## Search found 57 matches

Sun Mar 18, 2018 10:33 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Putting an Ice cube into a glass of water
Replies: 4
Views: 681

### Re: Putting an Ice cube into a glass of water

You do factor in the heat of fusion in these questions. You need to account for the heat required to melt the ice and then to raise the liquid water to the final temperature. Hopefully this helps.
Fri Mar 16, 2018 3:59 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Substitution Reactions
Replies: 2
Views: 141

### Re: Substitution Reactions

A substitution reaction occurs when an atom (or group of atoms) is replaced by another atom (or group of atoms). This kind of reaction is common in organic molecules with a leaving group and a nucleophile that replaces the leaving group in the reaction. Because Dr. Lavelle went over it in class, I w...
Wed Mar 14, 2018 9:31 am
Forum: General Rate Laws
Topic: Karen Leung Worksheet 7 Kinetics Part 1, #5 [ENDORSED]
Replies: 1
Views: 123

### Karen Leung Worksheet 7 Kinetics Part 1, #5[ENDORSED]

HI, when we are given a table of values like the one given in Karen Leung's Worksheet 7 titled Kinetics Part 1, how to we incorporate coefficients from the chemical reaction equation into our calculations? The equation given is 2A+B+3C=2D+F. The table given is the same as others we have seen in the ...
Tue Mar 13, 2018 8:17 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.57
Replies: 2
Views: 138

### Re: 8.57

Thanks Justin. I solved it the same way you did initially but upon checking the solutions manual it proposed a three step process and used Hess's Law to solve instead of ∆H values. Your explanation was very helpful, thank you!
Mon Mar 12, 2018 9:44 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.57
Replies: 2
Views: 138

### 8.57

How would we know to use oxygen in the proposed mechanisms for the hydrogenation of ethyne to ethane? I understand the process of Hess' Law used in the solutions manual but am confused on how the book knew to use oxygens in the separate steps.
Sun Mar 11, 2018 5:22 pm
Forum: Balancing Redox Reactions
Topic: Ranking elements
Replies: 8
Views: 351

### Re: Ranking elements

No, there is not a periodic trend we can rely on for this type of deduction. You need the reduction half reactions to determine which element in a reaction has a greater oxidizing or reducing power.
Sun Mar 11, 2018 5:20 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15.51? Molecularity?
Replies: 3
Views: 123

### Re: 15.51? Molecularity?

The rate law is determined by the slow step, which in this problem is NO+Br2-->NOBr2. The rate law based on this step in the process is rate=k[NO][Br2] and the molecularity is two because there is one molecule of each NO and Br2 colliding to create the intermediate NOBr2. Hopefully this helps!
Sun Mar 11, 2018 10:38 am
Forum: *Enzyme Kinetics
Topic: 15.67
Replies: 4
Views: 272

### Re: 15.67

In class on Friday, Dr. Lavelle showed us the Arrhenius equation, which is K=Ae^Ea/RT. This is another way of expressing the constant K and finding other components of the equation quantitatively. That is the equation used in this problem to find K.
Mon Mar 05, 2018 10:59 pm
Forum: General Rate Laws
Topic: 15.3
Replies: 3
Views: 146

### Re: 15.3

The unique rate is the instantaneous rate found using the ratios like (1/2)(dA/dt).
Sun Mar 04, 2018 6:05 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Rate Constant K
Replies: 4
Views: 198

### Re: Rate Constant K

The units for k change with the different orders of reactions we solve. Homework question 15.9 shows this concept in a question form nicely. Because the units for time and concentration are constant, the units for k must change when we use natural logs versus 1/concentration and so forth.
Sat Mar 03, 2018 6:40 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Proposing Reaction Mechanisms [ENDORSED]
Replies: 1
Views: 87

### Re: Proposing Reaction Mechanisms[ENDORSED]

That will not be on test 3, as we will only be tested on 15.1-15.6! We will go over that for the final I believe but you do not need to know that for the test this week. Hopefully that answered your question.
Sat Mar 03, 2018 12:49 pm
Forum: General Rate Laws
Topic: Unique Rate [ENDORSED]
Replies: 5
Views: 444

### Re: Unique Rate[ENDORSED]

The unique rate is the instantaneous rate of change at any time t during the reaction. Unlike the average rate, which takes into account all of the times at which the reaction proceeds and averages it, the instantaneous rate is specific to a time and therefore it generates a different k value. The g...
Tue Feb 27, 2018 12:20 pm
Forum: First Order Reactions
Topic: 15.23 (c)
Replies: 1
Views: 86

