## Search found 35 matches

- Sat Mar 17, 2018 1:12 am
- Forum: Administrative Questions and Class Announcements
- Topic: Derivations
- Replies:
**2** - Views:
**219**

### Re: Derivations

Dr. Lavelle said that he would not ask us questions saying "derive this" on the final, but it would probably be good to know how the derivations of the rate laws work just so you can get a better understanding of the equations as a whole.

- Sat Mar 17, 2018 1:04 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.33
- Replies:
**1** - Views:
**120**

### Re: 14.33

Could you indicate which equation you're referring to? They use the Nernst equation, which is in the textbook, as well as just a combination of Gibbs free energy from the half reactions to find the total Gibbs free energy.

- Sat Mar 17, 2018 12:58 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: qreaction and qcalorimeter
- Replies:
**3** - Views:
**420**

### Re: qreaction and qcalorimeter

So, to follow up on that question, is q

_{calorimeter}just another way of saying q_{surroundings}? Since when a reaction loses heat, the surroundings gain that heat, and vice versa?- Sat Mar 17, 2018 12:56 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.31
- Replies:
**1** - Views:
**117**

### Re: 14.31

As you can see in the solutions manual for part b), you are supposed to put the oxidated ion over the oxidating ion when it comes to writing out the half reaction for the anode. Since Pb4+ is the result of Pb2+ losing electrons (oxidation is loss), Pb4+ is oxidated, so you divide Pb4+ / Pb2+.

- Sat Mar 17, 2018 12:48 am
- Forum: *Nucleophilic Substitution
- Topic: Transition States
- Replies:
**3** - Views:
**792**

### Re: Transition States

To see how high the energy of the transition state is, just look at the graph of a reaction. The transition states are the peaks of each slope. Therefore, they are at the highest point of each step, so they have high energy.

- Sat Mar 17, 2018 12:39 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Adiabatic Process
- Replies:
**4** - Views:
**316**

### Adiabatic Process

What exactly does it mean when a process is adiabatic? And how does this affect the equations for internal energy and enthalpy?

- Sat Mar 17, 2018 12:32 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: qreaction and qcalorimeter
- Replies:
**3** - Views:
**420**

### qreaction and qcalorimeter

Will someone explain to me why it is that q

_{reaction}= -q_{calorimeter}? I don't understand why they are opposite of one another.- Fri Mar 09, 2018 3:08 pm
- Forum: Second Order Reactions
- Topic: Units [ENDORSED]
- Replies:
**5** - Views:
**392**

### Re: Units [ENDORSED]

As a general formula to calculate the units of the rate constant for any reaction order n, you can use the equation:

units of k = (L/mol)

units of k = (L/mol)

^{n-1}/ s- Fri Mar 09, 2018 3:02 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Chapter 15 #85
- Replies:
**2** - Views:
**168**

### Re: Chapter 15 #85

We went over something similar to this in lecture today. The proposed structure for the activated complex is going to be the lewis structures of the two interacting species, aligned with one another with respect to the bonds broken and the bonds formed. Dr. Lavelle demonstrated this in lecture when ...

- Fri Mar 09, 2018 2:33 pm
- Forum: First Order Reactions
- Topic: Calculating t using First Order Integrated Rate Law
- Replies:
**2** - Views:
**250**

### Calculating t using First Order Integrated Rate Law

If you were asked to find the time it takes for the concentration of a substance to decrease to, say, 50% of its initial concentration, how would you set the equation up? I understand that you would apply it to ln[A]t = -kt + ln[A]0, but I'm not sure what to do with the 50%, and I always mess up wit...

- Fri Mar 09, 2018 2:21 pm
- Forum: General Rate Laws
- Topic: Division in Rate Law
- Replies:
**3** - Views:
**208**

### Division in Rate Law

What is the order of the reaction if the rate law includes division, such as rate = k*[A]/[B]? Would it be zero due to the fact that [A] has a power of 1 and [B] has a power of -1? Explain how this works please!

