Search found 60 matches
- Thu Mar 14, 2019 11:17 am
- Forum: General Rate Laws
- Topic: k1 vs k2?
- Replies: 4
- Views: 1915
Re: k1 vs k2?
For a specific reaction at a specific temperature, there is only one rate constant k. However, if this same reaction took place at a different temperature, k would change. Note that there are different rate constant k's for different reactions.
- Thu Mar 14, 2019 11:10 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Arrhenius Equation
- Replies: 4
- Views: 486
Re: Arrhenius Equation
A as the pre-exponential factor/measure of the rate at which molecules collide is one of the Arrhenius parameters (the other being activation energy), which means it can only be found through experiments. I'd assume that we would be given the value of A if needed to solve a problem involving the Arr...
- Thu Mar 14, 2019 11:08 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Molecularity
- Replies: 5
- Views: 474
Re: Molecularity
As said above, molecularity is the number of reactant atoms/molecules/ions taking part in a specified elementary reaction. So yes, you would only look at the reactants to determine molecularity.
- Thu Mar 07, 2019 11:13 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Instantaneous vs Average
- Replies: 11
- Views: 1052
Re: Instantaneous vs Average
An average rate is taken over an interval of time. An instantaneous rate is taken at one specific instance/moment in time.
- Thu Mar 07, 2019 11:11 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Rate constant k
- Replies: 3
- Views: 475
Re: Rate constant k
K is a constant and does not change because k is characteristic for a specific reaction at a specific temperature. It follows that a different reaction, or even that same reaction at a different temperature, would result in a different value for k.
- Thu Mar 07, 2019 11:08 am
- Forum: Zero Order Reactions
- Topic: zero vs. first vs. second reactions
- Replies: 3
- Views: 336
Re: zero vs. first vs. second reactions
Note that a higher order also means that the reaction is less likely to occur/favorable. A second-order reaction depends on the interaction of two molecules at the same instant to form product, whereas a third-order reaction relies on the interaction among three molecules at the same time and would ...
- Thu Feb 28, 2019 6:36 am
- Forum: Balancing Redox Reactions
- Topic: Basic Solutions
- Replies: 3
- Views: 438
Re: Basic Solutions
In both acidic and basic solutions for redox reactions, you start with balancing the core element, and then move on to balancing the number of oxygens by adding H2O. Balancing the hydrogens after this is where the two differ: in acidic solutions, you simply add H+ to the side that needs more hydroge...
- Thu Feb 28, 2019 6:27 am
- Forum: Calculating Work of Expansion
- Topic: Calculating work
- Replies: 1
- Views: 494
Re: Calculating work
Specifically looking at expansion work, the system does work when expanding (-w, b/c it costs energy to do work and this lowers the internal energy of a system ΔU), but in the case of compression, the system is having work done on by its surroundings (+w).
- Thu Feb 28, 2019 6:23 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Delta H and Delta U
- Replies: 1
- Views: 243
Re: Delta H and Delta U
For constant pressure, ΔU = q + w, where q=heat and w=work. But we also know that q=ΔH, the change in enthalpy (for constant pressure and expansion work). (On another note, w can also be replaced with PΔV). So, since ΔU= q(aka ΔH) + w, when there is no work being done, ΔU=ΔH. Conceptually, for there...
- Thu Feb 21, 2019 11:41 am
- Forum: Balancing Redox Reactions
- Topic: sections covered on test
- Replies: 4
- Views: 451
Re: sections covered on test
My TA mentioned that Test 2 covers all topics on Gibbs free energy and electrochemistry up to but not including the Nernst equation, so any material from weeks 6 and 7 is fair game. I'm not sure if that corresponds exactly to those sections in the textbook, but just make sure they're covered!
- Thu Feb 21, 2019 11:30 am
- Forum: Student Social/Study Group
- Topic: Curve?
- Replies: 50
- Views: 6403
Re: Curve?
