CHEMISTRY COMMUNITY

Bryan Lau 3H

In this problem, we didn't get to decide. The equation was already given, so we just matched the oxidation / reduction reactions based on the overall reaction given.

Bryan Lau 3H

The problem tells you there's 50.0 mL of water, and 1mL = 1g => 50.0mL water = 50.0g water

Bryan Lau 3H

We consider tert and sec to be prefixes, while iso and neo to be part of the name (no "-"). I think it's just something we have to memorize. Just remember iso and neo are the only ones that are taken into account for ABC order.

Bryan Lau 3H

Hey Joseph! We have to know how to name a molecule with a functional group, double/triple bond, and a substituent. To answer your question, my TA said we won't be asked to name more than one functional group, unless you count double/triple bonds as another functional group.

Bryan Lau 3H

Because NO has 2 orientations: ON or NO, and oscillation between these two possible configurations contribute to higher residual entropy. BF3, on the other hand, has only 1 possible configuration. No matter how you rotate the trigonal planar, you'll get exactly the same configuration.

Bryan Lau 3H

A Cyclohexane is cis when e,a or a,e as in one is axial one is equatorial, with both facing the same direction, while trans is a,a or e,e, with both facing the opposite direction.

Bryan Lau 3H

NO because it has two possible configurations (ON or NO), while BF3 has only one possible, indistinguishable configuration (BF3 is the same no matter how you rotate it).

Bryan Lau 3H

Correct. Cis will always be a,e or e,a meaning 1 axial 1 equatorial, with both facing the same direction, while trans will always be a,a or e,e with each facing opposite directions.

Bryan Lau 3H

I think so, unless they are halogens.

Bryan Lau 3H

E in ethyl is before the M in methyl because the prefix isn't factored into alpha order. When there are multiple methyl's in a compound, you list the numbers of carbons to which they are attached and then say di/tri/tetra/methyl.

(Ex: 1,3,5 - trimethylalkane)

Bryan Lau 3H

This was a mistake made by the textbook, confirmed by Chem_Mod. You are supposed to use C :)

Bryan Lau 3H

For the quiz, #3) consider the reaction of nitrogen dioxide and fluorine gases (1) NO2 +F2 --> NO2F + F (2) NO2 + F --> NO2F How do you know which would be unimolecular, bimolecular, or termolecular? Is it determined by the reactions needed to make the products? I believe both reactions would be bi...

Bryan Lau 3H

Because .80V and .40V is the same as 0.80V and 0.40V, which is 3 significant figures Actually, I don't think you should consider the sig figs like this. 0.80V and 0.40V are still 2 sig figs. Rather, you should treat the sig figs like how I mentioned in the previous comment. Rules for sig figs are d...

Bryan Lau 3H

Because for addition the rules for sig figs are different. You round with respect to the number whose digits place ends furthest to the left. For example, .80 and .40 are both to the hundredths place, so the answer upon addition will be rounded to the hundredths place. Therefore, the answer is 1.20 ...

Bryan Lau 3H

At equilibrium, forward rate = reverse rate:

k[A][B] = k'[C][D] .

You can rearrange this to become:

[C][D] / [A][B] = k/k' .

We know that:

K = [C][D] / [A][B] .

Therefore:

K = k/k' , or kforward / kreverse

Bryan Lau 3H

So Quiz 2 covers the textbook only up to Ch15.10, but are we still responsible for the sections after 15.10 (for the final)? The Course Reader just says Read: Ch 15 (Omit 15.9), so I don't know if that means we should still read past Ch15.10 and just not worry about it for the upcoming quiz?

Bryan Lau 3H

It's because entropy and enthalpy are EXtensive properties. The more wood you burn, the more heat it will give off (enthalpy). The more of something you have, the more disordered it can be (entropy). On the other hand, E is an INtensive property. By definition, it doesn't matter how much of somethin...

Bryan Lau 3H

I don't know why, but it is okay to have both partial pressure and molar concentrations in the same calculation for Q in this kind of problem. So no, you don't have to convert them, as long as pressure is given in bar (approx. 1 atm), and concentration is given in M (mol/L).

Bryan Lau 3H

Would this help?:

http://physics.nist.gov/cuu/Units/units.html

If you scroll down you'll see stuff like Joules, heat capacity, entropy,etc.

