CHEMISTRY COMMUNITY

JonathanS 1H

Yes, we can use Q to determine delta G at any point in a reaction, but it will still be delta G rather than just G because Gibbs free energy refers to amount of energy available to do work, and if there is no change, no work is being done.

JonathanS 1H

K_w is the equilibrium constant for the autoprotolysis of water. Because this is equal to the concentrations of OH- times H+, it is also equal to Ka times Kb

JonathanS 1H

You can use PV/T due to the ideal gas law, but I believe you would need to multiply by a constant to get the units for entropy, J/K from atm.L/K.

JonathanS 1H

One defining quality of a catalyst is that it is not used up during the course of a reaction. In the mechanism, it will be includes, but on both sides of a certain step. Also, is the activation energy for a specific reaction is lowered, it is likely there is a catalyst involved in the reaction.

JonathanS 1H

The A is named the pre exponential factor because in the arrhenius equation, it is a constant that is before the exponential part of the equation.

JonathanS 1H

The Arrhenius equation allows us to model how temperature affects the rate constant of a certain reaction. If the reaction can be modeled by it, the plot of ln(k) vs. 1/T will be linear.

JonathanS 1H

The equilibrium constant K is equal to the concentration of products divided by the concentration of reactants of a reaction at equilibrium.

JonathanS 1H

You are correct. Microscopic reversibility is the assumption that the reaction mechanism does not change, but is only reversed when a reaction is reversed.

JonathanS 1H

The equilibrium constant, K has no units. If you mean rate constant, k, then the units are mol.L^-1.s^-1 for a zero order reaction, s^-1 for a first order reaction, and L.mol^-1.s^-1 for a second order reaction.

JonathanS 1H

A galvanic cell has a positive E^o_cell value, and therefore the redox reactions and flow of e- is spontaneous. An electrolytic cell has a negative E^o_cell value, so the flow of electrons requires an external source of current to cause the redox reactions to occur.

JonathanS 1H

Yes I believe your understanding is correct. When calculating E o of a cell, we use the natural E o of reduction values for E o cell = E o R - E o L . This applies even when the cell is unfavorable but you add a current to cause a redox reaction. Reduction always occurs at the cathode and oxidation ...

JonathanS 1H

We always want to use an inert metal such as Platinum because it is not involved in the chemical reaction, only in the transfer of e-.

JonathanS 1H

Another way to look at it is using the same process, but adding one step. After we balance using H+, we add an equal amount oh OH- to each side to cancel the charge. Then we combine the OH- and H+ to make H2O and remove excess H2O so that it is only on one side of the equation.

JonathanS 1H

Because you are given E^o for the whole reaction, we can use E^o = E_cathode - E_anode, all we need is the E value for the other half reaction and arrange the equation to give us the given E^o value.

JonathanS 1H

The oxidizing agent is the substance that causes oxidation in another substance. The same is true for reducing agents.

JonathanS 1H

The answer to both questions is that Hg (l) is the anode in this case. With a liquid anode, we include the liquid but since water is our solvent and not involved in either the anode or cathode, we do not include it in the cell diagram.

JonathanS 1H

If voltage is positive, we have a galvanic cell which creates a flow of e- from the anode to the cathode. If we have negative voltage, we have a flow of e- from cathode to anode, which means we do not have a working cell.

JonathanS 1H

I believe we know that Co is solid because it is in its natural state without a charge. With Ti 3+ and Ti 2+, thy have a charge, which means they were part of a dissolved salt, and are therefore aqueous. The only time we have ions as a solid are in salts.

JonathanS 1H

Under acidic conditions, we use the H+ ions while balancing the equation. Under basic conditions, we use the H+ initially, but balance the charge with H20 and OH- in order to have the reaction in a basic environment.

JonathanS 1H

When we balance a redox reaction with alkaline conditions, we initially balance it with H+ ions, but to make the reaction alkaline we add H20 and OH-, balancing the charge of the H+ since there are no H+ ions in an alkaline solution. If it is under acidic conditions, we use the H+ ions.

