CHEMISTRY COMMUNITY

Jack_Pearce_2H

It is easiest for chemists to look at linear graphs to determine what concentrations of ___ order reactions are at specific times. Thus, it is quick to determine the order of reactions based on the x and y axes. ln[A] v t is 1st order, 1/[A] v t is second order and [A] v t is zero order.

Jack_Pearce_2H

It boils down to that in an endothermic reaction, an increase in temp increases the rate constant of the forward reaction. In an exothermic reaction, the increase in temp increases the rate constant of the reverse reaction. Hope this helps!

Jack_Pearce_2H

So basically the slow step of a reaction is the step that determines the rate of the reaction because it is the "slowest" and thus determines how fast the reaction is. Kind of like the limiting reactant. Also you can identify it by seeing which rate law is closest to the primary rate law i...

Jack_Pearce_2H

It is actually not moles of gas! It is moles of electrons transferred and received in a redox reaction. Just balance the half reactions and then use the number in -nFEcell

Jack_Pearce_2H

Usually they will tell you which one is the slow step. If it is not given in the problem, the step whose rate law which resembles the primary rate law is the slow step. You can identify intermediates because they do not appear in the primary rate law.

Jack_Pearce_2H

Can somebody give a brief summary on the process to balance a redox reaction in a galvanic cell? I just want to make sure I have the right through process for how to approach these problems. Thanks!

Jack_Pearce_2H

Basically the salt bridge is an electrolyte (i.g. A salt dissolved in water to produce an anion and cation) that allows for electrons on the cathode to travel to the anode and back again. It also helps stabilize the electrodes by forming stable compounds in the anode and cathode.

Jack_Pearce_2H

Enthalpy should be heat flow at a constant pressure. Heat is the general temperature of the reaction and the thermal energy of the reaction. Heat itself is also not a state property and can be changed/is more fickle and harder to measure.

Jack_Pearce_2H

There are a lot of office hours and review sessions to attend. Check the 14B website for more but he also will probably send and email out early this week.

Jack_Pearce_2H

we should be done Wednesday. I would focus on the material after the last midterm but definitely review first midterm stuff because the final is cumulative.

Jack_Pearce_2H

You can usually check this through determining the reaction rates of each step/checking the activation energies of both. However a lot of other factors could contribute to this such as particle size and if they are spontaneous or not.

Jack_Pearce_2H

Sadly not, you have to look at the standard reduction potentials of the ions. Remember that better oxidation agents correspond to more positive reduction potentials of ions!

Jack_Pearce_2H

actually, when work is done on the system it is positive and when it is done by the system it is negative. I know it is a weird way to look at it but I like to think work is being "lost" by the system when it is being DONE by the system, and vice versa. When work is being done ON the syste...

Jack_Pearce_2H

It is actually E that equals 0 at Eq, not E(naught). This is due to the fact that concentrations of reduction and oxidation agents greatly affect the electrochemical potential of a cell. Bc of E=0, E(naught) will be equal to (RT/nF)(lnK).

Jack_Pearce_2H

Gibbs essentially is a state function that is used primarily in identifying if a reaction is spontaneous (deltaG<0) or non-spontaneous (deltaG>0). It also helps connect a lot of other thermodynamic properties

Jack_Pearce_2H

Okay so IU had trouble with this too. Basically all you have to do is switch the positions of Cl- and Pb on the anode side. Then on the cathode side, you don't have to put coefficients of 2AgCl and 2Ag so just leave them as AgCl and Ag. Also put Cl- on the cathode side directly to the right hand sid...

Jack_Pearce_2H

I've been really into KAYTRANADA at the moment. Listen to Caution, super good

Jack_Pearce_2H

Usually check the sign first. If it is>0 then it is spontaneous and if it is <0 it is non-spontaneous. From there the larger the abs value of the number is directly related to the more spontaneous or non spontaneous the reaction is relative to its original sign.

Jack_Pearce_2H

Yes, but do not forget temperature as well. Sometimes it is a bit overlooked but it also is determined by Final-Initial!

