Search found 30 matches
- Fri Mar 15, 2019 3:17 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Gibbs Free Energy of Formation
- Replies: 2
- Views: 486
Gibbs Free Energy of Formation
Is Gibbs Free Energy of Formation always zero for pure elements? Like in today's review problem, Cl2 had a Gibbs Free Energy of Formation = 0 - does this apply to any element?
- Fri Mar 15, 2019 3:15 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: H+ in Cell Diagram
- Replies: 3
- Views: 393
H+ in Cell Diagram
When do you include H+ in the cell diagram? Do you always put it in whenever there are H+ ions in a balanced reduction/oxidation half reaction?
- Wed Mar 13, 2019 2:59 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Rate Determining Step
- Replies: 9
- Views: 1146
Re: Rate Determining Step
As far as practice problems I've seen/done, they'll have to tell you the slow step in a mechanism, and depending whether the slow step involves an intermediate or not, you'll have to tailor your approach.
- Sun Mar 10, 2019 8:44 pm
- Forum: Balancing Redox Reactions
- Topic: 6K.3 part d
- Replies: 2
- Views: 286
Re: 6K.3 part d
You do the problem as you would any other redox reaction problem. Obtain the balanced half reactions, multiply the oxidation and reduction half-reactions by appropriate factors that will result in the same number of electrons present, and then add the half reactions. In this case, Cl2 just happens t...
- Sun Mar 10, 2019 7:30 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: solving for rates with charts
- Replies: 3
- Views: 409
Re: solving for rates with charts
And, once you solve for the reactant orders, you can solve for the rate constant by choosing a set of values from the table and plugging them in. Once you have the rate constant, you can plug in concentrations given from the table and the rate constant you solved for to find an unknown rate of a rea...
- Sun Mar 10, 2019 7:29 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: solving for rates with charts
- Replies: 3
- Views: 409
Re: solving for rates with charts
So, your rate law will be k = [A]^n x [B]^m. When given a chart, you have to look at two cases: one where [A] is changing but [B] is constant, and one where [B] is changing and [A] is constant. For whichever concentration is changing, use this formula - rate(2)/rate(1) = ([A(2)]/[A(1)])^n - with val...
- Sun Mar 10, 2019 7:25 pm
- Forum: First Order Reactions
- Topic: 7B.9a
- Replies: 1
- Views: 263
Re: 7B.9a
First calculate the concentration of A at 3 minutes.
[A(t)] = [A(0)] - (1 mol A/3 mol B) x [B(t)] = .015mol/L - (1/3) x .018 mol/L = .009 mol/L
The rate constant is then determined from the first-order integrated rate law.
k = (ln([A(0)]/[A(t)]) / t = ln (.015/.009) / 3 min = .17 min^-1
[A(t)] = [A(0)] - (1 mol A/3 mol B) x [B(t)] = .015mol/L - (1/3) x .018 mol/L = .009 mol/L
The rate constant is then determined from the first-order integrated rate law.
k = (ln([A(0)]/[A(t)]) / t = ln (.015/.009) / 3 min = .17 min^-1
- Sun Mar 03, 2019 5:35 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Rate Determining Step
- Replies: 9
- Views: 1146
Re: Rate Determining Step
The rate-determining step is the slowest step in a reaction mechanism. Because it is the slowest, it determines the rate of the overall reaction. In a reaction with a slow initial step, the rate law will simply be determined by the stoichiometry of the reactants.
- Sun Mar 03, 2019 5:31 pm
- Forum: Second Order Reactions
- Topic: first vs. second order
- Replies: 2
- Views: 472
Re: first vs. second order
In a first order reaction, there will be one reactant present in the rate law. For a second order reaction, you can either have a rate law with one reactant to the second order, or with two reactants both to the first order. I remember it by adding the exponents of all the reactants in the rate law ...
- Sun Mar 03, 2019 3:55 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Rusting Process
- Replies: 1
- Views: 253
Rusting Process
What is the chemical formula of the rusting process? I'm not understanding the book's explanation of it.
- Sun Feb 24, 2019 10:31 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Difference between Molar Entropy and Standard Residual Entropy
- Replies: 3
- Views: 558
Re: Difference between Molar Entropy and Standard Residual Entropy
Residual entropy is the measure of how many different arrangements a molecule can have. Molar entropy is a measure of a molecule's chaos/disorder.
- Sun Feb 24, 2019 10:28 pm
- Forum: Balancing Redox Reactions
- Topic: Salt Bridge
- Replies: 5
- Views: 458
Re: Salt Bridge
The purpose of the salt bridge is to maintain charge balance. As the electrons move from the anode to the cathode, electrons move from the cathode back to the anode to keep the flow of electrons constant. It ultimately maintains neutrality.
- Sun Feb 24, 2019 10:26 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: K>1
- Replies: 2
- Views: 280
Re: K>1
K is greater than one when there are higher concentrations of products than reactants at equilibrium. If K is equal to 1, the reaction favors neither the products or the reactants, meaning that their concentrations are the same.
- Tue Feb 12, 2019 10:00 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: HOTDOG Question 5A
- Replies: 1
- Views: 239
HOTDOG Question 5A
Can someone explain why we don't use the Cv for a monatomic ideal gas (3/2R) when calculating the change in entropy that accounts for the temperature change of the system? I have delta S = (3.73 mol)(8.314 J/Kmol)ln(348/323) = 3.48 J/K, which includes R but not the 3/2.