### 15.23 (c)

Can someone explain to me how we incorporate the coefficient of 2 from the reactant A into figuring out k? I got the correct answer for parts (a) and (b), but I am lost on which equation to use for part (c). Any guidance would be greatly appreciated!
Mon Feb 26, 2018 11:10 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: 5.13
Replies: 2
Views: 142

### Re: 5.13

I am assuming you mean 15.13 or else I would be doing the wrong problem! First, you have to convert the grams of reactants given into moles, and then from moles to their respective molarity values. You get these values to be 0.347 mol/L and 0.001 mol/L. Then, multiple these values and the given k to...
Sun Feb 25, 2018 11:49 am
Forum: General Rate Laws
Topic: rate of consumption sign
Replies: 2
Views: 122

### Re: rate of consumption sign

In 15.3, the entire rate law for the reactant should have a negative sign in front of it. However, because we find the change in concentration as final-initial, 320mmol-450mmol=-130mmol. When fully written out, rate=∆concentration/∆t=-(320mmol-450mmol)/20s=+6.5x10^-3. The negative signs cancel out t...
Tue Feb 13, 2018 10:07 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Calculating entropy with changing temperature and volume [ENDORSED]
Replies: 2
Views: 169

### Calculating entropy with changing temperature and volume[ENDORSED]

Hi, I am confused about what equation or set of equations to use when you have a problem with changing temperature and volume. An example of this would be from Karen Leung's worksheet titled 2nd and 3rd Laws of Thermodynamics, and it is the last part of question 6. Another balloon expands from 3.3L ...
Sat Feb 10, 2018 8:30 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Calculating Gibbs Free Energy [ENDORSED]
Replies: 2
Views: 137

### Re: Calculating Gibbs Free Energy[ENDORSED]

You will be able to tell which one you can use by the values you are given. If you have the ∆Formation for the reactants and the products given, then use that method for solving for ∆G. Ig you are given the proper values to calculate ∆G using enthalpy and entropy, then that method would be the best ...
Sat Feb 10, 2018 4:30 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: Residual Entropy
Replies: 1
Views: 167

### Re: Residual Entropy

Residual/positional entropy is the entropy calculated at a constant temperature, independent of the enthalpy value of the system. The other entropy, thermal entropy, incorporates the ∆H value because there is a change in temperature of the system. Boltzmann's equation does not account for temperatur...
Sat Feb 10, 2018 12:49 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: system vs surroundings
Replies: 6
Views: 327

### Re: system vs surroundings

The surroundings are everything else in the universe except for the system. So with you interpretation the system as just being the reaction, then the surroundings are equal to the entire universe minus that reaction system.
Fri Feb 09, 2018 9:52 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: How to calculate W [ENDORSED]
Replies: 6
Views: 298

### Re: How to calculate W[ENDORSED]

W is the degeneracy, or the number of ways of achieving a given entropy value. If you know the desired entropy, you can use the Boltzmann equation to plug in the value you are given along with Boltzmann's constant to solve for W. In this form of solving for W, you would have to use your previous kno...
Wed Feb 07, 2018 11:06 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Standard Conditions
Replies: 3
Views: 171

### Re: Standard Conditions

Standard conditions are 1.00 atm and 298 K, which is 25ºC.
Tue Feb 06, 2018 12:04 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.15
Replies: 4
Views: 178

### Re: 9.15

Using the equation ∆S=q/T, we can substitute ∆Hº for q, and the the appendix provides the value for that variable. Because the given ∆Hº is per mole, multiply the numerator by the number of moles in the sample to then produce the correct value. Hopefully this will give you a good jumping off point t...
Thu Feb 01, 2018 2:31 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.13
Replies: 3
Views: 145

### Re: 9.13

For this question in particular, you are given all of the information necessary to calculate the number of moles using the ideal gas law, PV=nRT. n turns out to not be 1.00 mol, so here the solution would be incorrect if 1.00 mol was assumed to be the value of n. I think a general rule would be that...
Wed Jan 31, 2018 3:35 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Entropy and Temperature
Replies: 3
Views: 125

