CHEMISTRY COMMUNITY

Angela Grant 1D

to add to your question, how can you tell which substance is a catalyst versus an intermediate when given a series of steps?

Angela Grant 1D

reactants are "-x" and products "+x" since the reactant is usually forming product. also you're usually given some reactant concentration and no product, so if you were to put "-x" under products you would get a negative concentration. hope this helps

Angela Grant 1D

you don't include H2O since it's a solvent

Angela Grant 1D

you put coefficients if it's in the equation. so if you have 2H+ as a product, you'd write "-2x." hope this helps

Angela Grant 1D

to add to your question, when do we use Cp vs Cv?

Angela Grant 1D

i'm not 100% sure, but i think sig figs still count. but i was told that for the midterm they wouldn't count for thermo questions. don't know if that still applies

Angela Grant 1D

I found this answer on an older, similar post:

"Breaking a diatomic molecule X2(g) requires energy to become 2X(g), and therefore the reaction is endothermic. This is true for other halogens as well"

Angela Grant 1D

H+ is aqueous so it's an active part of the reaction and is in the cell diagram but H2O is just a solvent so it doesn't get added into the cell diagram

Angela Grant 1D

^to add on, you have to subtract the new [A] from the initial [A]. so when you use the molar ratio to get [A] when [B] = .034 mol/L, you should get [A] = .068 mol/L. then you have to subtract from the initial: .153 - .068 = .085 mol/L. this is the [A] you should use for your computation

Angela Grant 1D

changing concentration wouldn't affect the reaction order. rate doubles because in a first order reaction, rate = k[A]. in a second order reaction, rate = k[A]^2, so if concentration were doubled the rate would increase exponentially, not double

Angela Grant 1D

rate of a reaction depends on the rate constant (k) and some other variable (we've been working with concentration)

Angela Grant 1D

we'd be given which steps are fast or slow, so we'd just have to know that the slow step is the rate-determining step

Angela Grant 1D

homogeneous reactions have reactants and products that are all in the same phase (gas, liquid, solid, aqueous). heterogeneous reactions include more than one phase in the reaction

Angela Grant 1D

for #4 on gibbs free energy, does it matter that the standard delta G is given in kJ instead of kJ/mol? i thought we'd have to divide by 2 to get the delta G per mole of NH3, but that didn't give me the right answer so i'm a little confused because plugging in the given value does give me the right ...

Angela Grant 1D

On #1 of the Gibbs to find the standard enthalpy and entropy of formation you have to do the sum of products minus the sum of the reactants using the numbers given right? For some reason I am not getting the same answer. the values are given per mole, are you multiplying them by the respective mole...

Angela Grant 1D

Madison Hurst wrote:In the Gibbs WS #4 how do you find T? or can you just assume its at 25 C?

yeah you just assume 25 C, so 298 K

Angela Grant 1D

it's similar to delta H / delta H^o or delta S / delta S^o, the ^o always refers to a standard state

Angela Grant 1D

^to add on, within each half-reaction in a cell diagram, you'd typically write in the order of what's being reduced/oxidized. so if Ni2+ is reduced to Ni, the cell diagram for this cathode would be Ni2+(aq)|Ni(s)

Angela Grant 1D

SydBenedict2H wrote:Will he ask us any acid base questions, or is he only asking Gibbs and electrochemistry?

acid base was covered in the midterm so it'll just be what's posted on his website

Angela Grant 1D

^to add, if the half-reactions have different numbers of electrons, you just multiply the reactions by whatever factors will make the electrons cancel out

Angela Grant 1D

yes, it's the number that cancels out when you balance a redox reaction by writing out its oxidation/reduction half-reactions

Angela Grant 1D

n is the number of electrons that cancel out when you write out oxidation/reduction half-reactions to get the balanced redox reaction

Angela Grant 1D

You'd use n = 2 electrons because when you split up the half-reactions you get:

Oxid.: Ce4+ + e- --> Ce3+
Red.: 3I- --> I(3)- + 2e-

Multiply the oxidation half-reaction by 2 to balance out the electrons and cancel them out, leaving you with two electrons getting cancelled.

