CHEMISTRY COMMUNITY

Scott Chin_1E

If I understand your question correctly, I think that all this means is that when you add up the cell potential of an anode (which has a positive value) and a cathode (which has a negative value), the cell will be not spontaneous (-E of the cell) and will not be a useful battery.

Scott Chin_1E

I think the stoichiometric coefficients are there to help us determine the rate law of the elementary step. It would give us the order of the reactant in the rate law.

Scott Chin_1E

What is the difference between an adiabatic system and an isothermal system? I know isothermal means that no heat is flowing in or out of the system (the system is isolated/insulated).

Scott Chin_1E

I understand the mnemonic device to determining which species is being reduced or being oxidized (OIL RIG or LEO GER). Let me clarify my question. In the question 14.25, they ask to list the elements Cu, Zn, Cr, and Fe according to increasing strength as reducing agents. I'm not sure how to determin...

Scott Chin_1E

Is there a mnemonic device or easy way to understand which species are stronger reducing agents or oxidizing agents?

Scott Chin_1E

Remember to use mnemonic devices to help you determine which is the anode and which is the cathode!

LEO - Lose Electrons Oxidation
GER - Gain Electrons Reduction

or

OIL - Oxidation is Lose (electrons)
RIG - Reduction is Gain (electrons)

REDCAT - Reduction is the Cathode

Scott Chin_1E

Yes, this is because we need to know how a change in concentration of one of the reactants will change the rate of the reaction. It is impossible to know the order of each reactant for every reaction without conducting experiments. We can come up with a proposed rate law, but it may not always be co...

Scott Chin_1E

"Everything else" meaning temperature and concentration held constant, correct?

Scott Chin_1E

In the case above, would molar mass be considered an extensive property since it's value is dependent upon the metal in question?

Scott Chin_1E

I know that the reaction that is the "slow step" will have the same reaction rate as the overall reaction.

Scott Chin_1E

I think Professor Lavelle used the example of diamond being thermodynamically favored to become a different product (though I'm not too sure what that was), however, we know that diamond does not indeed undergo this reaction because it requires a high amount of activation energy to overcome the ener...

Scott Chin_1E

I believe this is because for a zero order reaction, the rate at which the reactant is consumed is constant, thus the time it takes for the concentration of the substance to reduce by half would be dependent on the initial concentration. I could imagine that this explanation could almost be applied ...

Scott Chin_1E

It is definitely a lot harder to find the rate of a reaction when using products because the progress of a reaction occurs at an exponential decay, this, the reaction rate towards the end of a reaction occurs a lot slower than at the beginning of a reaction. Therefore, we using reactant concentratio...

Scott Chin_1E

But in a case in which you were unsure, you can always refer to the written reaction to determine which reactant is being oxidized and which is being reduced. This will help you find out which side is the anode or cathode ("redcat").

Scott Chin_1E

Another way of explaining this is that at higher temperatures, the particles have much more energy in them and will move more and theoretically "clash" with particles of the other reactant, thus accelerating the speed of the reaction.

Scott Chin_1E

Yes, make sure to determine which reaction is oxidation and reduction. This will help you in determining the order as the cathode (right) is associated with reduction and the anode (left) is associated with oxidation.

Scott Chin_1E

I think it's okay to assume initial rates because that's where we will be focusing our approximations with (because of the rate decreases as the reaction proceeds).

Scott Chin_1E

I think Professor Lavelle said that Graphite electrodes are used as well because they do not affect the reaction, but Platinum electrodes are the most common (even though they might be extremely expensive, go figure).

Scott Chin_1E

If this helps for comparison, a closed system would be like a sealed container, where heat can transfer in and out of the system but matter cannot (because the container is sealed).

Scott Chin_1E

Does Wmax usually occur when the system is at equilibrium?

Scott Chin_1E

This would especially help when applying Le Chatelier's Principle as exothermic and endothermic reactions can shift the equilibrium of a system as well as in Electrochemistry where voltage of a battery should always be positive (it would be a good way to double check that you identified the correct ...