Note that the syllabus says, "Each test and exam has a total score but is not assigned a grade. Only at the end of the class when the class average score (out of 500 points) is known are final grades assigned. This class does not use a curve." Your final grade will be determined at the end...
- Thu Feb 21, 2019 11:19 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy vs Entropy
- Replies: 4
- Views: 796
Re: Gibbs Free Energy vs Entropy
On that note, Gibbs free energy is important because it is used for looking at changes in a system with constant pressure and temperature. Most biomedical reactions occur at constant temperature and pressure, and \Delta G relates back to the spontaneity of a reaction. This is more convenient than lo...
- Thu Feb 14, 2019 11:18 am
- Forum: Calculating Work of Expansion
- Topic: Change in moles
- Replies: 3
- Views: 391
Re: Change in moles
You would only use it for substances in the gas phase since PV=nRT is the ideal gas law.
- Thu Feb 14, 2019 11:14 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Cp,m
- Replies: 5
- Views: 804
Re: Cp,m
Cp,m is the molar heat capacity (heat capacity divided by number of moles) of an ideal gas at constant pressure. Cv,m is the molar heat capacity of an ideal gas at constant volume.
- Thu Feb 14, 2019 11:11 am
- Forum: Calculating Work of Expansion
- Topic: S=0
- Replies: 12
- Views: 1780
Re: S=0
An entropy of zero indicates there is no disorder. An example of this would be in a perfectly ordered material. If a system can only be prepared one way (W=1), using S=k b ln(W), we have S=0, which indicates that the system could be a perfect crystal. This leads us to the third law of thermodynamics...
- Thu Feb 07, 2019 11:26 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Difference in Entropy Due to Temperature Change
- Replies: 2
- Views: 372
Re: Difference in Entropy Due to Temperature Change
This equation only applies for a constant temperature. We would expect disorder to increase when a system is heated, since the supply of energy increases the thermal motion of the molecules. To calculate the change in entropy when the temperature of a system is changed, we use delta S = C*ln(T 2 /T ...
- Thu Feb 07, 2019 11:13 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Negative q
- Replies: 7
- Views: 2778
Re: Negative q
A negative q signifies that heat is being lost/released from the system into the surroundings.
- Thu Feb 07, 2019 11:10 am
- Forum: Calculating Work of Expansion
- Topic: Ideal gas expansion
- Replies: 4
- Views: 422
Re: Ideal gas expansion
When an ideal gas isothermally expands or is compressed, it does so at the same (constant) temperature. Since there is no change in temperature, the gas molecules will continue to move at the same speed and there is no energy change, which is why the internal energy change delta U is equal to zero.
- Thu Jan 31, 2019 11:17 am
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Surroundings
- Replies: 11
- Views: 947
Re: Surroundings
The system + the surroundings make up the universe. If the system is a reaction, then everything else is the surroundings. Simply put, if something is not the system, it is a part of the surroundings.
- Thu Jan 31, 2019 11:15 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question on yesterday’s lecture
- Replies: 2
- Views: 320
Re: Question on yesterday’s lecture
For the 7th edition, this is still part of Chapter 4 since we are still covering outline 2. Professor Lavelle mentioned that we started with the latter half of outline 2 first, and now we're going back to the first half, which would be chronologically first in the textbook.
- Thu Jan 31, 2019 11:12 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Phase changes and enthaplies
- Replies: 3
- Views: 375
Re: Phase changes and enthaplies
Enthalpy is defined as the amount of heat released or absorbed at a constant pressure. You are correct in speculating that it is related to exothermic (releasing energy) and endothermic (requires heat) reactions, and that relates to phase changes. In a phase change, energy is required to break the c...
- Thu Jan 24, 2019 11:25 am
- Forum: Phase Changes & Related Calculations
- Topic: outline 3
- Replies: 1
- Views: 230
Re: outline 3
The homework we turn in each week should reflect the material that we are currently covering in class. So yes, next week's homework should be questions from Outline 3.
