## Search found 47 matches

Thu Mar 14, 2019 10:40 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Test 2
Replies: 6
Views: 735

### Re: Test 2

For the snowflakes, delta H is negative because the snowflakes lose heat, and delta S is negative because going from a liquid to a solid is less chaotic. Negative delta H minus a negative delta S makes a negative delta G. For the sublimation, delta H is positive because heat is going in, and delta ...
Thu Mar 14, 2019 5:04 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: 15.39 6th edition
Replies: 1
Views: 343

### Re: 15.39 6th edition

So first you want to find the concentration of [A] after the time period. Since they tell us that delta[B]=0.19M, we can multiply it by (1 mol A)/(2 mol B) to get delta[A]. 0.19 mol B/L x (1 mol A/2 mol B) = 0.095 M of A To get the final [A], simply subtract delta[A] from [A]not [A]final=0.15M-0.095...
Mon Mar 11, 2019 10:03 pm
Forum: First Order Reactions
Topic: Homework, 6th edition, 15.27
Replies: 2
Views: 111

### Re: Homework, 6th edition, 15.27

For parts c and d, you need to first find the value of k by using the half life formula. t1/2 = 0.693/k =355s, so k= 1.95x10^-3 s^-1 You can then plug this into the first order rate equation to find the time (t). ln[A]= -kt + ln[A]not c) ln(0.15)= -(1.95x10^-3 s^-1)*t + ln(1), so t=9.7x10^2 s d) ln(...
Mon Mar 11, 2019 9:57 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Test 2: Gibb's Free Energy Question
Replies: 1
Views: 148

### Test 2: Gibb's Free Energy Question

Does anyone know the answer to these and why?

1. The precipitation of snowflakes inside a cloud at -10 degrees Celcius and 0.839 atm. Is deltaG <0, =0, or >0?

2. The sublimation of dry ice in a warm room. Is delta G <0, =0, or >0?
Mon Mar 11, 2019 9:54 pm
Forum: General Rate Laws
Topic: Rate definition
Replies: 3
Views: 152

### Re: Rate definition

Rate constant is the term "k" itself, and it depends on the temperature, (T), frequency factor (A), and activation energy (Ea). Rate of reaction depends on both "k" and the concentration of reactants. For example, for a reaction aR --> bP, RATE= (-1/a)*(d[R]/dt)= (1/b)*(d[P]/dt) ...
Mon Mar 11, 2019 9:48 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Activation energy and k
Replies: 2
Views: 118

### Re: Activation energy and k

Looking at the equation: k= A * e^(-Ea/RT) = A / e^(Ea/RT) When activation energy (Ea) increases, the term "e^(Ea/RT) increases as well. Since that term is in the denominator, the bigger the denominator, the smaller the overall rate (k). The other factors that affect the rate, k, are the freque...
Sat Mar 09, 2019 6:46 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Calculating E naught
Replies: 5
Views: 273

### Re: Calculating E naught

You arrange it so that Enot is the most positive. So you should assign the half reaction with the larger Enot as cathode, and assign the half reaction with the smaller Enot as the anode. This way, Enot(cathode)-Enot(anode) will be the largest possible.
Tue Mar 05, 2019 1:41 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: E cell spontaneity
Replies: 5
Views: 274

### Re: E cell spontaneity

Since deltaG=-nFE, the reaction is spontaneous in the forward direction when deltaG is negative, which happens when E is positive. On the other hand, the reaction is spontaneous in the reverse direction when deltaG is positive, which happens when E is negative.
Tue Mar 05, 2019 1:38 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Concentration cells
Replies: 6
Views: 411

### Re: Concentration cells

In a concentration cell, Enot is always 0V, E is typically low voltage, and you always make the lower concentration the product ("it is in solution").
Tue Mar 05, 2019 1:33 am
Forum: First Order Reactions
Topic: order
Replies: 5
Views: 213

### Re: order

The order of a rate law is the sum of the exponents of its concentration terms. More specifically, the reaction order is the exponent to which the concentration of that species is raised, and it indicates to what extent the concentration of a species affects the rate of a reaction, as well as which ...
Thu Feb 28, 2019 4:25 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.9b
Replies: 1
Views: 104

### Re: 14.9b

Because the balanced half reactions are:
(6Fe^3+) + (6e-) ----> (6Fe^2+)
(2Cr^3+) + (7H2O) -----> (Cr2O7^-2) + (6e-) + (14H+)

So 6 moles of electrons are transferred.
Tue Feb 26, 2019 11:03 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Clarification on what temperature reaction will be spontaneous
Replies: 1
Views: 88

