## Search found 32 matches

- Fri Mar 16, 2018 11:17 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Review Session Question
- Replies:
**1** - Views:
**93**

### Re: Review Session Question

You need to have a solid or conductor to allow electron transfer. You need to add Pt(s) whenever there isn't a solid or other conductor already in the reaction on that side.

- Fri Mar 16, 2018 11:16 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Porous Disks vs Salt Bridges
- Replies:
**1** - Views:
**177**

### Re: Porous Disks vs Salt Bridges

They pretty much serve the same purpose; to stop charge buildup by allowing ion transfer. They should say which one is being used.

- Tue Mar 13, 2018 8:51 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Adsorption
- Replies:
**2** - Views:
**128**

### Re: Adsorption

Adsorption just means that the reactants are on the surface of catalyst, which shouldn't be confused with absorption.

- Tue Mar 13, 2018 8:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.57? I don't understand the way the solutions manual solved the problem?
- Replies:
**1** - Views:
**109**

### Re: 8.57? I don't understand the way the solutions manual solved the problem?

Since it's combustion, you would add the enthalpis of combustion together. Also, you would have to change the sign for C2H6 because it's a product.

- Tue Mar 13, 2018 8:15 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Cp and Cv
- Replies:
**2** - Views:
**115**

### Re: Cp and Cv

I think you don't need to really know too much about it because they give it to you on the constants and equations sheet. I just think you need to know that it's for a monatomic gas. Also, you should know that it increases for linear/diatomic gases [Cv =(5/2)R Cp=(7/2)R] and non-linear gases [Cv =3R...

- Thu Mar 08, 2018 12:23 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.23 part C
- Replies:
**3** - Views:
**153**

### Re: 15.23 part C

you can find [A] at 115 seconds by converting the [B].

[A] = 0.153 molA/L - [0.034 molB/L x (2 molA/1 molB)]

then you can use ln[A] = -kt + ln{A]o

[A] = 0.153 molA/L - [0.034 molB/L x (2 molA/1 molB)]

then you can use ln[A] = -kt + ln{A]o

- Tue Mar 06, 2018 7:51 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chapters on test?
- Replies:
**3** - Views:
**178**

### Re: Chapters on test?

It's 15.1-15.6

- Mon Mar 05, 2018 10:28 pm
- Forum: First Order Reactions
- Topic: 15.29
- Replies:
**3** - Views:
**225**

### Re: 15.29

part A: You have to change everything in terms of A, so when they give you the concentration of B after 3 minutes, to get A you would do [A]t=[A]o - [B]t(mol A/3 mol B) [A]t=0.015 mol A/L - (0.018mol B/L) x (mol A/3 mol B) [A]t= 0.009 mol/L t= 3 minutes Then you use ln[A]t= -kt + ln[A]o k=0.17/min p...

- Thu Mar 01, 2018 10:21 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order of the Reaction [ENDORSED]
- Replies:
**3** - Views:
**141**

### Re: Order of the Reaction [ENDORSED]

To find order you must use experimental data, like in lecture we used rates with different concentrations to calculate the order. You can't ever use the chemical equation to figure out order.

- Wed Feb 28, 2018 3:39 pm
- Forum: General Rate Laws
- Topic: 15.3 C
- Replies:
**8** - Views:
**500**

### Re: 15.3 C

When you calculate the change in concentration over the change in time, you get the rate of 2NO2, but you need the unique rate. This means that they're asking for the rate of one mole of NO2, so you divide the rate by 2.

- Tue Feb 27, 2018 9:37 am
- Forum: General Rate Laws
- Topic: Order of reactions?
- Replies:
**3** - Views:
**136**

### Re: Order of reactions?

It's easy to determine what n (the order) equals if you add up all the powers of the concentrations. Add the products' powers together and subtract the reactants' powers .

- Tue Feb 27, 2018 9:34 am
- Forum: General Rate Laws
- Topic: 15.9
- Replies:
**5** - Views:
**202**

### Re: 15.9

It's because for each one you need to determine what the units for k are. So the equation would be rate = k [(molA)/L]^n and you would have to divide both sides by [(molA)/L]^n to isolate k, then you cancel out the like terms (rate always equals mol/(Ls) ), then depending on the order, the n would b...