Angela Grant 1D

This question asks for the cell diagrams of given skeletal equations. For (b) Ce4+ (aq) + I- (aq) --> I2 (s) + Ce3+ , I understand that the cell diagram would need a Pt (s) on the cathode end where Ce4+ reduces to Ce3+, so it'd be Ce4+(aq), Ce3+(aq) | Pt(s). But the answer key also has a Pt(s) on th...

Angela Grant 1D

Jordan Lo 2A wrote:Can someone explain #4? Is this because q=0 when delta T=0, and w=0 when -PdeltaV=0?

When delta T = 0, the internal energy delta U = 0. Since delta U = q + w, if delta U = 0 then w = -q, aka the work done on the system is equal to the heat absorbed.

Angela Grant 1D

Can someone explain how to solve 10 and 11? For 11, I used the mol(cv)(ln(t2/t1)) and am not getting the correct answer. Can someone please outline the steps to solve? For 10, use delta S = nRln(V2/V1) For 11, you need to add up two different delta S's, one for the volume change and one for the tem...

Angela Grant 1D

Hannah Yates 1K wrote:Can someone explain 5c?

You use work = -nRTln(V2/V1). Isolate ln(V2/V1) and then solve for V2/V1 by taking e of both sides

Angela Grant 1D

can someone explain 6(d)? I tried using the additive equation for standard enthalpy of a reaction but I don't think I'm inputting the correct value for it or if that's even a possible way to get the answer

Angela Grant 1D

Kw implies that the solution is water since the expression represents the autoprotolysis of water

Angela Grant 1D

Minimizing the effect of changing volume is dependent on what side has more moles, so if both sides are the same then there's no shift that will minimize the effect. Aka it doesn't shift left or right

Angela Grant 1D

To add on, I don't think Dr. Lavelle would give us a borderline problem like that. If he did however, I'm pretty sure you still consider it negligible and add it.

Angela Grant 1D

The reaction responds in such a way that minimizes the effect of a pressure change, so increasing volume (decreasing pressure) in a system with more moles on one side will favor that same side.

Angela Grant 1D

We can disregard spectator ions when writing out equations with salts, but I think you should write out the entire equation for other scenarios to make sure you're noting the correct Kb / Ka values that correspond

Angela Grant 1D

∆Hfus = Hliquid- Hsolid ∆Hvap = Hvapor - Hliquid ∆Hsub = Hvapor - Hsolid Since sublimation goes straight from a solid to a vapor state, it'll have the same total change as the individual changes in fusion and vaporization put together. Therefore, you can combine them all to get ∆Hsub=∆Hfus+∆Hvap Hop...

Angela Grant 1D

Technically, it affects both since pressure and concentration are related by PV=nRT --> P=(concentration)RT <--> concentration=P/RT
But yes Kp is pressure and Kc is concentration

Angela Grant 1D

In addition to more products getting produced, the concentration of any other reactants in the reaction would decrease.

Angela Grant 1D

Yes, since Q and K are the same ratio ([products]/[reactants]) that means the [products] would have to decrease, or the [reactants] would have to increase, in order for Q to be less than K. Therefore, there will be more reactants yielding products so that the forward reaction is favored.

Angela Grant 1D

One reason that HF is a weaker acid than HI is because of the stronger bond - it is less likely to donate a proton

Angela Grant 1D

I don't think it's either actually, it's a neutral organic compound

Angela Grant 1D

First look at charge (greater charge corresponds to higher polarizing power) and then size (smaller atoms have higher polarizing power)

Angela Grant 1D

During one lecture Dr. Lavelle said any questions on ionic/covalent character would be pretty straightforward if we aren't given an electronegativity chart. So we can just figure it out based on how close/far apart the atoms in a molecule are on the periodic table.

Angela Grant 1D

what makes sigma bonds weaker than pi bonds?

Angela Grant 1D

Ashley P 4I wrote:Does anyone have a trick to memorizing all these things?

I made a big chart to study off of, you can find one online if you just search up VSEPR geometries

Angela Grant 1D

s-character is how much of a hybrid orbital is made up of the s orbital (ie sp = 50%, sp2 = 33%, sp3= 25%, etc.)

increasing s-character is basically decreasing the number of bonds, which allows for more space between bonds, therefore increasing bond angles.

