CHEMISTRY COMMUNITY

Lindsay H 2B

Ka is the equilibrium constant specifically for the dissociation of an acid. This means that if Ka is used, it means one reactant has to be an acid

Lindsay H 2B

Will we ever have to propose a reaction mechanism on our own on a test? In all the assigned homework problems about reaction mechanisms that I've found, the mechanism has already be given, but in some of the lecture examples it seemed like Dr. Lavelle wanted us to be able to propose a mechanism and ...

Lindsay H 2B

can someone explain the overall function of concentration cells? I'm having trouble with the very first question on Lyndon's (UA) practice midterm. How does the electron transfer occur?

Lindsay H 2B

I think for the purposes of this class, we're mainly going to use this equation to see how temperature relates to k. My TA said that we aren't really going to be using the frequency factor (A) here often cancels out in the way we use this equation. Page 642 of the textbook is helpful to see how to g...

Lindsay H 2B

This problem isn't asking for the activation of the reverse reaction though, it's asking for the activation of the forward reaction. Activation energy doesn't change with temperature. (you're right that the activation energy WOULD be different for the reverse reaction though, that's just not what th...

Lindsay H 2B

I've been under the assumption that k and k' are used to refer to the forward and reverse reaction rates for the reaction at a constant temperature, and that k1 and k2 are used to refer to the forward rate of the same reaction at two different temperatures, but the solutions manual for 15.61 and 15....

Lindsay H 2B

I don't think A has anything to do with the 0.59 in the solutions manual because A is eliminated when subtracting ln(k2)-ln(k1). However, I also don't understand why the 0.59 is there, could someone explain this part?

Lindsay H 2B

In this question, does it matter which is k and which is k'/which is T vs T' as long as k' and T' correspond to each other?

Lindsay H 2B

The reason Dr. Lavelle mentioned adsorption is because adsorption is the way that solid heterogeneous catalysts work. He said adsorption is sort of like a bunch of covalent bonds between the reactants and the solid catalyst on the surface of the solid (not literally covalent bonds, but similar conce...

Lindsay H 2B

I'm stuck on this question too. For the oxidation half-reaction, why is it Au(s)--> Au^3+(aq) + 3e- instead of Au+(aq)--> Au^3+(aq) + 2e-? Since the original reaction makes it seem like Au+(aq) is getting oxidized, not Au(s).

Lindsay H 2B

More of a technical than conceptual question, but I noticed that the solutions manual solves for Eº in problems 11, 13, and 15 even though cell potential is not asked for in any of these problems. 11, 13, and 15 are all along the lines of writing the balanced redox reaction from cell diagrams and vi...

Lindsay H 2B

I'm confused on this question too. Are both Br and O getting oxidized? Br seems to go from -1 in Br- to +5 in BrO3-, but according to this post, O3 is also being oxidized from -2 to 0.

Lindsay H 2B

O3(aq)+Br-(aq) -> O2(g)+BrO3-(aq) The solutions manual says oxygen is being oxidized from a -2 oxidation state in O3 to a 0 oxidation state in O2. I really don't understand what's going on in this reaction, since both O3 and O2 have neutral charges, and because the solutions manual also shows O3 gai...

Lindsay H 2B

Whenever working with the standard Gibbs free energy, does that mean temperature MUST be 298K? because question 11.21 asks for the standard Gibbs free energy and then part (a) says that T=1200 K.

Lindsay H 2B

Why do we need to find ∆G by using ∆H-T∆S instead of by using ∑∆G(products)-∑∆G(reactants)? I tried this with the ∆G values in the appendix but I got a different answer

Lindsay H 2B

In my notes from the lecture a few weeks ago on wednesday 1/24 I have written that ∆S=-nRTln(V2/V1) but the solutions manual and the book both use that equation without the negative sign. Can anyone explain why that is?

Lindsay H 2B

question 9.47 asks for ∆S, ∆Ssurr and ∆Stotal for an isothermal irreversible free expansion of an ideal gas. why is it that the solutions manual assumes ∆U=0 for this system?

Lindsay H 2B

More complex molecules also have more subatomic fundamental particles, and a higher number of particles always increases degeneracy (unless there is only 1 state in which case W=1). This goes for atoms too, atoms with more mass/higher atomic number have higher entropies

Lindsay H 2B

Individual atoms and molecules with more mass also have more subatomic particles and therefore greater complexity and higher degeneracy (W), which is why the more massive atoms have higher molar entropies than the smaller atoms.

Lindsay H 2B

This question asks for ∆S and ∆Stotal for 50.0 g H20 (l) at 20.0ºC mixing with 65.0 g H20 (l) at 50.0ºC at constant pressure in a thermally isolated vessel. Does ∆Stotal still refer to ∆S of the universe when the reaction happens in an isolated system?

Lindsay H 2B

I'm having a hard time understanding how it is that the change in entropy is lower at high temperatures because of ∆S=q/T, but that an increase in temperature causes an increase in entropy because of ∆S=nCln(T2/T1). Can someone explain the concept of the relationship between temperature and ∆S?

Lindsay H 2B

How does the number of available positions relate to energy?

Lindsay H 2B

For isothermal reversible expansion, w=-P∆V doesn't work because reversible expansion means P isn't constant, but instead changes very gradually. Instead, we use PV=nRT to substitute nRT/V for P and then integrate it in terms of V, which results in the equation w=-nRTln(V2/V1).

Lindsay H 2B

8.49: Oxygen difluoride reacts rapidly with water vapor to produce O2, HF and heat: OF2(g) + H2O(g) --> O2(g) + 2HF(g) ∆H= -318 kJ What is the change in internal energy for the reaction of 1.00 mol OF2? The solutions manual shows calculating the work portion of the change in internal energy by using...

