CHEMISTRY COMMUNITY

Steven Garcia 1H

In what cases are we allowed to add the cell potentials vs not allowed to?

Steven Garcia 1H

Does anyone know an easy way to remember/understand the electrochemical series?

Steven Garcia 1H

Does the pre-equilibrium condition only work if the early step is fast and the later step is slow? What if there's not a fast step before the slow step?

Steven Garcia 1H

Do we always ignore the reverse rxn of the slow elementary step ?

Steven Garcia 1H

I think the concentration is decreasing at a certain rate (with a certain slope) based on the reaction's k constant over time.

Steven Garcia 1H

The more strongly the rate constant depends on temperature, the more strongly the rate will depend on temperature (since the k constant is in the rate law)

Steven Garcia 1H

Hi! k is just the rate constant, whereas the whole expression represents the rate of the reaction. Another way to think about it is that the problem mentions the rate doubles as a result of doubling the concentration. Rate constants for a given reaction are CONSTANT, they do not change for a given r...

Steven Garcia 1H

When is it advised to use the pseudo-first order rxn as opposed to the second order rxn?

Steven Garcia 1H

For a zero order reaction, the most important thing to remember is that the rate is INDEPENDENT of the concentration of the reactant (provided there is some). Therefore, you can change the concentration all you want, but the rate will still remain constant. This is how you would be able to determine...

Steven Garcia 1H

Intermediates will be the product of one elementary rxn step and the reactant of the following elementary rxn step. For this reason, the intermediate does not appear in the overall chemical equation (we expect it to cancel with itself when adding the elementary rxn steps). However the concentration ...

Steven Garcia 1H

When given a chemical rxn and trying to determine if its spontaneous, are we allowed to ignored the coefficients of the given rxn when calculating the charge difference of the desired elements? and are we allowed to ignore the coefficients of the provided rxn when writing the corresponding half rxns...

Steven Garcia 1H

According to the solution manual, the cell diagram is shown without a solid conductor on the left hand side (anode side). I was curious as to why this is or if this was an error. The elements on the anode side of the cell diagram consist of the liquid Hg(l) and the aqueous solution Hg₂²⁺

Steven Garcia 1H

Hi, I'm a little confused.. how do we know that they are spectator ions? What gives that away?

Steven Garcia 1H

If you use molar convention, the amount in moles (n_r) is simply a pure number. In other words, think of the amount of moles as an integer without the unit of moles. The textbook briefly mentions molar convention.

Steven Garcia 1H

We are expected to know how to derive the Van't Hoff Equation so it is not explicitly on the formula sheet, but the two main equations we need to know in order to derive the Van't Hoff Equation are both on the formula sheet!

Steven Garcia 1H

Usually the element will be attached to one or more other elements forming a compound. The other elements might have a charge (e.g -2 for oxygen, +1 for hydrogen, and -1 for chlorine). Usually the compound with have a total indicated charge in the upper right hand corner of the compound. If the comp...

Steven Garcia 1H

I dislike these terms because of the confusion that they bring. The simplest way to think about it is that reduced species are oxidizing agents and oxidized species are reducing agents. Reduced species gain an electron, by pulling an electron from a specie, therefore the reduced species act as the A...

Steven Garcia 1H

Spontaneous usually means that delta G is negative, so you would expect the rxn in the forward direction to be favored (aka stable). An unfavorable rxn will have a positive delta G and therefore be unstable (products of rxn are unstable)

Steven Garcia 1H

Anything on Gibbs free energy or electrochemistry should be acceptable considering we officially finished thermodynamics today (2/20) in lecture.

Steven Garcia 1H

C(s, graphite) is the most stable element of carbon so that's why you would use C(s, graphite) as opposed to C(s, diamond). This can be confirmed if you look at the appendix of the textbook. The enthalpy of formation of C(s, graphite) is 0, which is what we expect for elements in their most stable f...

Steven Garcia 1H

For problems such as these, you will have to use the equation, standard change in Gibbs free energy= standard enthalpy of formation (of given compound) - T* standard change in entropy of rxn. First, set up a chemical equation that produces the given compound as the product. Make sure the elements (r...

Steven Garcia 1H

I also noticed this in the solution manual. I believe its a mistake. It should be written the way you have it. Since the volume is constant for that part of the problem, we would use the Cv value of 3/2*R in that equation

Steven Garcia 1H

Does constant volume always imply that w=0 ?

