## Search found 53 matches

Sat Mar 17, 2018 9:19 pm
Forum: Calculating Work of Expansion
Topic: Work with changeP and constant V
Replies: 4
Views: 363

### Re: Work with changeP and constant V

expansion work in a irreversible change = -P*deltaV, and when there is no change in V, deltaV = 0 so work = 0. This only holds true if we only consider expansion work.

expansion work in a reversible change = -nRT*(ln (Vfinal/Vinitial))
if vfinal = vinitial, then any expansion work also is zero.
Sat Mar 17, 2018 9:14 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: internal energy
Replies: 2
Views: 300

### Re: internal energy

If a system goes back its original internal energy, internal energy should be zero overall. I don't think you would have to flip the sign of any internal energy steps in this case, because the sign of internal energy would already be considered in your calculations of deltaU to make deltaU overall b...
Sat Mar 17, 2018 9:10 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: >> [ENDORSED]
Replies: 2
Views: 376

### Re: >>[ENDORSED]

I usually interpret >> as "much larger than" but I could be wrong.
Tue Mar 06, 2018 3:53 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Units for reaction rates
Replies: 9
Views: 312

### Re: Units for reaction rates

Reaction rate units concerning one substance are always Molarity/time.

But for k:
zero order is M/time
first order is 1/time
second order is 1/(time*M)
Tue Mar 06, 2018 3:48 pm
Forum: General Rate Laws
Topic: 15.5
Replies: 2
Views: 116

### Re: 15.5

You're correct, I think the answer key in the book does say that the rate for oxygen should be .44*3.
Tue Mar 06, 2018 3:39 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Midterm Question 5 (Step 2)
Replies: 2
Views: 263

### Re: Midterm Question 5 (Step 2)

I'm not really certain what you mean by internal energy being positive, but here's how I solved Q5: Because step one concerns an adiabatic process, q = 0, and consequently internal energy in step one = work. But the second step returns the internal energy to its initial state before step one happene...
Tue Mar 06, 2018 3:11 pm
Forum: General Rate Laws
Topic: Products having an effect on rate law
Replies: 1
Views: 148

### Re: Products having an effect on rate law

I believe the rate laws that are involved in 15.1-15.6 are supposed to calculate the instantaneous rate of reaction at certain quantities of reactant, assuming that no product has formed yet (because a product can undergo the reverse reaction and become the product, so the equation would have to be ...
Tue Mar 06, 2018 3:02 pm
Forum: General Rate Laws
Topic: Rate of formation/consumption
Replies: 3
Views: 575

### Re: Rate of formation/consumption

The rate of formation generally refers to one of the products (d[B]/dt, for example). The rate of consumption is the rate at which one of the reactants is being consumed at the beginning of the reaction. If you divided the rate of consumption and formation by the coefficients of the compounds the ra...
Tue Mar 06, 2018 11:52 am
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Isobaric vs. Isochoric [ENDORSED]
Replies: 2
Views: 709

### Re: Isobaric vs. Isochoric[ENDORSED]

Isobaric procedures occur at constant pressure and isochoric ones occur at constant volume. Usually when it's mentioned in a test question, it tells you what assumptions you should make and what kind of equations to use. Use Cp and Cv for each respectively when calculating heat capacity, and in isob...
Sun Feb 25, 2018 8:44 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: How to write the rate laws for homogeneous catalyst
Replies: 1
Views: 178

### Re: How to write the rate laws for homogeneous catalyst

A catalyst should be unchanged after the reaction that it facilitates occurs and is not consumed by the reaction. Because of this, it should not be treated the same as an intermediate, because intermediates are produced and consumed in the overall process of the reaction. That being said, a catalyst...
Sun Feb 25, 2018 8:34 pm
Forum: First Order Reactions
Topic: Reaction Order [ENDORSED]
Replies: 6
Views: 513

### Re: Reaction Order[ENDORSED]

I believe the reaction order does depend on the number of coefficients involved the reaction, but it more specifically depends on the coefficients of the reactants involved in the slow step.
Sun Feb 25, 2018 8:32 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Q
Replies: 7
Views: 399

