CHEMISTRY COMMUNITY

gabbi_r2C

there is a chart of bond enthalpies that we often reference when dealing with homework/textbook problems, but yes you would have to be given the bond enthalpy in order to calculate delta H.

gabbi_r2C

I think what they said above is right, we don't really have to worry about it. But to my understanding, the steady state method assumes that the concentration of an intermediate in the rate limiting step is constant and uses the fact that the derivative of a constant (or the rate of change of the co...

gabbi_r2C

everyone else pretty much said it, a galvanic cell consists of two different species in the cathode and anode (like Ag and Zn) while a concentration cell is the same species across (ie both Ag), and is driven by a concentration gradient rather than oxidation/reduction.

gabbi_r2C

I think it has mostly to do with how the catalyst changes the path of the reaction, but delta H is a state function and doesn't depend on the path taken. if you look at some of the graphs we did in lecture, its only the activation energy that is lowered with a catalyst, not the actual energy of the ...

gabbi_r2C

you typically look at elements that pretty much have constant oxidation numbers (like O being 2-, the only exception being when in H2O2, and H being +1) and then calculate the oxidation numbers of other elements in the compound based on what charge of the overall compound. ie, with Mn2O7, the charge...

gabbi_r2C

If you're given the order of the reactants, then the overall order of the reaction is the sum of those reactants' orders. ie, if [NO] is 2 and [H2] is 1, then the overall order is 3.

gabbi_r2C

I remember doing a problem like this with my TA and that was how we solved it. I think it is because you can think of the heating happening in one step and the expansion happening in another, since enthalpy is one of those propertied that depends on net change and not the path taken.

gabbi_r2C

I agree with everyone above. It is something that can be easily derived, but memorizing it would take much less time on the test, so after practice you should have it memorized.

gabbi_r2C

For question 9: generally speaking, rates are never negative. They're just the time it takes for a reaction to proceed, so having a negative rate would indicate negative time, which isn't possible--even when talking about the reverse reaction. For question 1: this is asking much less of you thank yo...

gabbi_r2C

I think you should know how to solve for the oxidation states in a given compound. If no compound or otherwise telling information is provided, then the oxidation number itself (if not H or O, since those are common) should be given.

gabbi_r2C

So this question is tricky because of how wordy and vague it is. But given that you have solid metal Al and gaseous O2, and knowing that Al only forms stable cations (not anions), Al must be oxidized. That leaves O2 to be reduced, and matches with how we know O forms stable compounds with 2- oxidati...

gabbi_r2C

I'm pretty sure Dr. Lavelle was referring to how delta Hº means the standard reaction enthalpy, and if you're not given any other information but standard values, you assume the reaction occurs under standard conditions. I hope this helps!

gabbi_r2C

Varsha was really detailed in her explanation, and the only thing I would add is to basically look at the slow reaction first and write the rate law that would accompany it (based on reactants and molar coefficients). If the slow reaction doesn't match the overall reaction and includes an intermedia...

gabbi_r2C

I struggled with this one too, wondering how you could have an "oxidation-reduction" reaction without the reduction, but that explanation helps a lot. So thank you!

gabbi_r2C

One way to think of it is if the slow step does not match what you would expect the rate be for the overall reaction, then you have to re-write it; the only way you can do this is by replacing the intermediate with an expression derived from the equilibrium expression.

gabbi_r2C

since H+-->H+, why don't you cancel them both out? since they appear in both reactants and products

gabbi_r2C

There won't always be intermediates in a reaction; there can sometimes be reactions completed in one step, which can be referred to as elementary reactions. You'll know what is an intermediate species because it later gets consumed in the reaction mechanism; ie, you will need a proposed reaction mec...

gabbi_r2C

I agree with Emma. The main difference between a concentration cell and galvanic cell is that a concentration cell has the same type of electrode/ion in both solutions, while a galvanic cell has different electrodes/ions in each solution. A concentration cell is powered by a concentration gradient d...

gabbi_r2C

I also was struggling with this one a lot, and turns out I just had messed up the order for the right side by accident, so your explanation helped me out a bunch!

gabbi_r2C

I agree with what everyone said above. First write the oxidation numbers of every atom to determine what is being reduced/oxidized, then separate the reactants into the following half reactions. It may seem a little confusing because of how the problem combines the products into a compound, but I th...