- Tue Feb 27, 2018 9:28 am
- Forum: First Order Reactions
- Topic: First Order Reaction Equations
- Replies:
**3** - Views:
**135**

### Re: First Order Reaction Equations

They are the same equation. The only difference is the way you write the equation that they are derived from. In this case for ln [A]t/[A]o = -kt, it is derived from ln[A]t - ln[A]o = -kt While the other equation comes from ln[A]o - ln[A]t = kt, which can be rewritten as -ln[A]t +ln[A]o =kt, if you ...

- Tue Feb 27, 2018 9:21 am
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Midterm 6A (Multiple Choice)
- Replies:
**4** - Views:
**318**

### Re: Midterm 6A (Multiple Choice)

Just in general, gases have the most freedom to move, then liquids and then solids have the least. There are bonds keeping the liquid together more, while for a gas they bounce off each other with more momentary interactions.

- Tue Feb 20, 2018 9:44 am
- Forum: Balancing Redox Reactions
- Topic: 14.3
- Replies:
**1** - Views:
**92**

### Re: 14.3

You have to compare it to the connected element. For example, oxygen is 2-, so for part a S2O3 2- , oxygen is 3x2=6 and there's still 2-, so S2 = +4 and S= +2 On the other side, oxygen us 4x2=8 and there's 2-, so S = +6. This means that it gave up electrons. Also, Cl2 has no overall charge, but ends...

- Tue Feb 20, 2018 9:33 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Concentration of Reactants and Products
- Replies:
**1** - Views:
**109**

### Re: Concentration of Reactants and Products

Standard reduction potential (E naught) is an intensive property, so it stays the same as long as it's a balanced equation. However, to find cell potential you must subtract that with (RT/nF)lnQ. This is where the concentrations are important because depending on the ratio of product concentration o...

- Mon Feb 19, 2018 12:54 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: E vs E naught
- Replies:
**3** - Views:
**283**

### Re: E vs E naught

Both E naught and G naught are at standard conditions, when everything is in their standard state. It's also used the same way as in using E naught to fine E as you would do G naught to find G, when the parts aren't in their standard state.

- Mon Feb 19, 2018 12:46 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: G=-nFE
- Replies:
**3** - Views:
**197**

### Re: G=-nFE

The only factor that changes the sign is E, so that's why knowing E can tell you whether it is spontaneous or not. n is simply the number of moles like in previous equations.

- Sun Feb 11, 2018 8:55 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.99: Enthalpy Values
- Replies:
**2** - Views:
**149**

### Re: 8.99: Enthalpy Values

Since they don't give you the enthalpy of formation of ZnCl2, you need to take it by parts. You need to use the one for Zn2+ and 2Cl-. Cl- is the same as HCl, hence the same number being used twice.

- Sat Feb 10, 2018 6:27 pm
- Forum: Balancing Redox Reactions
- Topic: 14.3
- Replies:
**1** - Views:
**151**

### Re: 14.3

Cl2 and Cl- don't have oxygen or hydrogen, so it's simply Cl2 + 2e- --> 2Cl-.

However, for MnO4- to Mn2+, there's the oxygen that is taken to form H2O.

MnO4- + 8H+ + 5e- --> Mn2+ + 4H20

However, for MnO4- to Mn2+, there's the oxygen that is taken to form H2O.

MnO4- + 8H+ + 5e- --> Mn2+ + 4H20

- Sat Feb 10, 2018 6:19 pm
- Forum: Balancing Redox Reactions
- Topic: 14.5
- Replies:
**4** - Views:
**184**

### Re: 14.5

I think you can reduce O3 to BrO3- becuase you need to oxidize Br-, so if the Bromine is losing electrons, it can't also be accepting it.