Steven Garcia 1H

Typically the product of combustion has water in its liquid state. I would stick with that as the norm unless the problem specifically states that the water produced should be in gas form.

Steven Garcia 1H

Hi! I believe it has to do with the fact that the system of a reversible process is continually adjusting in response to the external pressure. If the external pressure increases or decreases even a very small amount it will cause the piston to move in or out respectively at the expense of work. Whe...

Steven Garcia 1H

I don’t think we need to know the actual mechanics and details about the calorimeter. However, it’s important to know that the calorimeter essentially measures the change in temp of the solution/mixture inside the bomb calorimeter. The bomb calorimeter allows us to determine the q (heat transfer) of...

Steven Garcia 1H

Well depending on the system, energy could still be transferred in the terms of heat and work. Even though the pressure is constant, there’s still an opposing force which work may be required to act against. In regards to heat, heat transfer is usually always present with a change in temp. except in...

Steven Garcia 1H

Hi! Yes I think you’re thinking about it in the correct way. The system and surrounding concept is based on whatever the object of interest is. In the bomb calorimeter case, the solution or mixture within the calorimeter would be the system while everything outside the calorimeter would be considere...

Steven Garcia 1H

If I interpreted correctly, I believe that the standard enthalpy of formation is just a special case of the standard reaction enthalpy. The standard reaction enthalpy refers to the amount of heat released or taken up by a rxn when all the elements (reactants and products) are in their standard state...

Steven Garcia 1H

Are we suppose to round the x-value (change in concentration/pressure) in the ICE chart ? I noticed that in some of the textbook example, it rounds the x value and in other examples it doesn't round the x value. I was told that you don't and that you save the rounding for the very end (which would b...

Steven Garcia 1H

All the extra heat is being used up for the breaking of the bonds in order to cause the phase change. There is no heat contributing to the temperature during a phase change because all the heat is used up.

Steven Garcia 1H

Thanks! This helps a lot. Follow up question...how do you determine how exactly the weak compounds will dissociate? ex. How would you know that NH2NH2 becomes NH2NH3+? Thanks again!!! Hi! All derivatives of ammonia are weak bases so NH2NH2 is a weak base. Since it's a base, albeit weak, we would ex...

Steven Garcia 1H

If the compound has an H+ at its beginning or ends with a -COOH, then you can assume it is an acid. For bases, the compounds will usually contain a hydroxide ion, oxide ion, OR exist as some kind of derivative of ammonia (NH3). Strong acids include HCl, HI, HBr, HClO4-, HNO3, and H2SO4. You can esse...

Steven Garcia 1H

In this particular example, the system is being compressed, so the system should favor whatever side has the least amount of moles. Yet, the answer shows that neither side is favored. Why is that considering there are 6 moles on the left and 7 moles on the right of the reaction?

Steven Garcia 1H

Do you mean based on the quotient value? You can compare Q to R in order to determine which direction of a reaction is favorable. If Q<K, then this implies that there are less products and more reactants relative to K (think of the expression for calculating K, [products]/[reactants]). Therefore, mo...

Steven Garcia 1H

Yes, molar values and/or concentrations can go in the chart. However, whatever you decide to use, you should keep it consistent throughout the ICE chart. In addition, if using molar values it is common to convert the molar values into concentrations at the end to solve for the equilibrium constant.

Steven Garcia 1H

I believe it's because the availability of solids and liquids remains constant whereas the gases don't. The solids and liquids aren't as affected by volume like gases are.

Steven Garcia 1H

I noticed in a lot of the textbook example R was equal to 8.314 x 10^-2 L^- mol^- bar^- K^-

It has different variations in which it can be written depending on the desired units you need to cancel. I'm confident that the constant value will be provided on any exams.

Steven Garcia 1H

Yes! When the equilibrium is reached, the composition of the reaction mixture stays the same. Changes in the molarity and/or pressure occur before equilibrium is reached. However, at equilibrium, the composition stays the same precisely for the reason you mentioned. Hope this helps!

Steven Garcia 1H

Pumping inert gases should be the only way. The inert gases will not participate in any reaction with the gases present. If it were any other gases aside from the inert gases then we would expect the increasing pressure to favor the reaction towards either the reactants or products. Hope this helps!

Steven Garcia 1H

Will we be asked to compare the strengths of bases? (or will we be focusing on the strengths of acids)
If so, will we be provided with either a Kb or Ka value?

