CHEMISTRY COMMUNITY

Camille Marangi 2E

Conceptually, a lower molar entropy substance is just more ordered than another substance in comparison. For example H20 (s) has a lower molar entropy than say H20 (g). The particles in ice are more tightly and rigidly held together than in water vapor. You can also see this in the Boltzmann equatio...

Camille Marangi 2E

That reduction value is specific to the pH being 7 and should only be used for the problems that explicitly mention that the pH of the solution is 7. Otherwise you use the appendix Ecell values.

Camille Marangi 2E

If you could post the problem that would be a lot more helpful but usually if you have to use two trials that are not kept constant, you most likely will know the order of one of the reactants whose trials you are using. Then, because you know the order of one reactant you can see how much that spec...

Camille Marangi 2E

Usually, from what I have seen, if the slow step isn't the first step it has an intermediate in the rate law that you would form. So you find the rate law for the other step with the intermediate, solve for the concentration of that intermediate, then plug that new equation in for the intermediate. ...

Camille Marangi 2E

Capital K, the equilibrium constant, can also be found as rate constant of the forward reaction/rate constant of the reverse reaction or k/k'.

Camille Marangi 2E

Elementary reactions are the step by step sequences that make up overall reaction mechanisms. For example 3 elementary steps can add up to one overall reaction mechanism with intermediates, catalysts, etc.. The slow step of a mechanism determines the rate law and the overall reaction formula can be ...

Camille Marangi 2E

Mercury acts as a conductor when it is in a liquid state, so there is no need for an additional conducting metal.

Camille Marangi 2E

This is because of the equation: dG= dH-T*dS. The spontaneity still relies on the change in enthalpy as we see in the equation. If the reaction is exothermic and entropy increases, then no matter the temperature, we will see the forward process being favored as dG will be negative. In general, there...

Camille Marangi 2E

When at equilibrium, the change in gibbs free energy is zero due to there being no enthalpy or entropy change. So gibbs free energy is the same amount as whatever we started with. Say G=5 Kj/mol and dG=0 (equilibrium), then G is still 5 Kj/mole.

Camille Marangi 2E

For even more clarification, this would mean that the more free energy that the reaction has, the less stable it will be in general. More energy means less stability.

Camille Marangi 2E

It also depends on the sign of enthalpy- if the reaction is exothermic with a positive entropy, the reaction will without a doubt be more stable. If endothermic with a negative entropy the reaction will be less stable. Any variation in these signs are solely dependent on temperature to find the stab...

Camille Marangi 2E

As long as both the enthalpy and entropy are joules/mol, you can wait until after calculations to convert to KJ/mol.

Camille Marangi 2E

Some situations you have to use both and then add them together to find the total entropy change of the system- it all just depends on what is given. If the problem gives you 2 temperatures and 2 volumes and then asks for entropy change, chances are you'll be using both. However if there is just a t...

Camille Marangi 2E

Adiabatic just means there is no heat transfer in the reaction. So if the problem states that there is no heat transfer, then you know the system is adiabatic.

Camille Marangi 2E

Generally we are more interested in the system specifically gaining entropy or losing entropy. It just so happens that the majority of the time, the system tends to gain entropy and is therefore positive, meaning the surroundings must lose entropy and therefore the Ssurr must be a negative value. So...

Camille Marangi 2E

The possible orientations just depend on the atoms involved. For example, CH4 is a tetrahedral just like CH3Cl but only has one possible state to be found in and thus its degeneracy is seen as w=1^n. So this logic would follow with a trigonal bipyramidal- if the atoms attached to the central atom ar...

Camille Marangi 2E

Not that I've seen but Prof Lavelle will probably announce the sessions in lecture tomorrow and then they will be listed on the class website shortly after.

Camille Marangi 2E

The standard entropy of vaporization of benzene is approximately 85 J/Kmol at its boiling point. (a) Estimate the standard enthalpy of vaporization of benzene at its boiling point of 80 C. (b) What is the standard entropy change of the surroundings when 10. g of benzene, C6H6, vaporizes at its boili...

Camille Marangi 2E

Delta H is enthalpy, so it is just the measurement of change in heat *at constant pressure*. So if the reaction takes place at constant pressure, then delta H=q (enthalpy=q).

Camille Marangi 2E

How much heat is required to convert a block of ice of mass 42.30 g at -5.042 C into water vapor at 150.35 C? So I go through all the calculations (42.3*2.03*5.042+42.3*336.611+42.3*4.184*100+42.3*2259.23+42.3*2.01*50.35) and I end up with the answer 128 KJ but the solution is 132 KJ. Am I missing a...

Camille Marangi 2E

So 4E.9 from 7th edition reads as follows: Benzene is more stable and less reactive than would be predicted from its Kekulé structures. Use the data in Table 4E.3 to calculate the lowering in molar energy when resonance is allowed between the Kekulé structures of benzene. Am I supposed to assume a k...

