CHEMISTRY COMMUNITY

Andrea ORiordan 1L

Hello,

What will we be expected to know for the final with regard to functional groups? Should we be able to identify them in a complex molecule? Or should we just be able to name and draw them?

Thanks.

Andrea ORiordan 1L

For this class, I think we create a rate law from pre-equilibrium states. I believe we create a rate law from near-initial conditions rather than equilibrium conditions. If we have to account for a reverse reaction, I think it will be stated.

Andrea ORiordan 1L

Yes, you cannot tell which step is the slow step unless it is given or unless you are given the experimentally determined rate law. If you are only given the rate law, just match the rate law to whatever step it corresponds with. That step is the slow step.

Andrea ORiordan 1L

Rate laws involve concentrations, so you will need to use moles per liter or mmols per liter. However, some rates use seconds, some use minutes, some use hours. You just have to be consistent while solving an equation in terms of what units you use.

Andrea ORiordan 1L

I think it depends on the question. If its asking for units of K you could probably use mmol. Same with if its asking which order reaction it is. The question will probably specify-- I don't think you'll get points taken off if you use mmol vs mol.

Andrea ORiordan 1L

Rearrange ln[A]f = -kt + ln[A]0 and you will get kt= ln[A]0-ln[A]f. When you subtract ln[x]-ln[y] you can use log rules to write the expression as ln[x/y]. So ln([A]0/[A]f) = kt is the exact same equation, just written differently. You will get the same answer.

Andrea ORiordan 1L

Remember that [A] is the concentration of the reactants, so, as Clarisse said, ln[A] will decrease over time. The fact that we're able to get k from the graph of ln[A] for a first order reaction is actually super neat.

Andrea ORiordan 1L

An overall rate of reaction is not negative-- that would just be referring to the reverse reaction. A reactant's concentration is decreasing, so we use a negative sign to denote that. However, the product increases in concentration, and so we say that the overall rate of the reaction is positive. As...

Andrea ORiordan 1L

Do you mean the example where he discussed chromium plating? He was discussing how you can use a galvanic cell to plate metals onto other ones-- where a solution of chromium ions are reduced to form solid chromium that is plated onto the solid electrode.

Andrea ORiordan 1L

I don't know the question you're referring to in full, but in general, oxidation occurs at the anode and reduction occurs at the cathode. When looking at a reaction, identify the two half reactions to see which one is being oxidized (losing electrons) and which one is being reduced (gaining electron...

Andrea ORiordan 1L

I don't know the full question you're referring to, but I would assume that Pt is on the right because there are no solid or liquid state components to conduct the current, and you need an inert electrode. If there is no Pt on the left, I assume there would be a solid or liquid state compound to con...

Andrea ORiordan 1L

Yes, that is correct. The oxidizing agent (the one being reduced) is the one taking the electrons, so it is effectively oxidizing the other compound. The reducing agent (the one being oxidized) is the one releasing the electrons/giving them to the other compound, so it is effectively reducing the ot...

Andrea ORiordan 1L

It is conventional to write the anode on the left and the cathode on the right. If it is otherwise, the problem will usually specify.

Andrea ORiordan 1L

How Lyndon explained intensive versus extensive was as follows. Heat capacity is an extensive property because it depends on the amount of substance present. But specific heat capacity and molar heat capacity are extensive because they take into account the amount of substance present, and don't cha...

Andrea ORiordan 1L

It depends. If you're asking if internal energy is a state function, then yes. We can add up changes in internal energy to find the total change in internal energy-- it does not depend on the path taken. If you're asking about enthalpy, also yes. We see that in Hess's law, bond enthalpies, and entha...

Andrea ORiordan 1L

I'm not sure which problem you're referring to, but one thing to always double check is that your units for enthalpy and entropy are the same. Enthalpy is usually given in kJ, and entropy is usually given in J, which can sometimes lead to mistakes.

Andrea ORiordan 1L

For the volume change, I believe Dr. Lavelle meant that you cannot accurately measure the difference in volume between two beakers when one beaker is being poured into the other. In the same way, it is difficult to measure potential difference when a current is flowing from one beaker to the other. ...

