## Search found 35 matches

- Sun Mar 19, 2017 12:49 am
- Forum: *Cyclopropanes and Cyclobutanes
- Topic: energy barriers
- Replies:
**3** - Views:
**562**

### Re: energy barriers

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 :(

- Sat Mar 18, 2017 2:04 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Final 2013 Q4C
- Replies:
**2** - Views:
**302**

### Re: Final 2013 Q4C

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?

- Fri Mar 17, 2017 11:06 pm
- Forum: *Cyclohexanes (Chair, Boat, Geometric Isomers)
- Topic: axial/equatorial positions for multiple substituents
- Replies:
**2** - Views:
**358**

### Re: axial/equatorial positions for multiple substituents

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.

- Fri Mar 17, 2017 10:59 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Final 2013 Q4C
- Replies:
**2** - Views:
**302**

### Final 2013 Q4C

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?

- Fri Mar 17, 2017 10:57 pm
- Forum: *Cyclopropanes and Cyclobutanes
- Topic: energy barriers
- Replies:
**3** - Views:
**562**

### Re: energy barriers

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

- Fri Mar 17, 2017 10:36 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 2013 Final Q4 Part A
- Replies:
**2** - Views:
**270**

### Re: 2013 Final Q4 Part A

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.

- Fri Mar 17, 2017 2:13 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 2013 midterm q4
- Replies:
**1** - Views:
**247**

### 2013 midterm q4

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

- Fri Mar 17, 2017 1:23 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: final exam 2013 q3c
- Replies:
**1** - Views:
**197**

### final exam 2013 q3c

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?

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?

- Wed Mar 15, 2017 9:32 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Saying Thank You to Dr. Lavelle
- Replies:
**202** - Views:
**63405**

### Re: Saying Thank You to Dr. Lavelle

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

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

- Sat Mar 11, 2017 12:40 am
- Forum: *Haloalkenes
- Topic: 2.16 Intro to Ochem chapter 2
- Replies:
**3** - Views:
**439**

### Re: 2.16 Intro to Ochem chapter 2

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

- Fri Mar 10, 2017 11:49 pm
- Forum: *Haloalkenes
- Topic: 2.16 Intro to Ochem chapter 2
- Replies:
**3** - Views:
**439**

### 2.16 Intro to Ochem chapter 2

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.

- Wed Feb 22, 2017 5:03 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.67
- Replies:
**2** - Views:
**421**

### 15.67

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...

- Wed Feb 22, 2017 5:00 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.65
- Replies:
**1** - Views:
**250**

### 15.65

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 ...

- Mon Feb 20, 2017 1:36 am
- Forum: First Order Reactions
- Topic: 15.21
- Replies:
**1** - Views:
**423**

### 15.21

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...

- Mon Feb 20, 2017 12:09 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.17
- Replies:
**2** - Views:
**335**

### 15.17

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...

- Tue Feb 14, 2017 6:34 pm
- Topic: 14.37a
- Replies:
**1** - Views:
**200**

### 14.37a

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...

- Fri Feb 10, 2017 3:19 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.27
- Replies:
**3** - Views:
**246**

### Re: 14.27

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?

- Fri Feb 10, 2017 1:04 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.27
- Replies:
**3** - Views:
**246**

### 14.27

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?

- Thu Feb 09, 2017 9:12 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.29
- Replies:
**1** - Views:
**150**

### 14.29

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.

- Mon Feb 06, 2017 8:24 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15 - half reactions and cell diagrams for solubility equilibriums
- Replies:
**2** - Views:
**273**

- Mon Feb 06, 2017 7:28 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.21a - sign of cell potentials
- Replies:
**1** - Views:
**187**

### 14.21a - sign of cell potentials

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 -....

- Mon Feb 06, 2017 7:15 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: ∆G = -nFE
- Replies:
**4** - Views:
**325**

### Re: ∆G = -nFE

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...

- Mon Feb 06, 2017 5:52 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15 - half reactions and cell diagrams for solubility equilibriums
- Replies:
**2** - Views:
**273**

### 14.15 - half reactions and cell diagrams for solubility equilibriums

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?

- Mon Feb 06, 2017 5:27 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: ∆G = -nFE
- Replies:
**4** - Views:
**325**

### Re: ∆G = -nFE

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

- Mon Feb 06, 2017 5:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.13 [ENDORSED]
- Replies:
**1** - Views:
**182**

### 14.13 [ENDORSED]

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?

- Mon Feb 06, 2017 4:48 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: ∆G = -nFE
- Replies:
**4** - Views:
**325**

### ∆G = -nFE

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?

- Sat Feb 04, 2017 8:53 pm
- Forum: Balancing Redox Reactions
- Topic: 14.5a [ENDORSED]
- Replies:
**1** - Views:
**228**

### 14.5a [ENDORSED]

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-

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-

- Wed Feb 01, 2017 9:58 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Internal Energy
- Replies:
**2** - Views:
**221**

### Internal Energy

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

- Wed Feb 01, 2017 1:42 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Difference in molar entropy between complex and simple molecules
- Replies:
**1** - Views:
**263**

### Difference in molar entropy between complex and simple molecules

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...

- Tue Jan 31, 2017 6:39 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: HW problem 9.7
- Replies:
**1** - Views:
**201**

### HW problem 9.7

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.

- Mon Jan 30, 2017 11:05 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Internal Energy in Thermodynamics
- Replies:
**2** - Views:
**236**

### Re: Internal Energy in Thermodynamics

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).

- Mon Jan 30, 2017 11:00 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Equation S = q/T
- Replies:
**1** - Views:
**207**

### Equation S = q/T

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?

- Mon Jan 30, 2017 10:11 pm
- Forum: Phase Changes & Related Calculations
- Topic: Question 8.41
- Replies:
**2** - Views:
**223**

### Re: Question 8.41

That makes sense, thank you so much!

- Mon Jan 30, 2017 10:11 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Internal Energy in Thermodynamics
- Replies:
**2** - Views:
**236**

### Internal Energy in Thermodynamics

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 = ∆...

- Sat Jan 28, 2017 7:33 pm
- Forum: Phase Changes & Related Calculations
- Topic: Question 8.41
- Replies:
**2** - Views:
**223**

### Question 8.41

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...