Angela Grant 1D

A molecule's structure is an average of all resonance structures so the bond lengths would be an average

Angela Grant 1D

The VSEPR formula for seesaw is AX4E, so the 4 bonding groups and single lone pair result in <120 equatorial <90 axial

Angela Grant 1D

This is what my high school chem teacher taught us: "A molecule with a uniform distribution of electron density is nonpolar; and one with an asymmetrical distribution is polar. A molecule is nonpolar only if it has no lone pair electrons about the central atom and all groups attached to the cen...

Angela Grant 1D

Does anyone know how the grading scale works for Dr. Lavelle's class? Is it on a curve or does each letter grade correspond to a specific and fixed percentage range?

Angela Grant 1D

When you draw the Lewis structure, you can determine the number of bonding pair and lone pair electrons. You can then create a VSEPR formula which corresponds to a particular shape. For example, a formula of AX3 corresponds to a trigonal planar shape.

Angela Grant 1D

In the book the equation for electronegativity is given as the average of ionization energy and electron affinity. But in lecture Dr. Lavelle said we'd just be given the table of values if we need to use them on a test so we probably don't need the equation

Angela Grant 1D

The book states that most bonds are somewhere in between purely covalent and purely ionic. When depicting these bonds, how does one know when to depict a bond as ionic or covalent when it is in this in between state? The book uses a rough guideline in which a difference in electronegativity greater...

Angela Grant 1D

Can someone explain the relationship between number of electrons and polarizability?

Angela Grant 1D

I think as the electrons fill an orbital they take a +1/2 spin, then as the become paired they take -1/2 spin.

Angela Grant 1D

You are right about any atom with n=3 or higher being able to have an expanded valence shell. Dr. Lavelle mentioned that this applies to atoms in periods 3 and higher.

Angela Grant 1D

Electron affinity is the energy released when an electron is added to a gas-phase atom. You can think of it as how much an atom attracts an electron. On the periodic table, it is highest on the upper right (with the exception of noble gases since they have full valence shells).

Angela Grant 1D

I believe it's because the f block starts right after the s block at period 6, so it doesn't include the d block until after the lanthanide series ends.

Angela Grant 1D

The 3d orbital actually comes before 4s because it has lower energy, and the electron configuration is written from lowest to highest energy. For example, scandium's electron configuration is [Ar]3d(1)4s(2), not [Ar]4s(2)3d(1).

Angela Grant 1D

I believe for the ground state electron configuration of an atom, the convention is to fill orbitals with +1/2 spin and then go back to pair electrons with a -1/2 spin.

Angela Grant 1D

Remember that electron density is the probability of an electron being present in that region. So since nodal planes approach 0, there is no chance of an electron being found there (aka the electron density = 0).

Also, the number of nodal planes increases per orbital (s = 0, p = 1, d = 2, f = 3).

Angela Grant 1D

You have to know that the UV spectrum includes the Lyman series, in which n1 = 1. Using the given wavelength, you can find frequency. Then you isolate n2 in Rydberg's formula and plug in all your values to find the final energy level.

Angela Grant 1D

Where the orbitals lie also depend on their lobe orientation. For example the d-orbital can have lobes either on xy- yz- zx- planes, the x & y axes, or the xy-plane. So it varies within each orbital

Angela Grant 1D

Yeah that's fine, since we haven't finished covering the quantum world

Angela Grant 1D

It is possible! Take H2O for example. The formula can't get any simpler, so it is both the molecular and empirical formula of water.

Angela Grant 1D

When an atom absorbs a photon, the energy from the photon sends the electron to a higher energy level. The electron emits energy in the form of light when it returns to its ground state.

Angela Grant 1D

Since the textbook uses 6.022*10^23, it would probably be better to use that value so your answers can be more precise and line up with the textbook answers. In addition, it is a constant value so its sig figs are not dependent on other components of the problem.

Angela Grant 1D

Mass percentage is not necessary in this problem because you are already given the amount of each reactant with respect to the sample. You can use this information to set up ratios of mass of element per mass of sample (leaving you with grams), then divide by molar mass (leaving you with moles) in o...

Angela Grant 1D

In response to the most recent comment: no, it does not matter which order products or reactants are written in, as long as they are on their respective sides of the chemical equation (reactants on left, products on right)

So the equation can be written as C6H6 + O2-> H2O + CO2

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