CHEMISTRY COMMUNITY

jillian1k

The higher peak on rxn profiles is the peak of the rate-determining step. Slow steps are slow because they have to overcome such a high activation energy. This, in turn, makes the rate constant (k) smaller and therefore rate smaller (slower).

jillian1k

EMF or electromotive force is just the E of the cell, or the cell potential (E). It is the Ecathode minus Eanode. It's the cell's strength in producing electricity via redox reactions. It is never negative because the electrons only flow from anode to cathode. This is due to the fact that electrons ...

jillian1k

Q is the denotation for a reaction's concentration ratio when the reaction is not at equilibium. Q is [P]/[R]. Therefore, since the products are the cathode metal (s) and solution (aq) and the reactants are the anode metal (s) and solution (aq) (keeping in mind that solids and liquids are not includ...

jillian1k

Generally, unless otherwise specified, it's best to make your final answer units in moles. If you do the conversion it at the beginning, though, then it eliminates any possibility for unit mistakes later (ie what happens you're dealing with mmol and M). Rate is always M/s or mol/Ls. Concentration is...

jillian1k

d[A]/[A] in itself is not an equation. Also, because it doesn't have the variable for time (t) in it, it couldn't be a change in time. d[A]/[A] does come up when you're trying to find this change in concentration over time, though. Since a change in concentration over time is rate as a function of t...

jillian1k

The rate constant k is just an experimentally determined value for a specific reaction at a specific temperature (therefore different reactions have different rate constants). It is also independent of concentration or concentration change. Unless temperature (and/or activation energy) changes, k wi...

jillian1k

Why (conceptually) is the order of the rxn (n) more important than the rate constant (k) for explaining the mechanism of a reaction?

jillian1k

Is it likely that we'll have to integrate eqns or use them to explain differences b/w higher order reactions for problems on test 3? I know normally Dr. Lavelle shows us derivations and integrations to help us understand the background and foundation of a topic, not because it's what he's testing us...

jillian1k

I think the speed of a step in a reaction is just relative to the speed of all the steps in that reaction. By that I mean the fastest step in the reaction will be the 'fast' step by default, but that doesn't necessarily mean it's a 'fast' step in the grand scheme of things. I don't think there's any...

jillian1k

This is because the problem asks you to balance the half-rxns in basic solution. Therefore, you use H2O and OH-, not H+, to balance the reactions where O and H show up.

jillian1k

The oxidizing agent is the reactant that is oxidizing its fellow reactant by taking its electrons. Therefore, the oxidizing agent itself has a reduced product. The reducing agent is the reactant doing the reducing to something else by giving away its own electrons, therefore making it the oxidized p...

jillian1k

This is because Br2 is the only reactant, therefore, it is acting as both the reducing and oxidizing agent. So, when you add the two half-rxns, you have to add the Br2 from both reactions together to get the overall balanced (in terms of charge AND amt of each element) redox rxn. Technically you sho...

jillian1k

We know because the eqns for dS all come from dS=qrev/T. qrev is a greater value than qirev because q=-w for reversible systems. Since we know w is always greater in reversible systems than in irreversible systems, this difference translates to q in q=-w.

jillian1k

In step 1, acetone is in its liquid state, so you would use the molar heat capacity that corresponds to this phase. In step 2, liquid acetone in vaporized into acetone vapor at its boiling point. So, in step 3, acetone is now in the gas phase. Therefore, you have to use the corresponding Cp (since t...

jillian1k

dU=0 for an isothermal expansion or compression of an ideal gas only. This is a direct consequence of U = 3/2 nRT for an ideal gas. Therefore, for ideal gases, no temperature change=no internal energy change. Additionally, it's not really possible to expand/compress liquids or solids in the same way...

jillian1k

We know to use constant pressure over constant volume in this case because of what is given. They only give us the pressure of 1 bar. They say nothing about that pressure value changing or anything about a changing (or constant) volume, therefore we can assume they want us to use Cp for C in step 3....

jillian1k

This problem wants you to find the standard entropy of vaporization of water at 85 C reversibly. This takes much more work by the system than an irreversible vaporization of water at 85 C, which consists of one step and lots of energy lost as heat. To do it reversibly: 1.) reversibly heat the water ...

jillian1k

The number of moles on the reactants side is 4. The number of moles on the products side is 3. Therefore the total number of moles decreases by 1 mole somewhere in the reaction. Entropy is based partly on mass or moles that are in a substance. Generally, the more mass or moles present, the higher th...

