CHEMISTRY COMMUNITY

ZoeHahn1J

I believe adiabatic systems mean that no heat or matter is exchanged, only work, whereas isothermal is limited to heat.

ZoeHahn1J

Do you mean 14.9 part b?
The half reactions would be Fe3+ going to Fe2+ (reduction) and 2Cr3+ going to Cr2O7^2- (Cr is oxidized from 3+ to 6+).

ZoeHahn1J

Nope; if two things are in the same phase, they are separated only by a comma.

ZoeHahn1J

We know n=4 because that's the number of moles of electrons transferred when Fe is oxidized and O2 is reduced - split the reaction into two half reactions to check.

ZoeHahn1J

There are two reactants - Ag and Cl. The example is a little strangely formatted, but it gives us in the little diagram that the two electrodes are AgCl and Ag, and that the Ag electrode is oxidized, so it produces Ag+. The other electrode, AgCl, picks up electrons and produces Ag and Cl-.

ZoeHahn1J

I'm assuming that we are, since benzene has resonance in actuality, but whether benzene has resonance or not doesn't seem relevant to the solving of the problem, if that's a concern at all - the problem asks about irreversible work on a macro scale, so you don't have to worry about breaking resonant...

ZoeHahn1J

Ag is oxidized in the reaction (anode) and Br is reduced (cathode). We can tell which one is oxidized or reduced by how favorable it would be for each atom to either gain or lose an electron; since Br is very electronegative, we can assume that it would take an electron, and that therefore, Ag would...

ZoeHahn1J

I may be wrong, but I think water is not usually included in a cell diagram, since water and hydrogen tend to not be oxidized or reduced (especially as water molecules or H+ ions). Also, cell diagrams note both the anode and the cathode, as opposed to just one side of the reaction (ie, one half reac...

ZoeHahn1J

For a galvanic cell, you will have to flip the standard cell potential that will make E cell positive. There may be other reactions, however, where the question will give you the overall reaction and that reaction forces you to have a negative E cell.

ZoeHahn1J

That's the given heat absorbed when the oxidation of N2 to NO was completed in a bomb calorimeter in the question.

ZoeHahn1J

A concentration cell is just a galvanic cell, but the two electrodes are the same, just at different concentrations. Voltage is created as the concentration difference between the left side solution and right side solution decreases. In order for the less concentrated side to get more concentrated, ...

ZoeHahn1J

Right, but would there ever be a time when standard E is negative for a not galvanic cell, or when we would have to flip the reduction rxn for a galvanic cell?

ZoeHahn1J

Why can we assume that S in P4S3 is 2-? It's not like oxygen, since it's then oxidized to S^6+, and it doesn't make sense to me that S would be -2 (which is consistent with periodic table trends) but that P would have a +1.5 charge (not consistent with periodic table).

ZoeHahn1J

When calculating the standard cell potential of a reaction, do we always flip the oxidation half reaction? I know that for galvanic cells, Ecell has to be greater than zero, so we flip the larger value, but I'm not sure about non-galvanic cells.

ZoeHahn1J

You're correct that rate constant and rate are not the same thing; however, since this question asks for the factor by which the rate increases, and everything else stays the same, the factor between change in k and rate will be identical.

ZoeHahn1J

We are given that: K = 326 the attachment/loss of substrate to or from enzyme is 2nd order k of attachment = 7.4 x 10^7 L/(mol*s) and instructed to find the rate constant of the loss of substrate from the enzyme. Recall that K = k/k', if k is the rate constant of the forwards rxn (in this case, the ...

ZoeHahn1J

To solve 15.105, we are given: initial mass of drug = 150 mg half life = 4.5 hrs 1st order reaction and t= 2 hrs First, use the half life equation for a first order reaction to solve for the rate constant. Then, we can use the integrated rate law either a) solve for a ratio between the initial and f...

ZoeHahn1J

One could use spectrometry to measure the rate of very fast reactions since it tells us the absorption of a solution at very precise time units (ex: in stopped-flow technique), which can in turn be used to find the concentration of reactants/products at that time.

ZoeHahn1J

Since differential would be something like rate=k[A]^x[B]^y[C]^z, we can use it to: calculate the rate given k, concentration of reactants, and order of reactants calculate the rate constant k given rate, concentration of reactants, order of reactants etc. We cannot use it to calculate a new rate gi...

