CHEMISTRY COMMUNITY

Michelle_Tan_1G

Its caused by a collision between something (another molecule maybe?) and one of the atoms in the structure, which transfers the energy to change conformation. I'm pretty sure any calculations to do with it is beyond the scope of this course bc i dont remember learning anything about that :(

Michelle_Tan_1G

I'm confused. We already calculated decreasing (negative) reaction rates in part B. Why is it necessary to put another negative sign in the equation to make it pos, and what does O2 being a reactant have to do with it?

Michelle_Tan_1G

When you say adjacent to each other, do you mean on the same C atom? In that case, one would be axial and one equatorial. Otherwise, the more complex/heavier atom is put in the equatorial position. I think it's because it is more likely to be unstable/produce stronger repulsions.

Michelle_Tan_1G

They give the equation for average rate to be AVG RATE = -∆[O2]/∆t. Why is there a negative sign? Does an average rate have to be an absolute value?

Michelle_Tan_1G

Yes, any shift in conformation requires an energy collision with one of the structural atoms strong enough to overcome the energy barrier.

Michelle_Tan_1G

You look at the chemical equation and see that Fe (neutral charge) changes to 2Fe(OH)2 in the products, indicating Fe has a positive +2 charge. 2 e- are transferred, but there are 2 moles of Fe, so a total of 4 e- transferred, n=4.

Michelle_Tan_1G

How do you get n=4 e- transferred from the chemical eqn and info given?

Michelle_Tan_1G

This doesnt really require knowledge of the question. But the eqn they give you is

E = Ecell - .05916V/n * logQ

How do you get .05916? I think this is the eqn E = Ecell - RT/nF*logQ. I computed (8.314 J/K mol(298 K)/(n*9.6485e5 C/mol) = .025 V/mol. What am I doing wrong?

Michelle_Tan_1G

Dear Dr. Lavelle,

You've always been incredibly welcoming and supportive of each student in this class. I never felt afraid or judged when I asked you for help understanding the concepts after class. Thank you so much!

Cheers,
Michelle

Michelle_Tan_1G

ok, thanks! does that only apply to haloalkenes? In what case would you look at alphabetical order?

Michelle_Tan_1G

It asks you to name the haloalkene, which has a chlorine and fluorine atom attached to the end of a propene atom. However, when prioritizing which atom to list first (Cl vs F) do we look at atomic number or alphabetical order? The answer is 3-chloro-3-fluoro-1-propene.

Michelle_Tan_1G

The question says that a catalyst causes reduction of Ea from 125 to 75 kJ/mol. a) By what factor does the reaction rate increase at 298K, all other factors being equal? b) By what factor would the rate change if the reaction were carried out at 350K instead? I'm not sure how to go about solving thi...

Michelle_Tan_1G

The question says that for the reversible, one-step rxn 2A -> B + C, the forward rate constant k = 265 L/mol min and the reverse k = 392 L/mol min. The Ea for forward rxn = 39.7 kJ/mol and the reverse Ea = 25.4 kJ/mol. I found the equilibrium constant to be .676. How can I tell whether the reaction ...

Michelle_Tan_1G

The question says that a beta blocker is eliminated in a first-order process with a rate constant of 7.6x10^-3 min-1 at 37 degrees C. A patient is given 20. mg of the drug. What mass of the drug remains 5.0 hrs after administration? In this case, I used the first-order integrated rate law, but it as...

Michelle_Tan_1G

I have a question as to how to find the order of B. Knowing that C is zero order, I only compare A and B (using experiments 2 and 3, which have [A]o constant). My equation is 200^n/100 = 16/4.0. However, when I solve for n, I get a fractional number. I'm pretty sure I'm supposed to cancel 200 by usi...

Michelle_Tan_1G

Hey, I'm not sure quite how to write the equation for the reaction in a). It just looks like H2 + 2e- -> HCl to me, but that seems too simple. Otherwise, to solve problem, do we first calculate Eº of the reaction and then plug in given concentration to find Q? How do we translate 1.0 bar into concen...

Michelle_Tan_1G

That answer makes sense, but I'm still getting an answer of -.905 V while following those instructions. Is there something else I'm doing wrong?

Michelle_Tan_1G

According to the back of the book, the answer is -1.5 V, but I got -2.4 V. My work is below. What am I doing wrong?

Michelle_Tan_1G

When given two couples (in this case Co2+/Co and Ti3+/Ti2+) how do you determine which one gets oxidized and which reduced in a redox reaction? I assumed the one oxidized was the one with the more positive Eº (Co) but apparently Ti is oxidized according to the back of the book.

