CHEMISTRY COMMUNITY

Arjun_Anumula_3E

Catalyzed reaction makes use of another compound (that is not present in the overall equation) to help reduce the activation energy and speed up the rate of reaction. An example of this could be how ATP Synthase speeds up the conversion of ADP into ATP by helping increase the likelihood of favorable...

Arjun_Anumula_3E

Increasing temperature doesn't change the amount of gases in the mixture. While increasing temperature increases the total overall pressure in a mixture if volume remains constant, it does not change partial pressure like compressing the mixture would. Thus, it is only the equilibrium constant that ...

Arjun_Anumula_3E

Drawing a lewis structure can help. A molecule that lies in 1 plane (such as a benzene derivative) then can be rotated in different positions to find the possible number of microstates. Then, raise the number of microstates to the power of the number of particles to find W

Arjun_Anumula_3E

depending on the units of the y-axis, the slope can be used to find the rate constant.

For 0 and 1st order, k = -slope
For 2nd order, k = slope

Arjun_Anumula_3E

The value of the rate constants depend on the arrhenius equation. However, making changes such as adding a catalyst can increase both the reverse and forward rate by decreasing the energy of the activated complex

Arjun_Anumula_3E

A gas that has no interaction with other molecules and that move randomly. We also assume they take up no volume. In reality, they do not exist since everything takes up space and will interact with other particles at least a little bit, but it can be used to approximate

Arjun_Anumula_3E

How come on Achieve #15, the A value is much larger than 1?

Arjun_Anumula_3E

It isn't that k is negative, but rather the slope in the graphs where the plot would be linear shows a negative slope.

This is observed in the 0th and 1st order reactions and it is due to the nature of the integration that does not remove the negative sign from the initial -d[A]/dt.

Arjun_Anumula_3E

If the graphs are scaled properly with [A], ln[A], and 1/[A] vs time, then the graphs can be useful.

Positive linearity in the 1/[A] vs time graph indicates 2nd order
Negative linearity in the ln[A] vs time graph indicates 1st order
Negative linearity in the [A] vs time graph indicates 0th order

Arjun_Anumula_3E

The reaction rate may be given to use, or we may have to calculate it. Recall the first equation Dr. Lavelle showed us about reaction rate. The rate of formation/removal of a substance can be interchanged with the unique reaction rate: For the reaction: aA → bB + cC Unique rate = -1/a ( d[A]/dt ) = ...

Arjun_Anumula_3E

When trying to find the overall reaction order, it can be helpful to do multiple experiments to find n,m,l, etc. So k' just refers to the psuedo rate constant for rate law where there is only a noticeable change in one reactant. It can be useful to find the real rate constant after the overall react...

Arjun_Anumula_3E

Only in the first order reactions. Remember that both a 0th and 2nd order reaction require the initial concentration of the reactant to determine the half life. These observations are consistent with the integrated rate laws that Dr. Lavelle derived in class last week.

Arjun_Anumula_3E

What is the difference between using a different metal for an electrode than using the same one as the one in the reaction?

Arjun_Anumula_3E

This means that the equations 2Fe3+ + 6e- => 2Fe Fe3+ + 3e- => Fe have the same standard reduction potentials. This becomes useful when balancing more complicated redox reactions with different numbers of electrons needed or produced, because the reduction potential stays the same no matter what coe...

Arjun_Anumula_3E

Nernst equation is used to calculate the Cell Potential under any conditions and concentration ratios of product to reactant. As electrons from the oxidized metal flow toward the cathode, the Cell Potential will slowly decrease and will become 0 at which point the "battery is dead"

Arjun_Anumula_3E

At the Boiling point, the phases Co-exist and the formation of either full gas or liquid will not occur unless the temperature increases or decreases from this point.

Arjun_Anumula_3E

If you are given or solve for the Delta S of a system, then it must be used in conjunction with the gibbs free energy equation (ΔG = ΔH − T * ΔS) If ΔG is negative, then the reaction will be spontaneous If you are given or solve for the Delta S of the Universe, and determine that it is positive, the...

