of 1.00 bar and at constant temperature of 25°C?
The specific heat of water is 4.184 J/g °c and that of aluminum is .89 J/g °c. Will 1 gram of water or 1 gram of aluminum have the biggest increase in temperature upon the addition of 4.184 J? Explain.
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.
Classify each of the following as a path function or a state function.
iii) Distance traveled
At constant pressure, which of these systems do work on the surroundings? Check all that apply.
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)
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
When fuel is burned in a cylinder equipped with a piston, the volume expands from 0.255 L to 1.45 L against an external pressure of 1.02 atm. In addition, 875 J is emitted as heat. What is the ∆E? (101.3 J = 1 L • atm)
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
Which one of the following statements is FALSE?
A process is carried out on a system at constant external pressure. During the process 150. J of heat moves from the surroundings to the system, and the system expands, performing 400.J of work. The values for q, w, ΔU, and ΔH for the process are:
a. q = -150.J, w = 400.J, ΔU = 250.J, ΔH = -150.J
b. q = 150.J, w = 400.J, ΔU = 550.J, ΔH = 550.J
c. q = 150.J, w = 400.J, ΔU = 550.J, ΔH = 150.J
d. q = 150.J, w = -400.J, ΔU = -250.J, ΔH = -250.J
e. q = 150.J, w = -400.J, ΔU = -250.J, ΔH = 150.J
Oxygen gas at 34.5°C is compressed from 45.7 L to 34.5 L against a constant pressure of 0.987 atm. If the oxygen gas is defined as the system, how much work is done (in J) on the system? (1 L atm = 101.3 J)
A. 1.12 × 103 J
B. −1.11 × 101 J
C. 1.11 × 101 J
D. −4.55 × 103 J
E. −1.12 × 103 J
Which statement is true of a reaction in which ∆V is positive? Explain.
a. ∆H = ∆E
b. ∆H > ∆E
c. ∆H < ∆E
When 1 mol of a gas burns at constant pressure, it produces 2418 J of heat and does 51 of work. Identify ∆E, ∆H, q, and w for the process.
The heat of vaporization of water at 373 K is 40.7 kJ/mol. Find q, w, ΔE, and ΔH for the evaporation of 454 g of water at this temperature.
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 E system?
A rolling billiard ball collides with another billiard ball. That billiard ball (defined as the system) stops rolling after the collision. Identify the energy exchange and the sign of ΔE for the system.
A. work; - ΔE
B. work; + ΔE
C. heat; - ΔE
D. heat; + ΔE
Which statement is true of the internal energy of a system and its sorroundings during an energy exchange with a positive ΔEsys?
A. The internal energy of the system decreases and the internal of the surroundings increases.
B. The internal energy of the system increases and the internal energy of the surroundings decreases.
C. The internal energy of both the system and the surroundings decreases.
D. The internal energy of both system and the surroundings increases.
Which of the following is not a state function?
C. Internal energy
The internal energy of a system is the sum of all of its ___.
A. thermal energy and kinetic energy
B. potential energy and chemical energy
C. thermal energy and chemical energy
D. potential energy and kinetic energy
A typical breath is around 0.5 L, but to get 2 sig figs, let’s assume you breathe in 0.48 L. How much work is done when you exhale against an atmosphere pressure of 750 mmHg? Be sure to include a + or - sign in your answer.
Which set of signs for q and w represent a system that is doing work on the surroundings and losing heat to the surroundings?
a) -q, -w
b) +q, +w
c) -q, +w
d) +q, -w
e) None of these represent the system referenced above.
Which statement is always true of the internal energy of a system and its surroundings during an energy exchange with an endothermic value of ΔUsystem?
a) ΔHsurroundings > 0
b) The internal energy of the system and surroundings increase together
c) The internal energy of the system decreases while the internal energy of the surroundings increases
d) The internal energy of the system and surroundings decrease together
e) The internal energy of the system increases while the internal energy of the surroundings decreases
Calculate the amount of work done when 2.5 mole of H 2O vaporizes at 1.0 atm and 25°C. Assume the volume of liquid H2O is negligible compared to that of vapor. (1 L atm = 101.3 J)
1) -61.9 J
2) -6.19 kJ
3) 61.9 J
4) 5.66 kJ
5) 518 J
Which of the following depicts a situation where the least amount of work is done by a sample of gas? [1 atm = 101325 Pa]
a) A 20-L sample of gas expands to 500 L against a vacuum
b) A 1-L sample of gas expands to 30 L against a pressure of 2.22 atm
c) A 1-L sample of gas expands to 30 L against a pressure of 2.22 Pa
d) A 20-L sample of gas expands to 500 L against a pressure of 30 atm
e) Two or more are tied
A gas is allowed to expand, at a constant temperature, from a volume of 1.0 L to 10.1 L against an external pressure of 0.50 atm. If the gas absorbs 250 J of heat from the surroundings, what are the values of q, w, and ∆E?
1) A measure of temperature.
2) A measure of the change in temperature.
3) A measure of thermal energy.
4) A measure of thermal energy transferred between two bodies at different temperatures.
5) All of the above.
Which of the following statements concerning the first law of thermodynamics is/are true?
I) The internal energy of the universe is always increasing.
II) Internal energy lost by a system is always gained by surroundings.
III) The universe is an isolated system.
1. II and III only
2. III only
3. I and II only
4. II only
5. I and III only
6. I, II and III
7. I only
Heat absorbed by a system at constant volume is equal to
Which set of signs for q and w represent a system that is doing work on the surroundings and losing heat to surroundings?
a. – q, -w
b. + q, +w
c. – q, +w
d. + q, -w
e. None of these represent the system referenced above.
A gas at 25.00°C and 360.00 torr (760 torr = 1atm) expands from 25.0 L to 35.0 L. What is the value of the work done on or by the gas? (1 L•atm = 101.325 J)
a. 480 J
b. 3600 J
c. -3600 J
d. 1013.25 J
e. -480 J
What is the change in the internal energy (in J) of a system that releases 1000 J of heat and does 225 J of work on the surroundings?
Which is true when a gas expands isothermically against a constant pressure of two atmosphere (mark all that apply)?
a. The system does not work.
b. The gas releases heat.
c. Heat is absorbed by the gas.
d. The temperature of the gas decreases.
e. No heat flows.