Search found 51 matches
- Wed Mar 14, 2018 11:34 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: steady state and pre equilibrium
- Replies: 2
- Views: 572
Re: steady state and pre equilibrium
My TA told us that steady state will not be on the final because Dr. Lavelle didn't really cover it in class, and that we will only have to use the pre equilibrium approximation approach.
- Wed Mar 14, 2018 11:29 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: question 6 on test 2
- Replies: 2
- Views: 536
Re: question 6 on test 2
Once you write out the final redox reaction, only the reactants are considered oxidizing or reducing agents. The one that is being oxidized in the reaction is the reducing agent, and the one being oxidized is the reducing agent. Therefore you should write out the final equation before identifying th...
- Wed Mar 14, 2018 11:23 am
- Forum: General Rate Laws
- Topic: Test #3 Q3
- Replies: 3
- Views: 997
Re: Test #3 Q3
When you balance the equation, you should get AB4(g) + 2 C2(g) ---> AC2(g) + 2 B2C(g). Because the rate of the reaction is constant, we can use the equation rate = (change in concentration)/(change in time). You must multiply the rate by 2 because there are two moles of B2C produced for each mole of...
- Wed Mar 14, 2018 11:05 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Calculating Gibbs Free Energy
- Replies: 4
- Views: 661
Re: Calculating Gibbs Free Energy
You use difference of sums to find Gibbs free energy when you are using the Gibbs free energy of formation for the reactants and products. If you are given entropy, enthalpy, and temperature, then you would use deltaG=deltaH - TdeltaS.
- Wed Mar 07, 2018 11:10 am
- Forum: General Rate Laws
- Topic: Problem 15.37
- Replies: 3
- Views: 558
Re: Problem 15.37
Because they divide initial concentration by final concentration, you get a value of 10 in the ln and would therefore set ln(10) = k*t. If you divided final concentration by initial concentration instead, you would get a value of 0.1 in the ln, but you would have to set ln(0.1) = - k*t according to ...
- Tue Mar 06, 2018 8:37 pm
- Forum: General Rate Laws
- Topic: Coefficients affecting order & molecularity
- Replies: 2
- Views: 393
Re: Coefficients affecting order & molecularity
The orders do not have to do with coefficients. If you look at problem 15.17, for example, the reactants all have different coefficients, but that does not affect their orders. You simply look at the concentrations for each when determining order. In 15.23c, however, you do need to use the coefficie...
- Tue Mar 06, 2018 8:32 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.19c
- Replies: 6
- Views: 772
Re: 15.19c
The answer to part D is in mol*L^-1*s^-1 too, if that makes more sense.
- Tue Mar 06, 2018 8:31 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: 15.19c
- Replies: 6
- Views: 772
Re: 15.19c
I was confused at first too, and got the same answer as you. I figured that in the solutions manual they converted all the values from mmol to mol, so I did that and got 2.85 x 10^-12. However, in 15.17 they leave the values in mmol and I'm not sure which one they want us to do.
- Tue Mar 06, 2018 8:26 pm
- Forum: General Rate Laws
- Topic: 15.3
- Replies: 4
- Views: 604
Re: 15.3
Rate is similar to speed vs. velocity. Though velocity can be positive or negative because it has direction, speed can only be a positive value. The same goes for rate of a reaction, it can only be a positive value because it is simply how fast the reaction is proceeding.
- Tue Mar 06, 2018 8:23 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chapters on test?
- Replies: 3
- Views: 536
Re: Chapters on test?
My TA said that it covers up to #39 in the problem set.
- Thu Feb 22, 2018 3:37 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidation
- Replies: 3
- Views: 549
Re: Oxidation
I found a great website that tells you exactly how to determine oxidation states, as it would be too long to describe. I would check this out: https://www.chemguide.co.uk/inorganic/redox/oxidnstates.html
- Wed Feb 21, 2018 9:55 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15c
- Replies: 2
- Views: 324
Re: 14.15c
There is a typo in the solutions manual. It says on Dr. Lavelle's website that "The correct solution should have a 3 as the subscript for the nickel hydroxide reactant.