- Thu Jan 24, 2019 11:24 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: When to ignore change in Acid
- Replies: 5
- Views: 879
Re: When to ignore change in Acid
You can ignore the change in acid concentration when Ka < 10-3, or when the change is 10-4 and lower. We assume that the change in acid concentration is negligible because it is a weak acid and will not completely ionize as a strong acid would.
- Thu Jan 24, 2019 11:20 am
- Forum: Amphoteric Compounds
- Topic: amphoteric v amphiprotic
- Replies: 5
- Views: 1149
Re: amphoteric v amphiprotic
An amphiprotic substance, such as water, can both donate and receive protons while an amphoteric substance can act as either an acid or a base and can thus react with both acids and bases.
- Thu Jan 17, 2019 12:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium Constants
- Replies: 5
- Views: 382
Re: Equilibrium Constants
K itself is the equilibrium constant, but Kc specifies usage of concentrations in mols/L rather than Kp, which specifies usage of partial pressures (applicable to gases only).
- Thu Jan 17, 2019 11:33 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Q vs K [ENDORSED]
- Replies: 4
- Views: 540
Re: Q vs K [ENDORSED]
As mentioned above, K as the equilibrium constant is calculated when the reaction is in equilibrium. Q, the reaction quotient, is found when the reaction is at any point other than equilibrium- in other words, when the reaction is not at equilibrium. While K and Q are calculated the same way ([P]/[R...
- Thu Jan 17, 2019 11:18 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE tables
- Replies: 3
- Views: 287
Re: ICE tables
Molar ratios will affect the way in which you proceed with your ICE table since, for example, not all reactions use up all of the reactants to form a product. If, for instance, a balanced chemical reaction produces 1 mole of product for every 2 moles of reactant, the concentration of the reactant wi...
- Thu Jan 10, 2019 11:22 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Giving Qc or Qp when asked for Q
- Replies: 5
- Views: 629
Re: Giving Qc or Qp when asked for Q
As mentioned above, whether you need to find Qc or Qp will usually be indicated by the units of the given quantities. Moles/liter specifies Qc whereas bars, atm, ex. indicates Qp. I've found that some of the questions will specifically ask for you to find Qp (rather than Qc) if needed.
- Thu Jan 10, 2019 11:14 am
- Forum: Administrative Questions and Class Announcements
- Topic: ChemComm. Account
- Replies: 1
- Views: 233
Re: ChemComm. Account
As far as I know, you can use the same account as the one you used from 14A! I just made sure to update my profile so that it would reflect the right lecture number and discussion section for this quarter.
- Thu Jan 10, 2019 11:11 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: K=1
- Replies: 5
- Views: 386
Re: K=1
As mentioned above, when K=1, we can derive that the amounts of concentration of the products and reactants are equal at equilibrium. Such an occurrence indicates that there is equal stability in both the reactants and products and the reaction favors neither reactants nor products at equilibrium.
- Thu Dec 06, 2018 9:44 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Def
- Replies: 1
- Views: 351
Re: Def
Polarizability is the ease with which an electron cloud can be distorted. Polarizing power, on the other hand, is the ability of a cation to distort an anion's electrons. Highly distorted electrons are described as being highly polarizable.
- Thu Dec 06, 2018 9:39 am
- Forum: Lewis Acids & Bases
- Topic: Lewis bases and Coordinate Covalence
- Replies: 1
- Views: 278
Re: Lewis bases and Coordinate Covalence
By definition, Lewis bases only donate lone pairs of electrons (which are accepted by Lewis acids). A coordinate covalent bond is formed when two electrons are contributed by a single atom, so to the extent of my knowledge, you are correct.