- Tue Feb 12, 2019 8:52 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthalpies
- Replies: 8
- Views: 885
Bond Enthalpies
Just confirming - when using the bond enthalpies method, it's the sum of the bonds broken minus the sum of the bonds formed, right? This goes against the sum of products minus the sum of reactants in other forms of the equation, so I wanted to make sure I have it right. Thanks!
- Tue Feb 12, 2019 8:50 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: equations
- Replies: 1
- Views: 217
Re: equations
I believe you can use them in whichever situation - the variables that are constant in each case are the only thing that differ.
- Sun Feb 10, 2019 10:13 pm
- Forum: Calculating Work of Expansion
- Topic: Work Formula Involving Integral
- Replies: 3
- Views: 326
Work Formula Involving Integral
Is the integral that is used to describe "a sum of an infinite number of steps" just a way to obtain the w = -P x change of V equation? If not, in what situation would we use it?
- Sun Feb 10, 2019 5:08 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Reversible vs. Irreversible
- Replies: 2
- Views: 327
Reversible vs. Irreversible
This may be a really simple answer, but can someone explain the fundamental different between reversible and irreversible processes? I don't have it in my notes anywhere and I'm getting caught up on it. Thanks!
- Sun Feb 03, 2019 9:42 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Work Formula
- Replies: 5
- Views: 607
Work Formula
This is in reference to the work equation, where w = F x D = P x A x D = P x delta V. How does the third equality come from the second one?
- Sun Feb 03, 2019 4:48 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: 7th Edition, 4A.7
- Replies: 1
- Views: 254
Re: 7th Edition, 4A.7
The heat change will be made up of two terms - one term to raise the temperature of the copper and the other to raise the temperature of the water.
q = (300g)(4.18J/Cg)(100C-20C) + (400g)(.38J/Cg)(100C-20C) = 1.12x10^5 J = 1.12x10^2 kJ (with C being Celsius)
q = (300g)(4.18J/Cg)(100C-20C) + (400g)(.38J/Cg)(100C-20C) = 1.12x10^5 J = 1.12x10^2 kJ (with C being Celsius)
- Sun Feb 03, 2019 4:45 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Friday 2/1 Lecture Notes
- Replies: 2
- Views: 355
Friday 2/1 Lecture Notes
I was looking over my notes from Friday's lecture and had written down the diagram with C6H12O6 and 6O2 as reactants/6CO2 and 6H2O as products and H/heat given off increasing, but I don't really remember the context it was given in. Can someone explain?
- Sun Jan 27, 2019 8:33 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 4C.15 (7th Edition)
- Replies: 1
- Views: 207
Question 4C.15 (7th Edition)
Does anyone have any tips for figuring out this kind of question, where you're given the delta H of fusion and vaporization and heat capacities and have to match it to a graph? I'm just confused on how the information given is supposed to correlate to regions on the graph. Thanks!
- Sun Jan 27, 2019 6:49 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond Enthaplies
- Replies: 1
- Views: 206
Re: Bond Enthaplies
It's because most values are averages from measurements of many different compounds. The only time bond enthalpies are accurate is in the case of diatomic molecules, like N2, O2, or HCl.
- Sun Jan 27, 2019 6:47 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Preferable Method?
- Replies: 3
- Views: 342
Re: Preferable Method?
The only thing Dr. Lavelle said about accuracy was that using bond enthalpies was the least accurate method because the values found in the table are an average calculated from many different compounds. Because of this, I think the other two methods are preferred and can be used interchangeably.
- Sun Jan 20, 2019 4:15 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Strong/Weak Acids and Bases
- Replies: 1
- Views: 203
Strong/Weak Acids and Bases
For those who took 14A with Lavelle, is it encouraged to memorize different strong/weak acids and bases for the tests, or is it made known in the question? I just didn't have him last quarter so I'm not sure how much is given to us. Thanks!
- Fri Jan 18, 2019 3:58 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Question 6B.11 (7th Edition)
- Replies: 1
- Views: 230
Question 6B.11 (7th Edition)
My question is in regards to Part B, where it asks the mass of solid Na2O added to the original flask. I'm not understanding why you need to multiply by the molar ratio, 1 mol Na2O/2 mol NaOH, if we're just finding the mass of Na2O. Thanks for the help!
- Thu Jan 17, 2019 1:03 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Homework
- Replies: 5
- Views: 783
Homework
Do we still have homework due the day of the test?
- Sun Jan 13, 2019 3:09 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Module Part 2 Question
- Replies: 2
- Views: 255
Re: Module Part 2 Question
I set up my ICE table the same way you did and got the same values for the equilibrium constant equation, but when I plugged it into my calculator, I got 4.983. Is that an option? Maybe it was just a mistake in entering the values into the calculator.
- Sun Jan 13, 2019 2:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Question 5I.13 (7th Edition)
- Replies: 1
- Views: 112
Question 5I.13 (7th Edition)
My question is in regards to Part C of this problem, which asks us to use our results from Part A and B to determine which is more thermodynamically stable relative to its atoms at 1000 Kelvin, Cl2 or F2. I found both of the equilibrium concentrations for Cl2 and F2, and the eq. concentration of F2 ...
- Thu Jan 10, 2019 11:19 pm
- Forum: General Science Questions
- Topic: Chem 20A to Chem 14B
- Replies: 3
- Views: 807
Re: Chem 20A to Chem 14B
My roommate is in 20B right now, and her syllabus looks pretty similar to ours. Her class spends some time on reaction stoichiometry, which we did in 14A, but a majority of her content is thermodynamics and chemical equilibria, which is what we are starting right now. I'm sure you'd be totally fine ...