CHEMISTRY COMMUNITY

Kelsey Warren 1I

First you have to notice that when A and B are constant and C changes, the rate doesn't change so C must be 0 order. Then you can ignore C and just compare A and B when B and A respectively are constant to find their orders and the overall order.

Kelsey Warren 1I

For 0 order reactions, the y-axis is [A] to make a straight line graph. For 1st order, it's ln[A], and for 2nd order it's 1/[A]. You can figure out the units by always making the rate have units mol/Ls (or whichever concentration/time units you're given).

Kelsey Warren 1I

We learned the unique rate law as unique rate=-(1/a)(d[A]/dt), and you can set this equal to the same relationship for products if you make the 1/coefficient term positive instead of negative.

Kelsey Warren 1I

Both graphs really mean the same thing and can be about the exact same reaction, you're just plotting [A] vs time and then ln[A] vs time (which produces a linear graph from the exponential decay).

Kelsey Warren 1I

Something else to remember is that you can calculate k if the reaction rate, concentration of reactants, and the order of the reaction are known. Also, k is always positive and is the slope of the first order graph of ln[A] vs time.

Kelsey Warren 1I

[A] is just the concentration of the reactant A at any time t throughout the reaction, whereas [A]o is the concentration exactly at the beginning of the reaction.

Kelsey Warren 1I

If you're given a reaction and told that the products are thermodynamically stable, then the forward reaction will be favored. In contrast, if the reactants are thermodynamically stable, then the reverse reaction would be favored. (And vice versa for if you're told the products/reactants are thermod...

Kelsey Warren 1I

I don't have the 6th edition textbook so I don't know which problem you're talking about, but usually the cell diagram will include KOH (aq) if the reaction is taking place in a basic solution. So it basically just means that OH- is present in the solution and is part of the electron transfer or is ...

Kelsey Warren 1I

I think I understand what you're asking (?), and I'm pretty sure that what they give you depends on the compounds involved in the problem and what they're asking you to calculate in the problem. I'm sure that Dr. Lavelle will give us the appropriate oxidation numbers on any test/final problem.

Kelsey Warren 1I

I could definitely be wrong, but I think that delta G always equals the maximum energy available to do work because delta G takes into account entropy/temp changes/heat changes already.

Kelsey Warren 1I

The book uses the delta S values of the elements that make up the compound because the question is asking about their stability "with respect to the elements." So just use those delta S values to see which becomes less stable as temp is raised.

Kelsey Warren 1I

If delta G is negative, then the forward reaction is favored, which means that the reactants are unstable and will decompose into the products spontaneously.

Kelsey Warren 1I

q and U will always be in J or kJ, but delta H can be in J/mol or J/g or just J or kJ, etc., depending on what type of delta H it is in that problem.

Kelsey Warren 1I

Basically, you calculate the heat that was used to increase the temp of the 125g from -2.8C to 0C. Then, subtract that amount from the total energy and you'll have the amount of heat that was used to convert half the ice cream (62.5g) into liquid, so you divide that leftover heat by 62.5g to get the...

Kelsey Warren 1I

Yep, delta H is assumed to be in kJ/mol if it's not specified, and also it's assumed that if delta H is given, then the reaction is at a constant pressure.

Kelsey Warren 1I

Also, the pressure has to be constant in an open beaker because the pressure in the beaker would equal the atmospheric pressure, which is constant.

Kelsey Warren 1I

If the heat/energy value is really large (>10^3), then use kJ so that your numerical answer won't be so huge. If your heat/energy value is smaller, you can use J. It's really just dependent on the size of the numbers you're dealing with.

Kelsey Warren 1I

I'm not positive either, but I agree with what she said. Cv is for constant volume and Cp is for constant pressure.

Kelsey Warren 1I

Basically a calorimeter is a reaction vessel, so it acts as a container for a reaction to take place. It allows you to make measurements about temperature changes, volume, etc. that you can use to calculate enthalpy changes/heat capacities, etc.

Kelsey Warren 1I

Delta H naught is just delta H in standard/ideal conditions. You use them the same but one is for standard conditions.

Kelsey Warren 1I

Enthalpy is basically the heat possessed within a system, but we usually talk about enthalpy changes, which are the changes in heat within a system. Heat capacity is the heat required to raise the temperature of a substance by 1 degree C. So heat capacity has to do with heat entering/leaving a syste...

Kelsey Warren 1I

Ionized and deprotonated are basically the same thing when talking about acids and bases; these words just mean the amount of acid/base that has reacted divided by the amount of acid/base you initially started with (times 100).