Andrea ORiordan 1L

Hi all! Not really a question, but I thought I'd share this memorization technique I know. Dr. Lavelle said that Loss of Electrons is Oxidation (LEO) and that Gain of Electrons is Reduction (GER). He also mentioned that you could remember LEO the lion. I like to add GER to it so that it reads: LEO t...

Andrea ORiordan 1L

You cannot change whether or not a reaction is endothermic or exothermic (at least not in this course). If a reaction is exothermic, its reverse reaction will be endothermic. However, you can change temperature to make a reaction more energetically favorable. Given that ∆G=∆H-T∆S and reactions are s...

Andrea ORiordan 1L

A spontaneous reaction is one that is energetically favorable. The diagram of the hill was meant to show that increases in entropy are energetically favorable, just as decreases in enthalpy are energetically favorable. Gibbs free energy is what we use to determine whether or not a reaction is sponta...

Andrea ORiordan 1L

For the most part, oxygen and hydrogen gas do not exist in large enough concentrations near enough to each other in order to react to form water. I believe the textbook is referring to being able to safely store hydrogen and oxygen together without a reaction occurring, unless some outside source of...

Andrea ORiordan 1L

We use the first law a lot in class-- it applies to open and closed systems as well as isolated systems in the way that energy cannot be created or destroyed. If an open or closed system loses energy, that energy is transferred to the surroundings-- it doesn't just disappear. Hope that helps!

Andrea ORiordan 1L

All of the above are correct-- one more example I found helpful with regard to the difference between system and surroundings was the ocean analogy. If you add a bucketful of water to the ocean, it doesn't really change the volume or pressure.

Andrea ORiordan 1L

For this problem, you use the value from a table of enthalpies of combustion (I believe in Appendix 2A). The enthalpy of combustion for H2 is -286kJ/mol. Multiplying that by 14.7 moles, you get -4.20*10^3 kJ.

Andrea ORiordan 1L

Remember that it takes energy to break bonds, and energy is released when bonds are formed. Therefore, -6*518 + 3*837 = -597, which is the enthalpy of reaction.

Andrea ORiordan 1L

Also note that the ∆H of H2O is not -242 as a liquid. That is its gaseous state. To obtain the answer, -286 kJ/mol, you simply subtract the ∆H(vap)= 44.0kJ, to get -286kJ/mol.

Andrea ORiordan 1L

I believe he mentioned that we did not yet need them, but he might go over them in class on Friday. The equations are listed on the equation sheet, so we will likely have to know them at some point.

Andrea ORiordan 1L

I believe Michelle is correct, with the important distinction that q would equal -w or -q would equal w. The sum of the net work done on/by a system and the heat transferred to/from a system would have to be zero-- those two values would have to cancel each other out.

Andrea ORiordan 1L

You could feasibly write the answer either way as long as you included the correct number of sig figs. I believe it's just the author's preference.

Andrea ORiordan 1L

In discussing systems, we must discuss both the system itself and its relationship with its surroundings. An open system is characterized by its ability to exchange matter and energy with its surroundings. A closed system is characterized by its ability to exchange energy with its surroundings. An i...

Andrea ORiordan 1L

The main concept for that question is that having more electronegative atoms on an acid makes it stronger-- when a proton is released, the resulting anion is more stable as a result of the electronegative atoms stabilizing the negative charge left by the release of the proton.

Andrea ORiordan 1L

You use bis, tris, tetrakis, etc, if the ligand already contains a greek prefix (eg. ethyleneDIamine) or if it is polydentate (eg. oxalate).

Andrea ORiordan 1L

I don't believe so. Dr. Lavelle briefly went over it but said we wouldn't have to do it on this exam, at least.

Andrea ORiordan 1L

Liquids and solids are omitted from the equilibrium constant expression because, as you said, they do not change in concentration. Solids and liquids are virtually incompressible compared to gases. Pressure changes won't really affect the volume of a liquid or solid sample, whereas pressure changes ...