Scott Chin_1E

I think temperature is usually expressed in Kelvin as most of the conversions and constants are also expressed in Kelvin, so it makes it easier for calculations and as for Joules versus Kilojoules, q is usually found using heat capacity which is expressed in Joules.

Scott Chin_1E

Yes I believe it is because the molecule is monatomic.

Scott Chin_1E

No, there is no machine which we can use to measure DeltaG. The only way to get a value for deltaG is by calculating for it using

deltaG = deltaH - T(deltaS)

Scott Chin_1E

I know this might be a bit late, but in preparation for the upcoming midterm, I had trouble looking for problems dealing specifically with reversible expansion of ideal gases (ex: $w = -nRT(ln\frac{V_{f}}{V_{I}})$ )

Can someone point me to any good examples they found in the textbook? Thanks!

Scott Chin_1E

A quick correction to the post regarding the equation to determine DeltaG and thus spontaneity:
Instead of multiplying temperature and change in enthalpy and subtracting the product from entropy, it is actually

deltaG = deltaH - (temperature)(deltaS)

Scott Chin_1E

It would be useful to know the derivations for the purpose of understanding how we arrived at the 2nd Law of Thermodynamics, but hopefully Dr. Lavelle will clarify this before the midterm.

Scott Chin_1E

We can do this by using Degeneracy and as well as the fact that states of matter can “occupy” more positions when that matter has more entropy. For example, a substance in the gas phase has more entropy and thus can occupy more “positions” than a substance in the solid phase.

Scott Chin_1E

Reversible systems do more work because it’s is working against equal pressures on both sides (P internal = P external) and a pressure that will be constantly changing and thus will be pressing against a larger pressure than irresversible systems that work against a constant pressure.

Scott Chin_1E

Adding onto previous answers, I believe you will be using standard reaction enthalpies (from intermediate reactions) to calculate the enthalpy of formation (referring to the final reaction you wish to calculate for) in Hess's Law.

Scott Chin_1E

I think that it is acceptable to use fractions in the coefficients for Hess's Law because the final reaction is the sum of all the intermediate reactions, therefore, fractions of each reactant can exist in the intermediate reactions. Fractions may be useful in Hess's Law calculations because it may ...

Scott Chin_1E

I think this has something to do with equilibrium. If an reaction is at equilibrium, the the reaction is reversible and if the reaction is not at equilibrium, the the reaction is not reversible. Is there another, better way to explain this? Or is this correct? This is what I've gathered from the las...

Scott Chin_1E

I believe it is useful to know the specific heat capacity of water @ standard conditions because it is so commonly used and because it’ll make calculations much easier, but I’m sure for other molecules and substances the specific heat capacity will be given.

Scott Chin_1E

Intensive Properties refers to properties that do not depend on the size or amount of the substance while Extensive Properties do. For example mass and volume are extensive properties while density is not

Scott Chin_1E

You're correct in that the Triple Point is the point in which a substance can exist in all three states of matter. Even more so, this is the minimum pressure that is required for a substance to remain a liquid. You can think of this as a reference point in seeing phase changes, for example at pressu...

Scott Chin_1E

I think first looking at the reactants and products of the problem would be a good indication for whether the problem is a weak acid/base question, especially if they only give you K instead of K_{A} or K_{B} . If they give you K_{A} or K_{B} , I think its safe to assume that you are working with we...

Scott Chin_1E

Well we can't assume x=0 because then that would mean there was no change and we know that in a weak acid/base reaction, equilibrium is in constant motion. Instead, we assume x is a very small number, so much so that subtracting it from the initial concentration is negligible and wouldn't affect the...

Scott Chin_1E

A quick trick to hybridization is to find out the number of electron densities for a given atom that you want to find the hybridization for, then match it up with the appropriate hybridization which has the same number of orbitals. For example: An element with 2 electron densities = sp (s+p = 2) An ...

Scott Chin_1E

Exothermic means that the reaction will release heat as a product, therefore when you take heat away from this reaction, the reaction will continue to proceed towards the product to produce more heat to compensate for the fact that you've taken away heat from the products. Endothermic means that the...

