I feel like there’s gonna be a lot of vocab on this test, so here’s a vocab re-run:
Thermodynamics: study of energy and its transformations
Thermochemistry: aspect of thermodynamics that relates to chem reactions and energy changes involving heat
Joule (J): SI unit for energy; equal to 1 kg m^2/s^2
Calorie (cal): energy unit; amount of energy needed to raise temp of 1 g of H2O; 1 cal = 4.186 J
System: portion singled out for study
Surroundings: everything outside the portion being studied
1st Law of Thermodynamics: energy conserved
Internal energy: sum of kinetic and potential energies of all of its components
Endothermic: system gains heat (+)
Exothermic: system loses heat (-)
State function: property of system that is determined by specific condition of system (i.e. temp, pressure, etc.); value depends on the present system only, not the path it took to reach the present state
Enthalpy (H): function that accounts for heat flow, occurs at constant pressure
Hess’s Law: If reaction is in series of steps, H = sum of all enthalpy changes for each individual steps
Enthalpy of Reaction: H change that accompanies a reaction
CH4(g) + 2O2 -> CO2(g) + 2H2O(l) ΔH= -890 kJ
CO2(g) + 2H2O(l) -> CO2(g) +2O2(g) ΔH= 890 kJ
How much heat is released when 4.50g of methane gas (CH4) is burned in constant-pressure system?
Step 1: C2H2 needs to be in the products side, so you have to switch the first reaction; reaction switch means change ΔH sign only for that reaction, so instead of C2H2 + 5/2O2 -> 2CO2 + H2O ΔH = -1299.6 kJ, it’ll be
2CO2 + H2O -> C2H2 + 5/2O2 ΔH = 1299.6 kJ
Step 2: The 2nd and 3rd reaction can be left alone for now since C and H2 are needed in the reactant’s side. If you haven’t already figured out, you’re trying to get 2C + H2 -> C2H2 as the end result from those 3 reactions above.
Step 3: Now that you have all 3 of your “new” reactions, write them all together cause it’s neater/easier to see what happens that way.
2CO2 + H2O -> C2H2 + 5/2O2 ΔH = 1299.6 kJ
C + O -> CO2 ΔH= -393.5 kJ
H2 + 1/2O2 -> H2O ΔH= -285.8 kJ
Step 4: Since the end reaction needs 2 Cs, you have to take the 2nd reaction and multiply it by 2. Multiply even the ΔH by 2 as well.
2(C + O -> CO2 ΔH= -393.5 kJ) =
2C + O -> 2CO2 ΔH= -787 kJ
Step 5: Rewrite the 3 new reactions side by side again and cancel out the reoccurring things (*NOTE*: Since the 2 of the 3 O2s are in fractions, make the other one in fraction form too)
Isolated: The system cannot exchange energy or matter with the surroundings.
Closed: The system can exchange energy but not matter with the surroundings.
Open: The system can exchange both energy and matter with the surroundings.
Isothermal processes occur when the system’s temperature is constant, implying that the total internal energy of the system is constant throughout the process.
Adiabatic processes occur when no heat is exchanged between the system and the environment.
Isobaric processes occur when the pressure is constant.
State functions describe the system in an equilibrium state, including temperature (T), pressure (P), volume (V), density (ρ), internal energy (E or U), enthalpy (H), entropy (S), and Gibbs free energy (G). State functions are independent of path taken, but not necessarily independent of one another.
Process functions describe pathways taken from one equilibrium to another, including work (W) and heat (Q).
The standard conditions are defined as 25°C (298K) and 1 atm. Standard temperature and pressure (STP) is 0°C (273K) and 1 atm.
Under standard conditions, the most stable form of a substance is called the standard state of that substance.
The changes in enthalpy, entropy and free energy under standard conditions are called the standard enthalpy, standard entropy and standard free energy changes, symbolized by ΔH°, ΔS° and ΔG°.
Temperature (T) is related to the average kinetic energy of the particles of the substance whose temperature is being measured.
Heat (Q) is the transfer of energy from one substance to another as a result of their difference in temperature. The unit of heat is joule (J) or calorie (cal) for which 1 cal = 4.184 J.
The first law of thermodynamics states that the change in total internal energy is equal to the amount of heat transferred to the system minus the amount of work done by the system: ΔU = Q - W.
The process of measuring transferred heat is calorimetry. Two types of calorimetry include constant-presure calorimetry and constant-volume calorimetry.
The heat absorbed or released in a given process is: q = mcΔT, where c is the specific heat of the substance.
Specific heat is the amount of energy required to raise the temperature of one gram or kilogram of substance by one degree Celsius or one unit Kelvin. The specific heat of H₂O (l) is one calorie per gram per Celsius degree (1 c/g°C).
In constant-volume calorimetry, W_cal = 0 (process is isovolumic) and Q_cal = 0 (no heat is exchanged between the calorimeter and the rest of the universe).
Enthalpy is a state function so the pathway taken for a process is irrelevant to the change in enthalpy.
The change in enthalpy is equal to the heat transferred into or out of the system at constant pressure: ΔH_rxn = H_prod - H_reac.