Andrea ORiordan 1L

No, we won't get marked down to my knowledge. Usually he uses px, py, and pz when explaining specific concepts that require those notations to be understood, such as hybridization.

Andrea ORiordan 1L

The only way Kp is different than Kc is in the units. You can also convert Kp to Kc fairly easily, and therefore it makes sense that they represent effectively the same thing. Kp is used for simplicity in equations only utilizing gases.

Andrea ORiordan 1L

They specify the oxidation number with given roman numerals that so many transition metals have multiple oxidation states.

Andrea ORiordan 1L

Yes, the hybridization would be the same. Look at NH3 and NH4+ as examples, and you see that they both have sp3 hybridization. Hybridization does not depend on the number of lone pairs versus the number of bonded atoms- it only depends on regions of electron density, which can be either.

Andrea ORiordan 1L

We'll also have to know that the repulsion between a lone pair of an electrons and a bond is greater than the repulsion between two bonds- so knowing that something with a tetrahedral electron geometry will have bond angles of slightly less than 109.5 between the bonds is important too. But we won't...

Andrea ORiordan 1L

You kind of just have to experiment a little bit. There's also an easier way to figure out formal charge- you just count the number of electrons the bonded atom controls (count a bond as one electron and a lone pair as two) and compare it to the number of valence electrons it normally has. If it con...

Andrea ORiordan 1L

Radicals and Biradicals are extremely reactive and generally don't last very long in nature- for example, methyl on its own is so unstable that it cannot be stored.

Andrea ORiordan 1L

There's also an easier way to determine formal charge than to use the formula. You just see how many valence electrons that atom normally has, and compare it to how many electrons it has in the structure (a bond counts as one electron, a lone pair counts as two). If it has more electrons than it nor...

Andrea ORiordan 1L

This happens in periods 3 or higher, especially with p-block elements. This is because they have available d-orbitals that aren't filled with electrons that can be used in bonding. It results in a formal charge closer to 0, which makes the structure lower energy and more stable.

Andrea ORiordan 1L

Another notation style includes listing each sub-orbital (i.e. px, py, pz) with the number of electrons in that orbital (i.e. 2px1). That's also just a convention, and it's fine to just write p instead of separating it into 3 sub-orbitals.

Andrea ORiordan 1L

Electron shielding refers to the repulsion between two negative electrons, which decreases the nuclear charge/pull towards the nucleus that any one shielded electron "feels". The effective nuclear charge is the pull towards the nucleus that an electron "feels".

Andrea ORiordan 1L

Hello! Just a hypothetical question- if a large object (a dog, per say) slowed down to a very, very, very, very slow speed (think 10^-30 m/s), would it have observable wavelike properties? I'm aware that classical objects are made up of billions of particles and therefore you can't really calculate ...

Andrea ORiordan 1L

If we are expected to do calculations with other series besides the Balmer and Lyman series, the principle quantum number of the series (n) will probably just be given. I wonder how many series have actually been experimentally observed?

Andrea ORiordan 1L

Why do white bears dissolve in water? Because they're polar.

Andrea ORiordan 1L

To have your units correctly cancel, stick with the metric system's SI units while solving the equation. If the answer asks for velocity- say, in mph- you should complete the problem using the metric system & SI units, and then convert your answer to the desired units using dimensional analysis.

Andrea ORiordan 1L

Classical physics assumes than an oscillator (i.e. a charged particle) could oscillate at any energy, and therefore emit electromagnetic radiation at very high frequencies. Planck's hypothesis says that radiation of higher frequencies can only be generated when an oscillator of that higher frequency...

Andrea ORiordan 1L

Dr. Lavelle also said that he would go over the naming of compounds- and we'll probably go over oxidation numbers too. So don't worry if you don't automatically know the formula of calcium sulfide.

Andrea ORiordan 1L

For this upcoming test, we do not need to know how to write chemical formulas from their names (i.e. Nitric Acid is HNO3)- both the name of the compound and the formula (or just the formula) will be given to us. Dr. Lavelle said we would go over this in class at some point.

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