2Ni(OH)3 (s) + Cd(s) → Cd(OH)2 (s) + 2Ni(OH)2 (s)"
2Ni(OH)3 (s) + Cd(s) → Cd(OH)2 (s) + 2Ni(OH)2 (s)"
- Wed Feb 21, 2018 12:50 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Calculating K
- Replies: 3
- Views: 432
Re: Calculating K
You can use either of the first two equations, depending on what values are given. If given deltaG directly, then the first equation is easier to use. If only given standard cell potential, then the second equation is easier to use. The third can be used when the reaction is not in equilibrium.
- Wed Feb 21, 2018 12:41 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: appendix 2
- Replies: 2
- Views: 351
Re: appendix 2
Yes, I also think it was a typo because I'm pretty sure the solutions manual had Fe3+ instead of Fe2+.
- Wed Feb 21, 2018 12:40 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.37 C
- Replies: 2
- Views: 359
Re: 14.37 C
Because anode is always written on the left side of the cell diagram, you still have to calculate the Ecell = Ecathode - Eanode, which does give you a negative voltage. This tells you that the reaction is not spontaneous, and that the electrons would flow in the opposite direction.
- Mon Feb 12, 2018 4:15 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Example 9.5
- Replies: 3
- Views: 565
Re: Example 9.5
If you look at the bottom of page 325 under the example, it says to assume that argon is an ideal gas, so I am assuming that is why you would use argon's heat capacity.
- Mon Feb 12, 2018 4:10 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Pressure
- Replies: 3
- Views: 549
Re: Pressure
w=-p*deltaV, so pressure does in fact affect work.
- Mon Feb 12, 2018 4:01 pm
- Forum: Calculating Work of Expansion
- Topic: equilbrium = reversible
- Replies: 3
- Views: 532
Re: equilbrium = reversible
Reactions at equilibrium are considered reversible, as they go back and forth in order to reach equilibrium. Irreversible reactions, on the other hand, only proceed in one direction.
- Mon Feb 12, 2018 3:57 pm
- Forum: Calculating Work of Expansion
- Topic: q and temp increase
- Replies: 2
- Views: 384
Re: q and temp increase
Because q=n*C*deltaT, and neither moles nor molar heat capacity can be negative, a negative q would mean that delta T is negative. So yes, a negative q means that temperature is decreasing. A negative deltaU does not necessarily mean temperature is decreasing because deltaU=q+w. So even if q was pos...
- Fri Feb 02, 2018 1:37 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: entropy to surroundings
- Replies: 3
- Views: 489
Re: entropy to surroundings
The easiest way is that deltaS(total) = deltaS(system) + deltaS(surroundings), and deltaS(total) must be 0 for a system at equilibrium, so deltaS(system) = -deltaS(surroundings).
- Fri Feb 02, 2018 1:25 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Van't Hoff Equation
- Replies: 2
- Views: 1124
Re: Van't Hoff Equation
You start with deltaH - TdeltaS = -RTlnK. Then divide both sides by -RT to get (-deltaH/RT) + (deltaS/R) = lnK , or flip it around to get lnK = (-deltaH/RT) + (deltaS/R).
- Mon Jan 29, 2018 7:47 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy: State Function?
- Replies: 3
- Views: 473
Re: Gibbs Free Energy: State Function?
Also, gibbs free energy is equal to enthalpy plus entropy, which are both state functions. Thus, Gibbs free energy must also be a state function.
- Mon Jan 29, 2018 7:45 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: sign of entropy
- Replies: 11
- Views: 1687
Re: sign of entropy
Also, if delta entropy equals zero then the reaction is at equilibrium.
- Mon Jan 29, 2018 7:40 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs Free Energy [ENDORSED]
- Replies: 6
- Views: 1349
Re: Gibbs Free Energy [ENDORSED]
Delta G combines the values of delta H and delta S to fully determine whether a reaction is favorable or not.
- Mon Jan 29, 2018 7:38 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Entropy and Disorder
- Replies: 2
- Views: 473
Re: Entropy and Disorder
Entropy is the likelihood or probability that that a system will be in a particular state.
- Mon Jan 22, 2018 12:20 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.99 Negative Specific Heat Capacity
- Replies: 3
- Views: 479
Re: 8.99 Negative Specific Heat Capacity
Yes, the negative value is indicating that it is lost heat.
- Mon Jan 22, 2018 12:17 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.77 [ENDORSED]
- Replies: 2
- Views: 233
Re: 8.77 [ENDORSED]
The kekule structure of benzene suggests alternating single and double bonds, whereas the resonance structure contains six bonds that are between a single and double bond. The resonance structure is more stable, as all six bonds between the carbon molecules are the same.
