Search found 37 matches
- Sat Mar 16, 2019 2:38 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Review Session-Thermo-files- Wednesday-Q11-clarification [ENDORSED]
- Replies: 10
- Views: 1463
Re: Review Session-Thermo-files- Wednesday-Q11-clarification [ENDORSED]
For #11, why is Cv used for the change in temp instead of Cp? How do we distinguish when to use what?
- Mon Mar 11, 2019 6:36 pm
- Forum: Second Order Reactions
- Topic: error for 15.39 sixth edition
- Replies: 1
- Views: 216
Re: error for 15.39 sixth edition
The equation is always 1/[A] = kt+1/[A]initial. The coefficients are only taken into account when considering concentrations of reactants/products, not rate constants.
- Mon Mar 11, 2019 5:02 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate Law
- Replies: 2
- Views: 298
Rate Law
Why is it that for elementary steps, the rate law can simply be found using molecularity but this does not apply to the overall balanced reaction?
- Mon Mar 11, 2019 4:44 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Increasing reducing power
- Replies: 4
- Views: 1583
Re: Increasing reducing power
The E values we're given are reduction potentials, so the greater the value, the more likely it is to be reduced. Therefore, the species with higher reducing power are the species more likely to be oxidized, i.e. a smaller E value.
- Tue Mar 05, 2019 12:41 am
- Forum: General Rate Laws
- Topic: 15.9 6th edition
- Replies: 3
- Views: 402
15.9 6th edition
15.9 Express the units for rate constants when the concentrations are in moles per liter and time is in seconds for (a) zero-order reactions; (b) first-order reactions; (c) second order reactions.
What does it mean for a reaction to have zero-order?
What does it mean for a reaction to have zero-order?
- Tue Mar 05, 2019 12:38 am
- Forum: General Rate Laws
- Topic: Rate of Formation vs Unique Rate of a Reaction
- Replies: 2
- Views: 358
Re: Rate of Formation vs Unique Rate of a Reaction
The rate of formation is simply the rate of formation of a given product (or rate of consumption for a reactant). This is the value you get when you take the change in concentration over change in time, or d[R]/dt for the instantaneous rate. The unique rate of a reaction is when you take into accoun...
- Mon Mar 04, 2019 9:55 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 14.101 6th edition
- Replies: 1
- Views: 295
14.101 6th edition
14.101 In a neuron (a nerve cell), the concentration of K ions inside the cell is about 20–30 times as great as that outside. What potential difference between the inside and the outside of the cell would you expect to measure if the difference is due only to the imbalance of potassium ions? Does a...
- Tue Feb 26, 2019 12:10 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt(s) in cell diagram
- Replies: 4
- Views: 497
Re: Pt(s) in cell diagram
Shubham Rai 2C wrote:You would add Pt to the Ce4+ side of the diagram as it has no solids to conduct the charges. We use Pt when there are no solids in the half reaction.
the solutions show Pt(s) on both sides of the cell diagram though
- Tue Feb 26, 2019 12:07 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15b 6th ed
- Replies: 1
- Views: 183
14.15b 6th ed
14.15 Write the half-reactions and devise a galvanic cell (write a cell diagram) to study each of the following reactions: (b) H+(aq) + OH-(aq) --> H2O(l), the Brønsted neutralization reaction answer: (b) anode: 4 OH-(aq) --> O2(g) + 2H2O(l) + 4 e-; cathode: O2(g) + 4H+(aq) + 4e- --> 2H2O(...
- Mon Feb 25, 2019 11:59 pm
- Forum: Balancing Redox Reactions
- Topic: Water in Redox Reactions [ENDORSED]
- Replies: 3
- Views: 403
Re: Water in Redox Reactions [ENDORSED]
Karyn Nguyen 1K wrote:Don't we add water to basic solutions too?
Yes, we add water to basic solutions to balance out the oxygen as well. The difference with basic solutions is that you add the same amount of OH- as there are H+ to both sides of the reaction.
- Mon Feb 25, 2019 11:56 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Pt(s) in cell diagram
- Replies: 4
- Views: 497
Pt(s) in cell diagram
How do we know to add Pt(s) to the cell diagram?
example: 14.13 Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations:
(b) Ce4+(aq) + I-(aq) --> I2(s) + Ce3+(aq)
example: 14.13 Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations:
(b) Ce4+(aq) + I-(aq) --> I2(s) + Ce3+(aq)
- Tue Feb 19, 2019 9:38 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.83
- Replies: 1
- Views: 236
9.83
9.83 Hydrogen burns in an atmosphere of bromine gas to give hydrogen bromide gas. (a) What is the standard Gibbs free energy of the reaction H2(g) Br2(g) S 2 HBr(g) at 298 K? (b) If 120. mL of H2 gas at STP combines with a stoichiometric amount of bromine and the resulting hydrogen bromide dissolv...