Jack_Pearce_2H

Yes thats right. Only for systems at constant volume (delta V=0) is work equal to 0

Jack_Pearce_2H

A chicken parm hero would KILL right now

Jack_Pearce_2H

If Q<K then the reaction will eventually shift to the right in order to establish more production of products. If Q>K then the opposite occurs, the reaction will eventually shift to the left to establish eq. and produce more reactants.

Jack_Pearce_2H

all are considered spontaneous because they release heat, they might not occur quickly however despite the description of "spontaneous"

Jack_Pearce_2H

Irreversible reactions are usually quick and happen almost instantly. For example, an airbag expanding is irreversible expansion. Reversible is simply the opposite, very very slow equalization of pressures from an outside to an inside system. A soda can taken from low to high altitudes and eventuall...

Jack_Pearce_2H

For heat capacity questions you will usually use C but always double check with the specific heat of the problem (sometimes it can be different). For other applications though you will usually go from C to kelvin, especially in problems using ideal gas equations.

Jack_Pearce_2H

G° is just G but at STP (1atm and 298K). G is usually helpful though in the sense that it can be used in lab settings to determine the more "normal" free energy of the reaction.

Jack_Pearce_2H

q is the heat of the forward reaction and qrev is the heat of the reverse reaction. qrev is helpful in determining the heat of the reversible reaction at hand, usually associated with S and also the signs of the heat show if the reaction is reversible or not.

Jack_Pearce_2H

Not necessarily. Isothermal essentially means that there is no heat readily applied to the system. when delta U is 0 it either means that q and w is equal to 0 or -q=w

Jack_Pearce_2H

As seen above, yes that is the actual definition for work, but it is for reversible work (this will be stated in the problem). This can also be seen as w=-nRT(ln(Vfinal/Vinitial)). For irreversible work the usual equation is w=-p (external) (deltaV). Use these based on the information provid...

Jack_Pearce_2H

Okay so this is how I understood it. For constant V w=-P*deltaV so if there is no change in V (i.g. deltaV=0) then w=0 and if w=0, the internal energy (equal to w+q) would just equal q. This I believe does not hold true for constant pressure systems though. Hope this helps!

Jack_Pearce_2H

I used the PV=nRT equation, substituting the P, T, and V with the initial conditions of the problem Thanks! This was super helpful. However, which gas constant should I use? The one that uses J or atm? Thanks! Use 8.314 J/K*mol! You'll usually use this constant when doing thermo, use the 0.08206 co...

Jack_Pearce_2H

Because energy is being released prior to this scenario, the internal energy must be negative.

Jack_Pearce_2H

Closed systems do not allow for the transfer of matter in or out of said system. A volume change usually entails adding or removing gas (matter) so they closed systems are constant volume.

Jack_Pearce_2H

They actually cannot be negative! This is because it is the amount of energy REQUIRED to produce a change in a substance (usually temperature, hence heat capacity). There is a specific heat capacity for each compound/pure substance based on physical and molecular characteristics.

Jack_Pearce_2H

There are so many I would assume he would give you a chart for sure. It is always good to know though that any standard enthalpy of formation will be 0 for a diatomic gas involved in the reaction (i.g. 02) and eventually after practicing so many problems you will definitely pick up some common entha...

Jack_Pearce_2H

Your reasoning stands true! Because the noble gas is not involved in the reaction, it has no affect on the equilibrium shifts. Moreover, one could stand to reason that adding a noble gas would affect the pressure, and thus the equilibrium. This is true, but remember the volume of the chamber is not ...

Jack_Pearce_2H

You can approximate the solution (make X negligible) if the K value for the acid/base is at least 1.0x10^-4 or less than that. This method can be used for any setup (quadratic, cubic) but you should always test the approximation by taking the value, subtracting the x you achieved using approximation...

Jack_Pearce_2H

Usually if Ka2 is at least 1000x less thank Ka1, then it can be ignored and the approximation is valid. It is smaller in most polyprotic acids because the second/third/fourth hydrogen is harder to remove than the first, thus making it a weaker acid than its previous Ka1 counterpart and resulting in ...