Camille Marangi 2E

As far as from what TA's and Lavelle himself have told us we are supposed to wait until all calculations have been completed to round so we have the most accurate answer.

Camille Marangi 2E

Not that heat is necessarily an experimental reactant/ product in any way but the best way I can explain it is as follows: Endothermic- reactants + heat -> products Exothermic- reactants -> products + heat So if we are increasing the temperature, in the endothermic reaction we shift towards products...

Camille Marangi 2E

By increasing the partial pressure of NH3 you are in essence just increasing the reactants side. As seen in Le Chatelier's principle, the products would then increase to counter act this shift and the partial pressure of O2 would in fact decrease. I think you may be confusing this with the system be...

Camille Marangi 2E

So starting with the current concentration of OH- in the diluted solution you do pH+pOH=14 so pOH=14-13.25=0.75. Then 10^-0.75=[OH-] in the diluted solution. This comes out to be 0.178 M or 0.18 M with sig figs. From there, we have to work backwards to get to the original solution. So we know the co...

Camille Marangi 2E

Reactants just so happened to be favored in all of these because the reactants side had less stoichiometric moles than the products side did in each scenario. For example in part a, the reactants has 2 stoichiometric moles of O3 while the products has 3 stoichiometric moles of O2- therefore when the...

Camille Marangi 2E

I assume you're asking when it is ok to assume that 0.22-x can be simplified to just 0.22. Lavelle said it was fine to make that assumption as long as the equilibrium constant is at least 10^-3 smaller. So if there are 3 degrees of separation, then your initial value is at least 1000 times larger th...

Camille Marangi 2E

As far as I can tell, unless the question specifies which equilibrium constant it wants you're fine writing either. However it wouldn't hurt being prepared and knowing how to convert Kp to Kc and vice versa. Here's the conversion equation: Kp=Kc(RT)^n

Camille Marangi 2E

Dalton's partial pressure equation is Pt=Pa+Pb+... where Pt is total pressure of all gases in the reaction while Pa and Pb are individual gases- they can just be added up to find the total. You use it in many equilibrium scenarios where perhaps the total pressure is given so you have to solve for on...

Camille Marangi 2E

In 14A we studied quantum mechanics, (frequency, wavelength, speed of light sorts of problems), chemical bonds, molecular shape and structure, coordination compounds, and we started acids and bases which we will continue this quarter.

Camille Marangi 2E

I find that usually double displacement does the trick when determining which atoms will be a part of the salt. It is also helpful to know that a salt is just an aggressively ionic molecule, so whatever atoms are involved, just know the salt has to be ionically bonded together.

Camille Marangi 2E

In most cases there only ends up being one place for the radical/extra electron to be placed- I can't really think of a molecule where you have multiple options. Usually if we have say and nh3 ion, the extra electron goes on nitrogen because hydrogen can't physically take on another electron.

Camille Marangi 2E

As far as I can tell (as seen in my discussion today) you only need to write the abbreviation when naming compounds with di/ethylenediamine and you only need to know one of the columns and it seems Lavelle prefers the one with the asterisk next to it

Camille Marangi 2E

Yes a single electron can alter the shape as well. This is seen in radical lewis structures/ molecules and I think Lavelle assigned problems that use examples of this- I think there may be an NH3 radical formula in either the textbook problems or I have seen it in a review section. Either way, yes a...

Camille Marangi 2E

I believe atomic radii takes precedence over the ionic charges especially in terms of the row that the atom comes from. The larger in magnitude row on the periodic table, the larger the energy shell and thus nuclear effective charge becomes less effective as we go down the table. So the larger the a...

Camille Marangi 2E

As Lavelle explained in lecture on Monday, cisplatin is a well known chemotherapy drug. It forms a coordination compound with DNA and stops cell division through its chelating ligands. Chelates are complexes containing a ligand that forms a ring of atoms that includes the central metal atom and when...

Camille Marangi 2E

I don't think we really work much with transition metals in 14A so I'm not really sure there is a definitive answer for this. VSEPR seems pretty reliable though and it's all we know so up to this point I'd trust it.

Camille Marangi 2E

I would say yes because to have multiple isoelectric species, we must have ions that we can determine the electron densities for. Therefore we can still use the VSEPR model.

Camille Marangi 2E

The pi bonds can't be attached to hybrid orbitals due to their rigidity, just as Lavelle explained in his lecture last Monday. Also I think it's not that terminal atoms are most often not hybridized, rather that hydrogen isn't hybridized and it is often the terminal atom. Oxygen in this molecule has...

Camille Marangi 2E

An induced dipole moment occurs when there is movement in the electrons of a bond otherwise known as polarization that creates fleeting unbalanced distributions of electrons and thus small dipole moments, whereas in a regular dipole moment the sharing of electrons at all times is unbalanced due to o...

Camille Marangi 2E

Square planar molecular geometry occurs when there are 4 atoms and 2 lone pairs bonded around the central atom

Camille Marangi 2E

Soyoung Park 1H wrote:So what is exactly VSEPR model?