- Mon Jan 22, 2018 12:06 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Internal Energy
- Replies: 3
- Views: 565
Re: Internal Energy
It is like the hill analogy. The path you take to get to the top does not matter, as you only look at the change in elevation from your starting point to the end point. Therefore, internal energy is not concerned with the "in between", only the beginning and final states.
- Wed Jan 10, 2018 2:11 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Specific Heat Capacity vs Heat Capacity
- Replies: 5
- Views: 584
Re: Specific Heat Capacity vs Heat Capacity
Specific heat capacity is always constant for each respective substance, regardless of the mass present. Heat capacity on the other hand differs depending on the mass of the substance present.
- Wed Jan 10, 2018 2:06 pm
- Forum: Calculating Work of Expansion
- Topic: Work vs. Work Function
- Replies: 2
- Views: 478
Re: Work vs. Work Function
The work function is also the energy required to remove one electron.
- Wed Jan 10, 2018 2:02 pm
- Forum: Phase Changes & Related Calculations
- Topic: Temperature during Phase Change
- Replies: 6
- Views: 441
Re: Temperature during Phase Change
If you look at vaporization for instance, water is not able to increase in temperature past 100 degrees celsius, because after 100 it must be in its gaseous state. So even though more heat is being added, the temperature of the water evaporating remains at 100 degrees celsius and the heat that is be...
- Sat Dec 09, 2017 5:27 pm
- Forum: Trends in The Periodic Table
- Topic: chapter 2 #37 part d
- Replies: 1
- Views: 445
Re: chapter 2 #37 part d
You can think of it as the electrons in the p-orbital being more shielded by the electrons in the s-orbital, as it is closer to the nucleus. Therefore, the positive charge of the nucleus reaches the p-orbital electrons much less than s-orbital electrons, making the Zeff on the s-orbital greater than...
- Sat Dec 09, 2017 5:22 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Remembering VSEPR
- Replies: 3
- Views: 647
Re: Remembering VSEPR
You can memorize the VSEPR shape for each VSEPR formula (i.e. AX3E = trigonal pyramidal, or AX6 = octahedral). That way for every structure you draw, you can simply figure out its VSEPR formula and know which shape it denotes. Otherwise, drawing out the VSEPR structure should give you a clear idea a...
- Sat Dec 09, 2017 2:36 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: exception to nitrogen
- Replies: 2
- Views: 415
Re: exception to nitrogen
Also, you want to minimize the formal charges, so placing the lone electron on C will give you a formal charge of 0 for both atoms.
- Sat Dec 09, 2017 2:34 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: t-shaped
- Replies: 4
- Views: 588
Re: t-shaped
The bond angles for the atoms will only be slightly less than 90 degrees and slightly less than 180 degrees, not 120 as we do not count the angles between lone electron pairs.
- Sat Dec 09, 2017 2:25 pm
- Forum: Lewis Structures
- Topic: Lewis Structures
- Replies: 2
- Views: 306
Re: Lewis Structures
NO2 is a radical, meaning it does not have a full octet on the N atom, and N has one lone electron instead of a lone pair.
- Sat Dec 09, 2017 2:18 pm
- Forum: Limiting Reactant Calculations
- Topic: Limiting Reagent and finding amount produced
- Replies: 2
- Views: 506
Re: Limiting Reagent and finding amount produced
You would use the moles of the limiting reagent to find how many moles of your product are being produced, as the amount produced is limited by how much of the limiting reactant is present. This requires the use of molar ratios. So say your equation is 2A+3B->C. If you find A to be the limiting reag...
- Thu Dec 07, 2017 5:18 pm
- Forum: Administrative Questions and Class Announcements
- Topic: REVIEW WORKSHEET (CH 3 & 4) FALL 2017 [ENDORSED]
- Replies: 5
- Views: 1028
Re: REVIEW WORKSHEET (CH 3 & 4) FALL 2017 [ENDORSED]
On Dr. Lavelle's website it says that this review session is from 4:30-6:00 pm in LaKretz 110. Is it at this time or 4:00-6:30? Thank you!
- Wed Dec 06, 2017 7:03 pm
- Forum: Resonance Structures
- Topic: Delocalization of Charge
- Replies: 3
- Views: 436
Re: Delocalization of Charge
Delocalized charges occur when there is resonance, as the charge occurs in different spots due to the actual structure being a combination of all its resonance structures. It is therefore delocalized.