gabbi_r2C

Thank you, Can!! That helped SO much! I was wondering though, do we always assume the H2O we add is in liquid phase and the OH we add is in an aqueous phase? I can answer this: yes, we can typically assume H2O is in the liquid phase (rather than gas) because unless otherwise specified, we can assum...

gabbi_r2C

I've tried switching up the order of the elements on either side and get different, unhelpful error messages each time, so if someone did this question right, let me know.

gabbi_r2C

I keep getting similar wrong answers and error messages... I have the anode on the left side, so I'm not sure what the problem is.

gabbi_r2C

I was really struggling with this problem for the longest time, and that solution was really helpful. Can someone explain why you can add any amount of e- necessary to balance the charge, and why that can be greater than the difference in oxidation numbers between the Cl of Cl2O7 and ClO2-?

gabbi_r2C

I think Jonathan hit this one on the head in terms of an explanation. I always remember it as the anode gives off electrons, since I've associated anions with negativity, so in my head I track the movement of the negative charge. So to me, cathode is just the opposite of the anode. I hope maybe thin...

gabbi_r2C

I agree with Michael and broke down the problem into the same three steps. The reasoning behind this method is how no significant vaporization occurs below the boiling point, so you have to heat the sample to the boiling point first. This change causes an increase in entropy, so that's why you calcu...

gabbi_r2C

This one is sort of tricky; you have to consider what specifically they're asking. You don't need to calculate delta S for the liquids, and can use conceptual understanding instead: The question asked about the LIQUID samples of the GASES for which you solved for the standard entropy change of vapor...

gabbi_r2C

I think Faaizah nailed the explanation of this one. I used the same method to conclude the same answer; because no change in the system is necessary for the absorption of H2O in the SiO2, you can conclude the reaction must be spontaneous, therefore delta G is negative. In order for water vapor to be...

gabbi_r2C

I think irreversible expansions can be isothermal technically, so long as the surroundings contain enough heat to replace the heat lost from work for the expansion and the change in heat for the surroundings is not be very significant. But the very nature of an irreversible expansion does require a ...

gabbi_r2C

you'll know it's irreversible if the problem hints at things like "quick expansion" or if it says that a piston that was exerting pressure on the system is removed, or something along those lines. I think otherwise if nothing explicit was stated, we assume it's a reversible process, b/c th...

gabbi_r2C

This question should give you a table of values for Cv,m; you're given NO2, so if you draw out the VSPER model of NO2, you see that it has a bent shape, so use the value given for nonlinear molecules and plug this into q=mCdeltaT.

gabbi_r2C

I agree with everyone above. You don't need ICE tables, but that's how we organize all of the information into the formula we use to solve a question with a weak acid/base, since they don't dissociate completely. You can't directly calculate the pH given a concentration of a weak acid; you need to u...

gabbi_r2C

The first thing I see is you're using the wrong value for R: you use 8.314 J/(K*mol), when in the ideal gas law you're supposed to use 8.206*10^-2 L*atm/(mol*K). You can check this by writing out your units: they don't cancel. You're on the right track with solving for moles then plugging it into wo...

gabbi_r2C

You can use the equation we were given in class (S=KblnW) to derive a relationship: when W (degeneracy) increases, the value of the lnW also increases, and because Kb is a constant, that means S (entropy) also increases. The opposite is true for a decrease in W, making S decrease as well. So there i...

gabbi_r2C

We use the first equation (w=-PdeltaV) when pressure is constant; you can derive this from the integral equation Dr. Lavelle gave us in lecture. We use the second equation (w=-nRTln(Vf/Vi)) when pressure is not constant, as you cannot pull out P from the integral equation nor can you solve the expre...

gabbi_r2C

You can sort of draw a relationship between endo/exothermic behavior and bond enthalpies, but not necessarily bond strength; in terms of bond enthalpies, you can use the equation we learned (bonds broken - bonds formed), and if the bond enthalpies of the bonds broken add up to a number greater than ...

gabbi_r2C

I agree with above, and think that the external pressure is the pressure applied to the piston. External pressure doesn't necessarily need to be applied by a piston, like external pressure could be atmospheric pressure. It really just needs to be acting on the system.

gabbi_r2C

the above explanation is really accurate and concise. the only thing i would add is how irreversible systems use the w=-Pext(deltaV) while reversible systems use the -(integral of)PdV, which we will learn how to use later i think.