Bryan Lau 3H

Yup, by definition.

Bryan Lau 3H

To my understanding, it's because Br is heavier (79.904g) than F (18.998g). Heavier molecules tend to have more entropy.

Bryan Lau 3H

n=2 refers to the # moles of electrons transferred. You have: half rxn 1: 2e+F2 --> 2F- half rxn 2: 1e+Rb+-->Rb For the balanced full reaction, you have to multiple 'half rxn 2' by 2 and flip it so that the electrons cancel out. You'll see that the # moles of electrons that cancel out on both sides ...

Bryan Lau 3H

Why is H20(l) omitted in the cell diagram, on both the anode and cathode side?

Bryan Lau 3H

It's a mistake by the textbook. It is supposed to be deltaS=nCln(T2/T1).

Professor Lavelle gave us this link of all textbook mistakes:

https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_6Ed.pdf

Bryan Lau 3H

The negative means it is exerting/losing that amount of energy as work. Therefore, when determining which system does more work, you can compare their absolute values. -172 J means it is losing 172 Joules as work. -101 J means it is losing 101 Joules as work. The first system is doing more work beca...

Bryan Lau 3H

Enthalpy + for endothermic rxn - for exothermic rxn 0 for no heat transfer Entropy + for increasing disorder (Increase in temp/volume) - for decreasing disorder (condensation/freezing) 0 for ∆S(tot) for isothermal reversible expansion and for ∆S(surr) for isothermal irreversible expansion Heat (q) ...

Bryan Lau 3H

To my understanding, you use 5/2R or 7/2R only when you are dealing with gas molecules. In this case, you are dealing with liquid water.

Bryan Lau 3H

For 9.57, the solutions manual shows two possible solutions for ∆G of a reaction. One value is calculated via ∆G = ∆G(products) - ∆G(reactants). This value is -713.02 kJ/mol The manual then offers an alternative value of ∆G calculated using the relationship: ∆G = ∆H - T∆S. This value is -753.01 kJ/m...

Bryan Lau 3H

"Calculate the change in molar Gibbs free energy for the process NH3 (l) --> NH3 (g) at 1 atm and (a) -15.0 degrees celsius." To solve for ∆G, it seems like we just plug values in for ∆H, T, and ∆S from Tables in given in the textbook. My Question: Earlier in the textbook, we learned how t...

Bryan Lau 3H

In this question, you want to know which case has greater volume. In this question, greater volume = higher molar entropy because more volume allows the gas particles more space to act "disorderly" and with higher entropy. 1.0 mol Ar(g) at 1.00 atm has a higher molar entropy than at 2.00 a...

Bryan Lau 3H

For an isothermal irreversible free expansion: ∆S(surr) = 0 because w = 0, so q = 0, so q(surr)/T = 0. But ∆S(sys) ≠ 0. I understand w = 0 for free expansion, but then why isn't ∆S(sys) = qrev/T = 0 as well? Because isn't the system doing 0 work in expanding "freely?" In other words, why i...

Bryan Lau 3H

My first thought was to use C x ln(T2/T1) as well, so I'm also confused. For the example problem on page 325, the textbook tells us to use C x ln(T2/T1), so maybe some other people or Chem_Mod can tell us which equation they think we are supposed to use.

Bryan Lau 3H

I think the enthalpy value for a gas to a liquid is just the negative of the enthalpy value for a liquid to a gas.

Bryan Lau 3H

On page 347 of the textbook, Self-Test 9.18A asks us to "Confirm that liquid water and water vapor are in equilibrium when the temperature is 100. degrees Celsius and the pressure is 1 atm. Data are available in Tables 8.3 (Standard Enthalpies of Physical Change) and 9.2 (Standard Molar Entropy...

Bryan Lau 3H

I think there may have been an error in the question, which is why Professor Lavelle told us to omit this question.

Bryan Lau 3H

The question asks for the production of 1.00 mol H2. The equation given produces 3 mol H2, so you have to divide the equation by 3 to get just 1.00 mol H2. Two mol reactants becomes 2/3, and 4 mol products becomes 4/3. 4/3 - 2/3 = 2/3, and that's why ∆n = 2/3 :)

Bryan Lau 3H

Typically, larger/more complex molecules have higher molar heat capacities because they have more bonds. Bond vibrations absorb added energy (without increasing temperature), so the more bonds (larger/more complex molecule) the more bond vibrations to prevent change in temperature.