JonathanS 1H

Entropy is equal to the boltzmann constant times the natural log of (number of states)^{(number of atoms/molecules/objects)}.

JonathanS 1H

Entropy total is equal to 0 when there is a reversible reaction since the reaction is 'at equilibrium' throughout the whole time. Entropy of the surroundings is equal to 0 with an expansion against a vaccum.

JonathanS 1H

If you use the G = H - TS, with a - H and + s, you will have a spontaneous reaction. If you have a +H and -S, the reaction will be non-spontaneous. If both are positive, the reaction will be spontaneous if TS is greater than H. If both are negative, the reaction will be spontaneous if H is greater t...

JonathanS 1H

Because an irreversible reaction is spontaneous, we know that, according to the 2nd law of thermodynamics, the total entropy in the universe must increase. This tells us that the change in total entropy must be greater than 0.

JonathanS 1H

The Vant Hoff equation is used to relate the equilibrium constant of a reaction to the changes in enthalpy and temperature. It is used to find how a temperature change affects the equilibrium constant of a reaction.

JonathanS 1H

I guess my question is more of how does this equation prove that entropy always increases in a system.

JonathanS 1H

With U, we only consider it at specific points in time rather than considering how specifically the internal energy of a system has changed.

JonathanS 1H

Im pretty sure in this case that q rev = q p because we can consider phase changes to be both reversible and at constant pressure. An easier way to think of this is the transition from ice to water and vice versa under constant pressure because the system in this case is only having heat added to it.

JonathanS 1H

Could you please post the exact question number? I would like to confirm, but I believe it would be something like 2HCN --> 2C + H₂ + N₂.

JonathanS 1H

Hello everyone
Can someone explain the second law of thermodynamics in relation to the equation (delta)S = q_rev / T? I am confused as to what this equation means in a conceptual sense.

JonathanS 1H

I think the reason you assume constant volume is because it is in an insulated container which would be completely sealed and not able to expand, hence no change in volume.

JonathanS 1H

It does not matter because (delta)H=q_p as long as the system is under constant pressure.

JonathanS 1H

In a closed system, you can change the energy by adding heat and/or doing work on the system. I am less certain with the isolated system, but I am pretty sure there should be no net change in the total energy of the system.

JonathanS 1H

When we discuss the thermodynamic universe, we are referring to a system and its surroundings. If energy is transferred from a system to the surroundings, it remains in the universe. Energy cannot transfer from the universe or into the universe, because then there would just be more surroundings to ...

JonathanS 1H

Check Dr. Lavelle's website under outlines. The ones we are going through now are thermochemistry and thermodynamics.

JonathanS 1H

We can use -p(delta)v when we have constant pressure, because we can take of the p from the integral since it is a constant. In some way, we are always using the integral, but with constant pressure we can simplify it to make life easier. In the case of pressure not being constant, we must use the i...

JonathanS 1H

My textbook says the answer should be 4.90x10² which is the answer you have. I believe your answer is correct.

JonathanS 1H

When it states they are in the same state, that means there are no phase changes in the chemical reaction. The standard reaction enthalpy is the enthalpy of a specific reaction with standard conditions, such as 25 degrees C, 1 atm, and all products and reactants are in the same phase.

JonathanS 1H

We know to use the phase change from a liquid to gas, H_vap because that is what is demonstrated in the equation. We then use the H for breaking bonds because the Br₂ becomes 2Br.

JonathanS 1H

Using bond enthalpies is inaccurate because all enthalpies for everything except diatomic molecules are calculated averages, and not measured.

JonathanS 1H

The standard enthalpy of formation refers to the formation of a substance from it's most basic elemental components in their standard states. The case when it is 0 is when we consider elements being formed from their basic elements. The example in class was O 2 becoming O 2 . Because there is no cha...

JonathanS 1H

I'm not quite sure what you are exactly asking, but the endothermic/exothermic reactions refers to the reaction requiring or releasing heat. Endothermic reactions require heat to occur, while exothermic reactions release heat as a product.

