CHEMISTRY COMMUNITY

Justin Le 2I

You can tell by looking at the Lewis structure. In this case, both of them are on top.

Justin Le 2I

Thank you!

Justin Le 2I

I think the above post would be a great addition to the course reader.

Justin Le 2I

The numbering starts at the top of the double bond on the left size. Since this is an E molecule, the substituents have to be on opposite sides, which is why the chain looks like it is continuing to nine, but there are actually 8. Let me know if you are still confused.

Justin Le 2I

Thanks. Is isopropyl considered a functional group though?

Justin Le 2I

I don't understand why the numbering wouldn't start at the isopropyl group and then go counterclockwise. The numbers would then be 1,2,and then 4. The answer starts at the OH and goes clockwise so the numbers are 1,2,5.

Justin Le 2I

I would play it safe and remove the dash.

Justin Le 2I

When you draw it, it doesn't matter because the molecule is identical. Just rotate your paper 180 degrees and you'll get the other view.

Justin Le 2I

Adding onto Chem_Mod, the increase of entropy is why the free energy of the transition states will always be higher than the free energy of the reactants.

Justin Le 2I

I think we can assume that the reactions will be exothermic because they will usually be spontaneous. If the reaction were endothermic, it would not be spontaneous.

Justin Le 2I

Also, molecules with double or triple bonds are nucleophiles even though they may be neutrally charged because the double and triple bonds are electron rich regions.

Justin Le 2I

There is no fluorine in this problem, but I can tell you why it is trans. Each of the C of the double bond are sp2 hybridized and you look at what is connected to those C to see what you prioritize. On the left side, you prioritize H3CH2C because if you add up all the atomic numbers, it is higher th...

Justin Le 2I

Adding onto what Chem Mod said above, if you see CC then there is probably a triple bond and the structure is an alkyne.

Justin Le 2I

You can usually get the lowest numbering scheme if you start at the C with the most substituents to follow the number rule that Chem Mod said above. So here, yo u see that there are two methyl groups on a C so if you start there, you'll probably get the right answer.

Justin Le 2I

So in this case, does the Br add onto the central C?

Justin Le 2I

Double bonds are usually nucleophiles because their pi bonds are electron rich.

Justin Le 2I

I am confused how the molecule of 2-methyl-1,3-butadiene was found. I was looking for two double bonds that fit the 1,3 sequence but I couldn't. The answer key added a double bond onto the 3. Why does it do this and why is it right?

Justin Le 2I

In Dr. Lavelle's lecture, he said the dashed line of the intermediates for reaction profiles in the course reader should be lowered to the same line as reactants. However, the intro to ochem book and the answers for the homework problems do not do this. Which is right?

Justin Le 2I

Yes, you should put brackets and the double plus sign when writing out your activated complex. The double plus sign is used to indicate that there is a transition state. The Ar is technically in the reaction, but it appears in both the reactants and products. It gets canceled out which is why you do...

Justin Le 2I

It also helps to remember that activated complex diagrams are intended to show a middle transition phase. The reactants could continue to go on and form products (if they have enough energy and the right orientation) or go back and form reactants.

Justin Le 2I

Also, the increase in temperature will have a bigger impact on the larger activation energy. So if you have a reaction profile of an exothermic reaction, the activation energy of the reverse reaction will be more responsive to the temperature increase.

Justin Le 2I

Catalysts are consumed and then produced so there is no net change. Intermediates are produced and then consumed. So in chemical equations, you will see catalysts first appear on the reactants side and you will see intermediates first appear on the products side.

Justin Le 2I

Both techniques will get you the same answer but in this class, I think we are just going to use the pre-equilibrium approach that Dr. Lavelle went over in class.

Justin Le 2I

Since the slope is equal to -k and the slope is -3, then k = 3. k is always positive.

Justin Le 2I

k is also always positive because rates are always positive.

Justin Le 2I

Leave out H2O unless the problem explicitly says that water is not the solvent.

Justin Le 2I

The only part of this question I do not understand is part f. Can anyone explain please?