Michelle_Tan_1G

Hey, found a link that explains pt a!

viewtopic.php?t=5160

Michelle_Tan_1G

The textbook answer for 14.21a, which asks to find standard cell potentials, is +.75V. I used the equation Ecell = Eºreduction - Eºoxidation Ecell = Eº(Cu2+ -> Cu) - Eºoxidation(Cr3+ -> Cr2+) = (+.34) - (+.41) = -.07. The reduction cell potential for Cu is +.34 V, and the reduction Eº for Cr3+ is -....

Michelle_Tan_1G

While I did balance it, now I'm confused as to whether n refers to number of electrons transferred or moles, or whether it doesn't matter. Sorry...

Michelle_Tan_1G

I'm unsure how to write a half reaction/ cell diagram for a solubility equilibrium or a Bronsted neutralization reaction. Can someone explain to me what these terms mean in the context of cell diagrams?

Michelle_Tan_1G

But if you were given a reaction like in 14.9a: 2Ce^4+ + 3I- –> 3Ce^3+ + I-, how would you figure out n?

Michelle_Tan_1G

The cell diagram, in the solution, uses Pt electrodes instead of the oxidized solid (I2). I'm not sure when to add those in. Do we use Pt(s) as electrodes when there are no metals oxidized/reduced?

Michelle_Tan_1G

What does n in ∆G = -nFE represent? The book says that it is the amount n in moles that is transferred in a reaction. Does that just mean the difference in number of moles between the products and reactants? Why does that make sense?

Michelle_Tan_1G

Hi,

I'm not sure where to start when writing the reduction half reaction, in which O3 is oxidized from 0 to 2- (BrO3^2-). I just don't see a way to balance out Br.

My oxidation half reaction: 6OH- + Br- -> BrO3 2- + 3H20 + 6e-

Michelle_Tan_1G

Why does ∆U = 0 for the isothermal expansion of an ideal gas?

Michelle_Tan_1G

In the book it mentions that heavier atoms have more vibrational energy levels (because of allowed energies being closer together) and that results in a higher S. Can you clarify this statement, or explain how closer energy levels result in higher levels of vibration? Also, does the vibration result...

Michelle_Tan_1G

Problem 9.7 asks to calculate entropy change at constant pressure and volume. I'm pretty sure I need the heat capacity at constant pressure Cp. Do I use the values given in chapter 8 (ex. monoatomic atoms have a Cp of 5/2R) ? Cp is not given in the original problem.

Michelle_Tan_1G

Thank you, that makes sense! So w=-PV represents negative work done by expansion. When work is lost by expansion it is replaced by heat added into the system (in reversible reactions).

Michelle_Tan_1G

According to S = qreversible/T, when T increases, entropy decreases. However, I thought that at higher temperatures there was more disorder and therefore higher entropy. Can someone explain this to me?

Michelle_Tan_1G

That makes sense, thank you so much!

Michelle_Tan_1G

As I understand it, internal energy is equal to work + heat in the equation ∆U = q + w. However, there's a second equation that reads ∆U = ∆H - P∆V. Change in heat represents q, and P∆V = work of expansion. What is the reasoning behind the positive sign in ∆U = q + w changing to a negative in ∆U = ∆...

Michelle_Tan_1G

An ice cube is added to liquid water at given quantities and initial temperatures, and we are asked to find the final temperature. How do I factor in the enthalpy required to melt the ice cube? So far I have: qsys = qsurr qsolid = qliquid H(enthalpy of fusion) + nC∆T = nC∆T Is this the correct way t...

CHEMISTRY COMMUNITY

Search found 35 matches

Re: energy barriers

Re: Final 2013 Q4C

Re: axial/equatorial positions for multiple substituents

Final 2013 Q4C

Re: energy barriers

Re: 2013 Final Q4 Part A

2013 midterm q4

final exam 2013 q3c

Re: Saying Thank You to Dr. Lavelle

Re: 2.16 Intro to Ochem chapter 2

2.16 Intro to Ochem chapter 2

15.67

15.65

15.21

15.17

14.37a

Re: 14.27

14.27

14.29

Re: 14.15 - half reactions and cell diagrams for solubility equilibriums

14.21a - sign of cell potentials

Re: ∆G = -nFE

14.15 - half reactions and cell diagrams for solubility equilibriums

Re: ∆G = -nFE

14.13 [ENDORSED]

∆G = -nFE

14.5a [ENDORSED]

Internal Energy

Difference in molar entropy between complex and simple molecules

HW problem 9.7

Re: Internal Energy in Thermodynamics

Equation S = q/T

Re: Question 8.41

Internal Energy in Thermodynamics

Question 8.41