gabbi_r2C

i agree with everyone above. we will typically be given things like enthalpy values, bond enthalpies, heat of formation, etc for tests/midterms

gabbi_r2C

if work is negative, it means it is work done by the system (ie expansion); if work is positive, it is work done on the system (ie heating or compression).

gabbi_r2C

entropy is the measure of disorder in a system, known as delta S, like everyone above said. Things that increase disorder increase the number of moles, increase pressure, change phases (from liquid to gas and solid to liquid) etc.

gabbi_r2C

We so far have not learned how to use the integral and serves as just a conceptual understanding. I think we will use this equation to derive another, easier to use equation ( -nRTln(Vf/Vi) ) that we will use in calculations

gabbi_r2C

to find delta U, you have to use the eq nxC(v)xdeltaT; this is because you're given moles and temp and can solve for specific heat at constant volume aka C(v) using C(v)=C(p)-R. you use this because you're now focusing on the internal system, so the volume no longer changes.

gabbi_r2C

hi there, you can assume because of the fact that the system is open that the pressure is the same before and after because whatever gas is released from the reaction has an insignificant impact on the pressure of the universe. like with the ocean example, pouring a bottle of water in the ocean will...

gabbi_r2C

this can tell us whether a system is at constant volume (ie with a closed or isolated system) or when it is at constant pressure (ie with an isolated or open system), which can help in solving for internal energy and work of the system

gabbi_r2C

generally speaking, if the equilibrium constant is below 10^-3 or above 10^3, you can assume x is insignificant. to be safe, you can use 10^-4 and 10^4 instead.

gabbi_r2C

enthalpy is typically thought of as heat aka delta H, while entropy refers to the "disorder" of the system, aka delta S; we'll likely learn more about this later, but an increase in things like moles of gas or temperature increases disorder and thereby increases entropy.

gabbi_r2C

i agree with what has already been said, and that flat section of bond breaking is really just intermolecular forces (such as H-bonds for water) breaking, and the reason this part of the curve is flat is because of a change in potential energy, not kinetic (ie temperature)

gabbi_r2C

so the steam may not necessarily be at a higher temperature, but the fact that it has to go through a phase change (releasing potential energy that is converted to heat on your skin) in addition to the release of kinetic energy/heat (which is the only energy release of water) means that it releases ...

gabbi_r2C

yes water condensing on a surface is an exothermic reaction because the water vapor needs to release potential energy in order to form a liquid.

gabbi_r2C

a phase change is a change between solids, liquids, and gases; a solid-->gas (sublimation), gas-->liquid (condensation), etc. its a change in potential energy only, not temperature

gabbi_r2C

So Dr. Lavelle uses the derivative to describe the infinitely small changes in volume, hence the use of dV in the integral. The integral is used because of the infinite sum of these volume changes. Graphically, I think you can think of the derivative as how the slope of the tangent line at every poi...

gabbi_r2C

So because A is a solid, it does not exhibit any pressure on the system, so we don't count it when using the short-cut method of determining work. The moles that matter here is then 1 mol--->2mol, so the increase in pressure indicates that the system must do work on the surroundings.

gabbi_r2C

I agree with everyone before; if work is done by the system, then the system uses (ie loses) energy, so the value for work is negative--dont forget that energy is always in terms of the system, not the surroundings. Thus, work done on the system (by something else) is the opposite, and is positive b...

gabbi_r2C

Hi there! So Q and K are different because K is the reaction constant at equilibrium, and Q is the ratio of products to reactants at any point during the reaction (not usually at equilibrium; if it is, then Q will be equal to K). K can be measured at any temperature, but will change as temperature c...

gabbi_r2C

I agree with the previous responses. As a gas, H2O is not a solvent and thus has a measurable concentration change, so it must be included in the ICE table.

gabbi_r2C

I think this has to do with morphine being a very large molecule that makes it, as a base, fairly stable before receiving an H+ atom (having ot do with many electronegative O atoms distributing charge), so much so that it is more stable than its conjugate acid, while the opposite is true for BrO-.

gabbi_r2C

This is because tetrahedral compounds have atoms in different planes, so they will not add up to 360 or 180 degrees; the bond angles are 109.5 based on an experiment, but this is only because of the orientation of the atoms being 3D.