Arjun_Anumula_3E

At equilbrium, the concentrations of the products and reactants (the substances in the anode and cathode) are not changing. Equilibrium means that Delta G is 0.

This must mean that there is no net electron transfer, thus the cell potential (voltage) will be 0 V.

Arjun_Anumula_3E

As Dr. Lavelle explained on Friday's lecture, these batteries start out at a maximum voltage (most negative Delta G). However, as the reaction proceeds, the product to reactant ratio increases and thus the forward reaction that generates voltage becomes less favorable. So as Delta G decreases, so do...

Arjun_Anumula_3E

Since the Mn goes from a +7 to a +4, it is reduced and therefore an oxidizing agent. Since the S goes from a -2 to a 0, it is oxidized and therefore a reducing agent. Balance both the half reactions by adding the right number of electrons, water molecules, and H+'s to both reactions (only the Mn red...

Arjun_Anumula_3E

Yes it definitely can. If the temperature rises in a constant volume calorimeter or any sort of constant volume system, then by PV=nRT, both the pressure and temperature increases. Dr. Lavelle mentioned the equation for a constant Volume gas undergoing entropy change as change in entropy ∆S = n*Cv*l...

Arjun_Anumula_3E

Just to add on to what was said above, the equation should more specifically be ∆G°= -RTlnK. Make sure to know the difference between ∆G an ∆G°, because ∆G changes as the ratio of products to reactants changes during the course of a reaction.

Arjun_Anumula_3E

A spontaneous reaction means that the products are naturally favored in a reaction. When delta G is negative, this occurs

Arjun_Anumula_3E

Entropy is useful because it can help tell us whether a reaction is favorable. For example, we would think that if a reaction needs a great input of heat (endothermic) to proceed, we would think that it is not favorable. However, if the degree of disorder increases substantially enough or if the rea...

Arjun_Anumula_3E

Residual entropy is the difference between the non-equilbrium state and the crystal state of a substance at a temp close to absolute zero. Basically, a molecule that can exist in more than 1 state, like COF2, will have more residual entropy than BF3 (symmetries). Thermal entropy, however, is just th...

Arjun_Anumula_3E

Gibbs free energy is the maximum amount of non-expansion, reversible work that can occur inside of a thermodynamically isolated system at a constant temperature and pressure. A negative value means that the reaction is spontaneous, and a reaction at equilibrium has no change in gibbs free energy

Arjun_Anumula_3E

They are both related. Degeneracy is the specific number of ways you can achieve a certain state, while Entropy is calculated with the equation: S=kb ln(W). So while it is not a 1:1 relationship, as W increases - so does S.

Arjun_Anumula_3E

Although this doesn't occur very often in real life, the example of isothermal reversible expansion explains this concept very well. Since the process occurs in infinitesimal steps, all of the added heat to the system is converted into work. As mentioned in the previous reply, the internal energy st...

Arjun_Anumula_3E

I don't get why the volume basically doesn't change at all in the first step of an irreversible expansion. In my head, it makes sense for the system to expand quickly if the external pressure was quickly reduced.

Arjun_Anumula_3E

Could someone explain why the work for an irreversible process, like a constant pressure expansion, requires less work?

Arjun_Anumula_3E

I'm pretty sure we don't have to know the derivation of the equation, we just have to know how to use it. This might also include solving for W since Lavelle went over this in lecture, using:
W = (number of possible states)^number of particles

Arjun_Anumula_3E

In an isothermal reversible expansion, the reason the temperature stays the same is because it is a very slow process and all the heat that enters the system is converted into work that then expands the system. This is different from irreversible expansion where the temp decreases first. I was also ...

Arjun_Anumula_3E

I believe in cases of constant pressure, like the closed system example Lavelle gave on Friday's lecture, qp is equated to delta H, NOT delta H (rxn).

