Search found 50 matches
- Thu Mar 15, 2018 11:02 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Temperature for an ideal gas
- Replies: 2
- Views: 535
Re: Temperature for an ideal gas
Yes I think so.
- Thu Mar 15, 2018 10:52 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: delta U
- Replies: 2
- Views: 510
Re: delta U
Delta U in an isolated system when there is no heat change or work being done.
- Thu Mar 15, 2018 10:50 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Relationship between K and G
- Replies: 1
- Views: 894
Re: Relationship between K and G
Why is it that when K<1, G is + and when K>1, G is -? I know that this is because of the equation G = -RTlnK, but what does it mean conceptually? When K is less than 1, reactants are more favored in the reaction than products. Therefore, G is not spontaneous (positive) because spontaneous forward re...
- Tue Mar 06, 2018 10:25 pm
- Forum: General Rate Laws
- Topic: Integrating
- Replies: 2
- Views: 360
Re: Integrating
Lavelle has something on his website to help with integrating if you still need clarification!
- Tue Mar 06, 2018 10:16 pm
- Forum: Zero Order Reactions
- Topic: Units of k
- Replies: 12
- Views: 2038
Re: Units of k
If you are having trouble figuring out the units of k, write out the entire rate law with the units. You know that the rate must be in mol/Ls, so the units of k must cancel out the units of concentration of the reactants until you get mol/Ls.
- Tue Mar 06, 2018 10:09 pm
- Forum: General Rate Laws
- Topic: Writing rate laws
- Replies: 4
- Views: 581
Re: Writing rate laws
Also, when you have a zero order reaction, the rate is equal to the equilibrium constant k, so you don't need to include the reactant.
- Tue Feb 27, 2018 8:55 pm
- Forum: First Order Reactions
- Topic: Difference between 1st and 2nd order reactions
- Replies: 3
- Views: 21709
Re: Difference between 1st and 2nd order reactions
In a first order reaction, there will be one reactant present in the rate law. For a second order reaction, you can either have a rate law with one reactant to the second order, or with two reactants both to the first order. Two examples of second order reactions are rate=k[CO2]^2 or rate=k[CO2][H2O...
- Tue Feb 27, 2018 8:34 pm
- Forum: General Rate Laws
- Topic: 15.15 Rate Law
- Replies: 4
- Views: 1707
Re: 15.15 Rate Law
When the OH- concentration doubles, the rate increased by the same factor, and when the CH3Br concentration increases by a factor or 1.2, the rate increased by the same factor as well. Therefore, both reactants form a 1:1 ratio with the rate law, yielding the rate above. If, for example, when the co...
- Tue Feb 27, 2018 8:25 pm
- Forum: General Rate Laws
- Topic: Half life
- Replies: 3
- Views: 477
Re: Half life
You would first want to calculate the equilibrium constant (k) for the reaction, and then you plug that into the equation for half life.
- Thu Feb 22, 2018 8:02 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: 14.43
- Replies: 1
- Views: 348
14.43
In 14.43, why are we able to mix the concentrations and the pressures given to us when calculating Q?
- Thu Feb 22, 2018 2:11 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: K and Q at equilibrium
- Replies: 5
- Views: 714
Re: K and Q at equilibrium
K is equal to 1 at equilibrium, which would make Ecell 0 because there are no more electrons being transferred in the cell. I'm not sure whether Q is equal to 1 at equilibrium though.
- Wed Feb 21, 2018 10:51 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: E and Eo
- Replies: 2
- Views: 1204
Re: E and Eo
Eo is at standard state (25 degrees C and 1 atm) while E is the cell potential at anything other than standard state.
- Wed Feb 21, 2018 9:50 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: H+ in cell diagram
- Replies: 1
- Views: 246
H+ in cell diagram
In questions like 23 part a and c, why is H+ included in the cell diagram on the reduction side?
- Wed Feb 14, 2018 9:07 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: reversible system
- Replies: 9
- Views: 1154
Re: reversible system
Delta S total is zero when a process is reversible.
