CHEMISTRY COMMUNITY

Clara Rehmann 1K

One strategy is to experimentally determine what reactants are colliding with each other, since that is the most basic step of a mechanism. For example, if you are determining if A+B->C or A->C, one way to determine which mechanism is correct is by seeing if A produces C in the absence of B.

Clara Rehmann 1K

The difference in cell potential would be the difference between E and Eº, yes

Clara Rehmann 1K

Yes, mathematically you are also switching the Eº for each side.

Clara Rehmann 1K

I had to miss the second half of lecture on Friday - can someone quickly explain to me the relationship between catalysts and activation energy and how that is calculated?

Clara Rehmann 1K

This video https://www.youtube.com/watch?v=wGUKtJZJZUk helped me understand it a bit better - basically, electric current causes the "free" electrons to move from atom to atom of the conducting material.

Clara Rehmann 1K

I don't think so; or if so, the relationship isn't quantifiable

Clara Rehmann 1K

It possibly may have to do with the way molecules of these reactants interact with each other? I think we'd have to learn more about the mechanics of 0, 1st, and 2nd order reactions in order to understand the exact ways in which they differ, but I would guess that more of these molecules being in a ...

Clara Rehmann 1K

I'd assume you use it until the reaction reaches equilibrium, at which point A is no longer being converted into products.

Clara Rehmann 1K

I believe it depends on the order of the reaction.

Clara Rehmann 1K

Start with ΔG = -nFE and It = nF. Combining these equations, you get I= $\frac{ΔG}{-Et}$

Knowing that Eº = ~1 V, you can plug in values (ΔG = -10 x 10⁶, E = 1, t = 86400) and and solve for a solution in amps

Clara Rehmann 1K

Start with ΔG = -nFE and It = nF. Combining these equations, you get I= $\frac{ΔG}{-Et}$

Knowing that Eº = ~1 V, you can plug in values (ΔG = -10 x 10⁶, E = 1, t = 86400) and and solve for a solution in amps

Clara Rehmann 1K

Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode. Pt(s) | O 2 (g) | H + (aq) || OH - (aq) | O 2 (g) | Pt(s) Overall reaction: 4H 2 O(l) → 4H + (aq) + 4OH - (aq) Anode: O 2 (g) + 4H + (aq) → 2H 2 O(l)...

Clara Rehmann 1K

It seems as if most of the redox reactions that we are studying don't produce very high voltage - in my homework problems, voltage always is under about 2V. How do things that we use in our everyday lives (batteries, electronics) achieve such high voltage?

Clara Rehmann 1K

Clara Rehmann 1K

Standard potential follows the same trend as electron affinity - an element that holds onto its electrons more tightly will generate more energy when those electrons are taken away.

Clara Rehmann 1K

Today in lecture Lavelle mentioned that if there is a conductive material present in the cell, then both products and reactants can be in the same solution rather than being kept separate with a conductor and salt bridge in between (at least this is what I understood him to be saying; that could be ...

Clara Rehmann 1K

Regarding the question about the sun, you would know that there is heat exchange because the sun is causing the water to evaporate - it's transferring Joules of energy into the water to the point where the water reaches Hvap. As far as processes being exothermic or endothermic, for the sun the proce...

Clara Rehmann 1K

I think either is okay, but to be thorough maybe label them as "reduction" and "oxidation" or "anode" and "cathode."

Clara Rehmann 1K

Hi, I'm a bit confused as to deciding which value of R to use when calculating entropy - Lavelle's cheat sheet has a couple different values (based on changes in units). I try to base my choice of R off of the other units I am provided with when solving a problem, but that doesn't seem to always wor...

Clara Rehmann 1K

A reversible process does more work because it is working against a constant pressure, meaning the pressure in the system changes slowly. An irreversible process, on the other hand, involves fast pressure changes that limit the system's ability to do work. I found the graphs that Lavelle showed duri...

Clara Rehmann 1K

I think you are correct because Gibbs' free energy is temperature dependent, and thus the standard Gibbs' free energy should be at a certain temperature.

Clara Rehmann 1K

No matter how we're calculating enthalpy, we want the total "additive" energy in the reaction. When calculating enthalpy using bond enthalpies, I look it like this: breaking bonds requires energy, contributing positive values to my total (energy going in), while making bonds releases energ...

Clara Rehmann 1K

Monday's class gives a good baseline for some of Chapter 9's problems; I would do the ones you can given the equations Lavelle gave us and then fill in the rest of the seven with problems from Chapter 8. My understanding is that as long as it's relevant to what we're doing in class, we can do whatev...

Clara Rehmann 1K

In the original equation, we have 1 mol of S8. We want to calculate the enthalpy when there's 1.25 mol of S8. To get 1.25 mol S8, you multiply 1 mol of S8 by 1.25. Multiply the reaction's enthalpy by the same amount to get the enthalpy for 1.25 moles.

Clara Rehmann 1K

I assumed that the value was positive because we were looking at the energy from the perspective of the surroundings rather than the perspective of the system. Technically the TNT is losing energy by releasing it, but when observing that process from the outside, a positive amount of energy is being...

Clara Rehmann 1K

You are calculating the enthalpy of formation, which means you need to calculate the enthalpies for each bond broken and formed. Approach the calculation for C6H6 the same way you did parts (a) and (b) - in part (c), take into account the bonds in a standard structure, and in part (d), take into acc...

Clara Rehmann 1K

I think that unless stated, you assume standard pressure. Otherwise, there's really no way of knowing how to calculate it.

