CHEMISTRY COMMUNITY

Christine Wastila 1H

Lecture outline 4 says that we need to be able to explain how the collision model and activated complex model account for the temperature dependence of reactions.

Christine Wastila 1H

Remember also that emf represents the maximum potential difference.

Christine Wastila 1H

Either method will give you the same answer.

Christine Wastila 1H

This ratio determines the factor by which the catalyst increases the rate of reaction.

Christine Wastila 1H

Yes, AB is is the intermediate because it is produced in one reaction and consumed in another. It is completely consumed.
The overall reaction is: 2AC+B -> A₂B+2C. Therefore, A₂B is not formed using AB as a reactant.

Christine Wastila 1H

If a compound or element is both produced and used up, it is an intermediate regardless of how reactive it may be. Most intermediates are very reactive, which is why they are typically short-lived molecules that quickly convert to a more stable form. In other words, they are used up fast.

Christine Wastila 1H

Yes. Although test 3 only went to 15.6, the final will cover the rest (except 15.9).

Christine Wastila 1H

I don't think so. Rate is always a positive value, and since rate=k[A], then k cannot be negative.

Christine Wastila 1H

Yes because that would indicate that the species has undergone two half-lives.

Christine Wastila 1H

The test covers 15.1-15.6, which corresponds with homework problems up to 45 (and some integrated exercises).

Christine Wastila 1H

1st order reaction: t_1/2=0.693/k
2nd order reaction: t_1/2=1/(k[A]₀)

both can be found on the constants and equations sheet :)

Christine Wastila 1H

The other two responses are correct, but check the integrated exercises as well if you want some more practice.

Christine Wastila 1H

I think we would be given the steps and asked to analyze whether the proposed mechanism is valid (i.e. whether or not it agrees with the experimentally determined rate law). The example from lecture today is probably a good indication of what future problems will be like.

Christine Wastila 1H

I assume you mean in a chemical equation. In this case, since catalysts are neither reactants nor products, they are written above the arrow. If that's not given, I think it would have to be stated in the problem.

Christine Wastila 1H

They do not affect the rate law as written: rate=k[A]^order (i.e. you would not input different values than those given for concentration or oder because of the stoichiometric coefficients).

Christine Wastila 1H

Also remember that the rate law is determined by the slow step, so if the rate law is given, you know which one is the slow step because they have to correspond.

Christine Wastila 1H

We may not always be told explicitly, but it will be obvious by the information given. For example, if we are told the rate law, than we know which one is the slow step because they have to correspond.

Christine Wastila 1H

I saw on Dr. Lavelle’s website that test 3 will cover 15.1-15.6. Which homework problems does this material align with? I think I have an idea, but I just want to be sure I don’t leave anything out.

Christine Wastila 1H

Both of the other replies are correct, but I have one thing to add for clarity. The source of activation energy is usually heat, so reactions with a higher activation energy have a slower rate because it takes longer for the reactant molecule(s) to absorb enough heat in order to form products.

Christine Wastila 1H

If E is negative, then the process is called electrolysis, which is where an electric current is used to drive a non-spontaneous reaction. In a Galvanic cell, chemical energy is converted to electrical energy. The opposite is true of electrolysis.

Christine Wastila 1H

I think you would want to compare using two equations where the amount of electrons transferred is the same.

Christine Wastila 1H

Since you have different amounts of electrons being transferred, you want to find the least common multiple of those two coefficients. In the case of 3 and 2, the least common multiple is 6. So multiply the reaction with 3 electrons on the reactant side by 2 and multiply the reaction with 2 electron...

Christine Wastila 1H

The anode is on the left and the cathode is on the right. Solids are always on the outside (add Pt if none) and aqueous are closest to the salt bridge (I don’t think order matters if there are multiple).

Christine Wastila 1H

You don't necessarily have to draw out each form, but having a model of at least one of them will help you visualize how many different arrangements the molecule has. Drawing all of them can check your answer if you have time.

Christine Wastila 1H

The third law states that at 0K, the entropy of a perfect crystal is 0. Therefore, as T approaches 0, entropy also approaches 0. lim_T->0S=0.

Christine Wastila 1H

Just to add on, notation for the salt bridge would be ||

Christine Wastila 1H

9.13 is listed in the solution manual errors page on the class website. It should be C instead of R for calculating entropy change due to a change in temperature.

Christine Wastila 1H

The question asked if work was done. We know this is true for this situation because W=-PdeltaV. Since the volume of the balloon decreased, deltaV is negative, which shows that work was done on the system (the balloon).

Christine Wastila 1H

If given concentration or pressure, you could calculate Q and compare it to K if that value is given.

Christine Wastila 1H

That statement refers to the enthalpy of a reaction.

Christine Wastila 1H

Calorimeters are not isolated systems. Bomb calorimeters are able to exchange heat with the surroundings, therefore classifying them as closed systems.

Christine Wastila 1H

The Boltzmann equation relates entropy to the number of microstates. A microstate is one of many possible molecular arrangements for a given substance. More of these microstates indicate higher entropy.

Christine Wastila 1H

Just so you know, this won't be on the first test, and Dr. Lavelle said we would cover this topic in week 3.

Christine Wastila 1H

In terms of numbered problems, the only ones I see that have isothermal reversible expansion are 8.11 part B and 8.12 part B, so just omit those.

Christine Wastila 1H

The answer, 1626 kJ refers to how much work can be done by the fuel in the cylinder. You subtract 947 kJ because this amount of energy was lost to the surroundings as heat.

Christine Wastila 1H

Sara Varadharajulu wrote:I think what she means (as this was the same for me too), if you do you homework in pencil for example, get something wrong and correct it in pen, and turn in that marked up homework, you still get full credit.