JonathanS 1H

I am fairly certain that we will not be using that equation in this class, and if we do, I'm sure Dr. Lavelle will explain/derive it in lecture. I have never heard of that equation until it was mentioned by a student in class.

JonathanS 1H

Technically, we can use the equilibrium sign in almost every chemical reaction as there usually is a point of equilibrium. The reason we do not use it for strong acids and bases is because the concentration of reactant is so small, it is basically 0.

JonathanS 1H

To start off, you want to change all gram measurements into moles by dividing by the grams/mole of each substance because we are dealing with equilibrium. From the equation, we know that 2 moles of NH 3 are formed per mole of CO 2 formed, so we multiply our moles of CO 2 by 2 to get our moles of NH ...

JonathanS 1H

If you are asked specifically for molar concentrations, you would give the moles/Liter rather than the pH or pOH.

JonathanS 1H

In the description of the answer, it states that you can compare the equilibrium constants of the dissociation of F₂ and Cl₂ at a certain temperature. Because the K of Cl₂ is lower, it dissociates less and is therefore more thermodimanically stable.

JonathanS 1H

Because it is an increase in pressure by compression, the moles of each gas does not change, but the volume does. The reaction shifts right for a little bit, but the overall K value does not change.

JonathanS 1H

You can use partial pressures for ICE tables if you are given them and K_p. We also use partial pressures in the ideal gas equation and in calculation K_p.

JonathanS 1H

I believe we can because the Ice tables are used to help with the calculations of concentrations, and if we use K_p, we can, and have to, use partial pressures in the calculation.

JonathanS 1H

K = 1 is rare and difficult to achieve because it requires that the concentrations of both the products and reactants are equal to each other, which is fairly rare in most actual reactions.

JonathanS 1H

When a ligand binds to a central cation, it is chelating. Chelating is just another term for bonding used specifically for ligands.

JonathanS 1H

Myoglobin is a different compound from Hemoglobin. In Hemoglobin, four Heme groups rather than myoglobin groups.

JonathanS 1H

An acid or base is polyprotic if it can give off or accept multiple protons, respectively. NH₄⁺ is not polyprotic because once it donates one proton, the molecule becomes stable to the point that it will not donate another proton.

JonathanS 1H

Something is a Bronsted acid if it gives off a proton to water in solution. A bronsted base removes a proton from water in solution. A lewis acid accepts an electron pair, while a lewis base donates an electron pair.

JonathanS 1H

The coordination number is the number of ligands around a transition metal cation in a coordination compound. In this case, there are 6 CN^- groups around one Fe²⁺ cation, so the coordination number is 6.

JonathanS 1H

A molecule with resonance is a molecule that has delocalized bonding. Also, for acids and bases, if a central atom is attached to highly electronegative atoms, the delocalization of the charge will increase.

JonathanS 1H

The coordination number refers to the number of coordinate bonds surrounding a cation, while the oxidation number refers to the charge of a transition metal.

JonathanS 1H

Bronsted acids donate protons or lose H+ ions, and bronsted bases accept protons or gain H+ ions. Conjugate acids and bases are bronsted acids or bases, but correlate with a bronsted acid or base. A bronsted acid loses an H+ to become a conjugate base. A bronsted base gains an H+ to become a conjuga...

JonathanS 1H

Are the molecules in this equation in the aqueous state? If so, the NaOH acts as a base, and C0₂ acts as an acid, combining to create a complex salt and water. NaOH reacts with water to produce OH- and CO₂ reacts to form a bronsted acid, which donates a proton to OH-, creating H₂0.

JonathanS 1H

Coordinate covalent bonds occur when a covalent bond is created by the sharing of 2e- from a single atom with another atom without lone pair e-. As this is a covalent bond, it has the properties of a covalent bond.

JonathanS 1H

The roman numeral corresponds with the positive charge, or oxidation state, of the metal in a coordination compound. They should always be included in the name right after the metal.