- Tue Feb 19, 2019 7:55 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.57 units 6th edition
- Replies: 2
- Views: 303
Re: 9.57 units 6th edition
The calculations for deltaH consider the amount of moles when you multiply the deltaHf by the respective stoichiometric coefficients. So in this case, you would multiply the deltaHf of H2O2 by 2 to account for the fact that the reaction is decomposing 2 moles of H2O2. The end result is kJ/mol.
- Tue Feb 19, 2019 7:51 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.57 hmwrk 6th edition
- Replies: 2
- Views: 233
Re: 9.57 hmwrk 6th edition
*make sure you're using the correct deltaH values: H2O and H2O2 liquid
deltaH = deltaHf(H2O) + deltaHf(O2) - deltaHf(H2O2) = 2(-285.83kJ/mol) + 0 - 2(-187.78 kJ/mol) = -196.10 kJ/mol
deltaH = deltaHf(H2O) + deltaHf(O2) - deltaHf(H2O2) = 2(-285.83kJ/mol) + 0 - 2(-187.78 kJ/mol) = -196.10 kJ/mol
- Tue Feb 12, 2019 4:46 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22761
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Hannah Yates 1K wrote:Should number 4 on the midterm review be when delta T = 0?
Yes, I believe it should be deltaT=0 because temperature must be constant for q=-w.
- Tue Feb 12, 2019 4:44 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22761
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
How do you solve for the heat in #6a? Are we supposed to use standard enthalpy of formation or bond enthalpies? The standard enthalpy of formation of HI is not given and the bond enthalpy for C-O is not given.
*edit: nvm, I just realized change in internal energy is given.
*edit: nvm, I just realized change in internal energy is given.
- Tue Feb 12, 2019 12:57 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13 6th edition
- Replies: 1
- Views: 268
Re: 9.13 6th edition
On Lavelle's solution manual error pdf, it explains how to do 9.13. You assume ideal behavior and 1.00 mol N2 gas.
- Tue Feb 12, 2019 12:53 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: delta U [ENDORSED]
- Replies: 10
- Views: 2982
Re: delta U [ENDORSED]
With isothermal reversible expansion, work of expansion is being done by the system so energy is being lost. However, a heat reservoir also adds heat back into the system, so temperature remains constant. This is why q=-w.
- Tue Feb 12, 2019 12:49 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Delta U
- Replies: 1
- Views: 246
Re: Delta U
For irreversible processes, deltaU=q+w, so deltaU could be zero if q=-w. While it's possible, it's not ironclad like it is with reversible processes.
- Tue Feb 12, 2019 12:42 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22761
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Can someone explain how to do #3?
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ
I got deltaHrxn= 1mol(1560 kJ/mol) - [1mol(-1300 kJ/mol)+2mol(-286 kJ/mol)] = 3432 kJ
- Tue Feb 12, 2019 12:40 pm
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 22761
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
katherinemurk 2B wrote:Can someone explain how to do #1. I found the heat gained by the water but without the final temperature of the metal how would I find the specific heat??
The final temperature of the metal is the same as the final temp of the water.
- Tue Feb 05, 2019 4:25 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.57 (Sixth Edition)
- Replies: 3
- Views: 483
Re: Question 8.57 (Sixth Edition)
https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=17644 I found this thread that explains how to do the problem well. Essentially, you have to find the combustion reaction for each element of the given reaction and then use method 1 for solving for enthalpy to find the reaction enthalpy for the hy...
- Mon Feb 04, 2019 12:45 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 8.41 6th edition
- Replies: 2
- Views: 303
8.41 6th edition
A 50.0-g ice cube at 0.0 C is added to a glass containing 400.0 g of water at 45.0 C. What is the final temperature of the system (see Tables 8.2 and 8.3)? Assume that no heat is lost to the surroundings. For this question, I found the deltaH for the phase change from ice to water, the heat absorb...