Kelsey Warren 1I

Percent protonation is calculated the second way you mentioned, which is % protonation=concentration of conjugate acid / initial concentration of weak base x 100 (or the opposite if you start with a weak acid).

Kelsey Warren 1I

Yes, if they're just asking you to calculate the pH and there's an ion that does not take part in the reaction, then you always ignore that ion because it won't affect the pH.

Kelsey Warren 1I

Just to add on a little, the reaction quotient can be used to determine if the reaction will proceed forward or backward to reach equilibrium. For example, if the reaction quotient is smaller than the equilibrium constant, then the reaction will proceed forward to make more products to increase the ...

Kelsey Warren 1I

The equilibrium constant stays the same at equilibrium because it's a constant: it's a calculated factual value for that exact reaction at the given temperature. The ratio with the exponents is the equilibrium constant, so it will stay the same. And like Vincent said, the other ratio you mentioned i...

Kelsey Warren 1I

As far as the ICE table, sometimes they're helpful because they make a nice visual representation of what's happening in the reaction, but if you are given enough information to just use the K equation, that will work too (and it's faster).

Kelsey Warren 1I

Another way to think about it is that brackets are used when you're referring to a concentration and parentheses are used when you substitute in the actual numerical value of the concentration into the K equation. For example, if one of the compounds in your chemical reaction is NH3 and its concentr...

Kelsey Warren 1I

I think the book uses the word composition just to encompass both concentrations and partial pressures into its definition of equilibrium. So it's basically correct to say that that the compositions and concentrations in the mixture stay constant at equilibrium. You could also say the partial pressu...

Kelsey Warren 1I

I'm pretty sure that when you're dealing with gases, you use partial pressures to calculate K, so you'd use Kp. When you're using aqueous compounds, etc., so you're plugging in concentrations into your calculation for K, you'd use Kc.

Kelsey Warren 1I

NH4 can't be a base because it can't accept a proton or donate an electron. It can't accept a proton because the N already has 4 bonds each with one H, so neither the N nor the H's can make another bond. NH4 can't donate a base because there are no lone pairs of electrons and all single bonds. Becau...

Kelsey Warren 1I

Only a super acid or a super base would make the H+ concentration exceed .1M or less than 10^-14M, which correspond to ph 1 and 14. Because super acids and bases aren't as common as regular ones, values outside the regular pH range are rare.

Kelsey Warren 1I

I was a little confused about this too, but the aqua comes from the water ligand that is part of the complex (the OH2). Because the aqua comes before the cyano in alphabetical order, the aqua is the start of the name.

Kelsey Warren 1I

Another easy way to think about this is that increasing the strength of intermolecular attractions is like a bunch of molecules holding on tighter and tighter to each other. If they're holding on really tightly, it becomes very difficult for one to escape the others, therefore it takes more energy t...

Kelsey Warren 1I

Polarizability increases as electronegativity decreases and as atomic radius increases. Therefore, polarizability increases right to left and top to bottom.

Kelsey Warren 1I

Yes, the number is the period the element is in just because that's the period that the valence electrons (the ones actually doing the bonding) are in. Also, writing out the electron configuration is very helpful as they said above!

Kelsey Warren 1I

I think all we really need to know about the cause of hybridization is that it makes the molecule formed more stable and that a hybrid orbital interacting with the other atom's orbital makes the bond stronger/more balanced, whichever makes the most sense to you.

Kelsey Warren 1I

If a central atom has 6 regions of electron density surrounding it and they're all other atoms, then it will be nonpolar, but if there's one or more lone pair on the central atom then the shape is no longer symmetrical and could be polar.

Kelsey Warren 1I

H bonding is a form of dipole-dipole because the H has a slight positive charge and the O, N, or F has a slight negative charge, so yes they can.

Kelsey Warren 1I

An ionic bond is considered intramolecular within a molecule and intermolecular within an ionic solid. Since the metal completely gives one or more electrons to the nonmetal, the bond is much stronger than a H bond which is completely intermolecular and is only caused by the attraction of positive a...

Kelsey Warren 1I

Since a H bond is intermolecular and a covalent bond is intramolecular, a molecule can have a covalent bond and then share a H bond with another molecule of the same species. I guess the effect of that would be that the molecule with the covalent bond (like H2O) would be attracted to another H2O via...

Kelsey Warren 1I

I'm pretty sure you just draw one wedge because the wedge just represents the direction of the bond rather than the bond itself.

