CHEMISTRY COMMUNITY

Austin Ho 1E

You are correct that intermediates should not be in the rate law. However, what we are doing here is substituting the concentration of [HClO] for something that is known. In the first equation, we know that k 1 [ClO - ]=k 1 '[OH - ][HClO]. Thus, [HClO]=k 1 [ClO - ]/k 1 '[OH - ]. This is just then su...

Austin Ho 1E

Not sure what you mean? You don't need A at all in this problem. Given that ln(k₂/k₁)=E_a/R * (1/T₁ - 1/T₂), you just need to plug in values. Additionally, A is just the pre-exponential constant in the Arrhenius equation, so you could find it.

Austin Ho 1E

In general, yes, however, you should note that this is because they have the same units: Joules.

Austin Ho 1E

Negative orders occur when you have to take the product into consideration for a rate law. For example, in the textbook's example: rate = k[O 3 ] 2 [O 2 ] -1 , O 2 is a product of the reaction. The reason why O 2 is in the rate law is because its concentration is affecting the rate at which the reac...

Austin Ho 1E

I'm not really sure what manipulations you're talking about? All that matters in a 0th order half life is that t 1/2 =[A] 0 /2k. The manipulations you may be thinking about are how many half lives occur in a certain time span. You can find this by using the equation (1/2)^n=x, where x is the percent...

Austin Ho 1E

I'm pretty sure it's more of a conceptual thing. You just have to know how to identify a pseudo first order and the concepts about it, but there's not a whole lot of actual questions about them.

Austin Ho 1E

As stated above, I believe that there isn't really much of a strong relation between the two. Besides the fact they use the letter k and are constants, they relate two different things.

Austin Ho 1E

I believe that Dr. Lavelle said that we will not have to derive equations/formulas in general for tests. However, knowing the derivation is helpful not only for a conceptual grasp of how the equation was arrived at from previous understandings but also in case the equation is not provided.

Austin Ho 1E

I believe that, in general, the textbook won't be asking you questions about the half life of 0th order reactions. Therefore, the equation is not used, but is still provided so you know what it is in theory. This is like the half life equation of 2nd order reactions - it is provided, but you won't h...

Austin Ho 1E

I believe that, in general, the textbook won't be asking you questions about the half life of 2nd order reactions. Therefore, the equation is not used, but is still provided so you know what it is in theory.

Austin Ho 1E

Electrodes are simply the cells where electrochemical reactions take place. In a given example, an electrode refers to either the anode or cathode. Basically, an electrode in the cell (or half cell) where the electrochemical reaction happens.

Austin Ho 1E

This is probably an error in the solutions manual; you are correct in that the cell components with the same phase should be separated by commas, not |.

Austin Ho 1E

The reason why the cathode and anode are switched is because for this example they are writing the reverse cell potential and demonstrating how that would be setup. In general however, it should be anode||cathode, so you are correct.

Austin Ho 1E

The principle is that ΔG°= 0 for elements because we think about the reaction of their formation. For an element like Carbon, its formation reaction would simply be: C(s) -> C(s). Since both sides are identical (Carbon is identical to itself), they must both have the same ΔG°. Therefore, ΔG° for the...

Austin Ho 1E

I don't think Gibb's Free Energy is only for non-expansion work. Both systems with irreversible and reversible processes can do expansion work. In theory, ΔG=wmax, or the maximum amount of work that can be done is ΔG. In an irreversible expansion, w=-PΔV; pressure is constant while volume changes. T...

Austin Ho 1E

Additionally, for any reversible process ΔStot=0.

That's the only thing I can think of right now.

Austin Ho 1E

I believe that ΔSm is ΔS molar, and ΔSf is for ΔS of formation.

Austin Ho 1E

You should be able to distinguish whether a compound is thermodynamically stable (or unstable) by looking at its ΔG° of formation: if ΔG is positive, then its formation is not spontaneous; likewise, the reverse reaction is spontaneous, and it will decompose into its reactants. For a thermodynamicall...