### Re: Clarification on what temperature reaction will be spontaneous

You should solve for Temperature "Z" where deltaG=0 and explain that the reaction is at equilibrium at this point. Then to answer the question, you should write it as an inequality (ie: when Temperature> tempZ, the reaction is spontaneous in the forward direction). If you want to be extra,...
Tue Feb 26, 2019 10:58 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: n
Replies: 8
Views: 330

### Re: n

Follow up question: is n supposed to be (moles of electrons in reactants)-(moles of electrons in products) or is it (moles of electrons in products)-(moles of electrons in reactants)?
Tue Feb 26, 2019 10:53 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Midterm
Replies: 5
Views: 240

### Re: Midterm

Since the equilibrium constant is large, we can conclude that at equilibrium there are more products than reactants. Therefore, the products are more stable, while the reactants are more unstable in comparison.

https://lavelle.chem.ucla.edu/wp-conten ... am_ans.pdf
Thu Feb 21, 2019 10:04 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs Free Energy Decreasing
Replies: 2
Views: 148

### Re: Gibbs Free Energy Decreasing

You can take a look at problem 9.65 (6th edition). The solution manual states: "To understand what happens to deltaG as temperature is raised, we use the relationship deltaG=deltaH-T*deltaS. From this it is clear that the free energy of the reaction becomes less favorable (more positive) as tem...
Thu Feb 21, 2019 9:58 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs Free Energy
Replies: 4
Views: 151

### Re: Gibbs Free Energy

When deltaG is positive, it means that the free energy of the products is MORE than the free energy of the reactants. Therefore, an input of free energy is needed for the reaction to occur, so the reaction is NOT spontaneous. When deltaG is negative, it means that the free energy of the products is ...
Thu Feb 21, 2019 9:55 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Gibbs
Replies: 3
Views: 168

### Re: Gibbs

Gibb's Free Energy can be calculated through using enthalpy and entropy. deltaG=deltaH - T*deltaS When a reaction has a positive enthalpy (+deltaH): deltaG will be positive if entropy is negative (-deltaS) deltaG will be negative if entropy is positive (+delta S) and T*deltaS is greater than deltaH ...
Wed Feb 20, 2019 8:05 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Midterm Question
Replies: 5
Views: 432

### Re: Midterm Question

if K>>>10^3, then products are favored over reactants, and therefore products are more stable. if K<<<10^-3, then reactants are favored over products, and therefore reactants are more stable. if K=1, then products and reactants are exactly equally favored, and therefore are equally stable. for value...
Tue Feb 12, 2019 4:37 pm
Forum: Administrative Questions and Class Announcements
Topic: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!
Replies: 49
Views: 3946

### Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!

Can someone please explain why 3E is false? Because heat is required during melting or boiling (phase change transition), temperature of a sample can remain constant even though heat is being supplied. For example, if you have boiling water, it will remain at 100 C even though heat is being added, ...
Tue Feb 12, 2019 4:33 pm
Forum: Phase Changes & Related Calculations
Topic: #10 on Hotdog
Replies: 3
Views: 398

### Re: #10 on Hotdog

q(ice) = -q(water) FOR ICE: Melting: q=deltaH(fusion)= 6.01 kJ/mol x (25.0 g / 18 g*mol-1) = 8347.2 J Increasing Temp of Melted Ice: q=mCdeltaT= (25.0 g) x (4.184 J/K*g) x (Tf-273K) = 104.6(J/K)*Tf - 28555.8J FOR WATER: Cooling the Water: q= mCdeltaT= (265 g) x (4.18 J/K*g) x (Tf-298K) = 1107.7(J/K)...
Mon Feb 11, 2019 11:06 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Question 8.39 (Sixth Edition)
Replies: 4
Views: 196

### Re: Question 8.39 (Sixth Edition)

You have to break this problem up into two steps: 1. Find m * Heat of fusion 2. Find m * C * delta T When you find those two values, you add them up, and that is the total heat needed. Is it n*Heat of fusion or m* Heat of fusion? For the first part we use moles and the second part we use mass right?
Mon Feb 11, 2019 11:03 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Test Question
Replies: 2
Views: 128

### Re: Test Question

I had a different test, but generally speaking, by DECREASING the volume, you are INCREASING the pressure. With an increase in pressure, the reaction will tend towards the side with LESS moles. So you should count the coefficients (which tells you the number of moles) of the balanced equation on the...
Mon Feb 11, 2019 10:57 pm
Forum: Administrative Questions and Class Announcements
Topic: Midterm
Replies: 6
Views: 272