This is straight from the Outline 3 document:

Distinguish ∆U and ∆H and show how they are related. When q = qP then ∆U = ∆H - P∆V

Arjun_Anumula_3E

A state property just means that the path that you take to get from start to end does not affect its value. Energy, temperature, mass, density, etc. are all state functions because their values don't depend on the pathway to get there. However, if you look at work and heat, the specific path matters...

Arjun_Anumula_3E

I would say that a negative delta H always means that heat was given off. As we might learn later in this course, total the change in gibbs free energy could be positive when delta H is negative and vice versa. But yes, the sign on the delta H can indicate whether or not heat left or was introduced ...

Arjun_Anumula_3E

Since each of the 6 carbons is already attached to a hydrogen before, you don't need to worry about breaking and/or re-forming those bonds in your calculations. Drawing the structure as shown above on a problem like this definitely helps!

Arjun_Anumula_3E

Since the amount of energy that the copper gives off is equal to the amount of energy that the water takes in, you can set the equations up as so:

qw=-qc

(57)(4.18)(Tf-22) = -(20)(0.38)(Tf-100)

Solve for Tf, it should be between 22 and 100.

Arjun_Anumula_3E

Because it allows us to still perform calculations on the energy change within a system without having to know the specific amounts of moles or reaction equations. Take the first question from the week 3 discussion section. All we have to know is the heat capacity (in this case calculating it from a...

Arjun_Anumula_3E

All you need is an understanding of stoichiometry here. So they tell us that 358.8 kJ of Energy is absorbed/required per mole of this reaction. So for the first part, since there is 1 mol of s8/mol of rxn, we can just multiply the 358.8 kJ /mol rxn * 1.25 mol rxn to get 448.5 kJ. Do the same stoichi...

Arjun_Anumula_3E

Yes, you explained it pretty well. Irreversible processes do not return to equilibrium (e.g. turn from expansion back to compression) with only a small change of say the external environment. This would be the case in say an isothermal reversible expansion, where the outside pressure matches the ins...

Arjun_Anumula_3E

Knowing how much work was done on/by the system as well as how much heat was added or removed from the system help determine the change in internal energy, since Delta U = q+w. This is important because energy is constantly being transformed from one form to another, leaving certain systems and ente...

Arjun_Anumula_3E

Since the final product (liquid) has more energy than the reactants (solid), energy must have been added into the system. Hence, it is exothermic. However, I was kinda confused why the Delta H fus representing melting, because doesn't fusion mean bringing particles together??

Arjun_Anumula_3E

when more [H+] is introduced into the system, the solution becomes more acidic

Arjun_Anumula_3E

pH in a solution is not only determined by the acid dissociation constant. It's concentration also will affect the pH. It is very possible to have a weak acid that is just present in a very large concentration that then creates a very low pH. This would not be the case if we took a strong acid like ...

Arjun_Anumula_3E

catalysts can only affect the rate at which a reaction proceeds toward equilibrium, and as a result adding or removing it does not affects any values in the equilibrium equation. Only changes in temperature, concentrations of products or reactants, and pressure could cause a shift toward equilibrium.

Arjun_Anumula_3E

the cations in group 1 and 2 are conjugates of strong bases and so they would not reform bases in solution. That is why we consider them spectator ions. The conjugates of weak acids or bases, however, will react at equilibrium to change the pH. For, example, the CO3 2- in K2CO3 would reform HCO3- at...

Arjun_Anumula_3E

For every mole of N2O4 that is produced after the extra product is added, 2 moles of NO2 are used up. I thought of the reaction going in the opposite way, so I put N2O4 on the numerator and NO2 on the denominator of the equilibrium expression.

Arjun_Anumula_3E

When product is removed at equilibrium, the reverse reaction decreases and "doesn't push back" as hard. Thus, equilibrium shifts right as the reactants react to create more of the product that was lost.

Arjun_Anumula_3E

Basically adding heat will always cause an endothermic reaction to become more favorable (since adding heat helps get the required energy for the reaction). So when heat is given off in a reaction (exothermic) the reverse reaction will need be endothermic. That is why the reverse reaction, or formin...

