CHEMISTRY COMMUNITY

BNgo_2L

Intermediates are first seen as products and then are used as a reactant for another step. Catalysts are used first as reactants to speed up a mechanism and then reappear as products as they are never consumed in a reaction (used up and then reform). The overall reaction equation should only include...

BNgo_2L

Dr. Lavelle said that the pre-equilibrium approach is much simpler than the steady state approach, but it is less flexible. However, if done correctly, they will both have the same results.

BNgo_2L

Activation energy is determined by the amount of energy needed for the reactants to go to the transitional state, so on a graph, the activation energy will be the amount of energy from the level of the reactants to the peak (transitional state) of the graph.

BNgo_2L

You can determine the unit of the rate constants for each order reaction by using the rate laws to derive k.
(ex. second-order: mol/L s = k [mol/L][mol/L] --> k = L/mol s)

BNgo_2L

Dr. Lavelle said we would only be using the pre-equilibrium approach since it is simpler than the steady-state approach.

BNgo_2L

When delta G(naught) is positive then more reactants are present than products at equilibrium and vice versa when it is negative.

BNgo_2L

These coefficients were experimentally determined, so they are given. They're also on the equation sheet given to us during tests.

BNgo_2L

2.0^m comes from [rate 2's R]^m/[rate 1's R]^m after cancelling out all the constants --> ([rate 2's R]/[rate 1's R])^m = 2.0^m

BNgo_2L

The equilibrium constant disregards solids and liquids as the concentrations essentially do not change. Thus adding or removing water would not shift the equilibrium in either direction.

BNgo_2L

Also, 10^-3 < K < 10^3 means that neither the reactants nor products are favored.

BNgo_2L

H20 is involved in both redox half reactions. If you look in the appendix, you would use O2 + 2 H2O + 4 e2 --> 4 OH2 and O2 + 4 H1+ + 4 e2 --> 2 H2O. Also, for cell diagrams, the left is always the anode (oxidized) while the right is always the cathode (reduced).

BNgo_2L

You only add platinum/inert solid when the substance present cannot act as electrode itself (conduct). Hg can act as an electrode at its standard state, so you don't need platinum whereas NO/NO3 can't.

BNgo_2L

You would choose the direction and determine the cathode and anode in which the E(cell) is positive since the cell potential must be positive in order to be spontaneous.

BNgo_2L

That was an error in the solutions manual, so the actual equation should be Cl2(g) --> HClO(aq) + Cl-(g)

BNgo_2L

Given the cell diagram, you would write, balance, and combine the oxidation (anode) and reduction (cathode) half reactions in order to get the balanced redox equation. Balance the individual half reactions first and then balance both to each other by balancing the number of electrons transferred.

BNgo_2L

You can determine which compound is present more by determining Ka for C6H5CH2CH(CH3)NH3+. Since the Ka = [products]/[reactants] and the Ka for the compound is extremely small, you can infer that there is a higher concentration of C6H5CH2CH(CH3)NH3+ (reactant) at equilibrium.

BNgo_2L

Typically you can use the same equations for isothermal reversible and irreversible expansion. However when talking about isothermal reversible, q = -w and delta S(total)=0. For isothermal irreversible, delta S(surroundings)=0. All in all, how the equations are used depend on what's occurring in the...

BNgo_2L

In lecture today, what did Dr. Lavelle mean when he said that the max potential occurs when there is "very little current flows?" Furthermore, what does max potential in mean?

BNgo_2L

For the Van't Hoff equation, is entropy always constant? If not, in what instances would it not be constant?

BNgo_2L

For the equation, delta G = delta G (naught) + RTlnQ, what'sthe difference between delta G and delta G (naught)?

BNgo_2L

Monoatomic gases can occupy more space, so more disorder would occur than diatomic gases, which consist of two bonded atoms (more order).

BNgo_2L

It usually doesn't matter unless specified. However, the correct number of significant figures will determine if you write your answer in kJ or J.

BNgo_2L

If he didn't talk about the concept during lectures, it most likely won't be covered on the midterm.

BNgo_2L

To find delta H(vap), you need the energy required for vaporization per mole, so all you do is divide 4.76kJ by 0.579 moles.

BNgo_2L

You have a system consisting of 0.40 moles of an ideal gas contained in a 100.0L container at 1.0 atm. You just love chemistry to a fault, so you perform a series of steps to the system. First, you perform an isobaric compression of the container to 10.0L. Then, you pressurize the system to 10.0 atm...