- Mon Feb 04, 2019 12:29 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Question 8.3 (Sixth Edition)
- Replies: 2
- Views: 387
Re: Question 8.3 (Sixth Edition)
For part a, you want to find the work so you would use w=-PV. V=π(r^2)(h), so you would use the values given for the diameter of the pump and the distance the pump is depressed to find the volume. For part b, if work is being done on the system (the air in the pump), work is positive. If work the sy...
- Mon Feb 04, 2019 12:23 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.57 (Sixth Edition)
- Replies: 3
- Views: 483
Re: Question 8.57 (Sixth Edition)
I'm having trouble with this too... I did the sum of the standard enthalpy of combustion for products - reactants but I got +312 kJ instead of -312 kJ. I know it should be negative because it's a combustion reaction, thus exothermic, but that math doesn't match up.
- Mon Jan 28, 2019 6:23 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.3 6th Edition: Work being positive and negative
- Replies: 3
- Views: 298
Re: 8.3 6th Edition: Work being positive and negative
I still used 101.325 J/L*atm for 8.9. And then you have to convert it into kJ to solve for the change in internal energy.
- Mon Jan 28, 2019 6:19 pm
- Forum: Phase Changes & Related Calculations
- Topic: Work
- Replies: 5
- Views: 525
Re: Work
If work is done on a system, the value is positive. If the system is doing work, the value is negative.
- Mon Jan 28, 2019 6:16 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Method 3
- Replies: 2
- Views: 257
Method 3
What does it mean when the standard enthalpy of formation of one molecule is more negative than another? For example, in class the standard enthalpy of formation for CO2 was -394 kJ/mol and for CH4 it was -75 kJ/mol. Does this mean CO2 is more stable?
- Tue Jan 22, 2019 12:06 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: What does Kc have to be less than for you to estimate?
- Replies: 3
- Views: 283
Re: What does Kc have to be less than for you to estimate?
You can also double check by calculating the percent ionization. If it's less than 5%, the approximation is valid.
- Tue Jan 22, 2019 12:03 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: help with homework 12.57
- Replies: 2
- Views: 195
Re: help with homework 12.57
For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is the...
- Mon Jan 21, 2019 2:23 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Partial pressure vs pressure
- Replies: 2
- Views: 217
Re: Partial pressure vs pressure
With partial pressure, you treat the equilibrium composition the same as you would a change in concentration. If the partial pressure of something is increased on the products' side, for example, the reaction will shift towards the reactants. An overall increase in pressure usually is in response to...
- Mon Jan 14, 2019 11:08 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q vs. K
- Replies: 5
- Views: 476
Re: Q vs. K
K is the equilibrium constant while Q is the reaction quotient. So when you calculate Q, the reaction is not necessarily at equilibrium, though the equation to solve for Q and K are the same. You can then compare the values of Q and K to determine which direction the reaction will go.
- Mon Jan 14, 2019 10:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th edition 11.89
- Replies: 1
- Views: 158
Re: 6th edition 11.89
From the graph, you can see that B increases by 5, C increases by 10, and A decreases by 10. So you can think of it as +x, +2x, and -2x, if x=5. That's why the stoichiometric coefficients are 2A / B / 2C
- Mon Jan 14, 2019 10:47 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.77/11.87 6th Edition
- Replies: 1
- Views: 125
11.77/11.87 6th Edition
Why is an endothermic reaction? Shouldn't it be exothermic because enthalpy decreases/it is a catabolic reaction?
- Tue Jan 08, 2019 1:01 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7th Edition 5G.5 part c
- Replies: 2
- Views: 122
Re: 7th Edition 5G.5 part c
The balanced equation for decomposition is simply because in the (forward) reaction, the diatomic dissociates into two. The quantity of X2 and X in the flask is irrelevant for the equation.
- Tue Jan 08, 2019 12:52 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Concentrations
- Replies: 5
- Views: 438
Re: Concentrations
If I understand it correctly, products will be favored at equilibrium if you increase the amount of reactant when the reaction is at equilibrium in order to minimize the effect of adding more reactant; the forward reaction is favored, producing more product. Initial concentrations of reactants or pr...
- Tue Jan 08, 2019 9:45 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 6th Edition 11.7 part c
- Replies: 3
- Views: 239
Re: 6th Edition 11.7 part c
You also want to look at flask #3, when the reaction has reached equilibrium. I used the number of X2 and the number of X over the total number of molecules (17) for the mole fraction when calculating the partial pressures for X2 and X (mole fraction times total pressure) to find K. Hope that helps!