Arjun_Anumula_3E

There is almost always going to be a set of products that is more or less stable than the reactants. This is because unique configurations all have unique energies due to their arrangements and chemical makeup, so that's why its rare to find products with the same favorability of formation.

Arjun_Anumula_3E

Net ionic equations help us easily see what needs to go into the equilibrium constant equation. The spectator ions, if placed into the equation, would just cancel each other out.

Arjun_Anumula_3E

When Q<K, it means that the ratio of product/reactant is less than what it should be at equilibrium. So to increase this ratio to get to equilibrium, reactants must get used up to make products so the forward reaction would be favored in this instance.

Arjun_Anumula_3E

Hybridized orbitals overlap head on and cannot form pi bonds. Pi bonds can only result from un-hybridized p orbitals that do not allow free rotation and overlap side to side

Arjun_Anumula_3E

Thanks for the explanation. However I am still confused on how this would apply to atom like Helium. Even though it has multiple electrons, wouldn't the only available orbital be 1s? Because there are no other values of l in the first energy level right? "This degeneracy of orbitals with the sa...

Arjun_Anumula_3E

I believe in multiple electron systems, all orbitals cannot be degenerate due to electron-electron repulsion that distinguishes for example 2s from 2p at different energies. However, in single electron systems, without such repulsion, all orbitals will be degenerate. And as stated above by Darlene, ...

Arjun_Anumula_3E

Work function is just the energy required for an electron to be ejected from the metal by a photon. The overall equation is Photon Energy (hv) - Work Function = Kinetic Energy of electron (0.5mv^2)

Arjun_Anumula_3E

That's right, increasing the intensity simply increases the quantity of photons in a given beam of light, and it doesn't change the Energy of any single photon.

Arjun_Anumula_3E

Since the d-orbital is in the same principal energy level as a given s and p, leaking electrons from the d orbital to form bonds in period 3+ elements can create a molecule with a lower energy (more stable). This is because if a molecule was forced to place electrons into say the 4s level instead of...

Arjun_Anumula_3E

a pH can be be below 0 in the case of superacids, where the chemical potential of protons is higher than it is in pure sulfuric acid. This causes the mixture to behave as if there is greater than a 1.0 molarity of Proton, for example 1.1. Taking the negative log of this molarity would result in a ne...

Arjun_Anumula_3E

Strong acids are not given a Ka because they are assumed to almost fully dissociate in water. The Ka values would be so large (since the equilibrium would favor the forward reaction) that it would not be productive to compare the Ka's among strong acids. However, weak acids often favor the reverse r...

Arjun_Anumula_3E

Having both the chlorines on the same side of the platinum allows a coordination complex to form to the adjacent guanines on the DNA molecule. However, in the trans form, such a complex could not form. This is why the cisplatin is a much more effective chemo drug as it prevents DNA replication

Arjun_Anumula_3E

Similar to how Dr. Lavelle explained how cisplatin works with DNA, we may be given a new scenario where we have to apply concepts about coordination compounds. I'm guessing that many biological compounds form chelates (ligands which form rings that include the central metal atom), so it may be worth...

Arjun_Anumula_3E

Im pretty sure the answer is "triamminetrichlorocobalt(III)"

However, I thought that if there would be 2 of the same prefixes, then you would have to change one. Why wouldn't the first tri- change to tris- ?

Arjun_Anumula_3E

I believe we have to 'memorize them', or at least some of the common ones. Here is a list to them! https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.09%3A_Strong_Acids_and_Bases I think it's pretty easy to memorize that...

Arjun_Anumula_3E

A perfect example of an amphoteric molecule is water, because it could either gain a proton to become hydronium or lose a proton to become hydroxide. Another example, is HSO4-, which is an important buffer in biological systems. It is formed after hydrogen sulfate donates a proton, and the resulting...

Arjun_Anumula_3E

Take glucose and HCl. Both are polar molecules that have dipole-dipole interactions, and have hydrogens with partial positive charges. However, HCl separates (dissociates) into ions in solution, whereas individual glucose molecules stay together and are each surrounded by water molecules (they disso...