- Sat Feb 03, 2018 10:06 pm
- Forum: Van't Hoff Equation
- Topic: Final Van't Hoff Eqn
- Replies:
**2** - Views:
**322**

### Re: Final Van't Hoff Eqn

In lecture, I believe the reasoning was because the standard entropy is assumed to be constant and the same for both temperatures, so they would cancel out when you derive the equation. Also, since the R is a constant. If you look at the explanation above you'll see this. The enthalpies, however, do...

- Sat Feb 03, 2018 9:56 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.65
- Replies:
**6** - Views:
**230**

### Re: 9.65

Since everything naturally goes towards disorder or an increased entropy, that is why the most stable would be when entropy is highest. It takes more energy to decrease entropy, therefore meaning that any negative entropy of formation is unstable.

- Sat Jan 27, 2018 3:23 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: Calculating Degeneracy
- Replies:
**1** - Views:
**90**

### Re: Calculating Degeneracy

Yes that is the equation that we used in the example given in lecture.

- Thu Jan 25, 2018 7:33 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Reaction enthalpy involving H-H
- Replies:
**2** - Views:
**154**

### Re: Reaction enthalpy involving H-H

Bond enthalpy and enthalpy of formation are two different things. So the bond enthalpy is the energy for the bond between the two hydrogen atoms. However, the enthalpy of formation is for the standard state. In lecture, the example states that H2 -> H2 requires no energy because it's already in its ...

- Thu Jan 25, 2018 7:31 pm
- Topic: 8.49
- Replies:
**2** - Views:
**137**

### Re: 8.49

The question gives you you deltaH (enthalpy) which can be related to the change in internal energy. Delta H=Delta U+PDelta V. Then PDeltaV can be changed to deltanRT. You know the net change in moles would be one because the products has a total of # and the reactants has a total of 2, making the ch...

- Sun Jan 21, 2018 4:26 pm
- Topic: 8.65, where did that equation come from?
- Replies:
**3** - Views:
**159**

### Re: 8.65, where did that equation come from?

It's because the question states to find the enthalpy of formation for dinitrogen pentoxide (N2O5) from the two given equations as wells as the enthalpy of formation of nitric oxide (NO), which would be N2 +O2 ---> 2NO. That means that the equation of enthalpy of formation of N2O5 should be N2 + 5/2...

- Fri Jan 19, 2018 6:50 pm
- Topic: relationship between q and w
- Replies:
**2** - Views:
**141**

### Re: relationship between q and w

If q is positive then w is not necessarily positive as well. In 17, heat is absorbed making q positive, but work is being done by the system, which means that w must be negative. q and w are not automatically the same sign.

- Thu Jan 18, 2018 6:26 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Energy and matter?
- Replies:
**6** - Views:
**287**

### Re: Energy and matter?

Yes both matter and energy are important in open, closed, and isolated systems.

For open systems, they are able to transfer both energy and matter to the external surroundings.

Closed systems can transfer energy, but not matter.

While isolated systems cannot exchange energy nor matter.

For open systems, they are able to transfer both energy and matter to the external surroundings.

Closed systems can transfer energy, but not matter.

While isolated systems cannot exchange energy nor matter.

- Sat Jan 13, 2018 8:59 pm
- Forum: Calculating Work of Expansion
- Topic: Work (in general) [ENDORSED]
- Replies:
**4** - Views:
**177**

### Re: Work (in general) [ENDORSED]

In the book there is also an equation for work that relates the external pressure and volume.

w = -Pex x Delta V

w = -Pex x Delta V

- Sat Jan 13, 2018 3:45 pm
- Topic: Nonexpansion Work vs. Expansion Work
- Replies:
**1** - Views:
**111**

### Re: Nonexpansion Work vs. Expansion Work

Expansion work is work that changes the volume, while nonexpansion work is work that doesn't change the volume. I think that the meaning of constant volume and no nonexpansion work means that their is no other work done. That's why it is equal to q, because the transfer/change of energy is happening...

- Fri Jan 12, 2018 2:26 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heating Curve
- Replies:
**6** - Views:
**245**

### Re: Heating Curve

The steepness of the slope is related to the heat capacity. The steeper the slope, the lower the heat capacity is and vice versa.