CHEMISTRY COMMUNITY

Gurkriti Ahluwalia 1K

why do they multiply by the LCF in the solutions manual to obtain t-t'/t times t"

Gurkriti Ahluwalia 1K

Troy Tavangar 1I wrote:Does anyone know if any organic chemistry will be covered on the final?

just identify the functional groups. ketones, carboxylic acid, alcohol, ether, amine, and aldehyde:)

Gurkriti Ahluwalia 1K

"A catalyst is not consumed in the course of a reaction." I am confused as to why this is false. I understood that a catalyst is first found on the reactant side of the equation and then is later found as a product. Is it not consumed even though it goes from the reactants to the products...

Gurkriti Ahluwalia 1K

also, why do we set it up so that it is rate of the cat rxn/ rate of the uncat rxn?

Gurkriti Ahluwalia 1K

"A catalyst is not consumed in the course of a reaction." I am confused as to why this is false. I understood that a catalyst is first found on the reactant side of the equation and then is later found as a product. Is it not consumed even though it goes from the reactants to the products...

Gurkriti Ahluwalia 1K

it is consistent with products/reactants or y/x or final/intial^^

Gurkriti Ahluwalia 1K

think of "lowering the activation energy" as "it makes it easier for the reaction to occur" its like extra help for the products to undergo a reaction.

Gurkriti Ahluwalia 1K

in the solutions manual it tells you that the rate of the uncatalyzed reaction is .60. how did they know this??

Gurkriti Ahluwalia 1K

I agree: I don't believe one step equations have intermediates, only products and reactants. don't forget that while catalysts lower activation energy the energy of the products and reactants is still the same between a reaction with a catalyst and without. the reaction without a catalyst is lower b...

Gurkriti Ahluwalia 1K

I think its essential to know that the vant hoff equation relates temperature to the equillibrium constant and you get this by setting the two gibbs free energy equations equal to each other. always think critically about which information you are given and therefore which equation would be the best...

Gurkriti Ahluwalia 1K

intermediates get immediately used up, so in a sense its like they're not even there in the first place. their existence is temporary.

Gurkriti Ahluwalia 1K

let's say that the first step of a 2 step mechanism is the forward and reverse reaction, where both are fast and at equilibrium. and the second step is the slow step. does the first step also not affect the rate law? if not, then why? if anyone is curious im referring to problem 15.53 (c).

Gurkriti Ahluwalia 1K

Another main point to add to this is that the slow step must occur before other elementary steps in order for the overall reaction rate to only depend upon this slowest step; in other words, if there is another elementary step occurring before the slow step, this must be included in the overall rxn...

Gurkriti Ahluwalia 1K

ya this could have very well easily been a typo, because it makes sense that in step 1 the Cl- reacts with OH+ just like in the second step. I too treated it as an intermediate.

Gurkriti Ahluwalia 1K

if the diagrams are exactly the same, they are the same molecule. for simplicity sake, i started out by naming the entire molecule x or y or z so I would see where the intermediates are easily, and then when it asked me for the overall reaction or the actual molecule i wrote out the full version fro...

Gurkriti Ahluwalia 1K

since the question is still asking you for the rate law of each step, it is appropriate to include the intermediate molecule in the rate law since it is part of the reaction in step 2.

Gurkriti Ahluwalia 1K

can someone elaborate on what the solutions manual is talking about when we say that the second elementary reaction is fast and it does not affect the reaction order??

Gurkriti Ahluwalia 1K

why is the fraction subtracted from 1 when we asking for the time for the concentration to reach that value of the initial? subtracting implies that it is the remainder after the said fraction gets used up in the reaction.

Gurkriti Ahluwalia 1K

how do you get rid of the negative??

Gurkriti Ahluwalia 1K

a zero order reaction should have the same units as the rate, mol/L/s or mol x L-1 x s-1

Gurkriti Ahluwalia 1K

note that in solving for this the ln's DO NOT cancel, instead you use the inverse operation (e^) to get rid of the lns, and then divide both sides by A0 to get At=10% of A initial.

Gurkriti Ahluwalia 1K

i remember hearing something like "the more positive the positive the potential the more likely it will be reduced" in biology. is this the same principle perhaps??

Gurkriti Ahluwalia 1K

therefore, is it safe to say that you DO NOT calculate the order of C because it is independent of the rate?? in other words, only the orders of A and B matter because they affect the rate

Gurkriti Ahluwalia 1K

this reaction IS a first order reaction because the rates are directly proportional to each other

Gurkriti Ahluwalia 1K

shouldn't it be mol/l/s for zero order reactions??

Gurkriti Ahluwalia 1K

how are the units for k the same as the rate for a zero order reaction? if rate =k[A]^0 then wouldn't you divide both sides by mol/L to get s^-1=k? i know this is the answer to the units for k for a first order reaction, but i'm assuming for a zero order reaction they are the same because of the 0 e...

Gurkriti Ahluwalia 1K

how does one know to use the molar heat capacity rather than the specific heat capacity of water when calculating delta S??

Gurkriti Ahluwalia 1K

how do we know to use 5/2R for heat capacity?? what words or phrases in the question hint at this??

Gurkriti Ahluwalia 1K

in part c of this problem, why is it not acceptable to do 6 mol of C-C bonds? in other words, what does 3 mol CRC bonds mean, and why is this seprate from 3 mol C-C?

Gurkriti Ahluwalia 1K

can someone therefore explain why a isn't the correct choice also??

Gurkriti Ahluwalia 1K

in the solutions manual, to solve this table, the value given for electron affinity is subtracted instead of added like all the other values (with the exception of the energy used to form the compound). Is this because affinity for electrons requires energy from the compound to take in electrons? or...

Gurkriti Ahluwalia 1K

I have looked over multiple discussion questions on this question and there are two things I am still very confused about. First, how does this question assume there is no enthalpy of formation? Secondly, if we do apply Hess's Law using combustion reactions, how do we eliminate the appearance of Oxy...

Gurkriti Ahluwalia 1K

31A. If 3.607 x 10^-19 J is required to remove an electron with zero kinetic energy from a metal surface, what would be the longest wavelength light that could do this? So in other words, we are being asked to find the maximum wavelength where the photons will still have sufficient energy to remove ...

Gurkriti Ahluwalia 1K

as professor notes, the photoelectric effect detects electrons while atomic spectra detects the wavelengths/frequencies of light emitted. the photoelectric effect also completely removes the electron from the metal surface while in the atomic spectra experiments electrons typically remain bound to t...

Gurkriti Ahluwalia 1K

You can use Rydeberg's equation (1 step) or E=hv and then subtract the differences of the two (multistep). I believe this example is worked in the video module towards the end.

Gurkriti Ahluwalia 1K

sometimes, the limiting reactant is not always the one with the least value given. You should always use the ratios present in the chemical equation because although there may be less of one reactant you may need more than what's provided of the other. for example, if a ratio is 1:5 and you are give...

Gurkriti Ahluwalia 1K

I actually approached this problem a bit more practically in terms of limits. I tested the doublings in the formula MinitialVinitial=MfinalVfinal sort of like this: (.1M)(.010L)=(xM)(.020L) where the final concentration is .05 M. This is just the FIRST doubling. as expected, the concentration became...

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