- Thu Dec 06, 2018 9:28 am
- Forum: Ionic & Covalent Bonds
- Topic: Boiling point
- Replies: 5
- Views: 857
Re: Boiling point
Look at the intermolecular forces present: NH 3 has the capacity to form hydrogen bonds because of its H-N bond. Hydrogen bonds, which are the strongest intermolecular forces, also have dipole-dipole and London forces (in descending order of strength of the intermolecular forces). CH 4 , on the othe...
- Thu Nov 29, 2018 9:35 am
- Forum: Resonance Structures
- Topic: Labelling Resonance Structures
- Replies: 3
- Views: 508
Re: Labelling Resonance Structures
No, it is not necessary to label the atoms numerically in resonance structures. It was just used in the manual to emphasis the fact that resonance spreads multi-bond character across the overall compound.
- Thu Nov 29, 2018 9:32 am
- Forum: Polarisability of Anions, The Polarizing Power of Cations
- Topic: Ionic Character
- Replies: 3
- Views: 374
Re: Ionic Character
An electronegativity difference > 2 between two atoms indicates an ionic bond, while an electronegativity difference < 1.5 indicates a covalent bond. Generally, a greater electronegativity difference will result in greater ionic character.
- Thu Nov 29, 2018 9:25 am
- Forum: Naming
- Topic: coordination compound
- Replies: 1
- Views: 186
Re: coordination compound
By definition, ligands are and will always be Lewis bases that supply lone pairs of electrons (in other words, contribute all the electrons needed for a bond) in order to form coordinate covalent bonds with a central metal atom/ion that will be the Lewis acid. The central metal atom/ion bonded to li...
- Thu Nov 22, 2018 1:33 pm
- Forum: Hybridization
- Topic: Determining shape
- Replies: 3
- Views: 326
Re: Determining shape
Hybridization is the concept that atomic orbitals can be mixed to form new hybridized orbitals that helps us explain experimentally determined molecular structures. Another way to think of it is that hybridization acts as an explanation as to why structures undertake the shape that they do, for inst...
- Thu Nov 22, 2018 1:21 pm
- Forum: Ionic & Covalent Bonds
- Topic: Bond Variations
- Replies: 4
- Views: 454
Re: Bond Variations
Lone pairs weaken the strength of a bond because of electron repulsion - the lone pair electrons of an atom will repel the bonded electrons and such electron-electron repulsion will reduce the overall bond's strength.
- Thu Nov 22, 2018 1:12 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Does formal charge apply to the VSEPR Model?
- Replies: 6
- Views: 1789
Re: Does formal charge apply to the VSEPR Model?
Generally, formal charge is used to predict the most stable and therefore most correct structure, so it would be correct to assume that the most stable Lewis structure would be used to determine molecular shape via VSEPR.
- Thu Nov 15, 2018 9:36 am
- Forum: Octet Exceptions
- Topic: d orbital period 3
- Replies: 2
- Views: 1853
Re: d orbital period 3
As mentioned above, all elements in period 3 have access to their d-orbitals because their principal quantum number n=3. With n=3, the angular quantum number l can be 0, 1, and 2, which corresponds to s, p, and d-orbitals respectively. The d-orbital of sulfur simply isn't in use during its ground st...
- Thu Nov 15, 2018 9:29 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Fluorine and Electronegativity
- Replies: 2
- Views: 746
Re: Fluorine and Electronegativity
Iodine is a larger atom and therefore has more electrons than fluorine does. Because of iodine's greater number of electrons, there is a better likelihood/tendency of electron fluctuations to occur and create a dipole (greater polarizability). This greater likelihood of electron fluctuations/shifts ...
- Thu Nov 15, 2018 9:20 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Posted Grades
- Replies: 2
- Views: 290
Re: Posted Grades
It should be posted sometime soon, especially since we have all gotten our midterms back. It takes a little while for the grade to be posted on MyUCLA, maybe a few days after we've gotten our tests back, but it's never anything too long.