gabbi_r2C

I think in regards to HSO4- specifically, that it is not an amphoteric compound because its conjugate acid (if it were to accept a proton) is a strong acid, and thus will dissociate again; so HSO4-, if it does briefly accept a proton, will just dissociate again into HSO4- because of the strong acid ...

gabbi_r2C

So Bronsted bases are molecules that accept protons (aka H+ atoms), while Lewis bases are molecules that donate electron pairs. The real difference is how Bronsted acids/bases focus more on the movement of H+, while Lewis acids/bases focus on the movement of electron pairs. A molecule that qualifies...

gabbi_r2C

So electron diffraction is a phenomenon observed when a beam of electrons passes through a very small opening, and as a result of constructive/destructive interference (a property of waves only), a pattern of electron density is generated that shows high/low presence in unexpected places. ie, becaus...

gabbi_r2C

I agree with the responses above; acid rain is just a biological/environmental example of CO2 interaction with water in the real world, as this is an "applied chemistry" course of sorts, so the main purpose was just to demonstrate the usefulness of understanding acid/base interactions and ...

gabbi_r2C

Hi there! There are some simple things that he mentioned in the first lecture on acids/bases (so you can look back at that for reference; there was a pH chart listing various solutions) but some things that will be easy to remember are sodas being acidic, juices being acidic, soap being basic, water...

gabbi_r2C

Agreed, you should double the radius to get the diamteter to better represent the area/space the electron can travel around the nucleus.

gabbi_r2C

I agree with the above responses about which energy levels correspond with which type of EMR, but technically Lymann series is any drop (as far as 14A will have to deal with, that is) that goes down to n=1, Balmer is any drop down to n=2, and Paschen is any down to n=3.

gabbi_r2C

I agree with the answer above in regards to the original question. pOH is basically pH (ie -log[H+]) but instead of focusing on H+ concentration, we focus on OH- concentration. So pH and pOH are complimentary in that way, and always add up to 14.

gabbi_r2C

I agree with the response above. Potassium is essentially a spectator ion and can be left out of the equation because it does not affect the reaction at all. Fluorine interacts with water to form HF and OH-, and thereby raises the pH of the solution.

gabbi_r2C

To my knowledge, bent and angular are the same thing. And as far as the 2 vs 4 electron groups go for naming, there isn’t a discrimination. But for approximating bond angle, you have to take the groups of electron density into account.

gabbi_r2C

Like everyone else said, that is exactly right :) the stronger the acid, the weaker the covalent bond between H+ and A- (for HA acid).

gabbi_r2C

I typically just write out the Lewis structures and find formal charge. You can usually do formal charge (and sometimes even Lewis structures) in your head though, after enough practice :)

gabbi_r2C

I agree with the responses above. There isn’t a relationship between amplitude and frequency, but there is an inverse relationship between frequency and wavelength.

gabbi_r2C

I think it gives the number of nodes: l=0 is s (sphere shaped has no nodes), l=1 is p (peanut shape has one node), etc

gabbi_r2C

So when an acid is 100% ionized, it dissociates completely, and the moles of acid in the reactants directly relate to the moles of product form. ie, if 2 moles of HCl react, 2 moles of H+ and 2 moles of Cl- are generated. Only strong acids do this. Partial dissociation happens with weak acids, where...

gabbi_r2C

Typically it’ll be on the atom where it completes the octet. Atoms that commonly have radicals are N and C.

gabbi_r2C

The most basic way to determine hybridization is to look at the number of atoms bounded to the atom in question. 1 atom is just s, 2 is sp, 3 is sp^2, 4 is sp^3, etc. Hope this helps!

gabbi_r2C

Even in H2O you may see the lone pairs placed opposite each other, but the reason placing them next to each other is most accurate is because of the tetrahedral geometry the 4 regions of electron density in H2O mimic. Remember, lewis structures are 2D models of a 3D object, so what is drawn may not ...

gabbi_r2C

Because there are more lone pairs, which have a greater repulsion strength than bonding electrons, the lone pairs in CH2(2-) drive the bonding pair electrons farther away from the lone pair itself; this then means that the bond angle (between the two bonding electron pairs) is actually smaller in CH...

gabbi_r2C

Coplanar means that the atoms are in the same plane. ie, linear shape atoms are coplanar, while only two atoms of a trigonal planar atoms are coplanar.