Yes, this is what I meant :)

Christine Wastila 1H

I looked this topic up the other day and found this useful site: http://www.brighthubengineering.com/the ... processes/

Christine Wastila 1H

Cv denotes the heat capacity of a gas at a constant volume while Cp denotes the heat capacity of a gas at constant pressure. We have two values because, when heating takes place at constant pressure, some of the heat is used to do expansion work rather than raise the temperature of the system. For a...

Christine Wastila 1H

It's ok if you try one and get it wrong as long as you correct your answer. Last quarter, I would do my corrections in another color if needed, which earned me full credit.

Christine Wastila 1H

Lower values of pKa or pKb relate to stronger acids and stronger bases, respectively. So to answer your question, no. An extremely weak base would have a high pKb value.

Christine Wastila 1H

The equilibrium constant is only changed by temperature. However, the reaction quotient Q may be more or less than K depending on if the reaction is at equilibrium or not.

Christine Wastila 1H

This is more of a Lewis structure question. Why is the structure for SO2 drawn with a single bond on one of the oxygen + 6 valence electrons and a double bond on the other + 4 valence electrons rather than double bonds on both + 4 valence electrons? I have it with S in the middle + a lone pair and t...

Christine Wastila 1H

Whether or not something is an acid or base does not change no matter what definition you use, but the Bronsted/Lewis definitions focus on different characteristics. Bronsted acids donate protons while Bronsted bases accept them. Lewis acids accept electron pairs while Lewis bases donate them. Howev...

Christine Wastila 1H

I think if the total molecule has a charge, you should bracket it and indicate the charge like you would with any other Lewis structure.

Christine Wastila 1H

To summarize, the number of orbitals should match the regions of electron density. So if you have AX4 or AX3E, it would be sp3 (1 s orbital + 3 p orbitals=4 total) because there are 4 regions of electron density in both cases even though one of those regions is a lone pair in the latter example.

Christine Wastila 1H

To give you a quick answer:
single bond=1 sigma bond
double bond=1 sigma and 1 pi bond
triple bond=1 sigma and 2 pi bonds

Christine Wastila 1H

Be sure to make sure that the dipoles do not cancel, though. This can be determined by looking at the VSEPR model. For example, CO2 is nonpolar because of its linear form even though O is more electronegative than C. Since the dipoles are directly across from each other, they cancel.

Christine Wastila 1H

Look for sizable differences in electronegativity. For example, C-H bonds do not cause partial charges because there is not a large difference in electronegativity between C and H. However, a C-N bond or a C-Cl bond could cause polarity because N and Cl are more electronegative than C, causing a par...

Christine Wastila 1H

I think since Kp only applies to gaseous equilibria, changes in pressure would not cause changes in aqueous equilibria.

Christine Wastila 1H

Monodentate ligands only have one electron donor atom that binds with the central atom/ion. Thus, they are only attached at one point. Polydentate ligands are able to attach to the central atom/ion multiple times due to the presence of multiple electron donor atoms. In many cases, ligands are monode...

Christine Wastila 1H

Just to add on, ligands can be mono- or polydentate depending on how many donor atoms they have. A ligand with one donor atom can bind to the central atom once, making it monodentate. A ligand with multiple donor atoms can bind with the central atom more then once, making it polydentate.

Christine Wastila 1H

Q is the reaction quotient. By comparing this value to K, we can determine how the reaction will progress. If Q<K, the formation of products will be favored. If Q>K, the formation of reactants will be favored. If Q=K, then the reaction is at equilibrium.

Christine Wastila 1H

In response to the previous comment, yes, that is possible. A positive oxidation number represents how many electrons an atom can accept, while a negative oxidation number represents how many electrons an atom can donate.

Christine Wastila 1H

Just to add on, because lone pairs have a stronger negative charge than bonding pairs, they cause a higher degree of repulsion, which is why they appear to take up more space.

Christine Wastila 1H

I believe Dr. Lavelle said that the book uses a newer convention, but both are correct.

Christine Wastila 1H

There are two m values: ml and ms (l and s are subscripts). ml describes which orbital in a subshell an electron occupies and ms describes the spin. Depending on which orbital and which subshell the electron is in, ml will have one of these values. s: ml=o (one possible orbital) p: ml=-1,0,+1 (px, p...

Christine Wastila 1H

Effective nuclear charge is the net charge an electron experiences in a multiple electron atom. Shielding occurs when electrons in the inner shells prevent electrons in the outer shells from feeling the full positive charge of the nucleus. Thus, electrons closer to the nucleus have a higher effectiv...

Christine Wastila 1H

The 4s orbital is lower energy than the 3d orbital for Ca and K because they do not have electrons in the 3d state. However, after the 4s orbital is occupied and electrons enter the 3d state, 3d is then lower energy than 4s. For example, the electron configuration for Scandium is Sc: [Ar] 3d1 4s2.

Christine Wastila 1H

4s is lower energy than 3d for Ca and K because they do not have any electrons in the d block. After the 4s orbital is occupied and electrons enter the 3d state, 3d is then lower energy than 4s. That is why electrons in the 3d block are listed before those in 4s when writing the electron configurati...

Christine Wastila 1H

Just to add on, since the Lyman series have a shorter wavelength, they indicate that a greater amount of energy is being absorbed (because E=hv, v=c/wavelength). If you recall the model from lecture, this means the electrons are making a bigger "jump", such as from the first level to the s...

Christine Wastila 1H

You need to divide the work function by Avogadro's number (6.022x10^23) because the it was given in kJ/mol and the question asks about the energy needed to remove one electron from ONE sodium atom, not a mole of them. I got confused on that, too! Hope this helps.

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