Steven Garcia 1H

I understand that the CO bond is a triple bond, but I was curious why the Lewis structure can't be written as a double bond? I was able to write it as a double bond with no charges on either atom. Nonetheless, the CO triple bond lewis structure is preferred although it results in charges of +1 and -...

Steven Garcia 1H

The molecule CH2Cl2 is said to be nonpolar, which I understand in the sense that the four atoms attached to the central atom are not the same. However, what if the Cls are located opposite of each other... wouldn't their dipole moments cancel ?

Steven Garcia 1H

If the complex has a negative charge, we are suppose to add -ate to the end of the metal name. However, the textbook says that we add -ate to the end of the Latin stem of the metal (if metal originates from a Latin name). As a result, should we familiarize ourselves with the Latin stems?

Steven Garcia 1H

For figuring out denticity, you would probably need both the Lewis structure as well as the VSEPR shape. The Lewis stricture will allow you to know how many donor groups (groups or atoms with lone pairs) are in the ligand. The shape will allow you to see if the two donor groups in the ligand are clo...

Steven Garcia 1H

Kw is simply a constant that helps when calculating the pH. Kw has a value of 1.0x10^-14. You can also express Kw in the equation Kw=Ka*kb. This equation just goes to show that the product of the concentration of H3O+ ions (Ka) and the concentration of OH- ions (Kb) should always equal in 1.0x10^-14...

Steven Garcia 1H

Oxoacids, in particular, more readily lose H+ IF the resulting anion (the anion that results when H+ is released) is stabilized by electron withdrawing atoms. These electron withdrawing atoms, a.k.a atoms with greater electronegativities, delocalize and stabilize the negative charge of the anion. Ju...

Steven Garcia 1H

CHI3 has a higher melting point than CHF3 for reasons of polarizability (larger atom will increase London interactions), which I understand, but doesn't CHF3 have a larger dipole moment (thus stronger dipole-dipole interactions) than the CHI3 molecule because F is more electronegative than I? Does t...

Steven Garcia 1H

Why do the lone pairs of a molecule with let's say a VSEPR formula AX4E2 take the positions of the axial atoms, whereas the lone pairs of a molecule with a VSEPR formula AX3E2 take the positions of the equatorial atoms?

Steven Garcia 1H

Why is the ionization energy of oxygen lower than that of nitrogen ?

Steven Garcia 1H

Is there a way to know whether an H will be attached to an oxygen in a molecule or attached to the central atom? I noticed in some cases the H is attached to an oxygen in the molecule and other times the H is simply attached to the central atom.

Steven Garcia 1H

For this problem, wouldn't the SO3 molecule have a shorter bond length than SO2? The solution manual shows that they are approximately the same. Both the molecules only have double bonds but the SO2 molecule has a lone pair on its central S atom. Wouldn't the lone pair weaken the S-O bond in SO2 (th...

Steven Garcia 1H

The textbook mentions that "bonds between heavy atoms tend to be larger than those between light atoms because heavy atoms have larger radii than lighter ones". However, I'm a little confused with this idea. For example, if you go across a period, the atomic mass of the elements increase. ...

Steven Garcia 1H

If a molecule has a lone pair, can we predict that it will be a polar molecule? Or is this not always the case?

Steven Garcia 1H

The major significance is that an equatorial lone pair will repulse the other equatorial atoms more strongly. Therefore the angle between the atoms should be smaller than what is normally expected if there were just equatorial atoms (and no lone pairs). This is due to the idea that lone pair-atom re...

Steven Garcia 1H

Hi!

hybridization state of the central atom is the same thing as the hybrid orbitals of the central atom! Just a different way to phrase it I believe.

Steven Garcia 1H

If a molecule has more than four regions of electron density, how would one describe the hybridization of the orbitals (in other words if the electrons of the molecule occupy more than four orbitals) ?

Steven Garcia 1H

Does electron promotion and orbital hybridization go hand in hand? Or can any atom w/ s and p orbitals undergo hybridization ?

Steven Garcia 1H

Why is it acceptable for electron promotion to break the Pauli Exclusion Principle ?

Steven Garcia 1H

If a molecule has a single central atom and three bonds for three other atoms (therefore three regions of high electron concentration), how do we determine whether this molecule has a trigonal planar-, trigonal pyramidal-, or t- shape?