A positive ΔH_rxn corresponds to an endothermic process, and a ΔH_rxn corresponds to an exothermic reaction.
The standard enthalpy of formation ΔH°_f is the enthalpy change that would occur if one mole of a compound in its standard state were formed directly from its elements in their respective standard states.
The standard heat of a reaction ΔH°_rxn is the hypothetical enthalpy change that would occur if the reaction were carried out under standard conditions: ΔH°_rxn = Σ(ΔH°_f of products) - Σ(ΔH°_f of reactants)
Hess’s law states that enthalpy changes of reactions are additive. Hess’s law applies to any state function, including entropy and Gibbs free energy.
Bond dissociation energy is the average energy that is required to break a particular type of bond between atoms in the gas phase, given in kJ/mol of bonds broken.
Bond formation has the same magnitude of energy but is negative rather than positive, as energy is released when bonds are formed.
The enthalpy change associated with a reaction in regards to bonds is: ΔH_rxn = ΣΔH_broken + ΔH_formed = total energy absorbed - total energy released.
The second law of thermodynamics states energy spontaneously disperses from being localized to becoming spread out if it is not hindered.
Entropy is the measure of spontaneous dispersal of energy at a specific temperature: ΔS = Q_rev / t, in kJ/mol · K.
The second law claims that entropy of the universe is increasing:ΔS_uni = ΔS_sys + ΔS_sur > 0
The standard entropy change for a reaction is: ΔS°_rxn = ΣΔS°_prod - ΣΔS°_reac.
The change in Gibbs free energy is is the maximum amount of energy released by a process: ΔG = ΔH - TΔS
If ΔG is negative, the reaction is spontaneous.
If ΔG is positive, the reaction is nonspontaneous.
If ΔG is zero, the system is in a state of equilibrium where ΔH = TΔS.
The free energy change of reactions can be measured under standard state conditions to yield the standard free energy ΔG°_rxn.
The standard free energy of formation of a compound ΔG°_f is the free energy change that occurs when 1 mole of a compound in its standard state is produced.
The standard free energy of a reaction ΔG°_rxn is the free energy change that occurs when the reaction is carried out under standard state conditions: ΔG°_rxn = Σ(ΔG°_f of products) - Σ(ΔG°_f of reactants)
The standard free energy change for a reaction based on the equilibrium constant is: ΔG°_rxn = -RT ln K_eq, where R is the gas constant.
The greater the value of K_eq, the more negative the standard free energy change and the more spontaneous the reaction.
For a reaction in progress: ΔG_rxn = RT ln (Q / K_eq).
If Q < K_eq, the free energy change will be negative, so the reaction will spontaneously proceed forward until equilibrium is reached.
(Delta)U = Q + W - U: Internal Energy - Q: Energy Added to the system - W: Work done to o the system
Basically, energy cannot be created/destroyed. It always remains constant, even though it can change form from one form to another.
Say you add 2500J of Energy and do 1800J of Energy of work on a system. How much does internal energy (also called Enthalpy) change? (Delta)U = Q + W (Delta)U = (2500) + (1800) (Delta)U = 4300J of Internal Energy
What if you add 2500J of energy, but the system does 1800J of work? That means work is an OUTPUT, not an input. Therefore, it is negative. Outputs are negative. (Delta)U = Q + W (Delta)U = (2500) + (-1800) (Delta)U = 700J Internal Energy.
You can also have suck energy OUT of a system, which makes Q negative.
Isothermal: You do the process (changing volume/pressure) without changing the TEMPERATURE of the system.
Isobaric: You do the process (changing volume/temperature) without changing the PRESSURE of the system.
Isochoric/Isovolumetric: You do the process (changing temperature/pressure) without changing the VOLUME of the system.
Adiabatic: No heat (No Q value). The only thing you have is “Work” done on or by the system.
When using a PV Chart:
- Y Axis = Pressure - X Axis = Volume - Isovolumetric = Straight vertical line. - Isobaric = Straight horizontal line. - Isothermal = Curved line that is concave towards the origin - Adiabatic = It’s weird, but it’s a slightly more curvy version of Isothermal.
To find Work Done: Take the area underneath the reaction’s graph. So if you were measuring the work done by an Isothermal process, take the area underneath the curve (you have to know a bit of Calc to do that)
W = P x (Delta)V Work = Pressure x Change in Volume
(Delta)U = (3/2) n R (Delta)T Change in Internal Energy = 1.5 x (moles) (8.315J/mol) (Change in Temp)
Giancoli p448 Q 15-5 .25 moles of some random gas expands rapidly and adiabatically (no Q) against the piston. The temperature change is 750K. What is the work? (Delta)U = Q + W –> (Delta)U = (3/2) n R (Delta)T (3/2) (n) ® (Delta)T = (0) + W (3/2) (.25) (8.315) (750) = W Work = 2300J
Giancoli p448 Q 15-6 1 L water boils and makes 1671L of steam. how much is the internal energy? 1L of water is 1kg (1mL = 1g, that’s just special w/ water) HofVaporization: 22.6x10^5 Q = (mass) (HofV) Q = (22.6 x 10^5) J W = P x (delta)V –> We’re at atmospheric pressure, 100000 (delta)V = 671 L = .671 m^3 W = 100000 x .671 W = 1.7 x 10^5 J (Delta)U = Q + W –> You can do it from here.