Arjun_Anumula_3E

Whenever pH or acidity is mentioned, it always means that there is an accompanying solution. Without water or solution, these acids would be normal molecules that are not actively donating protons. It is only until acid comes into contact with water (such as pouring HCl onto your skin) causes the ti...

Arjun_Anumula_3E

In solution, an anion can be stabilized by ion-dipole interactions with electropositive hydrogens for example(-15 kJ/mol). This would reduce the energy of the anion, stabilizing it.

Arjun_Anumula_3E

The strength of Bronsted Acids is often determined by how strong the H-A bond is. For instance, a hydrogen bonded to the oxygen is much more easily ionized in solution when compared to say methylamine. If you were to view 3-D electrostatic potential maps of these molecules, the stronger acids will h...

Arjun_Anumula_3E

I'm pretty sure there isn't really a shape we assign to a central atom that only has 1 bond. If anything, there needs to be at least 2 bonded atoms to determine a proper angle.

Arjun_Anumula_3E

That is correct, the lower energy hybrid orbitals will make sigma bond(s) first, and the remaining un-hybridized p orbitals can make higher energy pi bonds.

Arjun_Anumula_3E

Similar to any other expanded octet atom, we would only be able to determine the shape if we knew which atoms and how many were bonded. For example, XeF2 would have 3 lone pairs and be linear, but XeF4 would have 2 lone pairs and be square planar. In summary, since Xe has no formal charge, match the...

Arjun_Anumula_3E

The axial atoms are the ones that are on the y-axis, and the equatorial ones are on the x-axis. This is similar to thinking about the axial atoms like the north and south pole, and the equatorial ones are the ones along the equator

Arjun_Anumula_3E

Yes, the hybrid model would be correct

Arjun_Anumula_3E

Can someone explain why the textbook says that NO2- is a radical (2C #1). I believe that it should be NO2 that is a radical, not NO2-.

Arjun_Anumula_3E

Unless there are 2 identical elements, there will almost always be some difference in electronegativity between atoms in a bond. However, in the case of C-H, that difference is so low that no significant dipole-dipole interactions ever arise. However, if the difference starts becoming higher than th...

Arjun_Anumula_3E

The more the difference between electronegativity there is, the more ionic character a covalent bond will have. Having a majority ionic character will cause the substance to be structured in more of a lattice, as the atoms will act more as ions rather than share electrons with a covalent bond. Havin...

Arjun_Anumula_3E

dipole-dipole interactions are found between 2 polar molecules (such as H-F and NH3), whereas dipole-induced-dipole is when a polar molecule casuses a non polar molecule to become polar to shifting electron density (H-F and O=O). Dr. Lavelle mentioned that they both have a strength of -2 kJ/mol

Arjun_Anumula_3E

Basically in molecules with stronger intermolecular forces, it becomes harder to melt or boil it. In a molecule like water (H20), the difference in electronegativity between the hydrogen and oxygen makes the dipole interactions much stronger than in H2S, and as a result the water molecules want to &...

Arjun_Anumula_3E

The structure with the least distribution of formal charge would be the most stable - also placing negative charges on more electronegative atoms will increase the stability of the molecule. However, in many cases there are many "equivalent" resonance structures. The example Dr. Lavelle me...

Arjun_Anumula_3E

So then how come Chlorine doesn't make 7 bonds? Is there even a maximum number of bonds an atoms can make?

Arjun_Anumula_3E

So yes, in general having O formal charge across a molecule makes it the most stable. However, if there happens to be charge in the molecule, it would be best to place negative charge on the most electronegative elements, and positive charge on the most electropositive elements. For example, it woul...

Arjun_Anumula_3E

Atoms with a larger radius would have lower electronegativity due to having a lower effective nuclear charge. Adding electron shells increases the shielding effect and prevents the protons in the nucleus from "holding on" as tightly to the valence electrons. So it would be harder to attrac...