- Wed Dec 06, 2017 7:02 pm
- Forum: Resonance Structures
- Topic: Delocalization of Charge
- Replies: 3
- Views: 436
Re: Delocalization of Charge
Delocalized charges occur when there is resonance, as the charge occurs in different spots due to the actual structure being a combination of all its resonance structures. It is therefore delocalized.
- Wed Dec 06, 2017 7:00 pm
- Forum: Sigma & Pi Bonds
- Topic: Delocalized Pi-Bonding
- Replies: 5
- Views: 888
Re: Delocalized Pi-Bonding
The easiest way to spot delocalized pi-bonding is in lewis structures that display resonance structures, as the pi-bond can move around and is therefore "delocalized." Just remember delocalized as not being local to one specific point in the structure.
- Wed Dec 06, 2017 6:54 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: Electron configurations of d block elements
- Replies: 2
- Views: 423
Re: Electron configurations of d block elements
Yes, it is just these exceptions, as it is more stable to have the d-orbital filled and only half of the s-orbital filled.
- Wed Dec 06, 2017 6:46 pm
- Forum: *Shrodinger Equation
- Topic: What does this even mean?
- Replies: 4
- Views: 784
Re: What does this even mean?
I believe that this topic was not heavily emphasized in class and we were not assigned any homework problems on it, therefore it most likely will not be on the final so I wouldn't stress about it too much.
- Wed Dec 06, 2017 6:27 pm
- Forum: Ionic & Covalent Bonds
- Topic: 3.29
- Replies: 2
- Views: 1035
Re: 3.29
Also, this is not an assigned homework problem so it most likely will not be on the final exam :)
- Mon Dec 04, 2017 12:39 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Units for K and Q [ENDORSED]
- Replies: 5
- Views: 2212
Re: Units for K and Q [ENDORSED]
The units used to calculate K and Q must be constant in the equation, so that the units will cancel out and leave K and Q without specific units, as K is a constant and Q is a ratio.
- Mon Nov 13, 2017 4:20 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Ions and Extra Electrons [ENDORSED]
- Replies: 2
- Views: 351
Re: Ions and Extra Electrons [ENDORSED]
It's easiest to treat ions like other molecules when determining lewis structures, by first adding up all of the valence electrons and either adding or subtracting the electrons (depending on whether it is a cation or anion) to the total number. Then just draw the structure so that the total number ...
- Sun Nov 12, 2017 4:29 pm
- Forum: Resonance Structures
- Topic: Understanding Resonance Structures
- Replies: 6
- Views: 981
Re: Understanding Resonance Structures
When comparing the bond lengths of two different molecules and one has resonance structures, you can determine its relative bond length by taking an average of the bond lengths within the resonance structure, which can then be compared to the bond length of other molecules. This is consistent with t...
- Sun Nov 12, 2017 4:23 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Length
- Replies: 3
- Views: 461
Re: Bond Length
Bond length has to do with electrons being shared, and can therefore be determined by the number of bonds. For example, a single bond has the longest bond length, and a triple bond has the shortest bond length.
- Sun Nov 12, 2017 4:21 pm
- Forum: Bond Lengths & Energies
- Topic: Bond Length
- Replies: 8
- Views: 3394
Re: Bond Length
You can determine the relative bond lengths by looking at the number of single, double, and triple bonds. Triple bonds have the shortest bond length, and single bonds have the longest. Because some structures may have resonance, the bond lengths can be averaged in order to compare it to other struct...
- Thu Oct 05, 2017 10:19 am
- Forum: Balancing Chemical Reactions
- Topic: Combustion Question
- Replies: 9
- Views: 1401
Re: Combustion Question
Yes, carbon dioxide and water vapor will always be produced in combustion, but depending on the reactants and due to conservation of mass, other products may be produced. In example H.19, nitrogen is one of the reactants, so it therefore must also be a product. That is why you would get carbon dioxi...
- Thu Oct 05, 2017 10:13 am
- Forum: Balancing Chemical Reactions
- Topic: Problem L35
- Replies: 3
- Views: 1051
Re: Problem L35
Yes, to approach this problem you must work backwards, using the moles of each product to find the moles of the reactant produced by the equation above it. Use the given moles of NaBr to find moles of Fe3Br8, which is the product of the previous equation above it. Use that to find the moles of FeBr2...