gabbi_r2C

If all four atoms have lone pairs, then no. The atoms will distribute evenly. If only one atom does then yes, and the bond angle between non-lone pair atoms will be smaller than 109.5, and the bond angle between the lone paired atom and non-lone pair atoms is larger than 109.5. A similar concept app...

gabbi_r2C

I think the most we need to understand about the Wave Function is conceptual; ie how it was used to generate a mathematical model (not actual) for the probability of electron density, and (because its 3D) its outputs generated the n, l, and ml quantum numbers. It thus can be synonymous with what we ...

gabbi_r2C

Yes, bent is the official shape name, no matter whether there are two lone pairs on the central atom or one.

gabbi_r2C

Molarity (moles of solute over liters of solution) is specifically referring to aqueous solutions, ie the ratio of solutes in a solvent, so no, it is never applied to solids or gases.

gabbi_r2C

No, I don't think so. I remember Dr. Lavelle saying as much a few times during lecture, but you should know their relative bond angles (ie trigonal planar has a larger bond angle than tetrahedral) and how adding lone pairs would affect those relative bond angles.

gabbi_r2C

I agree with everyone above. Lone pairs are their own regions of electron density :)

gabbi_r2C

I agree with the above. A ligand is basically just a Lewis base (electron donor), but specifically for transition metals.

gabbi_r2C

It sorta just depends on the number of atoms bonded around the central atom, and understanding that certain molecular shapes (ie tetrahedral, trigonal planar, and octahedral) have pairs of atoms in two different planes. As for which atoms are in those planes, it doesn't always matter, hense structur...

gabbi_r2C

I think the main point is to understand how VSEPR affects bond shape, like how having a lone pair on a central atom will affect the bond angle and where to put atoms and lone pairs around bonding atoms, but I don't think we have to memorize things like bond angles. Just know enough to be able to fig...

gabbi_r2C

I don't think it matters where you put d. I remember Dr. Lavelle saying in class you could do either way :)

gabbi_r2C

I think you're referring to isomers, and typically you have to rely on the way they give you the formula to know which structure, or other context clues they may be present in the question.

gabbi_r2C

Basically, a molecule made up of polar bonds will only be nonpolar if the dipole moments of each polar bond cancel, meaning they are of equal magnitude and opposite direction.

gabbi_r2C

Yeah, to be honest I don't think it matters. I remember Dr. Lavelle saying in class you could write it either way :)

gabbi_r2C

*lone pairs on the central atom. Sorry if that was confusing!

gabbi_r2C

Hi there! Typically a bent molecular shape is only established because of the presence of lone pairs, which creates an uneven distribution of electrons and makes the molecule polar. I think you can consider this also as the lone pairs do not have opposing dipole moments, and thus create a polar mole...

gabbi_r2C

I agree with the above two answers, but would add that the exceptions of Cu and Cr apply to all elements in their group as well.

gabbi_r2C

To be honest, I think that might just be a mistake in the textbook. The only other explanation I can think of would be that because 4d is in energy level n=4 and is full, it's paired electrons aren't considered valence because they aren't reactive, but even then that's not entirely true because we k...

gabbi_r2C

I agree with everyone above. Lewis acids accept a pair of electrons and lewis bases donate a pair of electrons.

gabbi_r2C

Hey John, Generally, an atom has high polarizability if it has a large atomic radius and many electrons, so based on the trends of atomic radius, polarizability increases down a group and decreases down a period. Atoms that have high polarizing power tend to be smaller and positive, so based on ioni...

gabbi_r2C

Hey Sofia! So "lone-bonding pair" and "bonding-bonding pair" refers to the type of repulsion taking place; for a lone-bonding pair, the repulsion is between a lone pair and a pair of electrons in a bond. A good example of this is NH3, where the lone pair on the nitrogen atom push...

gabbi_r2C

Yes you can multiply by 10^-2, as it is the same thing as dividing by 100 like you would in dimensional analysis; if you're not sure, this method is actually a very easy way to keep track of how units cancel out in a calculation, and to ensure you end up with the proper units at the end.

gabbi_r2C

Hi there! Seems like everyone is sort of in agreement about meters being used, and this is because you need to use meters to cancel or match with the units in the constant of the calculation, otherwise your numbers will be off.

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