BNgo_2L

To find the work done, you need to use delta n x R x T, and to find delta n, you would need to find the balanced equation and then calculate the total number of moles of gas in the products minus the total number of moles of gases in the reactants.

BNgo_2L

After finding the enthalpy of the reaction, you want the enthalpy of the reaction per mole of TNT, so you would divide it by 4. You would then proceed to multiply it by the density of TNT while also converting the answer to the correct units (kj/mol of TNT to kJ/g of TNT and cm^3 to L).

BNgo_2L

All the enthalpies of formation are found in Appendix 2A of the textbook. After finding the enthalpy for the reaction, you need to find how much energy is released per mole of TNT. The reaction requires 4 moles of TNT, but since you want the energy per mole of TNT, you divide the reaction enthalpy b...

BNgo_2L

In these types of problems, you only add and subtract the bond enthalpies of bonds that are broken and created. Therefore, since nothing happens to the C-H bonds, you do not have to account for their bond enthalpies.

BNgo_2L

It's easier to tell if you draw the lewis structure of the entire reaction. In this case, one O-H is broken, and one C-O and one C-H bond is created.

BNgo_2L

I think we went over something similar during lecture where Dr. Lavelle said the heat released by a reaction = - heat absorbed by the solution. The negative sign is just referring to the heat being released, which is negative. Heat released = - (heat absorbed), while - (heat released (-)) = heat abs...

BNgo_2L

An isolated system is a closed system that is insulated. Anything that has to do with bomb calorimetry is referring to an isolated system.

BNgo_2L

Since there is no work being done on the system, delta U = q, and q = deltaH x n, where n = moles. Therefore deltaU = -318 x 1.00 mol OF2.

BNgo_2L

We're still technically in the "Thermochemistry" section even though we started talking about thermodynamics. 4F is in the thermodynamics section, so no, we haven't talked about anything that will help solve 4F.

BNgo_2L

Enthalpy is the amount of heat absorbed or released during a reaction. Heat is the transfer of energy caused by a temperature difference that occurs in a system; and temperature represents the random motion of particles

BNgo_2L

ALegala_2I wrote:Does this mean that cooling an exothermic reaction will favor the products?

By "cooling," I assume you mean removing heat, so yes, cooling an exothermic reaction will remove the heat "on the product side" and will therefore favor the products (shift right).

BNgo_2L

You can also determine whether the reaction shifts left or right when volume is increased or decreased by determining if the concentration increases or decreases. When volume decreases, pressure increases and concentration also increases so Q < K and the reaction will shift right. Vice versa when vo...

BNgo_2L

We learned in 14A how to determine if a molecule is an acid or base. You can determine by its characteristics. Lewis acids accept electron pairs, so they tend to have a negative charge (Lewis acid), produce H3O+ (Arrhenius acid), and donate H+ (Bronsted acid). On the other hand, bases donate electro...

BNgo_2L

The quotient, Q, shows us the concentration of the reactants and products at any time while K shows us the unique concentrations for each when at equilibrium. Because reactions are in constant motion to reach equilibrium, you can use Q and compare to K to determine if the reaction will move forward ...

BNgo_2L

Stronger acids completely dissociate, so they will at equilibrium there will be more products (H3O+) than there will be reactants [Products]>[Reactants]. Thus, stronger acids will have higher Ka.

BNgo_2L

Liquids and Solids don't have/have minimal changes in pressures nor concentrations, so they are typically not involved in computing the equilibrium constant.

BNgo_2L

From what I've heard, Dr. Lavelle has a system that automatically checks whether we post 5 things each week. The TAs don't have any control over Chemistry Community.

BNgo_2L

His tests usually cover up to all the material up to the Friday of the week before the week before the test. So depending on if he goes over titrations this week, it may or may not be on next week's test. But, I assume that he will finish teaching all of acid and base equilibria, which includes titr...

BNgo_2L

Reactions of specific conditions have a unique equilibrium constant (unique equilibrium concentrations). Changing concentrations of any of the reactants or products will not affect K, because the reaction will occur in the direction that will return the reaction to its unique equilibrium concentrati...

BNgo_2L

Pressure only affects reactants and products when decreasing/increasing volume is involved. Decreasing the volume will increase the concentration of the gases in which [reactants]>[products] while increasing the volume will decrease the concentration of the gases in which [reactants]<[products]. Whe...

BNgo_2L

I believe Keq and Kc are the same. Kc seems to be just a type of Keq.

BNgo_2L

K is also the same as Keq, which can be specified as Kc when using concentrations or Kp when using partial pressures.