Kelsey Warren 1I

I was under the impression that the larger the molecule, there's more electrons and more electron shielding. The outer electrons of a large atom are less tightly held than for a smaller atom so they're able to be pulled by an attractive force more easily than for a smaller atom.

Kelsey Warren 1I

Polarizability is increased with increased size of atoms, number of electrons, and amount of negative charge (if you're talking about ions). Because larger hydrocarbons are larger and have more electrons, they are more polarizable. Their greater number of electrons experience more shielding and can ...

Kelsey Warren 1I

Yep, induced dipole-induced dipole is probably the most descriptive name for the intermolecular interaction but you can also use dispersion, VDW's, and London forces interchangeably.

Kelsey Warren 1I

I don't know how many details I can explain but I'll try! Basically what he means is that in real molecules, instead of a molecule clearly having two single bonds and one double bond (as shown by observed bond lengths), the molecule would have three bonds that were each "somewhere in between&qu...

Kelsey Warren 1I

I think he cares more about us knowing what each bond represents and the possible resonance for each molecule than drawing the bond lengths correctly. Also, the bond length differences for triple/double/single/etc. bonds are relatively small and would be extremely difficult to draw accurately on a t...

Kelsey Warren 1I

In its ground state, C has one electron in the px orbital and one electron in the py orbital and an unoccupied pz orbital. When an e- is added, it occupies the previously empty p orbital, which releases more energy than an additional e- in N pairing up with another in the px orbital.

Kelsey Warren 1I

Yes, electron shielding occurs in multi-electron atoms, so an electron far from the nucleus will experience less of the positive charge of the nucleus. This results in what is called an "effective nuclear charge." I don't know if we have to know that term but probably should be familiar wi...

Kelsey Warren 1I

The Cu and Cr are more stable when their electron configurations have a full (or half full) d shell. I think we just need to know these exceptions and what they look like rather than understand exactly why they occur.

Kelsey Warren 1I

Unpaired electrons are any electrons that don't have a another electron (with opposite spin) in their orbital. Draw out the Aufbau diagrams for each element using the rules from lecture and you'll be able to see which electrons don't have a buddy!

Kelsey Warren 1I

The equation is (usually?) used for an electron losing energy and consequently emitting a photon. Because the electron loses energy, its energy is "negative" while the energy for the photon is positive. Dr. Lavelle also said in lecture that the negative sign means that the bound electron h...

Kelsey Warren 1I

Like they said above, light doesn't have mass. A photon acts like a particle rather than a wave, but that "particle" is only used for visualizing discrete amounts of energy, 1:1 photon:electron interactions, etc. A photon has a velocity as well.

Kelsey Warren 1I

I agree with everything stated above I just wanted to clarify that 1 eV=1.6x10^-19 J, not the other way around. Hope this helps!

Kelsey Warren 1I

I'm not 100% sure, but it's derived from the equation in lecture today, E=-hR/(n^2). If you substitute E=hv, then you can divide both sides by h to get rid of Planck's constant, as in the equation you posted. Besides that, I'd guess that you would use it to find the frequency of a photon emitted fro...

Kelsey Warren 1I

I agree with the points they made above about the Bohr model vs. VESPR model and electrons jumping from ground states to excited states. I'm also wondering if when an electron absorbs such high frequency/high energy radiation that it approaches n=infinity and leaves the atom, does it leave the atom ...

Kelsey Warren 1I

You can't really measure "how fast" the kinetic energy of anything is. Kinetic energy is directly proportional to the velocity of the particle squared, so I guess you could say a physical sign of the kinetic energy of an ejected electron is the square root of its velocity.

Kelsey Warren 1I

Addressing the second part of your question, including the state of matter when writing and balancing equations has been very helpful for me in that I'm able to picture conducting the reaction in a lab, which we obviously aren't dealing with yet but we will in the future! It's also helpful when iden...

Kelsey Warren 1I

Another way to approach this is using dimensional analysis:

You have 1.56 mol reactant and you want to convert to mols of product, so multiply 1.56 mol reactant times (5 mol product/4 mol reactant). I think the last answer just got the 4:5 ratio flip flopped.

Kelsey Warren 1I

Hi Hannah,

I'm under the impression that the given pressure and temperature are not needed for calculating the amount of moles of carbon dioxide produced. I think that those come in handy when dealing with Ideal Gas Law problems, etc.

Hopefully it's true that we didn't need to use that information!

Search found 60 matches