ZoeHahn1J

Choose two experiments (rows) in which all but one reactant has the same concentration and the other changes. Compare the change in the reactant concentration to the change in the rate. If the reactant concentration doubled and the rate doubled, this must be first order for that reactant (ie, if the...

ZoeHahn1J

If the first step in a process is the slow step and its rate does not match the overall reaction rate, is it definitely an invalid proposed mechanism?

ZoeHahn1J

Why wouldn't we multiply the standard reduction potential by 3 before calculating the current? Are moles of electrons already accounted for in the current equation?

ZoeHahn1J

There are homework problems assigned that ask to find the half reactions and balance the equation (such as 4 and 11), which doesn't (as far as I know; please correct me if I'm wrong!) necessarily require one to find the standard reduction potential of that reaction. However, the standard potential i...

ZoeHahn1J

If there is an increase in V, work is being done by the system, so w will be negative. When work is being done on the system, there will be a decrease in V, so work will be positive.

ZoeHahn1J

Oh! Thank you so much!

ZoeHahn1J

When doing 11.111, I used dG of rxn 1 to find K of reaction one, multiplied K by 10 to get K of rxn 2, and then used that to find dG of reaction 2 (this gave the incorrect answer). However, I'm not sure of how to solve this even looking at the solutions manual; why would we add the two delta G value...

ZoeHahn1J

delta S is zero because change in entropy is a state function, and the final and initial volumes are the same. delta G is zero when a reaction is at equilibrium.

ZoeHahn1J

We can calculate the entropy of vaporization of water at a temperature lower than its boiling point by adding delta S due to increasing the temperature to the boiling point + entropy of vaporization of water at the boiling point + delta S due to decreasing the temperature of the water back to the or...

ZoeHahn1J

Correct me if I'm wrong, but I believe ideal systems can be isolated, open or closed since ideality is normally used to describe the gases in a system, as opposed to the system itself.

ZoeHahn1J

The number of states will either be given or found by looking at the number of different orientations/states that a molecule can be in.

ZoeHahn1J

Since delta G = delta H - T delta S, a system is favorable if: - delta G is negative - delta H is negative and delta S is positive - delta H is negative and delta S is negative and delta H is greater than T delta S - delta H is positive and delta S is positive and delta H is less than T delta S A sy...

ZoeHahn1J

If delta S and S are always equal to each other (S=S(0) + delta S = delta S), then why would we refer to klnW as S but qrev/T as delta S? Why not refer to both as S? Is there a difference?

ZoeHahn1J

Entropy must practically always be positive, since S approaches 0 as T approaches 0. The third law of thermodynamics holds that "The entropies of all perfect crystals approach zero as the absolute temperature approaches zero."

ZoeHahn1J

What's the difference between positional disorder and degeneracy? Is it the same thing?

ZoeHahn1J

In lecture today, Dr. Lavelle described that a gas expanding against a vacuum is reversible, and that as it expands, it does work against the piston (loses energy) that is then compensated by the heat (gain in energy). I'm a little confused, however, how could the gas be said to be doing work when i...

ZoeHahn1J

Replying to Claire: Irreversible systems are characterized by being more spontaneous than reversible ones. For example, in the model of a piston with gases on either side, an irreversible system could be when the pressure on the inside is substantially larger than the pressure on the outside, so the...

ZoeHahn1J

Delta U is internal energy, so delta U = q + w. Delta H is enthalpy (heat under constant pressure), so it doesn't account for work.

ZoeHahn1J

An open system is unique in that matter and energy (heat, work) can be transferred in/out of it (like a person, or a houseplant). Closed systems cannot exchange matter w/ surroundings, only energy (like an uninsulated tupperware - food won't get out, but the heat from the food will). Isolated system...

ZoeHahn1J

Calculating values for work, internal energy and solving reversible expansion problems will not be on the test!

ZoeHahn1J

How much does potential energy (as opposed to kinetic energy) of the atoms' positions in a molecule impact vibrational energy?

ZoeHahn1J

What determines a substance's molar heat capacity? From Cv,m = delta U/delta T, it appears that molar heat capacity would depend on the change in internal energy and temperature of a monatomic gas, but molar heat capacities are independent of temperature, so I'm unsure of what actually determines th...