Justin Le 2I

Adding on to what Neil said, don't confuse kinetics constants with equilibrium constants. When we did equilibrium in 14A, we used stoichiometric coefficients for the exponents. It makes sense to use coefficients when working with elementary steps because the order depends on the molecularity of the ...

Justin Le 2I

Remember that when you are working with kinetics, the exponent is not always equal to the stoichiometric power like we did for equilibrium constants (K and Q).

Justin Le 2I

For 14.47b, why do you think that rate =k*[NO2]^2*[O]? Neil's explanation is correct.

For 14.49, the answer is k[NO][Br2]. I think you might have read the solutions guide wrong.

Justin Le 2I

I think there is supposed to be a - sign on the Cl atom in 45. Does anyone else agree?

Justin Le 2I

If you look at the bottom of page 53 in the course reader (sorry, I have the 2014 course reader) or the second page in the kinetics section, you can see why it is better to use the fraction and to put the coefficient in the denominator. Otherwise, you would have to use improper fraction or a least c...

Justin Le 2I

A lot of it was experimental because the k value is experimentally determined for a certain temperature.

Justin Le 2I

The Lewis structures would be useful if you were calculating the enthalpy of reaction by using bond enthalpies, but we're not using that method here and it's also more inaccurate.

Justin Le 2I

The solutions manual makes this a little confusing because it says P_Br(g)^2 but it should actually be without the exponent because we are just solving for the pressure of Br(g)

Justin Le 2I

since deltaG = deltaH - TdeltaS, we can't determine if a reaction is spontaneous if we are simply given deltaS. We need more information to see if deltaG is less than 0

Justin Le 2I

They are equivalent and you will just use different coefficients when calculating H and G of reaction.

Justin Le 2I

The way I think of it is the concentration on the reactants side is decreasing and the concentration on the products is increasing. So the side with 1.0 M is decreasing and the side with 0.0010 M is increasing.

Justin Le 2I

It's usually better just to use RT/nF because you will get the exact answer and you can change the values if needed.

Justin Le 2I

Ka means a substance is dissolving in solution, so the equation will look like HF---> H+ + F-

Justin Le 2I

Usually any substances that are aqueous are closest to the double line. If you want to know what goes on what side, just look at the table of standard reduction potentials and use the one that has the elements that you have.

Justin Le 2I

Many of the problems that involve natural log have rounding errors. You'll be ok and you can always argue with your TA if you get it wrong.

Justin Le 2I

You can have the Ecell be negative. It just depends on what the question is asking.

Justin Le 2I

I think you can use either because they will both get you the same answer.

Justin Le 2I

There is nothing stopping the gas from expanding so if it's thermal energy increases then the volume will increase.

Justin Le 2I

You would add Pt to one of the sides if you need a metal electrode that can conduct the electrons. For example, if you have aqueous, gas, or liquid, then you would use Pt.

Justin Le 2I

It's also useful to think of it as Q = [P]/[R]. If you have more products, then Q>1 and then the reaction favors the products. It's better just to use the ratio and think about it intuitively then to memorize the different cases.

Justin Le 2I

You can probably get away with ordering it anyway, but some graders prefer oxidation/reduction.

Justin Le 2I

Don't forget to write out the states when you're doing the cell diagram. That will make it much easier.

Justin Le 2I

When I write cell diagrams, I write H+ and OH- just to be thorough but I leave out H20. Some problems require H+ or OH- to calculate pH, such as 43a.

Justin Le 2I

I think Hg (l) is an exception and can be an electrode because I see it being used as one in 19c and 43a.

Justin Le 2I

I don't think so because in redox reactions, the substance that is being oxidized is transferring electrons to the substance that is being reduced. There are no e- just lying around.

Justin Le 2I

Some of them are easy to memorize like H2 -> 2 H+ + 2 e-

Justin Le 2I

Did you make a typo when writing this out?
2H20+2e--->H2/2H2O+2e--->H2+2OH-

Justin Le 2I

I would use natural log because then you don't have to memorize a special number. You can just use the constants and you'll get a more accurate answer.