Bryan Lau 3H

Specific heat capacity refers to the amount of energy (Joules) it takes to raise 1 gram of a substance by 1 degree Celsius/Kelvin. Molar heat capacity refers to the amount of energy (Joules) it takes it to raise 1 mole of a substance by 1 degree Celsius/Kelvin. The question will prompt you to use on...

Bryan Lau 3H

You can arrive at (c) as the answer using process of elimination. (D) is incorrect because there are no horizontal lines to show ∆Hfus and ∆Hvap. (A) is incorrect because it indicates that ∆Hfus is greater than ∆Hvap, which is not true because the problem gives us the values of both ∆Hfus (10 kJ/mol...

Bryan Lau 3H

To my understanding, we add the negative because we are calculating the work done BY a system. Without the negative, we'd solve for 'x' joules, or kilojoules. However, when work is done BY a system (system expands), energy is exhausted (leaves the system). To illustrate that, we add the negative to ...

Bryan Lau 3H

To find the heat capacity of the calorimeter, you do not need to know the mass. The equation is qcal = Ccal∆T, and in this problem you are solving for Ccal. The only values you need to solve for this are the heat supplied to the calorimeter (qcal) and the change in temperature (∆T), both of which ar...

Bryan Lau 3H

I also had trouble with this problem. In accordance to what Chem_Mod advises, I assumed H2O was produced in its gas state. However, the solutions manual states H2O is produced in its liquid state, so the enthalpy of formation values used by the textbook and me are different. For this question, how a...

Bryan Lau 3H

That makes sense, but I'm still confused about the second value. Is the second value saying water has a second ∆Hvap at 25 degrees Celsius? I would think there's only one (@ 100 degrees Celsius).

Bryan Lau 3H

Table 8.3 on page 284 of the textbook states that the ∆Hvap of water is 40.7 (kj/mol). However, right under that value, in parenthesis, it says "(44.0 at 25 degrees Celsius)." I don't understand the meaning of this second value. Can someone please explain?

Bryan Lau 3H

The energy of the universe is constant because we look at the universe as an isolated system (given that the universe consists of its own system and surroundings). By definition, neither energy nor matter enters/leaves an isolated system because it "has no contact with its surroundings;" t...

Bryan Lau 3H

SubparChemist wrote:Could just be an error in the solution's manual, I was wondering the same thing Brian. The sig figs in the manual seem sorta wonky at time.

Okay, Thanks !

Bryan Lau 3H

I got 1.5 x 10^2 kJ for part (a) of this problem. The solution manual & textbook get 1.4 x 10^2 kJ, but it rounds twice, for q(water) and q(Cu). As a result, it gets 1.45, which is rightfully rounded to 1.4 for 2 sig figs. However, I got 1.5 x 10^2 kJ because I rounded (apply sig figs) only once...

Bryan Lau 3H

Yes, the enthalpy of condensation is the negative value of the enthalpy of vaporization, because ∆H of reverse process = -∆H of forward process.

Bryan Lau 3H

NO2 has a higher molar heat capacity because it is a larger/more complex molecule. NO2 has more bonds and thus more bond vibrations to absorb added energy without increasing in temperature.

Bryan Lau 3H

The quotations are just there to repeat, "Standard state for a...," which is stated right above the quotations ("Standard state for a gas:"). If you filled in the blanks, it would be: "Standard state for a" Sol'n, "Standard state for a" pure liquid or solid,&q...

Bryan Lau 3H

To my understanding, the temperature of the SURROUNDING increases in an enthalpy problem, for an exothermic reaction. Heat is being RELEASED by the system (-∆H). I think the increase from initial to final temperature refers to the temperature of the SURROUNDING, because that's what would absorb the ...

Bryan Lau 3H

Drawing the Lewis structures would definitely help you visualize the bonds being formed and broken. Of course you don't HAVE to draw the Lewis structures, but you would just have to be very very familiar with the reactants and products and mentally keep track of all the bonds being formed and broken...

Bryan Lau 3H

To my understanding, the standard enthalpy of formation for O2 and other elements in their most stable forms are zero because they are naturally occurring. In other words, it does not require any effort to form O2 if O2 already exists naturally; therefore, there is no change in heat for such a react...

Search found 53 matches