- Wed Feb 14, 2018 9:04 am
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Third Law Clarification/ Practice Midterm Question 3E
- Replies: 2
- Views: 450
Re: Third Law Clarification/ Practice Midterm Question 3E
Its true that the entropy of a perfect crystal is 0 at T=0 K. This is because perfect crystals only have one micro state, so using Boltzmann's equation and plugging in 1 for W, you get zero. However, most substances are not perfect crystals, so for these substances entropy is not 0 at T=0 K.
- Wed Feb 14, 2018 8:59 am
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: sign of delta S
- Replies: 1
- Views: 311
Re: sign of delta S
You want to look at the change as a whole and see if there are phase changes between reactants and products to determine the entropy.
- Sun Feb 11, 2018 2:44 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.13
- Replies: 3
- Views: 427
Re: 9.13
I believe it is a typo and I would trust the formula sheet.
- Sat Feb 10, 2018 8:06 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.15
- Replies: 3
- Views: 483
9.15
For 9.15 part a, the answer book gives the heat of fusion as -6.01 kJ/mol. How do we know that the heat of fusion should be negative?
- Wed Feb 07, 2018 10:49 pm
- Forum: Third Law of Thermodynamics (For a Unique Ground State (W=1): S -> 0 as T -> 0) and Calculations Using Boltzmann Equation for Entropy
- Topic: Molar Entropy
- Replies: 2
- Views: 359
Re: Molar Entropy
Larger molecules have more atoms, and therefore have more ways in which they can be arranged. Since entropy measures disorder, larger molecules have larger entropy.
- Thu Feb 01, 2018 6:55 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: units conversion for entropy
- Replies: 6
- Views: 1026
Re: units conversion for entropy
It doesn't matter what unit you use as long as you use the same unit for any other steps in the problem
- Thu Feb 01, 2018 6:49 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneous
- Replies: 14
- Views: 1925
Re: Spontaneous
Dr. Lavelle gave us guidelines to help determine when a reaction is spontaneous (deltaG is negative): Positive deltaS and negative deltaH = spontaneous at all temperatures Positive deltaS and positive deltaH = spontaneous at high temperatures Negative deltaS and negative deltaH = spontaneous at low ...
- Thu Feb 01, 2018 6:39 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework 9.13
- Replies: 2
- Views: 407
Re: Homework 9.13
On the class website, there is a correction to this problem that used Cv (5/2R)
- Tue Jan 23, 2018 10:39 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: units of enthalpy
- Replies: 1
- Views: 372
Re: units of enthalpy
The enthalpies of formation for the reactants and products are in kJ/mol because they depend on the number of moles of the reactant or product in the balanced chemical equation. When calculating enthalpy of a reaction from the enthalpies of formation of the reactants and products, you multiply the e...
- Tue Jan 23, 2018 10:32 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: relation between internal energy and enthalpy
- Replies: 1
- Views: 249
Re: relation between internal energy and enthalpy
When work is done by the system, the sign for work is negative. Therefore, by subtracting the negative amount of work from the change in internal energy (so basically adding), the enthalpy will be greater than the change in internal energy. When work is done on the system, the sign for work is posit...
- Mon Jan 22, 2018 11:07 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Bond enthalpies
- Replies: 1
- Views: 283
Bond enthalpies
In lecture on Wednesday of week 2, Dr. Lavelle made the comment that tables of bond enthalpies refer to breaking bonds in gases and therefore we need to add enthalpy of phase change for liquids and solids. I'm confused about what this means/how you would do this, can someone explain or give an examp...
- Tue Jan 16, 2018 9:08 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Heat Capacities
- Replies: 6
- Views: 342
Re: Heat Capacities
Molar heat capacities and specific heat capacities are virtually the same thing, just with different units. If you are given moles of a substance, you can easily convert to grams of the substance and vice versa.
- Tue Jan 16, 2018 9:00 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: K vs C
- Replies: 6
- Views: 543
Re: K vs C
If you are finding the difference from the final to the initial temperature, you can use C and K interchangeably because both will give the same difference. However, if you are using the ideal gas equation (PV=nRT) you want to make sure that your temperature is in K because using C instead of K will...
