CHEMISTRY COMMUNITY

Chris Lamb 1G

No class today.

Chris Lamb 1G

We learned about mechanisms earlier. Some enzymes (maybe even all enzymes) lower the energy by creating a different mechanism for the reaction. This new mechanism has a smaller energy barrier.

Chris Lamb 1G

We don't have a way to calculate A without using the equation we saw in class and on our formula sheet.

Chris Lamb 1G

I think Lavelle prefers reactants listed before products, but I don't expect this to be graded too strictly.

Chris Lamb 1G

These are ways to approximate the heat capacity of a gas without needing to be given experimental data.

Chris Lamb 1G

This will not be on test 3

Chris Lamb 1G

Lavelle stated in class that it would take a lot of chemistry knowledge to come up with an accurate intermediate on your own. He doesn't expect us to do that.

Chris Lamb 1G

The rate has to have units that match with the rate constant, k. k usually has s^-1 in its units, but there is a chance that a problem will give a constant with minutes or something.

Chris Lamb 1G

Some problems ask for answers in specific units, so it's important to read the question carefully.

Chris Lamb 1G

Although those are possible, we don't have to worry about them.

Chris Lamb 1G

Will we be talking about negative order, mixed order, or fractional order?

Probably not. We don't have to worry about that, unless he mentions it in class.

Chris Lamb 1G

I do not think that we will deal with non-interger reaction orders in this class.

Chris Lamb 1G

We assume that the reaction is going in the forward direction

Chris Lamb 1G

Each order produces a straight line when the y axis is set to a specific function of concentration.

Chris Lamb 1G

We haven't learned any convention for that specific order, but I would imagine that a rule likely exists for making very long reactions more organized. We don't have to worry about it for this class, though.

Chris Lamb 1G

This should be given in the problem.

Chris Lamb 1G

In the answer book, the cell diagram shows:
CD(s) | Cd(OH)₂(s) | KOH(aq) || Ni(OH)₃(s) | Ni(OH)₂(s) | Ni(s)

Why are there bars and not commas in between the solids, since they are the same phase?

Chris Lamb 1G

In the cathode of the cell diagram, Ce 4+ (aq) and Ce 3+ (aq) are both the same phase, so they require a conductor, Platinum. But why does the solutions manual put platinum on the anode side as well? I - (aq) and I 2 (s) shouldn't need a conducting metal, right? Is it a rule that we have to add the ...

Chris Lamb 1G

I would recommend knowing the derivations of the Van't Hoff equation and the work equations, just to be safe. He will not ask us to "derive this equation," but he may ask something that would require us to know how the equation is made.

Chris Lamb 1G

These values are all equivalent, they are just in different units. Depending on the units of the given information, these other values can be very helpful.

Chris Lamb 1G

That is just two ways of writing the same equation. They may look different, but the negative sign is just on different sides of the equation. You can use properties of logs to see why the k2/k1 becomes k1/k2.

Chris Lamb 1G

This is based off the equation $\Delta U=\frac{3}{2}nRT$ , which shows that with no change in temperature, there would be no resulting change in internal energy.

Chris Lamb 1G

∆S tot is the same as the entropy of the universe. According to the Second Law of Thermodynamics, the universe tends to shift towards disorder, so ∆S cannot be less than zero. Thus, a negative ∆S tot is not only nonspontaneous but also considered a forbidden process. This seems to contradict the en...

Chris Lamb 1G

Yes, but isn't ∆S(total) the entropy of the universe? Because the system + the surroundings = the universe. And the entropy of the universe can never decrease.

Chris Lamb 1G

For a nonspontaneous process, ∆Stotal < 0.
How is it possible for the total entropy to decrease? Since ∆S(total)= ∆S(sys) + ∆S(surr), doesn't that mean that ∆S(total) is the ∆S of the universe?

Chris Lamb 1G

The equation is not listed on the formula sheet. Does this mean that we need to know how to derive the equation for the midterm?

Chris Lamb 1G

I think we learned that at equilibrium, ∆G=0.
But the equation ∆G=-RTln(k) seems to contradict this, since it is calculating a nonzero ∆G value at equilibrium. Are these two ∆G values different?

Chris Lamb 1G

One reaction releases energy (like breaking ATP). This energy can be used to drive a reaction that is not favorable (like pumping ions across a barrier).

Chris Lamb 1G

Unlike energy, entropy can never be zero (unless you are at 0K).

Chris Lamb 1G

At 0K, there is no movement. The only disorder is resulting from position of the molecules. That is why residual entropy is positional entropy.

Chris Lamb 1G

We were sent an email saying "office hours are Wednesday's 2-4pm."

What about the Friday office hours? I am unsure if this is stating that office hours will only be on Wednesdays from now on or if this means that Fridays are staying the same and Wednesdays are changing slightly.