ZoeHahn1J

Why is heat not a state function? The explanation in the textbook uses both heat and work to show how it is not, but I'm a little unclear on the concept of heat alone not being a state function. Thank you!

ZoeHahn1J

Question 12.19 asks one to find the difference in the pH values of two solutions, one of which the concentration of HCl has been diluted to 12% of its initial value. To do this, we can use change in pH = (.12 x [HCl]o/[HCl]o) =.92. Could we use this same rule for weak acids as well?

ZoeHahn1J

Would anyone be willing to explain how the strength of an acid varies depending on the solvent? Also, how much does the solvent impact the strength of the acid relative to its polarity/electronegativity/number of oxygen atoms?

ZoeHahn1J

Thanks so much!
Also, would someone be willing to explain why across a period, a stronger bond = stronger acid, but down a group, stronger bond = weaker acid? I'm a little confused. Thanks again!

ZoeHahn1J

What are the factors in determining acid strength? How can we identify one acid as being stronger than another using periodic table trends?
Thank you!

ZoeHahn1J

When do we refer to iron as "ferrate" in the coordination compound name, and when do we use "iron"? I've seen it in the textbook both ways but I'm not sure when to do which. Thank you so much!

ZoeHahn1J

If I have a reaction with a reactant or product in an aqueous state, could I calculate Kp or only Kc? If I could calculate Kp, would I just exclude the aqueous reactant/product from the Kp calculation?
Thank you!

ZoeHahn1J

Think state of the reactant/product, but exclude solids and liquids from K calculations.
Hope this was helpful!

ZoeHahn1J

Is there a systematic way to name all coordination compounds? What are the steps? Thanks so much!

ZoeHahn1J

Oooh okay! Thank you so much!

ZoeHahn1J

Are bonded regions more electron dense/is that important to keep in mind?

ZoeHahn1J

I have in my notes that when calculating formal charge, the most electronegative element will be more stable with a negative charge. Why is this so/why not with 0 charge?
Thank you so much!

ZoeHahn1J

Does that apply for all row 3 atoms? That they can all accept more than 8 electrons? If so, can they all accept a total of 10, just like Sulfur? Thanks Any element that has an energy shell of n=3 or higher and is also a nonmetal can have an expanded octet. This is because they have access to the D ...

ZoeHahn1J

I'm not sure, but ClO2 being incomplete might have to do with the fact that Cl only has 7 valence electrons, so the total number of valence electrons is an odd number. Therefore, Cl must have one unpaired electron.

ZoeHahn1J

How common are diagonal relationships between elements in the periodic table? The textbook gives a few examples (metalloids, Li/Mg, Be/Al), but are there many more? Do most elements share chemical properties with the elements diagonal to them, or just certain ones? Do we only see diagonal relationsh...

ZoeHahn1J

As I understand it, elements with low ionization energies tend to be metals, since having low ionization energies allows the electrons to move around much more easily (ie, in a "sea" of electrons), which in turn allows for energy conduction in the element's solid state. Hope this is helpful!

ZoeHahn1J

I was wondering this too! The book says that electrons can spin counterclockwise (denoted by an arrow pointing up) or clockwise (denoted by an arrow pointing down). This article is interesting, too, since it's definitely difficult to think about electrons spinning like a baseball or a planet: https:...

ZoeHahn1J

Right - I like to think of it as photons as having energy. That is, you can calculate the energy of a photon, since photons are discrete units of light. Hope this was helpful!

ZoeHahn1J

When doing calculations with atoms with more than one electron (ie, homework question 1.25, which refers to sodium atoms), are we maintaining that each atom can emit only one photon, like with hydrogen? If so, why couldn't an atom having multiple electrons emit multiple photons?
Thank you!

ZoeHahn1J

The problem could ask you to: - find the percentage yield - find the theoretical yield (in moles or grams; "how much product in moles/grams is expected to form from x amount of reagent?") - find the reagent that is in excess Since all of these will require first finding the limiting reacta...

ZoeHahn1J

Ohhh I see! Thank you!

ZoeHahn1J

Hi! In part b of problem L39, it asks for the name of the oxide with the empirical formula (spoiler alert) SnO2. I'm a little unclear on how to find the proper name for it (tin (IV) oxide in the solutions manual, due to the +4 charge on Sn). Why, however, wouldn't tin dioxide be appropriate? Thank y...

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