- Thu Nov 08, 2018 9:49 am
- Forum: Interionic and Intermolecular Forces (Ion-Ion, Ion-Dipole, Dipole-Dipole, Dipole-Induced Dipole, Dispersion/Induced Dipole-Induced Dipole/London Forces, Hydrogen Bonding)
- Topic: Liquid and Solid Formations for Non-polar Atoms and Molecules
- Replies: 2
- Views: 302
Re: Liquid and Solid Formations for Non-polar Atoms and Molecules
As mentioned above, larger atoms and molecules have more electrons that are in higher energy levels and therefore further away from the nucleus, making them less tightly bound. In the context of molecules, the electrons are able to be more easily distorted and stronger intermolecular forces take pla...
- Thu Nov 08, 2018 9:30 am
- Forum: Bond Lengths & Energies
- Topic: Special Bond Length Rules
- Replies: 3
- Views: 446
Re: Special Bond Length Rules
As far as the content we have covered goes, I don't believe so. Just know that multiple bonds are stronger and shorter than single bonds, increasing in strength but decreasing in length from single, to double, to triple bonds. Also recall that lone pairs on neighboring atoms will repel one another a...
- Thu Nov 08, 2018 9:22 am
- Forum: Dipole Moments
- Topic: Dipole-Dipole vs Induced Dipole
- Replies: 5
- Views: 1069
Re: Dipole-Dipole vs Induced Dipole
Dipole-dipole interactions occur between molecules that have permanent dipoles. Induced dipole interactions occur when a molecule with a permanent dipole causes another molecule to have a dipole moment by distorting its electron cloud and thus giving it an uneven charge distribution, with one area b...
- Thu Nov 01, 2018 9:56 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal CHarge
- Replies: 5
- Views: 498
Re: Formal CHarge
As said above, formal charge is only assigned to an atom and is most stable with it is equal to zero. Make sure that you check the formal charge of each individual atom per each structure to determine which is the most stable. The total charge of a molecule is always one indicated charge and is foun...
- Thu Nov 01, 2018 9:46 am
- Forum: Octet Exceptions
- Topic: Max # of bonds
- Replies: 4
- Views: 465
Re: Max # of bonds
In theory, it could be possible but extremely unlikely. What matters is the stability of the structure as a whole rather than how many bonds an atom could form by accessing its d and f orbitals.
- Thu Nov 01, 2018 9:35 am
- Forum: Photoelectric Effect
- Topic: Rate at which electrons are emitted
- Replies: 2
- Views: 499
Re: Rate at which electrons are emitted
Hi! If by rate you mean the number of electrons emitted per a given time, the light's intensity would affect that rather than the energy. Intensity is linked to the number of photons, so a higher intensity would result in more photons. As long as the energy of each photon is greater than or equal to...
- Thu Oct 25, 2018 9:52 am
- Forum: Properties of Light
- Topic: Problem B9
- Replies: 4
- Views: 422
Re: Problem B9
Hi! 1.41x10^20 actually refers to the number of photons (emitted by the light in 2 seconds) instead of the amount of moles per 2 seconds. Because you already have the number of photons, you just need to convert it to moles- you already calculated the number of photons emitted in 2 seconds so you don...
- Thu Oct 25, 2018 9:32 am
- Forum: *Shrodinger Equation
- Topic: What do we need to know about the Shrodinger Equation? [ENDORSED]
- Replies: 1
- Views: 546
Re: What do we need to know about the Shrodinger Equation? [ENDORSED]
Hi! Professor Lavelle addressed during lecture that we won't be doing any calculations with Schrodinger's wave function equation. Conceptually, all you need to know is that the Schrodinger equation uses a wave function (psi)/orbital to describe an electron and its position in an atom. It uses the co...