Chris Lamb 1G

I thought it was odd that you are expected to assume one mole. It seems like something that you can't assume; you need to be given it? But it this a safe assumption for future problems?

Chris Lamb 1G

Boltzman's constant will also be provided on the formula sheet as 1.38x10^23 J/K

Chris Lamb 1G

Molecules with more "parts," such as electrons, protons, and other atoms bonded, have more possible orientations and vibrations than simpler molecules. This means they have more entropy.

Chris Lamb 1G

This depends on your TA, but I think we should be doing chapter nine problems this week, since all of our lectures this week will be chapter 9.

Chris Lamb 1G

I was under the impression that when calculating the standard reaction enthalpy, we use kJ/mol. For just regular reaction enthalpy, I would put kJ. But I would like to know if that is true or not.

Chris Lamb 1G

When doing combustion problems (such as finding the enthalpy of combustion), is the water produced in the liquid or gaseous state?
Problems usually don't give this information.

Chris Lamb 1G

I encountered the same issue. It is a mistake in the textbook.

Chris Lamb 1G

Does ΔU =0 in isothermal expansion?
Isothermal means ΔT = 0, so q=0.
But since there is expansion, doesn't that mean there must be work being done? Doesn't a change in volume guarantee that work is nonzero? I would think that ΔU = w in this situation.

Chris Lamb 1G

Here are the big keywords/hints for the q and w equations.

"Final Temperature" usually suggests using q=mCΔT, since ΔT is Tfinal - Tinitial.

Similarly, when a final and initial volume is given or mentioned, you are likely to use w = PΔV

Chris Lamb 1G

8.73 asks us to calculate the reaction enthalpy using bond enthalpies. In the answer manual, all of the answers are in units of kJ/mol. Normally, reaction enthalpies are just in kJ. Why are they in kJ/mol? Although bond enthalpies are in units of kJ/mol, we are multiplying them by the moles of bonds...

Chris Lamb 1G

We are considering it a closed and fully insulated thermos. Realistically, some heat will be lost, because nobody has invented a perfect insulation. But for the purpose of this problem, we are considering it 100% thermally insulated.

Chris Lamb 1G

I had the same issue. I really don't think it should be positive like the answer book says.

Chris Lamb 1G

There are equations where the moles of product is the same as the moles of reactant, so Delta n = 0. If the moles of gas is equal on both sides, does that mean that there is no work being done, since $\Delta n RT = 0$ ?

Chris Lamb 1G

There are different heat capacities for different states, so you need to know the waters's state. Also, you have molar heat capacity as well as specific heat capacity, and the one you choose depends on the units of your given information.

Chris Lamb 1G

Yes, heat capacity increases with molecular complexity.

Chris Lamb 1G

I see PV = nRT as fair game, because it was in Chem14A. However, he has not mentioned anything about this yet in 14B.

Chris Lamb 1G

[quote="Marina Georgies 1C" Sidenote- where is Dr. Lavelle's list of errors posted? I can't seem to remember.[/quote] It is a PDF on the 14B page titled "Solution Manual Errors" Here it is: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_6...

Chris Lamb 1G

Most of the diatomic elements are gasses at their standard states; however I₂ is a solid, and Br₂ is a liquid.

Chris Lamb 1G

The Acids and Bases topic was cut off in my 6-week summer session for Chemistry 14A. Is that topic necessary for 14B? I still plan on learning it eventually either way, as I will probably need it in the future.

Chris Lamb 1G

In the solution manual for question 55, one of the steps seems to break the rules. In order to get the O 2 molecules to cancel out, the solution manual multiplies one of them by 3/2. However, this coefficient is not multiplied to the rest of the equation; only the oxygen molecule was changed. I do n...

Chris Lamb 1G

The solutions manual and the textbook are the same as in 14A.

Chris Lamb 1G

So, this rule would only count for the naturally diatomic elements: H, O, F, Cl, Br, I, N?

Chris Lamb 1G

Why is the standard enthalpy of formation of O2 gas 0kJ/mol?
(This is from example 8.11 on page 296).

Chris Lamb 1G

Scientific notation is often used to remove any ambiguity about the number of sig-figs.

Chris Lamb 1G

You should be able to change the equations so that you can cancel things. For example, you could reverse the direction of one equation. Or you could multiply an equation's coefficients by 2. These changes would make it possible to cancel something that needs to be cancelled. I do not think there wil...

Chris Lamb 1G

The textbook says "At the melting point, the temperature remains constant, because all of the heat is being used to melt the sample."
So there is an energy barrier that must be overcome before a liquid could become a vapor.

Chris Lamb 1G

I am guessing that we are starting at page 278, at the section titled "Enthalpy." That would correspond to questions starting at 8.29.

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