Austin Ho 1E

This is a 3 part entropy calculation, because we need to (1) heat the liquid water to 100°C, (2) vaporize the liquid water into gas at 100°C, and finally (3) cool the water vapor down to 85°C. We need the specific heats to find ΔS for parts 1 and 3, using the heat for liquid water in part 1 and the ...

Austin Ho 1E

I'm in agreement with the statement above - we're probably not going to be asked for all three to change because that would be impossible to calculate with what we know about heat capacities for ideal gases.

Austin Ho 1E

If the ice is not already at 0°C, then you must heat the ice in addition to melting it (using ΔH of fusion). Therefore, we use the specific heat of ice to heat it to 0°C, then we melt it, then we heat the now liquid water using the heat capacity of water. Likewise, with the water, we use its specifi...

Austin Ho 1E

Not entirely sure what you're asking, but I'll give it a shot.

I think ΔG is 0 for pure reactants and products, e.g. graphite or a metal in its standard state.

Austin Ho 1E

I think in this question when they say C(s) they're specifying that it's solid carbon and not necessarily graphite. Therefore, its formation enthalpy is 0.

Austin Ho 1E

I believe that Boltzmann's Equation is used to find the residual positional entropy when the substance is at 0 Kelvin already.

Austin Ho 1E

Yes, that's pretty much it.

Adiabatic systems cannot have any heat transfer, so ΔU = w always. For isolated systems, there can be no transfer of energy at all, so ΔU=0 always.

Austin Ho 1E

I believe that whatever heat capacities, bond enthalpies, and enthalpies of reaction we will need will be provided to us like last year. Dr. Lavelle didn't specify that we needed to memorize anything so I think it's safe to assume that we won't have to for this test.

Austin Ho 1E

I'm not 100% sure either, but I think in this case since we're trying to find the enthalpy per density, we want a positive amount of enthalpy. To be honest I was a bit confused too but perhaps enthalpy per density can only be positive.

Austin Ho 1E

It does not always result in negative work. Remember that w=-Pext*delta V, thus work is only negative if the change in volume is positive. I.e., the work is negative (the system does work) if it expands. However, if the change in volume is negative, the work is positive as work is done on the system...

Austin Ho 1E

Another way to remember Heat vs. Energy is thinking about rectangles and squares. All squares are rectangles, but not rectangles are squares: likewise, all heat is energy, but not all energy is heat. If you understand the context of the question they're asking, for instance if you're asked to find c...

Austin Ho 1E

The way Dr. Lavelle is covering chapter 8 is by going through thermochemistry first and then thermodynamics. The way the textbook is setup, chapter 8 starts out with thermodynamics and then moves into thermochemistry. Therefore, if you want to follow how Dr. Lavelle is covering it, you should start ...

Austin Ho 1E

If a thermometer was an isolated system, it would not be able to interact with its surroundings, aka the mercury in the thermometer would not rise or fall according to different temperatures. Thus, we can rule out that a thermometer is an isolated system just on the principle that it works. Addition...

Austin Ho 1E

In short: yes, I think so. The charge on the outside lets you know the overall charge of the whole molecule, thus it should be kept into account when drawing a Lewis structure. For your example for instance, [HgF2(OH2))2]1+, I believe the oxidation number of Hg would be (V) since there are 4 negativ...

Austin Ho 1E

I'm not quite sure what you mean? Do you mean how an acid, such as HCl(aq), in water, is dissociated to H+(aq) and Cl-(aq)? Generally you don't have to write this unless it specifies a net ionic reaction or something. For a proton transfer equation, like for HCl(aq), you'd just write: HCl(aq) + H2O(...

Austin Ho 1E

Up to section 11.8, minus section 11.3. Basically everything in chapter 11 minus Le Chatlier's principle and changes in equilibrium.

Austin Ho 1E

There is only 1 lone pair on the Chlorine atom. In ClO2+, there are 7+12-1 electrons, or 18 electrons total. 2x6 electrons go on the oxygens, plus 2x2 electrons that bond the Oxygens to the Chlorine atom. That leaves us with 2 more electrons, or 1 lone pair on the Cl atom. Thus, since there are 3 re...