### Re: Entropy and Temperature

The term ∆S means the change in entropy, not just entropy itself. By increasing the temperature, the change in entropy is less than if the reaction were to occur at a lower temperature. The larger T value in the denominator makes this ∆S value less because as you said, the two are inversely related....
Wed Jan 31, 2018 10:24 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Relationship between ΔG⁰ AND K
Replies: 2
Views: 113

### Re: Relationship between ΔG⁰ AND K

The Kb that we are using here is not the same Kb as in acids and bases calculations. Here, we are using Boltzmann's constant, k = 1.38 x 10–23 J·K-1, to stand for Kb in free energy and entropy calculations. Hopefully this answers your question.
Tue Jan 30, 2018 8:36 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.11
Replies: 4
Views: 148

### Re: 9.11

I got the correct answer taking the natural log of (.500 atm/15 atm) and then multiplying it by the constant value R and 1.5 as the number of moles of neon. I will include the substituted values that I used below. Maybe it was a simple calculation error? ∆S=(1.5 mol)(8.314 J/Kmol)(ln(.500 atm/15 atm...
Mon Jan 22, 2018 1:26 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Internal Energy and Enthalpy [ENDORSED]
Replies: 2
Views: 172

### Re: Internal Energy and Enthalpy[ENDORSED]

∆U=∆H only when the work of the system is zero. As Dr. Lavelle was saying, this is almost always the case in biological systems because there is little to no work done on biological systems. In reference to the equation for determining the change in internal energy, ∆U=∆H+w, if w=0, then the equatio...
Sun Jan 21, 2018 2:54 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: isolated sytems
Replies: 4
Views: 172

### Re: isolated sytems

Another example of an isolated system could be a high grade thermos bottle like the question 8.1 shows us. The calorimeter is definitely the most popular and I would expect that to the the system used on a test dealing with isolated systems.
Thu Jan 18, 2018 3:27 pm
Forum: Administrative Questions and Class Announcements
Topic: discussion session tests
Replies: 1
Views: 129

### Re: discussion session tests

To my knowledge, tests are written by Dr. Lavelle, not the TAs. However, not all of the tests are exactly the same. They are all of equal difficulty but differing in the specific questions asked.
Thu Jan 18, 2018 2:19 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Internal Energy
Replies: 4
Views: 171

### Re: Internal Energy

Enthalpy is the amount of heat released or absorbed at constant pressure and it is written as H. When this H is changed between the products and the reactants in a chemical equation it becomes ∆H, meaning it is the change in enthalpy. This differs from the internal energy of the system greatly. The ...
Thu Jan 18, 2018 2:10 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Extensive v Intensive Properties
Replies: 2
Views: 140

### Re: Extensive v Intensive Properties

An intensive property is a property that depends on the amount of substance that is being measured. Extensive properties lack this substance measurement. We use the specific heat capacity as an intensive property because we can measure how much a a substance is involved in a calorimetry experiment a...
Thu Jan 18, 2018 12:51 pm
Forum: Phase Changes & Related Calculations
Topic: HW question
Replies: 5
Views: 205

### Re: HW question

Yes, that is all of the material that we have learned so far this quarter.
Sat Jan 13, 2018 6:58 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: When considering enthalpy (Δ H=q), is the letter "H" or "ΔH" Enthalpy?
Replies: 3
Views: 114

### Re: When considering enthalpy (Δ H=q), is the letter "H" or "ΔH" Enthalpy?

The delta just means that it is the change in enthalpy, so H is enthalpy. However, that H value is to as important to us as the change in that value which you already stated is equal to q. Enthalpy is the measure of heat released or absorbed by a system at constant pressure, and the delta just makes...
Sat Jan 13, 2018 6:22 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Intensive vs Extensive [ENDORSED]
Replies: 4
Views: 184

### Intensive vs Extensive[ENDORSED]

Why do we want to use extensive properties instead of intensive properties if possible? What is we are only given intensive to use for a calculation? Thanks!
Sat Jan 13, 2018 6:19 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy [ENDORSED]
Replies: 1
Views: 109

### Re: Enthalpy[ENDORSED]

Then, it would no longer be the enthalpy, it would just be the heat related or absorbed at the differing pressures. Enthaply has to be measured at a constant pressure.
Sat Dec 09, 2017 8:46 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Finding pH when given pOH
Replies: 1
Views: 121

### Re: Finding pH when given pOH

Yes that is exactly when you would use that equation:) The sum of pH and pOH is 14.
Sat Dec 09, 2017 3:55 pm
Forum: Naming
Topic: Ligand Suffix
Replies: 2
Views: 246