Arjun_Anumula_3E

For elements besides the ones Dr. Lavelle showed in class (Phosphorous and Sulfur), how are we supposed to know if say an element with d or f electrons would be more stable with 4 or 5 bonds? Is there some rule or concept that tells us why PCl5 is more stable than PCl4?

Arjun_Anumula_3E

Lewis Acids are compounds that accept electrons, such as boron in BF3. Lewis bases are compounds that donate electrons, such as F- donating 2 electrons to Boron in BF3. The resulting bond is a coordinate covalent bond, meaning that both electrons in the new B-F bond both came from the same species)....

Arjun_Anumula_3E

So yes, in general, a higher charged cation would have more polarizing power. However, if you were to compare to cations of the same charge (e.g. Mg 2+ vs Ca 2+), then the atom with the smaller atomic radius would have a higher polarizing power. This would be due to the fact that the smaller cation ...

Arjun_Anumula_3E

Having a negative formal charge on a more electronegative (and vice versa having a positive formal charge on the more electropositive atom) will be more stable. Since oxygen is more electronegative, it is thus more stable with a negative charge. This helps us determine which Lewis structures are mos...

Arjun_Anumula_3E

Tha Aufbau principle is also called the building up principle, and it means that electrons will (generally) fill lower energy orbitals before filling higher energy ones. For example, 2s would be filled before electrons start to fill 2p.

Arjun_Anumula_3E

Adding an electron to a filled valence shell would cause for the energy of the molecule to increase as the electron would have to occupy a new shell. This is why noble gases often do not react in nature due to the fact that they are very stable and electrons neither want to leave or join.

Arjun_Anumula_3E

The term delocalized means that the electron density of a particular double bond or charge is not bound to one electron and the lowest energy (most stable) form of the molecule involves that electron density being distributed across multiple areas of the molecule.

Arjun_Anumula_3E

Usually, we can calculate the valence electrons by looking at the s and p orbitals. So Dr. Lavelle just counts across a period and skips over the f and d blocks, to count the number of electrons that would be in the outermost shell.

Arjun_Anumula_3E

The gaseous state minimizes all other potential forces from interfering (such as the increased particle-particle interactions in solids and liquids), so that the Ionization energy can most closely depict how much energy is needed to remove one electron.

Arjun_Anumula_3E

H hat is generally used in higher level courses with more complex derivations, but as long as you know how to interchange the constants there should be no issue. H hat is the reduced form of planck's constant, with quantizes angular momentum. H hat = h/2pi

Arjun_Anumula_3E

Generally, a +1 would indicate px - However, I remember Dr. Lavelle stating that it was more important that we stay consistent with whatever is provided to us (it would be given to us on an exam) versus trying to remember what is what - since different books go by different rules.

Arjun_Anumula_3E

Spin states have to with the angular momentum of an electron. +1/2 means the spin is up, and if you use the right hand rule you'd see that the electron would be going counter-clockwise. And vice versa for the -1/2 spin

Arjun_Anumula_3E

After the lecture, I still have a question about 4s and 3d subshells. Does the 4s subshell have more energy than the 3d one? Some other online sources were saying otherwise so I'm not quite sure what is true

Arjun_Anumula_3E

An electron in an excited state will be in a different subshell or shell, so the configuration would be different. Although the total number of electrons would stay the same.

Arjun_Anumula_3E

They are temporary induced dipoles that occur in all molecules, but are most observed in non-polar molecules. This effect, caused by the rapidly shifting electrons in their electron clouds, creates temporary regions of partial positive charge and temporary regions with partial negative charge. These...

Arjun_Anumula_3E

effective nuclear charge of an element is defined as the atomic number of the element - the number of non-valence electrons. Going across across a row increases the number of protons but does not increase the number of shielding (non-valence) electrons. Thus there is a higher pull from the nucleus o...

Arjun_Anumula_3E

That symbol is used to relate 2 values that are proportional to each other. For example, in Planck's equation E=hv, the energy of a photon is directly proportional to the photon's frequency

Search found 107 matches