jillian1k

During lecture today, Dr. Lavelle discussed how the potential difference between electrodes is always at its maximum. He used the example of a volume change when pouring one beaker's solution into the other to illustrate this. Could someone please explain this concept to me? I don't think I got all ...

jillian1k

I know Dr. Lavelle said most of the time we'll be working with systems at equilibrium, but how will we know when a system is not at equilibrium unless it's stated? Would being given the boiling point of the system be one indicator of equilibrium always (since it's where the liquid and gas phase coex...

jillian1k

I was going over the notes from lecture on friday and came across the van't hoff eqn derivation. I'm confused about whether we're supposed to use this eqn -ln(K2/K1) = -(deltaH*/RT2) + (deltaH*/R) or this one -ln (K2/K1) = -deltaH*/R(1/T2-1/T1). If it's the latter, then my question would be why can ...

jillian1k

The example given asked at which temperature the reaction would spontaneously occur at 1atm. We know that for the rxn to occur spontaneously, it must have a negative deltaG (otherwise stated as deltaG<0). By stating deltaG as 0, we can substitute two known variables into the eqn for deltaG (deltaG=d...

jillian1k

The formula for bond enthalpies is easier to think about as deltaH=(sum of bonds broken)-(sum of bonds formed), not products-reactants. Not all chemical reactions have all bonds being broken (with none being formed) on the left and all bonds being formed (with none being broken) on the right.

jillian1k

Yes. In fact, it's often necessary to avoid assigning coefficients larger than 1 to products. For example, in the same rxn you gave, the other way to balance it is: 2H2 + O2 --> 2H20. However, this now multiplies the products (H20) by 2. Since reaction enthalpies given next to Hess's law rxns are th...

jillian1k

deltaHc refers to the enthalpy change of a combustion reaction. We know that a combustion reaction always consists of compound X + 02(g)-----> CO2(g) + H20(l). You just have to use the given compound for each deltaHc given in this problem to substitute for 'compound X' as I wrote, as well as balance...

jillian1k

I don't believe we have gone over this in class yet, but we still have a few lectures left before the tests on chapter 8 start. It's possible we will be if it's in any of the homework problems, but I'm sure Professor Lavelle will make it clear in class either when he teaches it or announces somethin...

jillian1k

Always assume so. It's not really worth the whole point lost for not having the correct sig figs so I'd say just try to make it a habit when doing homework to double check sig figs and units before boxing your answer.

jillian1k

deltaU is the internal energy of just the system. It takes into account heat (q) and work (w) of the system. This means that it is affected by changes in those values. More q transferred into system or more w done on system (or more of both) means there is more energy being put into the system and t...

jillian1k

It's a chemical change because there is a change in bonding that changes the identity/chemical composition of the substance. In graphite, each carbon atom is bonded to 3 other carbon atoms with one lone electron pair. Whereas in diamond, each carbon atom is bonded to the most number of carbons possi...

jillian1k

1 L•atm is equal to 101.325 Joules.
So, (1.48 L•atm)(101.325J/1 L•atm)=149.96 J (150. J with sig figs).
You could also do it using the ideal gas constant as long as you use the number with the correct units for this particular problem.

There's an example on page 274-275 using this conversion.

jillian1k

I think it's this way because Delta H fus and Delta H vap are both reactions that cause a physical change, therefore involving energy. Since, they involve energy, their units are expressed as such. But, the freezing and boiling points columns are just giving us those known, constant values for parti...

jillian1k

Ligands cannot be positively charged. The concept behind complexes/coordination compunds is that electron rich species (ligands) act as lewis bases and donate a lone electron pair to a lewis acid (the metal) to form a coordinate covalent bond. Multiple of these bonds combine to make a complex. There...

jillian1k

The periodic/group trends we've covered are atomic radius, ionic radius, ionization energy, electron affinity, and electronegativity. Also, I guess technically electropositivity, but this is just the exact opposite of electronegativity (and if I recall correctly, the only time it was used was for a ...

jillian1k

For Q and K, we only include gases and aqueous solutions. Therefore, in this problem, the only thing that will be included in Q is BCl3. Using the formula for Q ([C]^c[D]^d/ [A]^a[B]^b), we see everything is eliminated except for A so we're left with 1/A. As for the reason it's in pressure instead o...

jillian1k

From my understanding, only isomer b can form a chelating complex because it is the only isomer with ligands close enough to form the bond angle (109.5 because of tetrahedral shape on N in NH2) necessary to connect both of them to the same transition metal. This forms the ring of atoms from benzene-...