JonathanS 1H

Bronsted acids are proton, or H+ donators. Bronsted bases are proton acceptors. Conjugate acids are bronsted bases that have accepted a proton. Conjugate bases are bronsted acids that have donated a proton. Lewis acids are electron acceptors, while lewis bases are electron donators.

JonathanS 1H

Other than molarity calculations we did ad the beginning of the quarter, we have not done any other calculations with pH besides the ones you have mentions, so I assume so.

JonathanS 1H

Hydrogen bonding occurs between hydrogen bonded to a F, O, or N and a lone pair of electrons on one of those molecules. London dispersion forces occur between all atoms/molecules and are the same as induced dipole-induced dipole forces in which random e- movement causes areas that are partially char...

JonathanS 1H

Hydrogen bonds do not strengthen covalent bonds. Covalent bonds are between atoms, while hydrogen bonds are between molecules. They increase the boiling point of substances because they are another bond that must be broken to turn a substance into a gas from a liquid. The most notable interaction is...

JonathanS 1H

Ion-dipole occurs when an ion, such as Na+ interacts with a polar molecule, such as water. The partially negative oxygen will interact with the Na+ ion. Induced dipole-induced dipole occurs between all atoms/molecules and happens when random electron movement causes partially charged regions of the ...

JonathanS 1H

The lone pair e- for O are not drawn, but there are 2 lone pairs because O needs an octet. This means O will have 4 bonding regions, and thus a tetrahedral shape with bond angles close to 109.5 degrees.

JonathanS 1H

I don't think we need to know the specific bond angles, but we should know NH 3 bond angles are more than H 2 0 angles because we can look at how many bonding pairs there are versus how many lone pairs there are. The bond angles in H 2 0 are less than NH 3 because H 2 0 has more lone pair electrons,...

JonathanS 1H

Lone pairs are not bound to another atom, so the repulsion forces are not weakened by protons in another atom.

JonathanS 1H

The strength of London forces are dependent on the same factors as dipole-dipole bonds. If two molecules are more rod shaped, the bond will be stronger than if they were spherical. The closer the molecules are, the stronger the bond will be.

JonathanS 1H

Yes, larger differences in electronegativity cause polar covalent bonding. However, certain molecular shapes, such as tetrahedral, tend to not have a dipole moment because the dipole moment is actually a vector sum between all polar bonds in the molecule.

JonathanS 1H

The fluctuation in electron distribution causes certain regions of an atom to become partially charged. This partial charge in one atom has the ability to cause the same effect in a nearby atom due to the repulsion forces of two charges of the same kind. Another name for this random fluctuations of ...

JonathanS 1H

Fluctuating dipoles and electron fluctuation refers to the random movement of electrons around an atom/molecule. This allows regions to become partially charged, which in turn causes regions of another molecule/atom to become partially charged. This causes attraction known as van der waals/induced d...

JonathanS 1H

The same way you would any other atom. It helps to write out the e- configurations in order to do this, and the valence e- will be the number of e- in the outermost layer. Most transition metals have 2 valence e-. Is this because the d orbital tends to fill either halfway or completely full? Most h...

JonathanS 1H

If it is a true ionic bond, yes, because drawing a bond represents a covalent bond.

JonathanS 1H

A dipole moment occurs in covalent bonds between atoms of differing electronegativities. To draw a dipole moment, draw an arrow from the lower electronegative atom towards the higher electronegative atom. The vector sum of these arrows will be the overall dipole moment for the molecule.

JonathanS 1H

Atoms can have more than 8 e- when they have subshells greater than s or p. Atoms can use these orbitals to bond even if they do not have e- in them to begin with, but are in the third row of the periodic table or lower.

JonathanS 1H

To draw ionic compounds, draw the lewis structures of each ion individually, and label the charge of each and put them in brackets.

JonathanS 1H

The same way you would any other atom. It helps to write out the e- configurations in order to do this, and the valence e- will be the number of e- in the outermost layer. Most transition metals have 2 valence e-.