Justin Le 2I

I think part of the explanation is because ln ab = ln a + ln b. So it's like doing the concentrations and pressures separately and using log rules to combine them, which makes it easier on the calculator work.

Justin Le 2I

I think the best way to find Q in problems using the Nerst equation is writing the half reactions and then writing out the full reaction by adding the half reactions. This way, it is clear to you what is happening and you can simply plug in the products and reactants into Q.

Justin Le 2I

For this problem, why is n=1 instead of 2? I used 2 because of the standard reaction 2 H+ + 2e- -> H2.

Justin Le 2I

Also, I think the solutions got the 15 from the two significant digits after doing this: 1.52 V - 1.33 V

Justin Le 2I

For this question I got 1.1x10^2 because the voltage has two digits in the mantissa. Am I wrong?

Justin Le 2I

You will use the inert electrode such as Pt when you need a metal to carry the e- from the anode to the electrode. For example, if you have a gas and an aqueous, then you would use the inert electrode. You would also use it if you had all aqueous.

Justin Le 2I

I think one of the reasons why these problems have so many different answers is that there are multiple answers to making cell diagrams. I would include the OH- on both sides just in case.

Justin Le 2I

I have the same problem and I feel like the answer key is wrong because a nonaqueous state can be closer in the middle like in problem 13.b

Justin Le 2I

Bomb calorimeters are isolated from the rest of the universe, which makes it easier to measure heat that is released in a reaction. Since volume is staying constant in a bomb calorimeter, there is no work being done because expansion or compression cannot take place. So w=0. But heat can be transfer...

Justin Le 2I

For this problem, shouldn't the answer be 58.56 kj.mol-1, half of the answer given in the back? 58.56kj.mol-1 would be the delta G for the decomposition of one mole of HgO whereas 117.1 kj.mol-1 would be the answer for one mole of the reaction where the equation is 2 HgO(s) --> 2 Hg(l) + O2(g).

Justin Le 2I

Figure 7.16 on page 249 has a good visualization of the difference between reversible and irreversible processes. Note that the pressure in the reversible graph is not constant whereas the pressure in the irreversible graph is constant.

Justin Le 2I

You might also see an equation that looks like q=CdeltaT. A capital C means heat capacity, and the two that are listed in the first post are just different types of heat capacity.

Justin Le 2I

Another way to think of the equation is heat lost by system = heat gained by surroundings and vice versa. This logically makes sense provided that there is no heat being lost.

Justin Le 2I

I would recommend remembering the log rule where the exponent can be turned into a coefficient. If the equation is left as it is, your calculator might overflow like with one of the homework problems.

Justin Le 2I

Degeneracy can never equal 0. The lowest value it can get to is 1 which occurs at absolute zero because the system only has configuration without any kind of energy. Entropy, however, can equal 0 when w=1.

Justin Le 2I

You will know if you need to use a negative sign depending on if you're measuring the delta S of the system or the surroundings. Look at Example 8.12 on page 312-313 for an example of where the book uses both the positive and negative equation.

Justin Le 2I

I think the pictures on page 312 are also really helpful in understanding why the total change in S is zero for a reversible system and why deltaSsurr for irreversible systems is zero.

Justin Le 2I

What Martha said about heat capacity is specifically for monatomic gases.

Justin Le 2I

Adding on to the previous comment, if you notice that you know three of the values, then there is a good chance that you will have to use PV=nRT.

Justin Le 2I

For clarification, you don't need to convert Celcius into Kelvin because the size of the two units of temperature is the same.

Justin Le 2I

These two pathways are similar to when we use multiple reversible pathways to describe an irreversible pathway. We just take it in terms of one change at a time and depending on which you choose first, you will get a different answer.

Justin Le 2I

For this equation:
Delta H= H(vapor) - H(liquid)
Are you talking about the delta H of the entire heating process?

Justin Le 2I

In general, you don't use the bonds broken - bonds formed method unless the problem specifically says so because the other two methods are much more accurate. The values in the charts that give us bond enthalpies are just averages, which is why there are discrepancies between the bond method and the...