### Re: Midterm

Outline 3 is covered completely, Outline 4 is covered until problem 9.47 (after that the problems are on Gibb's Free Energy).
Mon Feb 11, 2019 10:53 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Deriving Formulas
Replies: 2
Views: 175

### Re: Deriving Formulas

I would assume anything covered in lecture is fair game, but it's probably smarter to focus on the application of formulas since the formulas themselves will be given on the constants and equations sheet.
Thu Feb 07, 2019 9:04 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Definitions of Heat Capacities
Replies: 2
Views: 150

### Re: Definitions of Heat Capacities

-Heat Capacity- heat required to raise the temperature of an object by 1C. -Specific Heat Capacity- divide heat capacity by the amount of substance present (g); heat required to raise the temperature of 1 gram of substance by 1C -Molar Heat Capacity- divide heat capacity by moles of substance (mol);...
Mon Feb 04, 2019 11:20 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: PdeltaV with Solids and Liquids
Replies: 3
Views: 132

### Re: PdeltaV with Solids and Liquids

Because you can't change the volume of solids and liquids, so delta V will be 0. Therefore, PdeltaV will equal 0, which makes it insignificant.
Mon Feb 04, 2019 11:18 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Finding Spontaneity of Reaction
Replies: 1
Views: 56

### Finding Spontaneity of Reaction

Towards the end of lecture on Friday Week 4, Lavelle talked about how just knowing delta U (or delta H) isn't enough to know if a reaction will occur and that we also need to know the entropy (S). Can someone explain how knowing delta U and S will tell us how the system will change? In other words, ...
Mon Feb 04, 2019 11:08 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Ideal Gas Internal Energy
Replies: 2
Views: 87

### Re: Ideal Gas Internal Energy

I believe he said we don't have to worry about deriving it! He didn't explain how he got 3/2 nRT.
Sun Feb 03, 2019 2:14 pm
Forum: Phase Changes & Related Calculations
Topic: Steam resulting in burns?
Replies: 6
Views: 283

### Re: Steam resulting in burns?

Steam burns causes more severe burns than water burns. This is because heat is required during melting or boiling (phase transition), and therefore the temperature of the sample REMAINS CONSTANT even though heat is being supplied. When you get burned by water, since it is not in a phase transition, ...
Sun Feb 03, 2019 2:05 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Midterm Exam Content
Replies: 4
Views: 176

### Re: Midterm Exam Content

On the website it says under announcements that "Midterm covers: Chemical Equilibrium, Acid and Base Equilibria, Thermochemistry, Thermodynamics to the end of entropy."
Wed Jan 30, 2019 12:35 am
Forum: Phase Changes & Related Calculations
Topic: 101.325 J/(L . atm)
Replies: 2
Views: 70

### Re: 101.325 J/(L . atm)

It comes from the gas constants (this is provided on the list of constants). Gas constant, R = 8.314 (J)/(K*mol) = 8.206 x 10^-2 (L*atm)/(K*mol) Since the values equal each other, 8.314 (J)/(K*mol) / 8.206 x 10^-2 (L*atm)/(K*mol) equals to 1. Cancel out the units and you get 101.316 (J)/(L*atm), whi...
Mon Jan 28, 2019 9:18 pm
Forum: Ideal Gases
Topic: Q and K
Replies: 13
Views: 554

### Re: Q and K

Q is called the "reaction quotient," and can be calculated anytime during the reaction. K is called the "equilibrium constant", and can be calculated only when the reaction has reached equilibrium. For a specific reaction (at a constant temperature and pressure), you can get diff...
Mon Jan 28, 2019 9:11 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Calorimetry
Replies: 5
Views: 245

### Re: Calorimetry

To add on, Constant P. Calorimetry (I guess this is the "normal" kind) gives you the enthalpy values. Both types are used to measure specific heat capacity.
Mon Jan 28, 2019 9:08 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: qp
Replies: 3
Views: 129

### Re: qp

I believe qp is equivalent to deltaH (change in enthalpy), and it represents the heat at constant pressure. qp=n*Cp*T where n is in moles, Cp is the molar heat capacity, and T is the change in temperature. I think he's going to talk continue to talk about this in Wednesday's lecture, but if anyone c...
Thu Jan 24, 2019 11:32 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: PV= RnT
Replies: 2
Views: 199

### Re: PV= RnT

This will allow you to convert the partial pressure of gas to molar concentration.
PV=nRT
P=(nRT)/V
P= (n/V)*RT
P= Concentration *(RT) and Concentration= P/(RT)
R and T are normally constant for these problems.
Tue Jan 22, 2019 10:21 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: What Happens to a Reaction if Inert Gas is Added?
Replies: 7
Views: 630