### Re: Q

Q and K are written with the same scheme that we learned last quarter. for reaction aA + bB ---> cC + dD, K = ([C]^c * [D]^d)/([A]^a * [B]^b) In a galvanic cell, use the equation for the full reaction (combine the half reactions) to get your products and reactants. The substances that are favored to...
Sun Feb 18, 2018 8:54 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reactant and Product
Replies: 1
Views: 111

### Re: Reactant and Product

Are you referring to concentration cells? If so, there are some explanations for that here:

viewtopic.php?f=142&t=20974
Sun Feb 18, 2018 8:45 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Q
Replies: 7
Views: 399

### Re: Q

Q is similar to K (products/reactants) but the amounts of products and reactants used in calculating Q represent the reaction when it is not at equilibrium and is instead in the process of reaching equilibrium. It allows us to compare the current ratio of products and reactants at a certain point in...
Sun Feb 18, 2018 8:39 pm
Forum: Balancing Redox Reactions
Topic: Addition of H3O+ or OH-
Replies: 4
Views: 517

### Re: Addition of H3O+ or OH-

Usually, the question should tell you whether the reactions are occurring in a basic or acidic environment.
Sun Feb 11, 2018 11:32 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.13 - Use of Cv
Replies: 2
Views: 162

### Re: 9.13 - Use of Cv

I think the two calculations that consider entropy increase as a result of temperature change and entropy increase due to volume change should be done entirely separate from each other. Therefore, since we already took entropy change from volume into account in a prior calculation, we assume that th...
Sun Feb 11, 2018 11:21 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Standard molar entropy of cyclopentane vs 1-pentene
Replies: 2
Views: 235

### Re: Standard molar entropy of cyclopentane vs 1-pentene

I assumed that cyclopentane has a lower entropy than 1-pentene because cyclopentane has to conform to a specific shape due to its carbons bonding in a closed ring structure, while 1-pentene is a straight molecule with two free ends.
Sun Feb 11, 2018 11:13 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Non-spontaneous
Replies: 4
Views: 313

### Re: Non-spontaneous

If a reaction is non-spontaneous, its deltaG is positive and therefore the formation of products is not favored. This means that it's favored for the reactants to stay reactants instead of undergoing the reaction, and if there were a great number of products under certain conditions, the reaction wo...
Sun Feb 04, 2018 8:10 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Delta U (Concept Clarification)
Replies: 6
Views: 298

### Re: Delta U (Concept Clarification)

viewtopic.php?f=130&t=26453

Sun Feb 04, 2018 7:50 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Question about Change in Gibb's Free Energy
Replies: 4
Views: 211

### Re: Question about Change in Gibb's Free Energy

I believe deltaG can be used to represent the total amount of energy that is released or absorbed into or by a system during a chemical reaction. This energy released can be used to perform different functions, hence deltaG representing the amount of energy free to do work. For example, if we said A...
Sun Feb 04, 2018 7:41 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Question about "Real Processes"
Replies: 3
Views: 189

### Re: Question about "Real Processes"

A process such as a reversible expansion is not seen in real world scenarios due to two reasons. First, expanding a gas reversibly involves incrementally decreasing the pressure by infinitely small amounts to make the gas expand, which would take forever to do. Secondly, all heat in a system expandi...
Sun Jan 28, 2018 1:56 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Reversible vs Irreversible
Replies: 2
Views: 190

### Re: Reversible vs Irreversible

In the case of a reversible expansion, the system does the maximum possible amount of work. All heat is converted into work, which models a perfect engine, something that doesn't exist. A reversible expansion is 100% efficient and would not be found in real life, to the best of my knowledge. I suppo...
Sun Jan 28, 2018 1:47 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Maximum Entropy
Replies: 1
Views: 60

### Re: Maximum Entropy

At equilibrium, the system is no longer trending towards a spontaneous change, and by consequence it has reached its maximum entropy state. (Entropy increasing can drive processes like diffusion, at which the system wouldn't be at equilibrium.) If an isolated system at equilibrium were to decrease i...
Sun Jan 28, 2018 1:35 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Gas Expansion [ENDORSED]
Replies: 3
Views: 259