jillian1k

Both names mean the same thing, it's just that cyanido is the new IUPAC name convention. In the notes, Dr. Lavelle tells us that when naming anionic ligands, -ide endings become -o. He gives the example "CYANIDE---> CYANO". I don't think using the names (cyano vs. cyanido) interchangeably ...

jillian1k

Based on my understanding, knowing the coordination number of a complex tells you how many ligands/bonds a complex or coordination sphere contains. This amount influences the overall size of the coordination sphere and can show the bonding capacity of the central metal atom. It can also tell you the...

jillian1k

This was the most recent announcement on Dr. Lavelle's website: "Week 9: Test 4 covers material covered from Chapters 4, 17, and Ch 11.1 to the end of 11.8 (but Omit 11.3, and pages 452-453 as stated in the syllabus)." We've finished chapter 4 and 17. We will be covering chapter 11 this up...

jillian1k

Also, do the 5 modules on chemical equilibrium cover all the material covered in week 8?

jillian1k

Have we finished covering chapter 4? I know the likely answer is yes, but I just wanted to double check since I've been using a course reader from a previous year that covers through the end of the chapter (which includes molecular orbital theory, MO energy diagrams, etc.).

Thank you!

jillian1k

I just wanted to clarify which shapes we need to know. Dr. Lavelle went over most of them in class, but the textbook has a few more that we did not go over. Would it be correct to assume we only need to know the information about the linear, bent/angular, trigonal planar, tetrahedral, trigonal bipyr...

jillian1k

Is electropositivity just the opposite of electronegativity? Would it literally be the opposite periodic/group trend of electronegativity with it increasing down a group and right to left across a period? There was a problem in the chapter 3 homework about this and I just want to make sure I underst...

jillian1k

The uncertainty in the velocity will be smaller. This is simply because the mass of a baseball is larger than the mass for an electron. Mathematically, both the coffee cup and the electron will have the exact same calculations with exact same values for ΔP and ΔX. The point where the calculations ch...

jillian1k

Hi! I also had the same question when I did the problem. I found an explanation in the textbook on page 47 in the paragraph below "THINKING POINT." I also found something on this website that worded the explanation differently: http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html D...

jillian1k

Since Cl-is an anion, it's ionic radius will be larger than the radius of its parent atom Cl. Also, since K+ is a cation, its ionic radius will be smaller than the radius of its parent atom K. The radii for Potassium and Chlorine therefore aren't equal. They actually both change size in a way that m...

jillian1k

What homework questions pertain to Schrodinger's Equation? I'm just wondering what exactly we're supposed to know about the equation and how to apply it. Since we didn't do any examples in class, I'm not sure if we're just supposed to understand the concepts behind it and where it came from, or if t...

jillian1k

We use noble gases because they are the most stable elements. This is true because they have the maximum/ideal number of valence electrons and are less likely to react with other elements.

jillian1k

Do we need to know each of the specific shapes/models of the orbitals for each state? Or are we just supposed to know generally how many nodes and lobes each one has?

jillian1k

I just wanted to clarify something about wave-like properties. Can large objects, like cars, demonstrate wavelike behavior even if we can't detect it? I'm confused because part of the notes in the course reader reads that "all matter has wavelike properties, but is only noticed for objects (lik...

jillian1k

For the Balmer series, must the energy levels of e- increase or decrease only in intervals of 2? For example, can it only go from n=2 to n=4, but not from n=2 to n=3? I'm just wondering because the book says the reference state for the Balmer series is n=2.

jillian1k

I know you were probably looking for an answer from Professor Lavelle, but I thought I'd answer to give the information I know in the meantime. For all the tests (excluding the midterm and final), we will only be given 40 minutes in our discussion classes to complete the whole test. Prof. Lavelle al...

jillian1k

You would use the equation E(photon)-E(energy to remove e-)=E(excess), where E(photon)=hv, E(energy to remove e-) is also known as work function, and E(excess)=Ek=(1/2m)(v^2). 1.) Calculate Ek=(1/2m)(v^2) using the velocity(v) of e- given in the problem and the constant mass of an e- (9.109 383 × 10...

jillian1k

The question I ran into a problem with is: A solution is prepared by dissolving 55.1 g of KCl in approximately 75 mL of water and then adding water to a final volume of 125 mL. What is the molarity of KCl(aq) in this solution? I know that I have to use the formula Mf=(MiVi)/Vf. I also know I need to...

jillian1k

so just to be clear, it doesn't matter how many sig figs we use at any step in the problem as long as there is the correct amount in the final answer?

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