Clara Rehmann 1K

To add on to what other posters have replied, standard enthalpy of formation requires that reactants must be at their most stable states.

Clara Rehmann 1K

Yes, three posts/replies per week.

Clara Rehmann 1K

Are the terms "energy" and "heat" interchangeable? And, if so, how does the measurement of temperature relate to the measurement of energy applied to/released by a substance?

Clara Rehmann 1K

Because NO2 has more bonds and atoms, it has a higher heat capacity. The additional matter means that there is greater ability to contain energy, i.e. heat.

Clara Rehmann 1K

Is there a trick for telling if a molecule will give off a proton or accept one? It's sometimes difficult to tell, especially when there are a lot of H's already in the molecule.

Clara Rehmann 1K

If all molecules have dipole moments, how do you tell when the dipole moment is great enough to create a polar molecule without having access to specific numbers?

Clara Rehmann 1K

Yes - hybridization refers to regions of electron density so lone pairs contribute to it. The C atom in CH3 for example is sp3 hybridized because it has four regions of electron density (three in its bonds with H atoms and one as a lone pair).

Clara Rehmann 1K

Electron arrangement refers to the position of electrons as well as the position of atoms in a molecule. Molecular shape essentially ignores extra electrons and just looks at the atoms in the molecule - for example, H2O is "bent" rather than tetrahedral because you look at just the atoms, ...

Clara Rehmann 1K

There's more O2 in the same size space (.1 mol in one L vs. .5 mol in one L), thus the second container will always have a higher concentration of O2.

Clara Rehmann 1K

A coordination number refers to the number of bonds formed rather than the number of ligands, so if a ligand is polydentate it will be contributing [number of bonds] to the coordination number rather than just the one ligand.

Clara Rehmann 1K

Notice that you can see which molecules have decayed within the flask. Start with counting out how many molecules have split apart!

Clara Rehmann 1K

Sigma bonds only have one field of electron sharing, so the atoms involved in a sigma bond can rotate independently of one another on that sigma-bond axis and still remain bonded. However, because pi bonds have two areas of electron sharing, no rotation is allowed: If an atom tried to rotate on one ...

Clara Rehmann 1K

Lone pairs are a little more electronegative than bonded pairs, so they "push away" a little more as well. This results in the angle of the bonds being smaller, since they are being pushed together by the electronegativity of the lone pairs.

Clara Rehmann 1K

Dipole moments represent the charge difference in a molecule. The arrow points toward the δ- atom. If the dipole moments do not cancel each other out (ie, if the arrows do not point at each other) then the molecule is polar because there is a charge difference across the molecule.

Clara Rehmann 1K

If you know the molarity of the stock solution, you can use the equation MiVi = MfVf, with Mf being .650M and Vf being .25 L.

Clara Rehmann 1K

That sounds about right. The most important thing to note is that atomic orbitals are three-dimensional areas where the probability of finding an electron is above a certain threshold. I'm not sure that just writing out the wave equation will be enough; you should demonstrate that you understand tha...

Clara Rehmann 1K

You can predict the nature of the bond based on the elements in it and how electrons need to be shared. If both can fulfill the octet rule (or whatever rule they need to fulfill) by just giving/receiving an electron, then it's probably an ionic bond, but if it looks like electrons need to be shared ...

Clara Rehmann 1K

We haven't learned the equation for potential energy, but we know that it is proportional to (meaning it changes at the same rate as) (Q1)(Q2)/r where the Q's are the charges and r is the distance between the two. Because r is in the denominator, as it increases (the charges get farther apart), the ...

Clara Rehmann 1K

Jessica is right - there are more exceptions than Cr and Cu, but those are the only two we learned about. Having 5 or 10 electrons in the d orbital is more stable than having 4 or 9, so any d-block element that ends up with 4 or 9 electrons in the d orbital will actually have 5 or 10 (and then just ...

Clara Rehmann 1K

Nodal planes are similar to nodes in that no electrons are found in that space. The difference is that a node is just a single point, while a nodal plane is a "sheet" (remember, we are dealing with 3D space) through which electrons don't travel. Think of it as just a really long/tall node,...

Clara Rehmann 1K

Ψ^2 is the wave function squared, meaning that all values of the wave function are positive. It equals the probability of finding an electron in that space, since electrons behave like waves.

Clara Rehmann 1K

Does the question supply any other information? Also, does it specify that it is referring to the photoelectric effect?

Clara Rehmann 1K

The 150.6 kj is required to remove electrons from a mole of sodium. The question is asking the amount of energy needed for just one sodium atom, so some conversions need to happen - find the amount of energy required for just the one atom. Hope this helps you piece it together!

Clara Rehmann 1K

Electrons hop up between discrete energy levels when they are excited by a photon. What connects them to spectral lines is that the "hopping" of an electron from one energy level down to a lower energy level releases a photon of a certain energy level and wavelength. The energy level, and ...

Clara Rehmann 1K

1. Find the number of moles of sodium carbonate added (2.111g / molar mass) 2. Find molarity of the solution (M = n / V) - don't forget to convert 250.0 mL to L! a) 1. Convert mmol of Na+ to mol 2. Knowing that there are 2 Na+ molecules in each sodium carbonate molecule, the molarity of Na+ in the s...

Clara Rehmann 1K

Rounding to the nearest even number (eg. 2.5 -> 2, 5.5 -> 6) prevents there from being a significant shift up in the quantities you have. If you always always always round 5 up, especially with a big data set, you'll end up with the average of all your numbers being shifted up. Rounding to the neare...

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