JonathanS 1H

H, He, Li, and Be all want to fill only the 1s orbitals, only wanting 2 e- total. As for the atom with the least ionization energy being the central atom, the lower the ionization energy, the more the atom will hold on to its e-. This means that it will be more likely to bond with multiple atoms and...

JonathanS 1H

These numbers represent the observed length of each bond. The delocalization of e- means that the e- is not around one specific atom, but around the entire molecule as a whole.

JonathanS 1H

No, for certain atoms with d-orbitals, there can be more than an octet surrounding a specific atom. The example we went through in class that shows this is the sulfate ion, SO₄^2-.

JonathanS 1H

The Quantum number that describes the orbitals in a subshell is m_l. The formula for this is l, (l-1),..., -l.

JonathanS 1H

Yes, I believe that all atoms with a d-orbital or higher have the ability to be an exception to the octet rule. These orbitals do allow for more than 8 electrons around a singular atom while drawing Lewis structures.

JonathanS 1H

I believe you always put the closest noble gas with atomic number less than the atom you are referring to in brackets to save time and space while writing e- configurations.

JonathanS 1H

Hello everybody,
Is it safe to assume that for the purposes of this class, we will only be referring to the Heisenberg Indeterminacy equation in one dimension for $\Delta x$ ? Would there be a case in which we used this equation in three dimensions (x,y,z)?

JonathanS 1H

I assume you are asking why these electron configurations are they way they are. I understand that the electron configurations drop a 4s e- because it is easier (takes less energy) to have more stable shells such as 3d^5 or 3d^10. This is the example for Cr and Cu, but it also applies further down t...

JonathanS 1H

The way I understand the exceptions is that it takes more energy to have 2 e- in the 4d orbital than to have one in each sublevel evenly distributed (hence Cr: [Ar] 3d^5 4s^1). This also applies for copper in which the 3d shell is full.

JonathanS 1H

If an electron is in a plane, it is aligned with a certain plane (x,y,z). I'm not quite sure what you are asking for, but I hope this helps.

JonathanS 1H

Electron diffraction has to deal with electrons having wavelike properties, and atomic spectra deals with light as a wave and quantized levels of electron energy. I am not quite sure about black body radiation though.

JonathanS 1H

Although it would be helpful to read all chapters in the quantum world, I believe you would be okay to read the chapters that are associated with homework problems. this will not only help you with doing the homework, but also gain a better conceptual understanding of the quantum world for tests, etc.

JonathanS 1H

No, we did not discuss it in depth in class, only mentioned it briefly. I believe we will be discussing it tomorrow. You can double check the outlines on the website to be sure.

JonathanS 1H

I believe the correct answer would be the photoelectric effect for a few reasons. The first being that only certain frequencies of light are able to eject electrons from the metal. If light only had wavelike properties, the total amount of energy (including amplitude and frequency) would eject elect...

JonathanS 1H

I believe it will depend on the number of electrons in the orbitals. If there are two in a specific orbital and one is removed, the other will remain in that orbital. If there is only one in a certain orbital and it is removed, the next electron will be housed in the orbital before. Also, I believe ...

JonathanS 1H

I believe the reason for this is back in the idea that energy cannot be created or destroyed. When an electron goes from a higher to a lower state, there has been a decrease of energy, but that energy must go somewhere. In the case of electrons it is released as light.

JonathanS 1H

So they are a type of particle though?

JonathanS 1H

Hi everyone,
I noticed that on certain homework problems, such as E.9, we are not given the molecular formula of a compound, but rather the name of the compound. Does anybody know if this will be the case on tests, or will we be given the formula as well as the name?

JonathanS 1H

I could be wrong, but I believe that units are not used because of the convention for atoms and formula units. Because there is no unit besides "atom(s)" or "formula unit(s)" that could be used, you can leave them off. I'm not 100% sure about that though.

JonathanS 1H

Hi everybody,
I have a few questions about photons that came up during lecture last week. Are photons a type of particle like electrons/neutrons/protons? Or are they what we use in order to describe light with the properties of particles? I'm just getting a little confused...

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