Steven Garcia 1H

Just to clarify, we can assume the bond length will more than likely always be inversely proportional to the bond strength ? (increased bond length--> reduced bond strength)

Steven Garcia 1H

The equation for the potential energy of London interactions is -(alpha1*alpha2)/r^6. The concept that the potential energy increases with increasing polarizability of the molecules involved makes sense to me. However, for highly polarizable molecules, we would expect for them to have a larger radiu...

Steven Garcia 1H

Why isn't the Lewis Structure for ClO2 shown with its lowest formal charge? The formal charges shown in the manual can be further reduced to all 0.

Steven Garcia 1H

Oh thank you, I understand now!

As a follow up, was the initial configuration of Ag 4d^10 5s^1 because it used the other s orbital electron to fill up the d orbital ?

Steven Garcia 1H

Hi! If a Lewis structure for an atom can be written two different ways having the same formal charge, is it ideal to use the Lewis structure that let's say has a triple bond compared to another Lewis Structure of the same atom (with the same formal charge) that has a double bond instead? An example ...

Steven Garcia 1H

Hi!

I'm confused on when to separate the Lewis structures of polyatomic species? I noticed for exercise 39 in Ch.3, some of the Lewis structures of the atoms/ compounds are separated from others? When do we NOT separate them and when do we ?

Hope this isn't too confusing. Thanks!

Steven Garcia 1H

Hi! Why is the electron configuration for Ag+ [Kr] 4d^10 I originally thought it would be 4d^8 5s^2. I thought adding the s orbital electrons to the d orbitals (in attempt to fill the d- orbitals) only applies for certain cases such as Cu and Cr? Would this example be considered an exception as well...

Steven Garcia 1H

How is the valence electron configuration for Group 5 in the periodic table (n-1)d^5 ns^2 when none of the elements in group 5 have this configuration in the periodic table? I originally thought the answer would be (n-1)d^3 ns^2, which I think would make more sense when comparing to the given config...

Steven Garcia 1H

I'm curious as to why barium has a greater ionization energy than cesium? I figured that the two valence electrons in barium, compared to the one in cesium, would provide a greater electron-electron repulsion; therefore making it easier to remove an electron from barium.

Steven Garcia 1H

If two or more elements are isoelectronic (that is they have the same electron configuration), how do you determine which element has the smallest or largest ionic radius ?

Steven Garcia 1H

I understand that the 4s- orbitals get filled before the 3d- orbitals because the 4s- orbitals have lower energy levels. However, I'm confused as to why some of the elements in the periodic tables (specifically the elements in period 5 of the d-block) have s^1 instead of s^2? Don't the 4^s-orbitals ...

Steven Garcia 1H

Hi Shirley! If you're referring to the configuration, then there is a convention. It goes 1s^2, 2s^2, 2p^6, 3s^2, 3p^6. After this, it gets a little confusing. The 4s- orbitals actually come before the 3d- orbitals because 4s- orbitals have a slightly lower energy. We prefer lower energies first bec...

Steven Garcia 1H

Hi Ricky! You're partially correct in your thinking. The s-electrons are very close to the nucleus and this plays a role in the shielding that occurs. S-electrons penetrate through the inner shells of the atom. Because p-electrons, d-electrons and so on don't penetrate these inner shells like s- ele...

Steven Garcia 1H

If something has an extremely small wavelength such as 6.94x10^-38m, would we say that it has wavelike properties ? Why or why not?

Steven Garcia 1H

A bound electron can be an electron in the energy state n=1, the grounded state. A bound electron is still relatively close to the nucleus. Whereas an energy state of n=4 is farther away from the nucleus. The farther away from a nucleus an electron is, the higher energy state it's at.

Steven Garcia 1H

The change in energy results from an electron going from one energy level (for instance, a higher energy level farther from the nucleus) to a different energy level (for instance, a lower energy level closer to the nucleus). As a change like this where an electron goes from a higher to lower energy ...

Steven Garcia 1H

I noticed that the answer in the solutions manual for problem L.39 part b shows tin (IV) oxide. I was curious what does the roman numeral IV represent? What's the importance of including it?

Steven Garcia 1H

After I changed the mass percentages of each element into masses, I then converted these masses into moles by dividing the masses of each element by their corresponding molar masses. In the end, I ended up with a ratio of 2.95:1:5.91 for Na:Al:F. Is it safe to round these ratio values to whole numbe...

Steven Garcia 1H

For the question being asked, I think it's safer to use the actual ratio of atoms present. The question states "in what atom ratios are the atoms present in VANILLIN" so I would assume that you would write the actual number of atoms present in that compound. In this particular case, the em...

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