CHEMISTRY COMMUNITY

Tiffany 1B

Catalysts aren't really consumed in the reaction since they are usually formed in a later state so that they can be used again and again. And I think the equilibrium isn't affected because both the forward and reverse rates are affected by the catalyst equally, since creates a new pathway. Hope that...

Tiffany 1B

Chapter 19 and 20 are on O chem, so you might want to check those chapters. I think chapter 20 would be your best bet. Hope that helps.

Tiffany 1B

Hi,
I had a similar problem with this concept, but I think this has to do with the fact that the more positive cell potential (1.69>1.40)wants to be reduced more, so it becomes the cathode where the reduction occurs.. Hope this helps.

Tiffany 1B

Hi,
I'm a little confused on how the Ea corresponds to whether the reaction is endothermic or exothermic. Can someone please explain this to me?
Thanks.

Tiffany 1B

Oh my haha. You don't see that everyday.

Tiffany 1B

Hi, One of the easiest ways to determine this is the complexity of the molecule. In this case, butane has much more complexity. Hope that helps

Tiffany 1B

Hi,
I think this is bc what youre finding is delta H rxn so per mole of product or whatever youre measuring, if I remember correctly. I think they said they wouldnt mark off points for using just kJ vs kJ/mol tho.. Hope that helps.

Tiffany 1B

Yeah, I think you should just go with given equation. If you convert, you're adding an additional step where you could potentially introduce error.

Tiffany 1B

Yes, exactly what Beza was saying. You compare the two rate laws and their initial rates. Although the solutions manual does say 2&3 for the order of B you would actually use 1 and 3. So to solve for B it would look something like rate3/rate 1= [A]^l[B]^m[C]^n/[A]^l[B]^m[C]^n where everything wo...

Tiffany 1B

Hi, So I think the difference in these has to do with the way it is integrated, using the second order differential rate law (-1/a* d[A]/dt = k[A]^2) as starting point then separating variables and integrating using the power rule (which is in his integration rules tab on the website) you get -[A]^-...

Tiffany 1B

If Avagadro calls, tell him to leave his number.

Tiffany 1B

What is the chemical formula for banana?

BaNa^2

Tiffany 1B

Hi,
So I believe this is because the equation itself accounts for the sign change, but as Ava said you can change the sign and then simply add them (instead of subtracting them). Hope that helps.

Tiffany 1B

Test 2 covers Chapter 14.1-14.11 so basically all of Electrochem. Hope that helps

Tiffany 1B

Why did the chemist coat his shoes with silicone?
-He wanted to reduce his carbon footprint.

Tiffany 1B

Yeah, Q is the concentrations of products over reactants just like K, except Q represents any point in the reaction while K is specifically at equilibrium.

Tiffany 1B

The equation for an ideal gas is U=3/2*nRT. When you're calculating the change in internal energy, the number of moles and the gas constant are constant values and that means the change in internal energy is dependent upon temperature change. But in an isothermal expansion temperature doesn't change...

Tiffany 1B

Yeah, since it's a state function the 'path' it took to get from initial to final doesn't matter so you could use the reversible path and that equation to solve for the irreversible. I think the book goes over this idea in more depth in the first few pages of chapter 9 as well. Hope that helps.

Tiffany 1B

Hi,
Yeah so since at boiling point the particles don't have a higher tendency to go to liquid over gas or vice versa, you could just imagine some particles as gas and others as a liquid at this point. Hope that helps.

Tiffany 1B

Hi, So I think the first one is used for any constant temperature while the standard Gibbs free energy is used when both the products and reactants are in their standard states at one bar and the temp(which I think is) 298K. Other than that I'd say just using what you're given is main way to tell. H...

Tiffany 1B

What type of dog do chemists have?
Laboratory Retrievers

Tiffany 1B

I think this would be because the heat is contained within the system.

Tiffany 1B

Yes, bc delta G takes into account both enthalpy and entropy, the two factors Dr. Lavelle said were needed to determine spontaneity. If delta G is negative, then it's spontaneous. Hope that helps.

Tiffany 1B

Hi, So the difference of when to use those two equations is whether or not the pressure is constant. If the pressure is a constant than you would use the w=-p delta V. If the pressure isn't a constant you would use the w=-nRT ln(v2/v1). Dr. Lavelle showed us how to derive them on the 24th I think if...

Tiffany 1B

Yeah, definitely check the syllabus and the TA and UA hours as well because if you cant do Dr. Lavelle's hours there are a some that would work for your schedule.

Tiffany 1B

Yeah so with spontaneous processes the entropy does increase so I think you would probably be able to assume that a positive delta S means spontaneous. Although Gibbs free energy might be a more definite way...

Tiffany 1B

"Thermodynamics just gets me so q-ed"
-Frustrated chem students everywhere

Tiffany 1B

DamianW, Something being in its "most stable form" refers to it being in its 'most pure/most stable' state, which I like to think of as how you would normally find it; an example of this would be oxygen which we know is all around us in the form of O2. In this particular problem, we know t...

Tiffany 1B

Hi. So I think the difference between the R value of 0.08206 or (0.0821) and the R value the book used is the pressure unit. The R=8.314 is using kPa vs the R= 0.0821 is using atm Straight from my constant sheet from last quarter: R = 8.314 L ∙ kPa ∙ K−1∙ mol−1 R = 0.08206 L ∙ atm ∙ K−1∙ mol−1 Hope ...

Tiffany 1B

Yeah those are the only ones I have written down as well. Carbon as graphite, I2 as a solid,Br2 as liquid and the others diatomic ones(N2, F2,O2 Cl2) as gases.

Tiffany 1B

Yeah I think so because molar heat capacity is given by Q=nC Delta T so the higher number of moles, the greater the heat capacity you should get/expect.

Tiffany 1B

Yeah exactly what Daniisaacson2F was saying. Using the bond enthalpies was just an alternative to the first method w/ Hess's Law. The bonds that were broken were the C=C bond and the H-Br bond and the ones formed/ rearranged were the C-C, C-H, C-Br bonds. Then he used the bond enthalpy values(knowin...

CHEMISTRY COMMUNITY

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Re: Catalysts

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