Justin Le 2I

After you solve for the delta H of that reaction, you will get your answer in kJ. You can then decide if you have to add a per mole to your answer depending on what the problem is asking.

Justin Le 2I

Also, for this problem there needs to be two separate pathways because doing both at the same time is too difficult. It is easier if we split them up into two parts, one where the pressure changes and one where the volume changes. This is similar to when we have to separate irreversible changes into...

Justin Le 2I

The size of the molecule doesn't affect the heat capacity of atoms or molecules, the complexity of the atom or molecule does. On page 255 there is a table that gives the molar heat capacity at constant volume and constant pressure for three types: atoms, linear molecules such as CO2 and nonlinear mo...

Justin Le 2I

Also, be careful about which R you pick because they have different units. Look at the units in your calculations and pick the one that fits.

Justin Le 2I

For these kinds of problems, one way to check if you have the right answer is to see what units your answer is and see if it makes sense. In the context of this question, the units kJ/year would be most appropriate so you would try and get your final answer with that unit.

Justin Le 2I

Just a note:
I think 2H2 (g) + O2 (g) --> 2H2O (g) is better written as H2 (g) + 1/2 O2 (g) --> H2O (g) because enthalpies of vaporization and fusion are given for just one mol of the product.

Justin Le 2I

Yes, delta E and delta U are used interchangeably. It just depends on the book and personal preference.

Justin Le 2I

Yes you could use that equation, but I think the equation that the solutions manual is easier to understand and you don't have to memorize. Enthalpies of vaporization and fusion are always given in kJ/mol, so if you use the first equation then the units make sense.

Justin Le 2I

Oh sorry the question is 7.57, not 5.57.

Justin Le 2I

The solutions guide says that the final answer is in kJ.mol-1, but I think that it should be in kJ because you multiply the final answer by 1.00 mol. Any other opinions? Thanks.

Justin Le 2I

In an isothermal process \Delta U=q+p and since \Delta U=0, q=-w. The way of it is when work is being done on a system, the system must be losing heat at the same time and vice versa. Otherwise, the temperature and thus the internal energy would change. Check out this link for more clarification: ht...

Justin Le 2I

To add onto the previous point, each axis (x,y,z) contributes 1/2RT of translational energy. That is why the molar internal energy of a monatonic atom is 3/2RT.

Justin Le 2I

Chem mod also answered the question here:
viewtopic.php?f=76&t=4999

Justin Le 2I

You will be using kj most of the time because problems typically ask how much heat is required or released. kJ/mol is the unit used for standard enthalpies, which you can see in the tables throughout the chapter or the appendix. Just think about the question is asking for and see if you need to use ...

Justin Le 2I

It takes more energy to break down the intermolecular forces between solids and liquids than for liquids and gases. In liquids, molecules are moving around a little more, but in gases, the molecules are moving a lot more. That's why the heat of vaporization is so much larger than the heat of fusion.

Justin Le 2I

Two examples of path functions are heat and work, because they describe how the system reached equilibrium, not the system at equilibrium.

Justin Le 2I

It might be easier to understand if you move the negative sign, so it would look like this:
-heat lost by metal = heat gained by water

Justin Le 2I

The problem doesn't specify that there is a constant external pressure so I don't think we can use w=-P $\Delta$ V.

Justin Le 2I

Lavelle has also said in another post on Chemistry Community that when writing the formula, the anion ligands should be named before the neutral ligands.

Justin Le 2I

Evan Huang 3L wrote:It can also be a complex with an overall positive charge, for example

[Ni(en)₃]³⁺

has the name: Triethylenediamine Nickel (III) ion

The name should actually be tris(ethylenediamine)nickel(III) ion because en is bidentate and there are three of them.

Justin Le 2I

When writing out the formula, you write the cation transition metal first, then the anion ligands, then the neutral ligands.

Justin Le 2I

Gina Young is correct. My previous comment should only apply to complexes.

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