### Re: Ligand Suffix

There are two different naming systems used for ligands. They can both be found on the Naming Coordination Compounds sheet that Dr. Lavelle has posted on his website. Ligands with the -ido endings come from the IUPAC form of the name and the -ido is interchangeable with -o. Just stick to one form wh...
Tue Dec 05, 2017 3:13 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: HF vs HCl [ENDORSED]
Replies: 3
Views: 204

### Re: HF vs HCl[ENDORSED]

HCl is stronger than HF because of their size, so in turn that means that it is indeed because of their bond length. Fluorine atoms are smaller than chlorine atoms and that makes the bonds between the hydrogen atom and the fluorine atom shorter and stronger than that of a hydrogen and a chlorine ato...
Tue Dec 05, 2017 3:04 pm
Forum: Identifying Acidic & Basic Salts
Topic: HClO2
Replies: 1
Views: 242

### Re: HClO2

This molecule HClO2 is an acid, so it releases an H+ proton in solution. After that dissociation occurs, you would be left with a molecule with an O as the central atoms when it really should be a Cl. Physically, the much more stable arrangement of the molecules has the two oxygens bonded to a centr...
Sun Nov 26, 2017 3:33 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Post Assessment Equilibrium Part 2 #30
Replies: 2
Views: 209

### Post Assessment Equilibrium Part 2 #30

30. A mixture of 2.5 moles H2O and 100 g of C are placed in a 50 L container and allowed to come to equilibrium subject to the following reaction: C(s) + H2O (g) ⇌ CO (g) + H2 (g). The equilibrium concentration of hydrogen is found to be [H2] = 0.040 M. Calculate the equilibrium constant Kc of this ...
Sun Nov 26, 2017 3:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Reactions and their Reverse Reactions
Replies: 1
Views: 107

### Re: Reactions and their Reverse Reactions

I will use the example of an equilibrium equation A+B<-->C. The Kc for this forward reaction is Kc=C/A*B. The reverse reaction of this equilibrium system is C<-->A+B. In this reverse reaction, Kc=A*B/C. The two Kc equations have the numerator and the denominator flipped because the reactants and the...
Tue Nov 21, 2017 5:36 pm
Forum: Naming
Topic: 17.29 Cobalt vs. Colbaltate
Replies: 3
Views: 164

### Re: 17.29 Cobalt vs. Colbaltate

Cobaltate is used when the compound is an ion with a net negative charge while cobalt is used when the molecule has a net positive charge. Other TMs like iron can become ferrate when they have a net negative charge. Hope this helps.
Sat Nov 18, 2017 11:22 pm
Forum: Naming
Topic: Roman Numerals
Replies: 3
Views: 433

### Re: Roman Numerals

In terms of naming compounds with transition metals in them, the oxidations of transition metals is never zero (I think) and therefore you would never have a negative Roman numeral. The Roman numeral corresponds to the TW, so as the TM oxidation states are positive, the Roman numerals are also posit...
Sat Nov 18, 2017 10:13 am
Forum: Polarisability of Anions, The Polarizing Power of Cations
Topic: Methanol polarity
Replies: 2
Views: 193

### Re: Methanol polarity

When you draw out the lewis structure for the molecule, you can see that the dipoles do not cancel out. Unlike CH4 where the bonded atoms are all the same, methanol has three Hs and one OH. Therefore, the dipoles will not cancel because there are differences in their strengths.
Wed Nov 08, 2017 10:27 pm
Forum: Balancing Chemical Reactions
Topic: Net Moles of Gas
Replies: 5
Views: 254

### Re: Net Moles of Gas

Once the equation is balanced, it becomes 4C4H10+26O2-->16CO2+20H20. There are 30 moles of reactants and 36 moles of products. The difference between total moles of products and total moles of reactants gives us the net moles of gas produced (36-30=6). Therefore, the answer is 6 moles of net gas pro...
Wed Nov 08, 2017 10:20 pm
Forum: Trends in The Periodic Table
Topic: 2.93
Replies: 3
Views: 207

### Re: 2.93

There is an error in the solutions manual, so this is what Dr. Lavelle has posted on his website in reference to this error: 2.93 In the picture, it shows A (smaller atom) + B (larger atom) --> C (larger ion) + D (smaller ion) The solution manual says that A=Na and B=Cl, and it references Figure 2.2...
Sat Nov 04, 2017 10:33 pm
Forum: Electronegativity
Topic: Electronegativity related to solubility
Replies: 2
Views: 177