Jack_Pearce_2H

You can ignore it only when the second Ka value (Ka2) is at least 1/1000 smaller than the first Ka value. Also, depending on the acid, yes there most definitely can be K(n) protonations based on how many hydrogens the acid has (H(n)A).

Jack_Pearce_2H

Shifts in equilibrium ultimately result from changes in the concentrations of the products or reactants. Because the reaction does not have a noble gas in its products or reactants and the volume of the chamber is not changing, the equilibrium concentrations of the products and reactants will remain...

Jack_Pearce_2H

Pressure does not really affect liquids or solids much, so shifts in equilibrium based on the pressure of a container would only apply to gasses.

Jack_Pearce_2H

I think what he means is that to make sure the products will be favored in the reaction, since the reaction is endothermic, there must be additional heat added to the system in order for it to shift to the right and thus form more products. Hope this helps!

Jack_Pearce_2H

Usually, yes, if not try to use the ideal gas equation to solve for it!

Jack_Pearce_2H

The enthalpy of vaporization is higher than that of fusion, so more energy is released when steam goes to water than when water loses heat and stays in the same phase. Also, steam first has to lose heat if it is above 100 C and then phase change, again transferring more heat and causing larger burns.

Jack_Pearce_2H

Basically this happens when the water on the surface of a cooler object comes in contact with an extremely hot object (like lava lol) and the water does not have enough time to heat up and vaporize all at once so just the water at the very surface of the object turns into vapor, protecting the objec...

Jack_Pearce_2H

I always check real quick to make sure it is valid but I don't even bother with approximations if the K value is greater than 1.0x10^-4

Jack_Pearce_2H

Percent ionization is just the percent an acid or base is ionized compared to the original concentration of the acid or base. It's important though because it allows one to start to show if the approximation is valid and show if an acid or base is strong or weak.

Hope this helps!

Jack_Pearce_2H

Hi!

Use the -x when subtracting from the initial concentration (product or reactant, depends on the problem) given in the problem and add a +x for the molecules produced by the R or P. Add a coefficient of nX when the R or P is (...)^n.

Jack_Pearce_2H

I would make sure you have all the right equation with all the proper parameters. Make sure you do the equation twice by twice with the (+) and (-) sign after the -b term and before the square root.

Jack_Pearce_2H

Use 8.206x10^-2 when you convert all the units of pressure/temperature/volume in PV=nRT to atm/Kelvin/L, respectively.

Jack_Pearce_2H

Use an ICE box always for weak acids. Acids will either be strong or weak usually so for strong you can assume 100% ionization, then just use ICE boxes for the weak!

Jack_Pearce_2H

You would need to know whether the reaction is endo or exothermic in order to answer this.

Jack_Pearce_2H

Yes! Different weak acids could result in different percent ionizations above 5%, but this usually applies to weak acids that are regularly known. If the professor does not give you an acid you know is weak, the approximation could be above 5%

Jack_Pearce_2H

It will eventually reach K yes and q is the initial value measured before k is established

Jack_Pearce_2H

You can solve for n(moles of the substance at hand) and divide it by V(volume of the substance) to obtain the molarity of the material! Typically either n or V is given in the problem or you can solve for it using another equation, so you just have to solve for one of the unknowns. Hope this helps!

Jack_Pearce_2H

H20 can be omitted from the ice stable but if it is in gaseous form it typically needs to be included

Jack_Pearce_2H

You just exclude it from your evaluation of Kc. For example if you had Ag(s) in your reactants for the forward reaction, it would not be included in the denominator of the Kc.

Jack_Pearce_2H

They're always supposed to be in Kelvin, not Celsius.

Jack_Pearce_2H

Kc usually deals with equilibrium regarding concentrations and K is a more general term for the eq. constant. Hope this helps!

Jack_Pearce_2H

Essentially it means the liquid has strong intermolecular bonds (i.e. LDF's, h-bonds, dipole dipole interactions)

Jack_Pearce_2H

A salt is an ionic compound formed from an acid base neutralization reaction. Stable salts are usually composed of a metal (from the base) and a nonmetal (from the acid) however in equilibrium reactions, other salts can form. Water is also created in the neutralization reaction.