### Re: Gas Expansion[ENDORSED]

A gas can still expand without internal energy changing because deltaS, the maximum number of positional states (entropy) of a system, is going up. If we assume that the gas is expanding into a vacuum at constant temperature (deltaU is zero), its expansion is still favored because deltaS is positive...
Sun Jan 21, 2018 7:52 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Intensive vs. Extensive Properties
Replies: 4
Views: 222

### Re: Intensive vs. Extensive Properties

I guess you could discern intensive and extensive properties by imagining a block of the substance you're examining, and then cutting it in half. If the characteristic you're trying to figure out is intensive or extensive changes, then it's extensive. If I had a block of water and (hypothetically) I...
Sun Jan 21, 2018 7:38 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Exercise 8.19
Replies: 1
Views: 208

### Re: Exercise 8.19

We need to calculate the temperature rise for copper and water separately because they have different specific heats. The specific heat gives us the amount of energy needed to raise a substance based on mass and temperature, so that's what we'll use: Copper: 500g*(.38 J*C^-1*g^-1)*(100 degrees C - 2...
Tue Jan 16, 2018 2:39 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Expansion Work vs. Nonexpansion Work
Replies: 2
Views: 417

### Re: Expansion Work vs. Nonexpansion Work

Part of your question has actually been already answered here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=75&t=25593 The equation w = - (external pressure)(change in volume) actually only applies to one of the expansion work scenarios given by the book. This equation applies to what the...
Tue Jan 16, 2018 2:05 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: HW 8.21
Replies: 2
Views: 146

### Re: HW 8.21

Given that the final temperature of the metal will be less than the initial, the calculated energy lost by the metal will be negative. And given that the final temperature of the water will be more than its initial, the calculated energy gained by the water will be positive. So before modifying the ...
Tue Jan 16, 2018 1:55 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Determining sign of q
Replies: 2
Views: 125

### Re: Determining sign of q

If a system loses energy in the form of heat, q should be negative. But if a system absorbs heat, q should be positive.

When water changes from a liquid to a gas, water molecules absorb heat to break the attractive intermolecular forces between them, so q in this reaction is positive.
Tue Jan 16, 2018 1:52 am
Forum: Administrative Questions and Class Announcements
Topic: Discussion post due dates
Replies: 2
Views: 235

### Re: Discussion post due dates

If you are in Chemistry 14B, we have to do 3 comments per week now, but Chem 14A students will do 2 comments a week (Sorry about this ambiguity, I can't really tell what class you're in) I'm guessing that the week counts as Monday - Sunday, but I will ask my TA tomorrow to be sure. In one of my disc...
Sat Dec 09, 2017 4:28 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: [H30+] and [OH-] for strong acids/bases
Replies: 1
Views: 231

### Re: [H30+] and [OH-] for strong acids/bases

When a strong base (or acid) is dissolved in water, they dissociate completely. Therefore .10 M HCl means .10M H+. However, watch out for strong acids/bases that have more than one H+ or OH- ion for every molecule. .10M Ba(OH)2 is actually .20 M OH - because there are 2 molecules of OH- for every mo...
Sat Dec 09, 2017 4:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Changing Kc
Replies: 2
Views: 596

### Re: Changing Kc

You are correct that K for a reaction only changes when temperature changes. However, when we halve all the stoichiometric coefficients for the reaction, we are actually changing the reaction we are using in our calculations. 2SO2 + O2 ---> 2SO3 is different than SO2 + (1/2)O2 ---> SO3. Therefore, t...
Mon Dec 04, 2017 1:37 am
Forum: Lewis Structures
Topic: Midterm Question: Lewis Structure for HOCO [ENDORSED]
Replies: 6
Views: 1261

### Re: Midterm Question: Lewis Structure for HOCO[ENDORSED]

While it is true that this formation would result in formal charges of zero for every atom, the accepted answer is still the one listed on the midterm solutions. There are two tip offs that let you know this is the case: First, note the way the molecule is written. The molecule you bring up actually...
Thu Nov 30, 2017 9:42 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.7 Part C
Replies: 2
Views: 187