Lindsay H 2B

the minus sign makes it so that the equation represents energy in relation to the system. When work is done on the system (i.e. system is compressed), then ∆V is negative, and (-P∆V) becomes positive. This means that when work is done on the system, the system gains energy. Alternately, when the sys...

Lindsay H 2B

I like to think of it as how when something is compressed (like the air in the bicycle pump in problem 8.3) it means work is being done on the system, so the system gains energy (positive work). In an opposite scenario if the air in the bicycle pump were allowed to expand, the energy for the expansi...

Lindsay H 2B

when calculating the work using w=-P∆V, what units are best to use for pressure and volume in order to get work in joules?

Lindsay H 2B

question 8.19 has to do with heating up a copper kettle in order to heat up the water inside. In this situation, is the kettle considered part of the system or the surroundings?

Lindsay H 2B

I think it implies that the element would begin in its most stable form

Lindsay H 2B

for any reaction with a given reaction enthalpy, is the enthalpy of the reverse reaction always just found by changing the sign of the forward reaction?

Lindsay H 2B

The textbook uses q for heat and H for enthalpy

Lindsay H 2B

Example 8.6 in the book and question 8.31 both ask for a change in energy at (a) constant pressure and (b) constant volume. Does this mean that only part (a) is considered finding the enthalpy, since the enthalpy definition Dr. Lavelle gave us in lecture was the change in heat at a constant pressure?

Lindsay H 2B

ex 12.3 c is to write the proton transfer equilibria for H2PO4^- and identify the conjugate acid-base pairs. in the case of H2PO4^- reacting with H2O, it seems like this ion could behave either as a base (by accepting a H+ from H2O and resulting in H3PO4), or as an acid (by donating a H+ and resulti...

Lindsay H 2B

The conjugate acid is usually the initial base plus H+; and the conjugate base is the initial acid minus H+. I think this makes them easier to identify in a chemical equation

Lindsay H 2B

What are the units normally used when calculating Kp? or are the units omitted like they are for calculating Kc?

Lindsay H 2B

I think both. I don't know if we'll specifically be asked to write the formula just from a lewis structure but it's probably useful to be familiar with. One of the ch 17 exercises gives you the name and tells you to write the formula

Lindsay H 2B

Toolbox 17.1 (p 741) says that if the ligand already contains a Greek prefix or if it is polydentate, then use the prefixes bis-, tris- etc instead of the Greek prefix, but is there any way to tell whether the "bis" in the name means that there are two ligands, that the ligand is bidentate...

Lindsay H 2B

I understand that in order to be polydentate, a ligand needs to have more than 1 lone pair to donate to the central metal atom, but does this mean that ligands ALWAYS donate ALL their lone pairs? In other words, just because a ligand has two lone pairs, does it necessarily have to donate both of the...

Lindsay H 2B

pi bonds have 2 areas of overlapping e- density because the p-orbitals overlap side-by-side, so if the atoms in a pi bond were to rotate, then the bond would break. today in lecture, Dr. Lavelle drew the pi bonds like two figure 8s on top of each other, so that both lobes of each pi bond overlapped.

Lindsay H 2B

Why is it that when there are five regions of e- density but 2 of them are lone pairs on the central atom (such as ICl3 in ex 4.9a) both lone pairs are in the equatorial positions rather than the axial positions? I thought that the point was to keep lone pairs as far apart as possible, and if 2 lone...

Lindsay H 2B

the bond angle can determine whether an atom can, must, or cannot have lone pairs because the given bond angle in this exercise suggests the number of regions of e- density around the central atom. for part (a), the 120º bond angle means 3 regions of e- density around the central atom but since ther...

Lindsay H 2B

***typo, it's Ga^3+ and Tl^3+; sorry!!

Lindsay H 2B

can anybody explain why according to the solutions manual, in question 3.5 the ground-state electron configuration from Ga^3+ is [Ar]3d^10 instead of [Ar]3d^8 4s^2, and also why the ground-state configuration for Tl is [Xe]4f^14 5d^10 instead of [Xe]4f^14 5d^8 6s^2?

Lindsay H 2B

Which of the following increase when an electron in a lithium atom undergoes a transition from the 1s-orbital to a 2p-orbital? a) energy of the electron. b) value of n. c) value of l. d) radius of the atom. Compare your answers to the answers of ex 2.1 (2.1 asks essentially the same thing about a hy...

Lindsay H 2B

it does matter; when writing electron configurations for electrons in the d block, 3d should be written before 4s

Lindsay H 2B

basically, the number of electrons in a subshell can be up to twice the number of orbitals in that subshell, since 2 electrons can fit in 1 orbital.

Lindsay H 2B

The fact that 3d comes before 4s can be confusing because the d-block on the periodic table doesn't begin until the 4th period, so try to think about it in the context of the angular momentum quantum numbers instead of the literal rows on the periodic table. If n=4 then l can equal 3 which correspon...

Lindsay H 2B

When a question asks "how much energy is required to remove an electron from a metal surface?" is it asking us to find the threshold energy or the energy of the photon?

Lindsay H 2B

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6x10^3 km.s-1. (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50x10^16 Hz. How much energy is required t...

Lindsay H 2B

Sodium Bromide, NaBr, which is used to produce AgBr for use in photographic film, can be prepared as follows. Fe + Br2 --> FeBr2 FeBr2 + Br2 --> Fe3Br8 Fe3Br8 + Na2CO3 --> NaBr + CO2 + Fe3O4 What mass of iron, in kg, is needed to produce 0.250 t of NaBr? Note that these equations must first be balan...

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