BNgo_2L

Maya Beal Dis 1D wrote:When would we need to specify and use Kc as opposed to just using Keq?

I believe we should use Kc any time we use concentrations to calculated Keq.

BNgo_2L

Ipsita Srinivas 1K wrote:Ok, just to clarify, K_c and K_p are just types of K_eq, and you label the equilibrium constant depending on whether K_eq is calculated using partial pressure or molar conc. ?

Yes, Kc and Kp are both types of Keq, and Keq is labelled based on whether you use concentration or partial pressures.

BNgo_2L

The reaction quotient demonstrates the relative amounts of reactants and products at any given particular time whereas the equilibrium constant demonstrates the favored direction of the reaction when the forward and reverse reaction rates are at equilibrium.

BNgo_2L

Strong acids tend to have weaker bonds because when bonds are more easily broken, protons are more easily given off, which is a characteristic of strong acids.

BNgo_2L

Lewis bases also tend to have lone pairs to donate while lewis acids don't. Furthermore, the F- anion has a negative charge so it will tend to donate a lone pair to stabilize it.

BNgo_2L

Lewis bases typically have a lone pair in its structure to give off while lewis acids typically don't.

BNgo_2L

He did discuss the concept that is used to solve this problem using these formulas. It was from the Dec. 2nd lecture when he discussed the relative strength of acids.

BNgo_2L

Lewis acid and Bronsted acid can be the same. When an acid donates a proton (bronsted), sometimes the lone pair from the bronsted base is given to the acid, which also makes it a lewis acid.

BNgo_2L

NH2OH is polar. The dipole moment distributions are not balanced, so there is an overall dipole moment.

BNgo_2L

Yes, they are. The bonds of a resonance structure are partial bonds so they are all the same length. The length is the average of the bond lengths of all the possible resonance structures.

BNgo_2L

The number of hybrid orbitals are consistent with the number of electron densities of the molecule determined using its VSEPR model. In hybrid orbitals, the max number of orbitals of each shape are s:1, p:3, d:5, etc. The orbitals are then filled using the number of electrons in the atomic orbital o...

BNgo_2L

Sigma bonds allow atoms to rotate whereas pi bonds are created by overlaps so the bound atoms can't rotate. When the atoms of a pi bond rotate, the pi bond breaks. Moreover, sigma bonds are always the first interactions between atoms, so that also makes it more stable than pi bonds.

BNgo_2L

In resonance structures, the number of sigma and pi bonds are not affected because resonance only affects the location of the electrons on the molecules.

BNgo_2L

Smaller cations have higher polarizing due to the minimal amount of electrons between the nucleus and the outer shell. When there is less electrons, the positive charge of the nucleus is able to distort the electron cloud of anions more easily.

BNgo_2L

Kellylin_4D wrote:What is this equation? Was it in lecture or the textbook? Any help would be great please.

Dr. Lavelle talked about it during the lecture last Friday I believe. All I understand is that it has to do with the energy contributed when two atoms or molecules are attracted to each other.

BNgo_2L

bonding-bonding pair < lone-bonding pair < lone-lone pair

BNgo_2L

VSEPR models center around the electrons/electron pairs of the central atom.

BNgo_2L

A molecule is polar when then there is a clear dipole moment, meaning there is one end that is distinctly partially negative and the other end that is distinctly partially positive. A molecule is nonpolar when there is no dipole moment or when the dipole moments cancel each other out by the same par...

BNgo_2L

Yes, dissociation energy is the same as bond enthalpy since they're both the amount of energy stored in bonds.

BNgo_2L

Typically radicals are formed by breaking bonds, so wherever the bond is broken is where the single lone electron will be. Other than that, I think they're usually placed at the central atom, such as in methyl and hydroxyl.

BNgo_2L

High polarizability and high polarizing power contributes to more covalent character because the electron cloud is being distorted towards the cation and therefore the electrons are being "shared" like in a covalent bond.

BNgo_2L

Polarizability increases as the size of the anion increases. The larger electrons cloud, the more susceptible it can be distorted since the electrostatic attraction of the outer electrons being reduced.

BNgo_2L

Polarizability, which is a characteristic of anions, increases down and left as the size increases. Polarizing power, which is a characteristic of cations, increases up and to the right as the size decreases.

BNgo_2L

Yes, the main demonstration of the photoelectric effect is to explain that light doesn't act like a classical wave when it is emitted as a particle (photon).

BNgo_2L

I think the central atom is usually the one with the least electronegativity. Like in CCl4, the carbon atom is the central atom and has the smaller electronegativity.