Austin Ho 1E

What Rita said is correct. The difference in electronegativities are more important than the values of electronegativity themselves. Also, you need to understand molecular shape to understand dipole moments: look at how 'balanced' the differences in electronegativity are. For example, in CCl4, the t...

Austin Ho 1E

I'm not sure what context 'phase' is being referred to, but in the context of equilibrium, I think it's talking about the physical state. Since we only consider reactants and products in concentration or gas phase to contribute to the equilibrium constant, only R's & P's that are (aq) or (g) are...

Austin Ho 1E

You should be expected to find bond angles for any future VSEPR problems. It's mainly memorization but honestly it's not very much since you just need to know the bond angles of the 'parent' structures. For AX2, Linear, all derivations will start with 180, for AX3, Trigonal Planar, all derivations w...

Austin Ho 1E

What Johann said is correct. Hybridization only occurs in bonding because it allows certain elements to bond more than you'd expect them to be able to due to the positions of the electrons in their orbitals, such as Carbon. Carbon only has 2 unpaired electrons in its p orbital, but if you use sp3 hy...

Austin Ho 1E

In general, since the noble gases already have stable electron structures, they do not 'want' another electron; thus, their electron affinities are always positive. Since the electron affinity is positive, it requires energy to push an extra electron onto the stable core of a noble gas. I'm not enti...

Austin Ho 1E

Hey guys, Recently I've come across 2 different lewis structures for SO2 - one with the Oxygens double bonded to the Sulfur, and the other with 1 Oxygen double bonded and 1 Oxygen single bonded. According to formal charge, we should use the first one with both Oxygens double bonded, but I saw someti...

Austin Ho 1E

Electronegativity describes the tendency of an atom to pull electrons toward itself within a species. This is related to the effective nuclear charge - in the molecule H2O for instance, Oxygen has higher electronegativity than the Hydrogens, and the covalent bond is shared unequally as Oxygen pulls ...

Austin Ho 1E

Valence shell electrons are written out following the closest previous noble gas; for Copper it would be [Ar]3d104s1. The part we care about is the 3d104s1. Note that we have a 4s1 and not a 4s2; this is one of the exceptions you need to know for this class (including Chromium). I won't go into deta...

Austin Ho 1E

Effective nuclear charge increases as you move right across a period and it decreases as you move down a group. You were correct in this assumption. However, for these isoelectric atoms, we know they all have the same valence electron configuration; in this case, for F-, Na+, and Mg2+ they would all...

Austin Ho 1E

Part (c) asks about the wavelength of light required to remove the electron at the speed given (3.6*10^3 km/s). This is important, as you must use the photoelectric effect equation that relates Kinetic Energy to the energy of the photon and the work function. The equation gives us: 1/2*mass of the e...

Austin Ho 1E

I believe that atomic/emission/absorption spectrum all refer to the same thing. The way I think of it is that electrons in a Hydrogen atom, for instance, start out at the 'ground state' or lowest possible energy level, n=1. They are then excited by a photon of light that corresponds to the energy di...

Austin Ho 1E

When calculating wavelength of light you actually use E=hc/v, where h is Planck's constant, c is the speed of light, and v is the frequency that you are given. You would not use E=hc/lambda because lambda is the wavelength of light, thus you would not need to calculate it. When using the first equat...

Austin Ho 1E

Hi all, Just finished reading Chapter 1 and I still feel a bit confused about the Schrodinger Equation and Wave Functions. From my understanding, wave functions are substituted since we can't use trajectories to describe where electrons will be (due to their particle-wave duality), and that the squa...

Austin Ho 1E

When finding the excess, you must find which reactant is limiting first. Once you've done this, you can then determine which reactants are in excess; i.e. there is some left over that isn't used in the reaction. An easy way to find how much excess is left over is to convert all of your reactants int...

Austin Ho 1E

Whenever I think about molar mass I always remember that molar mass is how many grams 1 mole of that substance would be. I.e., 1 mole of Carbon-12 is 12.01 grams. Molecular weight is essentially the same thing as molar mass (as mentioned above) but instead uses amu. All that really matters is that y...

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