The VSEPR model stands for valence shell electron pair repulsion theory and it's basically model used in chemistry to predict the geometry of individual molecules from the number of electron pairs surrounding their central atoms

Camille Marangi 2E

The larger an atom is, the more electrons it has. The more electrons the atom has, the greater likelihood of depolarization of said electrons. This polarization is what creates dispersion forces between atoms or molecules. Therefore, the larger the atom/molecule is, the greater polarization ability ...

Camille Marangi 2E

It's negative in that it has high electronegativity and thus a high propensity for electrons. The higher the electronegativity of an atom the more electrons are attracted to it.

Camille Marangi 2E

As far as I know resonance is only in reference to delocalized bonding pairs, not lone pairs but we have yet to get to hybridization which might provide more insight on this

Camille Marangi 2E

It is most likely a matter of sig figs. One of the givens is 2.4 x 10^21 photons and that only has 2 sig figs so that means the problem's answer will have 2 sig figs (750 as zero in the last digit of a whole number is not a sig fig).

Camille Marangi 2E

I think both explanations revolve around the same concept in regards to oxygen's ionization energy. Oxygen will have lower ionization energy than nitrogen and fluorine because it will be more stable by losing one electron to reach the half-filled state much like your TA said and the electron repulsi...

Camille Marangi 2E

I think the appropriate strategy is just to remember that atomic radii decrease across a period and increase down a group. So for this specific issue, with each additional period down the periodic table an energy level is added and the atomic radius grows by an entire shell. Elements on the left sid...

Camille Marangi 2E

Normally you use the uncertainty in diameter unless the givens in the problem state otherwise as seen in the exercise you mentioned. To answer your second question, yes the uncertainty in position would be exactly as you stated because that's how the problem presents that given. Hope this helped.

Camille Marangi 2E

Why does Boron's lewis structure show it as being perfectly stable without a full octet of electrons when paired with 3 Fluorine atoms? (Seen in BF3 if that wasn't clear)

Camille Marangi 2E

Oxygen has a lower electron affinity because it has a smaller atomic radius than sulfur and thus the electrons experience significant electron-electron repulsions. Here's a link to a photo that shows the electron affinities of elements in the top right of the periodic table. https://i.stack.imgur.co...

Camille Marangi 2E

'l' is the orbital angular momentum quantum number. What this means is that it can tell us the subshell that an electron is lying in. 'l' can equal n-1 and any value below that. So 'l' could potentially be 0,1,2,...n-1 depending on whatever principle quantum number the problem is giving you. Pro tip...

Camille Marangi 2E

So the original equation for Heisenberg's uncertainty principle is actually (delta)p(delta)x = (1/2)ħ just as the solution's manual says. However there is a difference between ħ,an H-bar, and h, Planck's constant. 'ħ' can be reduced to planck's constant as seen as: h/2π. Multiply (1/2) and h/2π toge...

Camille Marangi 2E

I believe the cone-shaped nodal plane are both the planes in the d-orbital. These cone-shaped planes only occur in one orbital of the 'd' type, where there are 5 in total.The specific orbital that it occurs in is the 3dz^2. All the rest have regular planes (square-shape as we previously have seen). ...

Camille Marangi 2E

The way I answered this problem was to use the equation λ=h/mv. So we have Planck's constant as h=6.626x10^-34 J.s, velocity is given as 2.75x10^5 m/s and then you would do this equation twice, once using the mass of the proton (m=1.673x10-27 kg) and again using the mass of the neutron (m=1.674x10^-...

Camille Marangi 2E

I believe in this scenario the work function and threshold are interchangeable. So your equation can also look like this: (E=hv) - threshold energy = (Ek= 1/2me-ve^2) or exactly like the one you posted. Threshold energy is the amount of energy it takes to remove an electron from a metallic surface, ...

Camille Marangi 2E

Hi so I am currently stuck on the HW question 1B.15 which is as follows: The velocity of an electron that is emitted from a metallic surface by a photon is 3.6x10^3 km/s (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency...

Camille Marangi 2E

Chlorine is a diatomic element. Diatomic elements are pure elements that form molecules consisting of two atoms bonded together because they are extremely unstable alone due to their *almost* full valence shell. There are 7 of these diatomic elements and the easiest way I find to remember them is to...

Camille Marangi 2E

The 1 atm and 25 degrees C is given information to ensure that the reaction is done at STP (standard temperature and pressure) to show that the gases in this reaction act ideally and guarantee that all calculations/measurements are consistent- basically no further action is necessary when this info ...

Camille Marangi 2E

It is just one extra step compared to parts a and b. 50.0mg means 50.0 milligrams of sodium carbonate and is also equivalent to .05 grams of Na2CO3. From there you convert the grams to moles using the molar mass, 105.988 g/mol, which will give you 4.72 x 10^-4 moles of sodium carbonate. Then, like i...

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