BNgo_2L

Elements with expanded electron shells (with electrons more than s2p6), lewis acids and bases, Group 13 elements, and radicals (compounds with unpaired electrons) are all exceptions to the octet rule.

BNgo_2L

Yes, the ionic bond has more or less covalent character depending on the anion's polarizability and the cation's polarizing power. It has less covalent character when there is the anion is less polarizable and/or the cation has high polarizing power. It has more covalent character when there is high...

BNgo_2L

Ionic bonds have covalent properties because the electrons of the anion enter the bonding region between the cation and anion as the cation polarize the anion.

BNgo_2L

Resonance structures are just different versions of the actual structure. This means that the resonance hybrid has a blend of all these structures. As a result, the bond lengths of the actual structure is a blend of both the single and double bonds thus creating equal bond lengths all around.

BNgo_2L

More charges means positive formal charge.

BNgo_2L

Dr. Lavelle told us in class that elements have expanded valence shells when they can orbitals that can accommodate more than the usual octet, so I think the exception applies to elements that go beyond s2p6.

BNgo_2L

The formal charge of each atom always adds up to the charge of the molecule.

BNgo_2L

Negative formal charge is determined when there are more lone and shared electrons than the valence electrons. This is basically the same as having more electron attraction, known as electron affinity. So all in all, negative formal charge is paired with the atom with higher electron affinity.

BNgo_2L

If the energy of the light equals the wave function (threshold), electrons will still be emitted, but there wouldn't be any excess energy (kinetic energy).

BNgo_2L

The Rydberg can be only used for calculating the change in energy between quantum levels for light emitted by H-atoms. DeBroglie's wave equation is used to determine if a particle with momentum has detectable wave properties.

BNgo_2L

Ground state electron configuration means the lowest energy electron configuration. This configuration follows all the all the principle when determining electron configuration that we learned in class. I assume that an excited state electron configuration means that there are electrons misplaced in...

BNgo_2L

Electron density is basically the probability of finding an electron in the region in question. Because electrons tend to be found in higher electron density, a region with zero electron density has probability of close to zero of finding an electron in said region.

BNgo_2L

Doubling up before occupying each state with one electron would cause asymmetry and thus make the atom unstable. By occupying each state before doubling up, the atom becomes more symmetrical and is thus more stable.

BNgo_2L

The gaps between the energy levels decrease as the levels increase because as the levels increase, they get farther from the "gravity" of the nucleus, which means that higher energy levels require less energy to remove the electrons from said level. Vice versa, lower energy levels require ...

BNgo_2L

The extent of the change in energy corresponds to the slope of the electric field. Decreasing the frequency widens the wavelength which then decreases the slope/extent of change. As for the others: Speed is constant Wavelength is inversely proportional to frequency Energy is directly proportional to...

BNgo_2L

In the photoelectric effect, intensity has little to do with the intensity of the light. Because the energy emitted in this experiment doesn't act like a wave, high intensity/amplitude does not eject electrons and produce kinetic energy from the metal surface unless the E(photon) is greater than or ...

BNgo_2L

After converting the wavelength of the spectral line to the energy it emits, you have to use E=E(final)-E(initial) to figure out the n values.You're already given the n value for the final energy as UV follows the Lyman series stating from any n value, the electron must return to ground state at n=1...

BNgo_2L

The particle part of the wave-particle duality derives from the characteristics of photons as they act as bundles of electromagnetic radiation when an electron emits it.

BNgo_2L

(1) Find the moles of each reactant. Converting the mass to moles will allow easier comparison between different reactants. (2) Use molar ratios in order to calculate the required moles needed for the reaction to occur. (3) Compare the required moles to the available moles (from step 1) to determine...

BNgo_2L

Given that we're only reviewing the fundamentals right now, it's not necessary to indicate how much energy was used nor how much was produced. But I think later on when we learn different concepts such as thermodynamics, we'll have to indicate the energy (heat) input/output in a combustion and other...

BNgo_2L

In order to find the moles needed to produce a certain product or needed to react with another reactant, you should always compare molar ratios using the limiting reactant. This is because the limiting reactant is always the first to run out, so it determines how much product can be produced or how ...

BNgo_2L

P4 isn't considered when finding the limiting reactant of the second reaction between O2 and P4O10 because P4 isn't its own reactant in the second reaction.

BNgo_2L

Yes, you need to use the information in the first reaction to solve the second reaction. After finding the limiting reactant of the first reaction, find the remaining moles of the excess reactant as well as the moles of P4O6 produced. You can then use that information to find the limiting reactant, ...

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