CHEMISTRY COMMUNITY

Jakob von Morgenland 2C

I highly doubt we would be asked to regurgitate memorized molecules on the final. Dr. Lavelle said he wants us to understand functional groups visually, meaning you should most likely be able to decipher which function groups are present in a given molecule.

Jakob von Morgenland 2C

I do not think so, I think the functional group is just the C bonded to the N, the rest can vary.

Jakob von Morgenland 2C

If at a constant pressure, deltaU = deltaH - Pext(deltaV).

Jakob von Morgenland 2C

I'd like to point everyone's attention to page 621 in the textbook, there is a possibility for orders to be negative numbers (you can see the example that they provided of the decomposition of ozone to oxygen to understand it better). So yes, that would be the right rate law, according to the textbo...

Jakob von Morgenland 2C

Using these plotting methods, a first order reaction would have a straight line because the rate is directly proportional to the concentration. If you were to plot a second order reaction using similar techniques, it would not be a straight line because a second order reaction's rate is proportional...

Jakob von Morgenland 2C

Reaction intermediates are molecules that appear as the products in a step and then again as reactants in a later step (AKA they get crossed out when writing the summation of the reaction).

Jakob von Morgenland 2C

Yeah you should definitely know all of the derivation/integrations that we do in class. Many (if not all of the ones you need to know) are on the constants and equations sheet already.

Jakob von Morgenland 2C

You can determine the molecularity by looking at the number of molecules that participate/are used in the rate-determining step/rate law.

Jakob von Morgenland 2C

Yes, that is a correct statement to make.

Jakob von Morgenland 2C

I would assume you would use the one that cancels out the charge on the ion. For this, since it has a 2+ charge, I would use the one that has 2 electrons in it.

Jakob von Morgenland 2C

I think it flows from cathode to anode because if it's a negatively charged electrolyte, that implies that the reaction is non-spontaneous due to a negative voltage. Because spontaneous reactions are spontaneous when the electrolyte flows from anode to cathode, non-spontaneous would just be the reve...

Jakob von Morgenland 2C

They will most likely provide all standard reduction potentials that you will need for the test.

Jakob von Morgenland 2C

I would assume you can't use that formula since the pressure is changing. Therefore, I believe you cannot solve for enthalpy of the reaction due to the fact that the pressure is not constant (therefore q =/= deltaH).

Jakob von Morgenland 2C

I believe Dr. Lavelle will generally put the anode on the left and the cathode on the right. However, if you are unsure, look at the standard reduction potentials chart and determine which one is being reduced and which one is being oxidized from those values. Always try to make the voltage positive.

Jakob von Morgenland 2C

Don't forget to multiply your deltaHfus of ice by the amount of moles of ice you have (50/18). Other than that, looks good!

Jakob von Morgenland 2C

It's based on the degeneracy of the two respective molecules. The trans molecule has less possible equivalent energy states/positions of its elements than the cis model. If you draw out the two molecules and move the elements around while preserving their cis/trans conditions, you can see the cis ha...

Jakob von Morgenland 2C

The equations sheet that we received for the first test has all the entropy formulas you would need to solve any problem given; in addition to those listed above, there's the nCln(T2/T1), and the formula for standard entropy of formation.

Jakob von Morgenland 2C

Sounds like the question is not worded in the best manner. I think that's technically compression work, since it the system is being compressed.

Jakob von Morgenland 2C

In a perfect crystal, there are no imperfections, and in other words, a perfect crystal has no disorder (so the entropy will equal 0).All other objects have some sort of imperfection and disorder, so the entropy only theoretically approaches 0.

Jakob von Morgenland 2C

∆G° is considered the standard free energy change of a reaction, while ∆G is considered the free energy change of a reaction. Remember, standard free energy change implies that the reactants and products are in their standard states at 1 atm and generally 25 celsius.

Jakob von Morgenland 2C

If deltaU=0, then deltaU=0=q+w, or q=-w. As you said, q would equal 0 if the temperature is constant, so sub that in for q, and you would get q=0=-w, or w=0. This is assumed for and ideal gas in an isothermal process generally.

Jakob von Morgenland 2C

It would most likely be given in that scenario.

Jakob von Morgenland 2C

Since w=-PdeltaV, we make it negative on both sides to simply substitute the value in for work. And we know that if we make one side negative, we have to make the other side negative as well, so that's why it's -deltanRT.

Jakob von Morgenland 2C

Yes, 0 degrees Celsius is the temperature at which ice melts, and 100 degrees Celsius is the temperature at which water vaporizes. At these temperature values, you need to calculate the enthalpy of melting/vaporization and incorporate that into your total enthalpy amount.

Jakob von Morgenland 2C

I don't think the work needs to be negative if the given q is positive; I believe it is not a causal relationship ("if q is positive, then w must be negative", etc.). If you take a look at 8.13, you can see that this problem gives a negative deltaU value and a negative (but not as large) d...

Jakob von Morgenland 2C

I believe the value assigned is the enthalpy of the reaction, which is a state function, and is different from heat. Also, joules and kilojoules are units of energy, which gives way to understanding that we are tracking the change in energy of the function, which is the enthalpy.

Jakob von Morgenland 2C

In Friday's lecture, we discussed standard reaction enthalpies, which occur when the reactants and products are in standard states and the pressure is 1atm. We also talked about standard reaction enthalpy for formation, which we can subtract that standard reaction enthalpies of formation of the reac...

Jakob von Morgenland 2C

The internal energy is the system's total storage of energy; heat is the amount of energy transferred; enthalpy is a measurement of energy (also has to be held at a constant pressure in order to be a state function).

Jakob von Morgenland 2C

Since they are interchangeable, I do not believe it matters which notation you use. However, since Dr. Lavelle has been using deltaH almost exclusively in lecture so far, I take it that we would mainly use deltaH.

Jakob von Morgenland 2C

I believe it has something to do with the fact that pressure (P) is considered a state property. Because it is dependent on a state property, Qp itself becomes a state property; however, Q itself is not a state property.

Jakob von Morgenland 2C

When we calculated the individual bond enthalpies for the bonds that we broken in the reaction compared to the bonds that were formed, we found that the bond enthalpies of the bonds being formed were larger than the bonds broken. Since the bonds being formed are given negative enthalpies and the bon...

CHEMISTRY COMMUNITY

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