### 11.7 Part C

Question 11.7 states that the initial pressure exerted by 11 molecules of the gas X2 is 0.10 bar. I assume that X2 decomposes according to the equation X2 --> 2X. At equilibrium there are 12 molecules of X and 5 of X2. The solution manual states that the equilibrium constant of this equation is equa...
Sun Nov 26, 2017 10:30 pm
Forum: Ideal Gases
Topic: Ideal Gas Law
Replies: 3
Views: 216

### Re: Ideal Gas Law

The criteria that I gave for an Ideal Gas (no intermolecular attractions and no kinetic energy is lost in the collisions between molecules) was not actually every requirement that a gas needs to be ideal. Another requirement of Ideal Gases is that the individual molecules of the gas do not have volu...
Sat Nov 25, 2017 9:17 pm
Forum: Ideal Gases
Topic: Ideal Gas Law
Replies: 3
Views: 216

### Re: Ideal Gas Law

The Ideal Gas Law is used to analyze gas samples under the assumption that gases are "ideal gases," which have two criteria: 1) no intermolecular attractions and 2) no energy is lost in the collisions between molecules. No gas is an ideal gas, and the Ideal Gas Law becomes less accurate as...
Sun Nov 19, 2017 7:20 pm
Forum: Lewis Structures
Topic: Midterm Question: Lewis Structure for HOCO [ENDORSED]
Replies: 6
Views: 1261

### Re: Midterm Question: Lewis Structure for HOCO[ENDORSED]

That figuration would not have the lowest formal charge possible. The oxygen on the left now bonds with carbon twice, so its formal charge is now 6 - 5 = 1 e-. The oxygen on the right now also has the formal charge 6 - 5 = 1 e-, (assuming it has the lone pair and the one radical electron) while the ...
Sun Nov 19, 2017 6:15 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: AX2E2
Replies: 2
Views: 458

### Re: AX2E2

The molecule is not linear because both lone electron pairs have more repulsion than electrons from atoms bonded to the central atom, and there are four regions of electron density around the atom. Keep in mind that the "bent" shape is not entirely flat; the regions of electron density are...
Sun Nov 19, 2017 6:07 pm
Forum: Lewis Structures
Topic: Midterm Question: Lewis Structure for HOCO [ENDORSED]
Replies: 6
Views: 1261

### Re: Midterm Question: Lewis Structure for HOCO[ENDORSED]

The H atom should have a single bond with the oxygen, which in turn should have a single bond with the carbon. The carbon atom should have a double bond with the remaining oxygen. In this case, HOCO is a radical because all the valence electrons of the atoms add up to form an odd number. The remaini...
Mon Nov 13, 2017 5:52 am
Forum: Trends in The Periodic Table
Topic: Electron Affinity Exception
Replies: 3
Views: 1088

### Re: Electron Affinity Exception

This exception remains true for most elements in Group 15, but as you go down the group it becomes less pronounced. By the time you reach Bismuth the exception no longer holds true. It appears that when atoms in group 15 have a much larger number of protons and electrons (are at much higher energy l...
Wed Nov 08, 2017 2:10 pm
Forum: Trends in The Periodic Table
Topic: Atomic/Ionic Radius [ENDORSED]
Replies: 2
Views: 216

### Re: Atomic/Ionic Radius[ENDORSED]

You are correct that Br is smaller than Na. However the Br atom discussed in this question is actually an ion with an extra electron. Adding another electron to Br increases its radius by a large amount. This is because the electrons now experience a reduced effective nuclear charge and are not draw...
Sun Nov 05, 2017 11:44 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity [ENDORSED]
Replies: 6
Views: 690

### Re: Electron Affinity[ENDORSED]

There does seem to be many exceptions when following electron affinity down a period, but theres a trend that can be very loosely applied: electron affinity increases across a period and decreases down a group. There are explanations for some of the exceptions, though there are a lot of said excepti...
Sun Nov 05, 2017 11:28 pm
Forum: Octet Exceptions
Topic: Octet Exception Question [ENDORSED]
Replies: 3
Views: 615