- Tue Jan 16, 2018 9:00 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: K vs C
- Replies: 6
- Views: 543
Re: K vs C
If you are finding the difference from the final to the initial temperature, you can use C and K interchangeably because both will give the same difference. However, if you are using the ideal gas equation (PV=nRT) you want to make sure that your temperature is in K because using C instead of K will...
- Tue Jan 16, 2018 8:53 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: What does R stand for?
- Replies: 13
- Views: 3312
Re: What does R stand for?
Going off of what everyone else said, there are several different gas constants so its important to use the one that ensures the correct units cancel out.
- Tue Jan 16, 2018 8:51 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Calorimeter
- Replies: 10
- Views: 1019
Re: Calorimeter
Calorimeters are used to find unknown variables in the q=mCΔT equation, such as specific heat or energy of the reaction.
- Fri Jan 12, 2018 2:40 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.3
- Replies: 2
- Views: 250
Re: Question 8.3
Yes, since the equation to find work is w=-PdeltaV, we know we need to use the volume of the pump which we can assume to be a cylinder. The equation we use for volume would vary by problem and the container we're working with.
- Fri Dec 08, 2017 10:41 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Strength of acids and bases
- Replies: 3
- Views: 288
Strength of acids and bases
Does electronegativity have anything to do with the strength of acids or bases?
- Thu Dec 07, 2017 12:25 pm
- Forum: Ionic & Covalent Bonds
- Topic: Strength of ionic v covalent bonds
- Replies: 2
- Views: 350
Re: Strength of ionic v covalent bonds
Ionic bonds have higher dissociation energy that covalent bond because the polarity of ionic compounds makes it easier for them to break apart in solution. For example, NaCl is highly dissociative in water because of its polarity.
- Sat Dec 02, 2017 4:04 pm
- Forum: Amphoteric Compounds
- Topic: 12.17 [ENDORSED]
- Replies: 9
- Views: 1700
12.17 [ENDORSED]
State whether the following oxides are acidic, basic, or amphoteric: a) BaO b) SO3 c) As2O3 d) Bi2O3
How do you tell what these are, particularly how do you know when they're amphoteric?
How do you tell what these are, particularly how do you know when they're amphoteric?
- Fri Dec 01, 2017 10:24 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Changes in K
- Replies: 4
- Views: 347
Re: Changes in K
There is a temperature associated with every K value, which is what allows us to use the K value for all reactions at a given temperature. But, when the temperature changes, we must use the K value associated with that new temperature.
- Fri Dec 01, 2017 10:22 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Exothermic vs. Endothermic?!
- Replies: 5
- Views: 4539
Re: Exothermic vs. Endothermic?!
Sometimes it helps me to think of heat as a reactant or product. If heat is added to the reactant side (endothermic) then more products will be produces, and if heat is added to the product side (exothermic) then more reactants will be produced.
- Fri Dec 01, 2017 10:20 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Exothermic vs. Endothermic?!
- Replies: 5
- Views: 4539
Re: Exothermic vs. Endothermic?!
Sometimes it helps me to think of heat as a reactant or product. If heat is added to the reactant side (endothermic) then more products will be produces, and if heat is added to the product side (exothermic) then more reactants will be produced.
- Tue Nov 21, 2017 8:12 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Unequal Reactant
- Replies: 3
- Views: 544
Re: Unequal Reactant
When there are more reactants than products, the equilibrium sits to the left. This is because the denominator of the K expression is larger than the numerator, which leads to a K value that is less than one meaning that the equilibrium sits to the left.
Re: HCN
hydrogen cyanide, because CN is a compound known as cyanide.
- Mon Nov 13, 2017 10:17 pm
- Forum: Hybridization
- Topic: Hybrid orbitals
- Replies: 2
- Views: 245
Re: Hybrid orbitals
To add to this, if the shape of the molecule is linear, the number of hybrid orbitals is 2 (sp) since there are 2 regions of electron density. If a molecule is trigonal planar, the number of hybrid orbitals is 3 (sp^2) since there are 3 regions of electron density. It helps to draw the Lewis structu...
- Mon Nov 13, 2017 10:10 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: 4.23
- Replies: 1
- Views: 200
Re: 4.23
In the answer book, there is a not with this problem that says that other Lewis structures are possible and more stable, like the ones you are talking about, but that they don't change the shape of the molecule. You're right about the double bonded structures being more stable, but I think for the s...