### Re: Electronegativity related to solubility

The electronegativity values help us to see what type of bond is formed between elements. This in turn can help us see what types of molecules will dissociate in water and become soluble. Compound bonded together ionically tend to be soluble in water due to them being salts. Those molecules fully di...
Fri Nov 03, 2017 11:00 pm
Forum: Ionic & Covalent Bonds
Topic: Electronegativity and Covalent Bonds
Replies: 2
Views: 128

### Re: Electronegativity and Covalent Bonds

When the difference in electronegativity of two bonding elements is greater than 2, then the bond is going to be ionic in character. If the difference in electronegativity is less than 1.5, then the bond is going to be covalent. The above explanation did a good job explaining why this occurs.
Wed Nov 01, 2017 9:30 pm
Forum: Trends in The Periodic Table
Replies: 4
Views: 214

When the electron jumps from the 2s orbital to the 2p orbital, it occupies a shell further from the nucleus, resulting in a higher atomic radius. It transforms to be more similar to the atoms in the p block that we know are larger than the s block atoms.
Fri Oct 27, 2017 10:14 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Electron affinity
Replies: 4
Views: 230

### Re: Electron affinity

The number of valence electrons greatly affects the electron affinity of an atom. An atom is much more likely to gain an electron from another atom when their outer valence shell is almost filled because they are looking to reach the ideal state of having a full outer shell. Because the nuclear char...
Fri Oct 27, 2017 10:08 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Post-Module Question #14
Replies: 2
Views: 206

### Re: Post-Module Question #14

In addition to the answer above, by looking at the Heisenberg equation that we have, we see that (deltaP)(deltaX)=h/4pi. deltaP and deltaX are inversely related, meaning that as one value increases, the other value decreases.
Sun Oct 22, 2017 6:26 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: XYZ [ENDORSED]
Replies: 4
Views: 241

### Re: XYZ[ENDORSED]

The two systems are actually very similar. For example, the p level has the capacity to hold 6 total electrons within it, and more specifically two per subset. There are three subsets within the p orbital, bringing the total amount of electrons tat can fit up to six which is what we would expect. Th...
Sun Oct 22, 2017 4:42 pm
Forum: DeBroglie Equation
Topic: Wavelike Properties
Replies: 3
Views: 349

### Re: Wavelike Properties

Also, in lecture it was determined that the general rule of thumb number to be used to detect if the object has wavelike properties is determined by the value of lambda. Anything that has a wavelength less than 10^-15 generally does not have wavelike properties.
Fri Oct 13, 2017 9:50 pm
Forum: Photoelectric Effect
Topic: Question 34 from Photoelectric Effect Post Module Assessment
Replies: 1
Views: 170

### Re: Question 34 from Photoelectric Effect Post Module Assessment

Hi Kelly, After finding the threshold energy in #33 to be 7.22x10^-19, you must look back at the equation hv(aka E)-phi=1/2mv^2. Phi is 7.22x10^-19 and because E=hc/lambda, you can calculate the energy of the photon to be 1.02x10^-18. Then, plug those two numbers into the above equation and you will...
Fri Oct 13, 2017 9:35 pm
Forum: Photoelectric Effect
Topic: Post Assessment #28 [ENDORSED]
Replies: 4
Views: 254

### Re: Post Assessment #28[ENDORSED]

Also be aware that the value in in kJ not just J, so you need to also do that conversion
Thu Oct 05, 2017 8:13 pm
Forum: Limiting Reactant Calculations
Topic: Limiting Reactant Calculations Post-Module Assessment
Replies: 2
Views: 218

### Re: Limiting Reactant Calculations Post-Module Assessment

The temp and pressure numbers are just denoting that the reaction is taking place at a standard temperature and pressure, meaning all of the products and reactants will behave ideally:) Long story short, the reaction is happening in a totally normal state so it can be solved without accounting for a...
Thu Oct 05, 2017 8:10 pm
Forum: Limiting Reactant Calculations
Topic: M11
Replies: 5
Views: 619

### Re: M11

I didn't use math to get O2 as the excess reactant, I just reasoned it out because the problem saw that if there is enough oxygen, the process can be further carried out in a secondary reaction, so thank you very much for elaborating on that! I would have been so lost on how to solve for it specific...