### Re: What Happens to a Reaction if Inert Gas is Added?

Increasing the pressure by introducing an inert gas has no effect on the equilibrium composition. However, compression of a reaction mixture at equilibrium tends to drive the reaction in the direction that reduces the number of gas-phase molecules.
Mon Jan 21, 2019 7:07 pm
Forum: Bronsted Acids & Bases
Topic: Sig Figs
Replies: 4
Views: 308

### Re: Sig Figs

My TA said he will be lenient on the first test in regards to significant figures, but it probably depends on how your TA grades. Lavelle posted two links under "Math Assistance" on his website about significant figures that you can review. https://lavelle.chem.ucla.edu/wp-content/supporti...
Mon Jan 21, 2019 7:03 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Acid Strength
Replies: 3
Views: 133

### Re: Acid Strength

To add on to the post above, the concept that the stronger the acid, the weaker its conjugate base, and the stronger the base, the weaker the conjugate acid is known as conjugate seesaw. It's also helpful to remember that Ka x Kb = Kw and pKa + pKb = 14.
Mon Jan 21, 2019 6:55 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: weak acids/bases
Replies: 2
Views: 310

### Re: weak acids/bases

Weak acids have Ka<10^-3 and weak bases have Kb<10^-3. Between two Ka and two Kb, the one with the bigger number is stronger. Since Ka x Kb = Kw, they are inversely related, which is why a strong base is paired with a weak conjugate acid and a strong acid is paired with a weak conjugate base. This i...
Tue Jan 15, 2019 11:35 pm
Forum: General Science Questions
Topic: Rounding/ Significant Figures
Replies: 3
Views: 396

### Re: Rounding/ Significant Figures

I think I counted wrong. I forgot that numbers like 20.0 is only 1 sigfig and not 3. Thank you for your help!
Tue Jan 15, 2019 10:40 pm
Forum: General Science Questions
Topic: Rounding/ Significant Figures
Replies: 3
Views: 396

### Rounding/ Significant Figures

I did not have Lavelle before and was wondering about how we should round our answers. I noticed that the answers in the answer manual for the 6th edition textbook often don't follow the rules of significant figures. For example, let's say there is a given value of 0.385M in the problem. I would ass...
Mon Jan 14, 2019 10:09 pm
Forum: Ideal Gases
Topic: Homework for week 2 [ENDORSED]
Replies: 10
Views: 2399

### Re: Homework for week 2[ENDORSED]

The problems from acids & bases are more relevant to what is currently being covered in lecture, but you can do problems from chemical equilibrium too since we're still doing ICE tables.
Mon Jan 14, 2019 10:05 pm
Forum: General Science Questions
Topic: molarity question
Replies: 2
Views: 182

### Re: molarity question

Just remember that pure water has the neutral pH of 7, so as the comment above states, [H3O+] and [OH-] should always be equal. If they are not equal, the solution will either be more acidic (higher concentration of H3O+) or more basic (higher concentration of OH-).
Mon Jan 14, 2019 9:58 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Are Either Products or Reactants Favored?
Replies: 3
Views: 85

### Re: Are Either Products or Reactants Favored?

As long as you explain that products are favored since K>1 and add/clarify that it isn't strongly favored since K<10^3, it should be correct.
Tue Jan 08, 2019 6:08 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Using Net Ionic Equation
Replies: 1
Views: 110

### Using Net Ionic Equation

Quoting the 6th edition textbook on page 427, it says "when a reaction involves fully dissociated ionic compounds in solution, the equilibrium constant should be written for the net ionic equation, by using the activity for each type of ion." I was wondering how you would know if there are...
Tue Jan 08, 2019 5:56 pm
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: Spontaneous reaction
Replies: 4
Views: 167

### Re: Spontaneous reaction

To add on, the products of a spontaneous reaction are generally more stable than the reactants. Therefore, a spontaneous reaction can occur without any additional input of energy, and as said in the reply above, it normally releases energy.
Tue Jan 08, 2019 5:50 pm
Forum: Ideal Gases
Topic: Equilibrium Constants
Replies: 4
Views: 214

### Re: Equilibrium Constants

This will depend on whether you are changing the coefficient of a reactant or product. Let's say the equilibrium equation has aA and bB as the reactants, and cC as the product, with the lowercased letters representing the coefficients. Therefore, K=(partial pressure of C)^c/(partial pressure of A)^a...