## 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: 837

### 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: 90

### 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: 93

### 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: 406

### 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: 142

### 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?
Tue Mar 05, 2019 12:38 am
Forum: General Rate Laws
Topic: Rate of Formation vs Unique Rate of a Reaction
Replies: 2
Views: 117

### 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: 113

### 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: 146

### 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: 97

### 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: 133

### 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: 146

### 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)
Tue Feb 19, 2019 9:38 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.83
Replies: 1
Views: 113

### 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: 111

### 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: 82

### 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
Tue Feb 12, 2019 4:46 pm
Forum: Administrative Questions and Class Announcements
Replies: 179
Views: 12557

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
Replies: 179
Views: 12557

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.
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: 116

### 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: 624

### 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: 74

### 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
Replies: 179
Views: 12557

Can someone explain how to do #3?
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
Replies: 179
Views: 12557

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: 219

### 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: 102

### 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: 156

### 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: 219

### 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: 125

### 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: 221

### 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: 101

### 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: 134

### 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: 64

### 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: 91

### 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: 230

### 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: 71

### 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: 50

### 11.77/11.87 6th Edition

Why is $X_{2}\rightleftharpoons 2X$ 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: 63

### Re: 7th Edition 5G.5 part c

The balanced equation for decomposition is simply $X_{2}\rightleftharpoons 2X$ 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: 263

### 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: 107

### 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!