Scott Chin_1E

When the problem has has K, it's usually implicit that they are referring to K_{c} or K_{p} . You would just have to pay attention to the units in which the initial concentration/pressures are given. If they give the initial numbers with the units M (molarity) you will know that the problem is refer...

Scott Chin_1E

(Greek prefix) + Ligand names in alphabetical order (that are bonded to the TM) + TM cation + (Roman Numeral) + (if there are anions within the coordination complex we will add a -o) + (if there are anions bonded to the coordination complex) the anion name + (Greek Prefix) + Hydrate. For example: [C...

Scott Chin_1E

As mentioned above, hybridization is the fusion of two different orbitals (the s and the p) which enable such atoms such as Carbon to form more than just 2 bonds as it would be expected to make. Instead, the "hybridization" of the s and p orbitals further show that Carbon will form 4 bonds...

Scott Chin_1E

It would be helpful to know that Halogens and elements like Oxygen and Sulfur tend to have high electronegativity values than that of metals. This is because metals tend to lose electrons to form cations while nonmetals tend to want to gain one, two, or three more electrons to complete their octet b...

Scott Chin_1E

A simple way to distinguish the differences in shape between polar and nonpolar molecules would be if the shape were "symmetrical" or not. Keep in mind this a simple an easy was to distinguish, but not always true and there are exceptions. However, it does apply to a lot of the shapes.

Scott Chin_1E

Because Hydrogen does not take an octet (as mentioned above) and it forms a 1+ ion (thus taking a max of 2 valence electrons), Hydrogen will be more willing to complete its unstable 1s shell rather than lose it (because it can't otherwise it wouldn't be an element). And because of this character, it...

Scott Chin_1E

Oxygen is in the middle of the structure because it has a higher first ionization energy value. This means it is less likely to give off an electron, and more likely will hold onto its electron and thus draw other electrons towards it. The electronegativity value for Chlorine means that it will more...

Scott Chin_1E

Also, metals are known to form ionic bonds due to their tendency to give off electrons and bond with nonmetals (who more often accept electrons due to their higher electron affinity values).

Scott Chin_1E

"Delocalized electrons" speak to the fact that electrons have wave-like properties as well as particle-like properties. Delocalized essentially means that the electrons are in constant movement around the atoms involved in bonding rather than in rigid "orbitals" around a give ato...

Scott Chin_1E

I think the exceptions have to do with Oxygen's electron configuration. Given its configuration, 1s^{2}2s^{2}2p^{4} and Nitrogen's 1s^{2}2s^{2}2p^{3} , Nitrogen's completed half shell of the p subshell makes it slightly more stable than oxygen's (which is has one more electron). Nitrogen's relative ...

Scott Chin_1E

I know for a fact that we will have to know the conversions between micro (10^-6 m), nano (10^-9 m), and picometers (10^-12 m) because of their relation to wavelength. First it is important to realize that the conversions for these three measurements decrease by 1000 (10^-3) beginning with 10^-6. A ...

Scott Chin_1E

In reference to chromium and copper, I think the professor is going to go over this exception this week.

Scott Chin_1E

Generally, according to Hund's Rule, electrons in the same subshell occupy different orbitals with parallel spin due to their repulsion to each other. So more commonly, you can fill in electron configurations first with the spin up electrons until all orbitals are filled, then fill in with spin down...

Scott Chin_1E

I also believe that in Atomic Spectroscopy, the goal of the experiment was to determine at what energy levels of photons, will electrons of specific metals become excited to then release the energy. This experiment for Atomic Spectroscopy was more of a verification of the shell model for atoms while...

Scott Chin_1E

The equation E_{k}=hv-phi , should be used when trying to find out the kinetic energy of an ejected electron. Also, you may rearrange the terms to solve for hv in which this would be the energy of the photon released. Notice that E_{k} is also equal to \frac{1}{2}(m_{e^{-}})(v_{e^{-}}^2&...

Scott Chin_1E

I believe the reason why the solution manual is trying to prove the amount of doublings that would need to occur before there is only 1 atom of substance X is that after 70 doublings, the solution would essentially have 0 atoms of the substance, thus showing any more doublings past 70 doublings woul...

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