Jack_Pearce_2H

Ka is just a term to measure the strength of an acid. It is proportional to pH in that the lower the pKa (-log(Ka)), the stronger the acid. He will elaborate more in 14B but basically it is how much an acid (strong or weak) will dissociate in water.

Jack_Pearce_2H

Yeah that will be the case. I think instead of pKa though it would be given in the problem to be pKb. But yes your logic does hold true, it would be neutral.

Jack_Pearce_2H

Im pretty sure the central atom has to be a metalloid or similar to a metalloid (i.g. S,P,C). Br is a strong halogen so it does not have the capability to be amphoteric.

Jack_Pearce_2H

I don't think you need to know how it was derived but definitely know that ph=(-log([H+]) and that [H+]=10^(-pH)

Jack_Pearce_2H

both are the same thing. a pH of 1 for example simply means the molecule disassociates with hydrogen almost completely, releasing the hydrogens into solution and thus the low pH

Jack_Pearce_2H

amphoteric is a molecule that can act both as an acid AND a base, while amphiprotic is a molecule that can both accept and donate electrons. Based on the reaction at hand and other compounds/molecules introduced associated, the amphoteric/amphiprotic molecule will act either as a proton donator/acce...

Jack_Pearce_2H

S04-2 would be the conjugate base in the reverse reaction in that the compound acts as a bronsted base and can accept protons. In the forward reaction you are indeed right though, H2SO4 would be the bronsted acid and donate protons

Jack_Pearce_2H

A lewis acid is an electron pair acceptor and a bronsted acid is a proton donator. Not super sure on a rule about if the terms can be interchanged but it doesn't help to check based on the question.

Jack_Pearce_2H

Hi everybody. I just had a quick question on how to determine if a Pi bond is perpendicular or parallel a plane. I believe this was like number 18 on Sapling and I was confused on the premise behind it. Could somebody please explain this? Also, with regard to the orbitals themselves, why a Pi bond w...

Jack_Pearce_2H

I think of it all going back to thermo. It is most favorable to have the orbitals mix this way (hybridization) and allow for bonding in the most energetically favorable (so the lowest energy state). Moreover, it is the hybridization process that essentially allows bonds to form in the first place by...

Jack_Pearce_2H

I see you logic, but actually no it would not. central atoms with LP only affect the molecular geometry so no need to think of the accessory atom LP's as affecting the molecular geometry :)

Jack_Pearce_2H

They most definitely do! in terms of the magnitude of which molecular geometry is affected by other regions of electron density it goes as follows, LP-LP, LP-atom, atom-atom. Lone pairs are not shared and essentially belong to the atom which the LP is assigned, which is why affect vsper so much.

Jack_Pearce_2H

usually called bent/v-shaped. I would specifically identify it by saying the AX2E2 general formula has <109.5 degree bond angles and usually rests around 104.5 degrees

Jack_Pearce_2H

coplanar means for atoms in a molecule to be on the same "level" per say. for example, if you hold you hand flat on a table, the hand is on the same "plane" as the table. However, if you make you hand perpendicular to the table, it is now "out of plane". the same logic ...

Jack_Pearce_2H

Not necessarily. aqueous acids and bases disassociate in water so I would more closely relate it to an ion-ion interaction. Depending on structure, acids and bases may have the capability of hydrogen bonding, but simple interactions would not be.

Jack_Pearce_2H

Actually no! This is a common mistake I honestly do make too. ALWAYS check out the dipole moments! They will give the best representation of whether a molecule is polar or nonpolar. If there are too many dipole moments to draw out (sometimes it is time consuming really), look at the groups attached ...

Jack_Pearce_2H

honestly I usually just check out the dipole moments to help visualize if the molecule will be polar or non polar. If all dipole moments offset one another=nonpolar. If they are leaning or pushing to one side, they are polar. an example would be cis- and trans-dimethylcyclohexane

Jack_Pearce_2H

Hey there! For me, I like to first notice if there are any positive or negative net charges on the molecules. If they are negative, they are lewis bases and positive as lewis acids. If there are no net charges, I like to draw out the lewis structures and see if there are any unpaired electrons or at...