### Re: Octet Exception Question[ENDORSED]

I looked online for an answer and this is what I got: "However, some of the third-period elements (Si, P, S, and Cl) have been observed to bond to more than four other atoms, and thus need to involve more than the four pairs of electrons available in an s2p6 octet. This is possible because for ...
Mon Oct 30, 2017 12:21 am
Forum: Octet Exceptions
Topic: Octet Rule [ENDORSED]
Replies: 1
Views: 1856

### Re: Octet Rule[ENDORSED]

I think the octet rule is usually used when drawing lewis structures. This is the idea that bonding atoms generally end up with their electrons in the electron configuration of a noble gas; as in, they all have 8 electrons in their valence shells. The octet rule does have exceptions, however. It doe...
Mon Oct 30, 2017 12:14 am
Forum: Conjugate Acids & Bases
Topic: Acids and Bases [ENDORSED]
Replies: 8
Views: 765

### Re: Acids and Bases[ENDORSED]

Weak acids create "strong" conjugate bases because their dissociation reaction (the reaction of it releasing an H+ ion) is not favored. If you take HNO2 (nitrous acid) as an example, the reaction for HNO2 to produce a conjugate base would be as follows: HNO2 = H+ + NO2- In this case, NO2- ...
Sun Oct 22, 2017 4:33 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Test 3 [ENDORSED]
Replies: 7
Views: 2479

### Re: Test 3[ENDORSED]

I can't say for sure but the test schedule says that Test 3 will cover "new material up to October 27," while the midterm, which is cumulative, says "all material up to November 3." If I had to go off this I'd say Test 3 isn't cumulative AND covers any material featured in the le...
Sun Oct 22, 2017 4:26 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Electron density distribution?
Replies: 2
Views: 281

### Re: Electron density distribution?

Nodes are areas in orbitals where there is a zero percent chance of finding an electron. Orbitals like the p orbital are drawn as dumbbell shaped because they represent spaces around the nucleus in which electrons are more likely to appear. (This is all represented by the wave function). Nodal plane...
Mon Oct 16, 2017 11:10 pm
Forum: Properties of Light
Topic: Lyman and Balmer Series Significance
Replies: 4
Views: 645

### Re: Lyman and Balmer Series Significance

I'm not sure about that. We should either ask if we need to know them or memorize them to be safe.
Mon Oct 16, 2017 11:06 pm
Forum: Properties of Light
Topic: 1.3
Replies: 5
Views: 434

### Re: 1.3

This is what I wrote to explain choice c on another person's question about 1.3: From the textbook: Light is a form of electromagnetic radiation, which consists of oscillating (time-varying) electric and magnetic fields that travel through empty space at about 3 * 10^8 m/s, or at just over 670 milli...
Sun Oct 15, 2017 7:32 pm
Forum: Properties of Light
Topic: Homework 1.3
Replies: 2
Views: 185

### Re: Homework 1.3

From the textbook: Light is a form of electromagnetic radiation, which consists of oscillating (time-varying) electric and magnetic fields that travel through empty space at about 3 * 10^8 m/s, or at just over 670 million miles per hour. This speed is denoted c and called the speed of light. Visible...
Sun Oct 15, 2017 7:19 pm
Forum: Properties of Light
Topic: Lyman and Balmer Series Significance
Replies: 4
Views: 645

### Re: Lyman and Balmer Series Significance

Here's my idea of a question you might get that would involve the series. If course I have no way of knowing exactly what will be asked on future exams but here's my idea: The Balmer Series and the Lyman Series both correspond to certain types of electromagnetic radiation. The Balmer series includes...
Thu Oct 05, 2017 10:27 pm
Forum: Limiting Reactant Calculations
Topic: M3
Replies: 3
Views: 928

### Re: M3

Hi Essly! Since CaCO3 is the only reactant in the equation, it has to be the limiting reactant.
Wed Oct 04, 2017 10:42 pm
Forum: Balancing Chemical Reactions
Topic: H1 // Book Problem [ENDORSED]
Replies: 7
Views: 532

### Re: H1 // Book Problem[ENDORSED]

I believe you're correct. Adding an oxygen atom to the right side of the equation would be adding a product that isn't produced by the reaction in the first place, so now the modified equation isn't even a representation of the original.

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