- Fri Mar 09, 2018 2:10 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Order less than 0
- Replies:
**4** - Views:
**313**

### Re: Order less than 0

Yes, a reaction order can be negative. This means that the concentration has an inverse effect on the rate of reaction.

- Fri Mar 02, 2018 2:22 pm
- Forum: First Order Reactions
- Topic: Pseudo First Order Reaction
- Replies:
**5** - Views:
**590**

### Re: Pseudo First Order Reaction

A pseudo first order reaction is just another way of saying a second order reaction but under more ideal conditions in order to simplify your calculations. If the concentration of one reactant is in great excess, you may assume that it is constant, allowing it to be ignored so that the rate therefor...

- Fri Mar 02, 2018 2:15 pm
- Forum: Zero Order Reactions
- Topic: Units of k [ENDORSED]
- Replies:
**13** - Views:
**459**

### Re: Units of k [ENDORSED]

I know this question has been answered many times, but the simplest response is just to do dimensional analysis and multiply everything out. The units of k will be whatever makes the equation work out, so they differ according to the order of the reaction.

- Sun Feb 25, 2018 6:52 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies:
**7862** - Views:
**1100756**

### Re: Post All Chemistry Jokes Here

I would tell another chemistry joke, but it appears that all the good ones Argon.

- Sat Feb 24, 2018 7:41 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Calculating n in the Nernst Equation
- Replies:
**2** - Views:
**451**

### Calculating n in the Nernst Equation

I have a question about the variable n in the Nernst Equation. I have read that it correlates to the number of electrons transferred across the reaction. However, how is this calculated when there are 3 electrons added on the reactants side of the equation and two added to the products side? Do you ...

- Tue Feb 13, 2018 5:35 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Practice Test, Question 3A
- Replies:
**2** - Views:
**122**

### Re: Practice Test, Question 3A

The question specifies that you have 125 grams of Krypton gas and 9.00 grams of Helium gas. Therefore, you may use the molar masses provided by a periodic table to calculate the number of moles. The molar mass of Krypton (Kr) is 83.80 g/mol and the molar mass of Helium is 4.00 g/mol.

- Tue Feb 13, 2018 5:31 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Problem 9.35
- Replies:
**1** - Views:
**98**

### Re: Problem 9.35

As specified in the exercise, Container A contains monoatomic gas, and gas molecules are more free to bounce off of one another and create disorder compared to atoms that are "bound together." If an atom is more free to move around, it has greater entropy.

- Tue Feb 13, 2018 5:02 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Enthalpy of fusion [ENDORSED]
- Replies:
**4** - Views:
**252**

### Re: Enthalpy of fusion [ENDORSED]

Can someone please explain the answer to this question in more detail? I don't understand why we use q=mcdeltaT + deltaH. Also, I plugged the values into this and got an incorrect answer.

- Fri Feb 09, 2018 1:58 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Exercise 11.19
- Replies:
**2** - Views:
**109**

### Re: Exercise 11.17

My mistake, I meant to say Exercise 19.

- Fri Feb 09, 2018 1:36 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Exercise 11.19
- Replies:
**2** - Views:
**109**

### Exercise 11.19

In 11.19(b), I noticed that, when finding the equilibrium constant at 25 degrees C for the oxidation of carbon monoxide, 2CO(g) + O2(g) = 2CO2(g), the solutions manual says to subtract the entropy of formation of 2CO from the entropy of formation of 2CO2. In 11.19(a), the solutions manual just keeps...

- Thu Feb 08, 2018 11:49 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Exercise 9.19
- Replies:
**2** - Views:
**163**

### Exercise 9.19

I was wondering if anyone could explain exercise 9.19 for me? More specifically, why does the solutions manual describe that you must heat the water to 100 degrees and then cool it back down to 83 degrees Celsius when finding the standard entropy of vaporization of water at 85 degrees C? I don't und...

- Wed Jan 31, 2018 11:25 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Define Entropy
- Replies:
**4** - Views:
**175**

### Re: Define Entropy

In summary, entropy is equal to the amount of disorder. Gibbs Free Energy is equal to the enthalpy minus the entropy times the temperature. We can use Gibbs Free Energy to determine whether or not a reaction is spontaneous, as a negative value of G will show spontaneity.