- Wed Nov 08, 2017 3:37 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Beryllium Octet Rule
- Replies: 6
- Views: 10741
Re: Beryllium Octet Rule
To add to this, you can tell that beryllium will now have a full octet because of formal charge. In BeCl2, Be has 2 shared bonds, making its formal charge 0, and Cl has 6 unshared electrons and 1 shared bond, making the formal charge of both Cl atoms in BeCl2 0. This proves to be the most stable str...
- Thu Nov 02, 2017 5:07 pm
- Forum: Trends in The Periodic Table
- Topic: first and second ionization energies [ENDORSED]
- Replies: 3
- Views: 623
first and second ionization energies [ENDORSED]
According to my notes from lecture, removing the second electron is always harder (second ionization energy is larger), but what about with elements like Mg? Wouldn't the first ionization energy for Mg be larger than the second since Mg has a full s orbital, therefore making it harder to remove the ...
- Wed Nov 01, 2017 11:58 pm
- Forum: Lewis Structures
- Topic: electron configuration
- Replies: 6
- Views: 772
electron configuration
When writing electron configuration, should the d orbital electrons of lower energy come before the s and p orbital electrons of higher energy? For example, in the electron configuration for gallium, should the configuration be [Ar]3d^10 4s^2 4p^1 or [Ar]4s^2 3d^10 4p^1 ?
- Sat Oct 28, 2017 5:04 pm
- Forum: Ionic & Covalent Bonds
- Topic: Chemical Bonds Lewis Structure: Octet Rule Violation [ENDORSED]
- Replies: 3
- Views: 2004
Re: Chemical Bonds Lewis Structure: Octet Rule Violation [ENDORSED]
This exception is like the one we talked about in class when we discussed radical exceptions to the octet rule, which is when there is a compound with an unpaired electron.
- Mon Oct 16, 2017 9:59 pm
- Forum: DeBroglie Equation
- Topic: λ=c/v VS. λ=h/mv
- Replies: 9
- Views: 12891
Re: λ=c/v VS. λ=h/mv
I think you are correct that you want to use λ=c/v when dealing with a photon of light, and λ=h/mv for other particles. What also helps is that when you are given a velocity in the problem, you know you must use λ=h/mv, but if you are not given a velocity, you can assume that the velocity is the spe...
- Sat Oct 14, 2017 9:07 pm
- Forum: Properties of Light
- Topic: How to use the Rydberg Formula? [ENDORSED]
- Replies: 6
- Views: 1287
Re: How to use the Rydberg Formula? [ENDORSED]
You want n1 to be the energy level that the electron drops to, and n2 to be the energy level that the electron drops from, so n1 will be the smaller n value and n2 will be the larger n value.
- Sat Oct 14, 2017 9:02 pm
- Forum: Properties of Light
- Topic: Units for HW Problem 1.23 [ENDORSED]
- Replies: 2
- Views: 398
Re: Units for HW Problem 1.23 [ENDORSED]
In this problem it makes sense to convert the wavelength to picometers because one picometer is equal to 1x10^-12 meters. Since the wavelength was calculated as 8.8237x10^-12 m, 8.8237 pm expresses this wavelength in the simplest form.
- Fri Oct 13, 2017 8:23 pm
- Forum: Properties of Electrons
- Topic: Wave Properties of Electrons [ENDORSED]
- Replies: 4
- Views: 610
Re: Wave Properties of Electrons [ENDORSED]
To add to that, diffraction occurs because of the different wavelengths of the waves passing through the crystal. Since electrons passing through a crystal show diffraction, this proves that instead of being a uniform stream of energy, electrons behave like waves.
- Fri Oct 13, 2017 8:16 pm
- Forum: Properties of Electrons
- Topic: Energy Level Change
- Replies: 5
- Views: 713
Re: Energy Level Change
I think that an excited electron in a hydrogen atom will always drop back down to the first energy level because that is where electrons are most stable, especially since hydrogen only has one electron and that electron exists naturally in the first energy level.