Jack_Pearce_2H

Hi!
Hybridizations are basically the mixing of electron orbitals. They help to stabilize a molecule and are also inherent to the physical properties of atoms and orbitals!

Hope this helps clarify that for ya

Jack_Pearce_2H

Honestly, I would just draw them out! Usually it is pretty easy to see if there are a lot or a little by playing with the double bonds, accessory atom location or if you can make new double bonds by messing with the formal charges!

Jack_Pearce_2H

The bond had to occur between a H that was not bonded to a carbon and one of the following atoms: F,O,N. Hope this helps!

Jack_Pearce_2H

Bond character just refers to how covalent or ionic a bond is (usually a covalent bond in a molecule) this is given through the bond length and difference in electronegativity between the two atoms at question :)

Jack_Pearce_2H

I like to take a look at double bonds, if there can be movement of them or reassembly between two atoms in the molecule, and if accessory atoms can be switched around! Usually this is all there is to consider :)

Jack_Pearce_2H

I had the exact same question. just a memorization thing really, know that F, O, H, N can make hydrogen bonds assuming H is not bonded to a carbon! Hope this helps!

Jack_Pearce_2H

Fluorine has the least electron shielding due to a low number of inner shells (thus a higher effective nuclear charge) and a HUGE affinity for electrons due to its 7 valence electrons. This makes a single fluorine atom intrinsically unstable and in need of an 8th electron to complete the octet, givi...

Jack_Pearce_2H

Yes Hydrogen bonding only occurs between F, O, or N. C-H and H-C hydrogen bonds do not occur because the dipole moment is too weak for there to be a hydrogen bond with any hydrogen connected to a carbon.

Jack_Pearce_2H

They are all lewis bases yes. But honestly it is good to double check anyways really quick just to make sure. Just know also not all lewis bases are CC bonds too!

Jack_Pearce_2H

Try to get it as close as possible to 0! The closer it is the more stable, but if you really cannot get it to 0 thats okay too :)

Jack_Pearce_2H

diople dipole interactions are constituted as van-der-waal attractions or london dispersion forces so they are inherently weaker than hydrogen bonds, but may or may not be more prevalent in a molecule than hydrogen bonds so cumulatively they could be larger. Individually however, no.

Jack_Pearce_2H

They totally can spin! But just like the heisenberg uncertainty principle, if we could efficiently determine what specific spins electrons had, they would change states. Thus I believe they cannot conform to specific spins, especially when bonded, but most definitely can possibly have the same + or ...

Jack_Pearce_2H

This is a special case! NH4 is inherently more stable with 4 bonds formed but usually the coordinate covalent bond (the lone pair you mentioned) goes unbonded as in NH3. just depends on the molecule i believe :)

Jack_Pearce_2H

You can check on the periodic table as well. Just look at the respective element's (s) and (p) block and count how many s and p electrons are contained in that element. For example, carbon is the 2nd element in its 2p block. this means carbon has 2s electrons and 2p electrons, giving it 4 valence el...

Jack_Pearce_2H

Hi!

What I did as use the Bohr Frequency equations [v=Rh((1/n2^2)-(1/n1^2))] to solve for v then use E=hv to solve for the energy. Make sure to pay attention to the values so they make sense logically! Hope this helps :)

Jack_Pearce_2H

It's usually just an arrow unless otherwise stated (i.e. if the questions says that another electron with the same first three quantum numbers and an Ms of +1/2 is in an orbital, you know the other electron will have -1/2).

Jack_Pearce_2H

J x Sec for planks constant. m/s for velocity and meters for position (x)

Jack_Pearce_2H

should be the third period down. But honestly it depends on which element youre talking about because some are specifically expanded octets and others aren't

Jack_Pearce_2H

yes and depending on if it is a double or triple bond, you will have different numbers of pi bonds (1 and 2 respectively)

Jack_Pearce_2H

UA sessions help sooooooo much! if you had trouble with your discussions questions they help you with that as well

Jack_Pearce_2H

you can post or reply to get participation points right?

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