- Wed Jan 31, 2018 11:18 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacity
- Replies:
**3** - Views:
**170**

### Re: Heat Capacity

Yes--at equilibrium, the heat of the object must equal the heat of its surroundings.

- Wed Jan 31, 2018 2:44 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.5
- Replies:
**5** - Views:
**172**

### Re: 9.5

In addition, when using the equation deltaS = qrev/T, you'll find that a lower temperature (in the denominator) will result in a greater overall entropy deltaS, and vice versa. So, 200K will result in a greater deltaS (deltaS = 200) than 800K (deltaS = 50). They are inversely proportional. To solve ...

- Wed Jan 31, 2018 2:36 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: 9.33 (a)
- Replies:
**3** - Views:
**276**

### Re: 9.33 (a)

If you think back to how Dr. Lavelle explained it in class, entropy is equal to the amount of disorder. Atoms in a gaseous state have more space to move around and bounce off one another compared to liquid and solid states. Therefore, a greater ability to bounce off of one another allows for more po...

- Wed Jan 31, 2018 2:27 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Why should deltaG=0?
- Replies:
**4** - Views:
**404**

### Re: Why should deltaG=0?

The above responses are correct. To elaborate further, solving for G = 0 will give us the boiling point at which the liquid turns into a gas. The question asks us to find the temperature at which the vaporization of Br2 is spontaneous. Since we don't have enough information to solve for any negative...

- Wed Jan 31, 2018 2:21 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Class Example
- Replies:
**6** - Views:
**209**

### Re: Class Example

With the information that we had, we were forced to solve for G = 0, which gives us the boiling point at which the liquid will become a gas. However, since it must be a gas, anything greater than 333K will be correct for this equation. Any higher temperature will result in G having a negative value,...

- Fri Jan 26, 2018 3:52 am
- Forum: Phase Changes & Related Calculations
- Topic: Homework Exercise 8.43 and Phase Changes
- Replies:
**1** - Views:
**111**

### Homework Exercise 8.43 and Phase Changes

Exercise 8.43 mentions the connection between values such as heat of fusion and heat of vaporization, as well as heat capacities for different phases, and relates them to the appearance of phase change graphs. Can someone refresh me on how to read and interpret heating curves, and how to relate cert...

- Fri Jan 26, 2018 3:43 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Homework Exercise 8.53
- Replies:
**1** - Views:
**176**

### Homework Exercise 8.53

Exercise 8.53(b), when referring to a temperature increase in the calorimeter assembly, asks to calculate the internal energy change, deltaU, for the reaction (deltaU = q). At the end of the calculations, you find the number of moles and the value for C*deltaT separately, and you are supposed to use...

- Wed Jan 24, 2018 10:58 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Formation of a Cation from an Atom
- Replies:
**3** - Views:
**417**

### Re: Formation of a Cation from an Atom

At first glance, I would say that it is an exothermic reaction because you are removing electrons from the atom, therefore releasing energy.

- Wed Jan 24, 2018 10:55 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Homework Problems
- Replies:
**4** - Views:
**289**

### Re: Homework Problems

Is there anyone reading this who has Mackenzie as their TA? Does anyone know her preference?

- Wed Jan 24, 2018 10:51 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies:
**7862** - Views:
**1100756**

### Re: Post All Chemistry Jokes Here

heheheheheheheh

- Wed Jan 24, 2018 10:42 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Homework Problems
- Replies:
**4** - Views:
**289**

### Homework Problems

When we turn in our 7 homework problems this week, are they supposed to be from Chapter 8 or Chapter 9, or both? I think I remember hearing Dr. Lavelle mention that we were currently in Chapter 9, but we've also done some of Chapter 8 this week.

- Sat Jan 20, 2018 7:43 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies:
**7862** - Views:
**1100756**

### Re: Post All Chemistry Jokes Here

The base is under a salt.