- Thu Oct 25, 2018 9:25 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Spherical Polar Coordinates
- Replies: 1
- Views: 227
Re: Spherical Polar Coordinates
Hi! I wouldn't worry about knowing the polar coordinates for the test. Professor Lavelle didn't address them during lecture either so I wouldn't be too worried. As long as you know the concepts behind Schrodinger's equation (which is what the spherical polar coordinates refer to), you'll be more tha...
- Thu Oct 18, 2018 10:01 am
- Forum: Properties of Electrons
- Topic: Nodal Plane
- Replies: 5
- Views: 899
Re: Nodal Plane
Hi! A nodal plane is basically a two-dimensional plane that indicates the electron density distribution is zero, owing to the shape of the orbital. In other words, there is zero chance of finding an electron in a nodal plane. For instance, an s-orbital is spherically shaped and has a uniform electro...
- Thu Oct 18, 2018 9:50 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Orbitals
- Replies: 3
- Views: 236
Re: Orbitals
As mentioned above, a nodal plane essentially indicates an area where the electron probability density is zero. In other words, there is zero probability of finding an electron in a nodal plane. Nodal planes also depend on the shape of the orbital - for instance, an s-orbital is spherically shaped w...
- Thu Oct 18, 2018 9:41 am
- Forum: Photoelectric Effect
- Topic: Pre Assessment Module Question #13
- Replies: 2
- Views: 336
Re: Pre Assessment Module Question #13
I agree, I think the experiment needs to be performed underneath a vacuum so that the electrons can be detected. Without one, electrons displaced with a lower amount of kinetic energy might not necessarily be detected otherwise.
- Thu Oct 11, 2018 9:55 am
- Forum: Significant Figures
- Topic: Module: Molarity #15
- Replies: 6
- Views: 1271
Re: Module: Molarity #15
I'm not sure either, but I refer back to the given information of the problem when determining the number of significant figures in my answer. You shouldn't have more significant figures in your answer than was present in the information with the least number of significant figures in the question. ...
- Thu Oct 11, 2018 9:45 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Molar mass
- Replies: 3
- Views: 4142
Re: Molar mass
You will use only the subscripts in the chemical formula to calculate molar mass rather than using the stoichiometric coefficients that are present when a chemical reaction is balanced. Stoichiometric coefficients are looked at to get the molar ratios of the reagents and products of a chemical react...
- Thu Oct 11, 2018 9:38 am
- Forum: Properties of Light
- Topic: Photoelectric effect
- Replies: 4
- Views: 238
Re: Photoelectric effect
The energy of an individual photon (calculated by E=hv) must be greater than or equal to the work function (energy required to remove an electron from the metal) in order to displace an electron. E photon -E work function/energy to remove an electron =E kinetic energy of electron , so a photon with ...
- Thu Oct 04, 2018 9:37 am
- Forum: Significant Figures
- Topic: M11 Sig Figs
- Replies: 4
- Views: 415
Re: M11 Sig Figs
I've noticed that a lot of the solutions in the back of book have different numbers of significant figures than given in the info in the problem. I would go back and use the number of sig figs in the info given to you. In this case, your answer would be rounded to 3 sig figs instead of 2 as shown in...
- Thu Oct 04, 2018 9:29 am
- Forum: Limiting Reactant Calculations
- Topic: Significant figures in textbook 7th edition [ENDORSED]
- Replies: 5
- Views: 359
Re: Significant figures in textbook 7th edition [ENDORSED]
I typically go back to using the number of significant figures as provided by the info given in the problem. It's possible that using some conversion factors in dimensional analysis might give you additional significant figures during calculations, but I would ultimately refer back to the original i...
- Thu Oct 04, 2018 9:25 am
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Problem E7 converting atoms to moles
- Replies: 4
- Views: 370
Re: Problem E7 converting atoms to moles
Yes, you need to divide by Avogadro's constant in order to convert from atoms to moles. And yes, you can submit questions from the sixth edition, just make sure that it's clearly written somewhere on your homework (alongside the chapter and homework number) before you turn it in :)