Second Law of Thermodynamics:
This is the law about Spontenaiety (spelling’s bad, sorry). Basically, if it’s spontaneous in one direction (hot to cold) it won’t be spontaneous in the other direction. (cold to hot) For example: A teenager’s room will spontaneously get messier. It’s just the way it works. But when does a teenager’s room spontaneously get cleaner?
Engines use the fact that heat flows from hot to cold to their advantage. What happens: Heat energy goes into the engine, and some of it comes out of it as mechanical work - it moves the pistons and stuff. But some of it comes out as a lower heat energy, because the engine isn’t perfect and can’t use ALL of the energy.
Q (high) = Q (low) + Work High Input Heat = Work + Low Output Heat
e = W/Q (high) efficiency = Work / High Input Heat
e = 1 - (Q (low) / Q (high)) efficiency = 1 - (High Energy Input/Low Energy Output) Sometimes they ask for efficiency as a percent - just x100
This guy was important. He tried to make engines more efficient, and realized that there’s a “perfect” engine (no such thing in real life, but he found out what it SHOULD do)
Perfect efficiency Equation:
e = 1 - (T (low) / T (high)) efficiency = 1 - (Temp of output/Temp of input)
Kevin Planck Statement: No one can make a perfect machine where Heat is transformed completely to Work and Energy.
Refrigerators: They want to do the opposite of engines - taking energy from low temperatures and taking in Work in order to make higher energy. There can’t be a perfect fridge either.
CP = Q (low) / W Coefficient Performance (basically efficiency) = Low Energy Input / Work
CP = Q (low) / ( Q (low) - (Q (high)) Coefficient Performance = Low Energy Input / (Low Energy - High Energy)
CP (ideal) = T (low) / ( T (low) - T (high)) Perfect Coefficient Performance = Low Temp of Input / (Low Temp - High)
CP of a HEAT PUMP (basically a backwards fridge) = Q (high) / W
Entropy is basically the randomness or disorder of a system. We already talked about it a bit earlier.
(delta)S = Q / T Change in Entropy = Heat Energy / Temperature (KELVINS ONLY)
In order to find changes in entropy, you must first find the change in HEAT. This often uses q = mc(delta)T or q = m(HoVaporization or Fusion)
And another thing to keep in mind - Entropy can only ever increase or stay the same. Entropy (randomness) never goes lower. Just like how the teenager’s room never gets cleaner.
Things tend to get messier, or more disordered.
And this is where I stop for today - I have to do some Macro later. Hope you got some of my blabbering.
Today was probably the most productive school day I have ever had - I finished Looking for Alaska like a good Tumblr user should, and one of the quotes stuck with me through all my classes: “Entropy increases. Things fall apart.” When I got to APChem, we were, of course, learning nothing other than entropy, which is the science behind why there is so much chaos and things ending and forming new, simpler versions of the sum of their parts. My teacher said the main law in all thermochemistry is that energy cannot be created nor destroyed, so I asked how energy came to be in the first place.
“Well, that’s the million dollar question.”
“So that’s it? No one knows?”
“Well, the big-bang theory is what most scientists assume to be true, but when religion comes into play…you can’t just create something from nothing. Someone had to jump start it all.”
This reminded me of an earlier discussion in class about spontaneous reactions, and how a ball rolling down a hill is spontaneous because there is no force needed to make it start moving - gravity just does it’s thing and the ball rolls. A couple of smart-asses who didn’t really get the concept started arguing that since someone had to put the ball at the top of the hill, it wasn’t truly spontaneous - it started the reaction just as much as shoving it would have.
“So, God put the ball at the top of the hill then?”
For student teaching I created a thermochemistry unit, in accordance with the Indiana State Standards. I’ve linked my lesson plans below (which include suggestions), along with any handouts. I also presented my students with this Unit Guide.
Energy and Its Forms This is a short lecture/ discussion walking through and “unpacking” some of the information they gathered in the Thermochemistry Guided Discovery. handout
Heat Flow This activity is a small scale lab where students mix different salts with water. This lab was done previously to discuss exo/endothermic reactions but I made them do it again using the words: system, surroundings, molecular motion, energy, and heat. handout
Gen chem II final tomorrow. Acid/base, kinetics, equilibrium, electrochemistry and thermochemistry….sigh. And when we ask our prof what would be on the test he simply said “everything”. At least gives us some guidance on what to focus on…
Please, I need all the prayers I can get! I’m beyond nervous! :\
On one hand it seems everyone got a D on the thermochemistry exam and that sucks. On the other hand I get to say, “Yeah, we got the D!” and laughing and throwing our hands in the air, a group of people related by nothing but a crumbling academic life get to cheer “We got the D!!” And that’s what life is about man.