## Search found 30 matches

Sat Mar 10, 2018 7:54 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Molecularity and Coefficients
Replies: 3
Views: 332

### Re: Molecularity and Coefficients

Molecularity is basically just the number of molecules that come together in a reaction. A way to find this is by just adding all of the stoichiometric coefficients of the reactants in an elementary reaction. So 2A would be bimolecular, 3A would be trimolecular, and A would be unimolecular. And with...
Sat Mar 10, 2018 7:42 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Equations: ln Q vs. log Q [ENDORSED]
Replies: 5
Views: 697

### Re: Equations: ln Q vs. log Q[ENDORSED]

ln and log are not interchangeable. Using ln will give you a different answer than using log. Be sure to apply whatever function the equation uses. As Sahil mentioned, you technically could use ln or log instead of the one in the equation but you would need to use the conversion factor of ln(X) = 2....
Sat Mar 10, 2018 7:35 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.49
Replies: 2
Views: 228

### Re: 8.49

The reason the 8.314 value is used is because of its units. The units of the 8.314 value is Joules/(K*mol). The units of the 0.08206 R that you mentioned is L * atm / (K * mol). Since the question is asking for the internal energy, the answer should be in the units of joules or kJ. The same answer c...
Fri Mar 09, 2018 11:25 pm
Forum: First Order Reactions
Topic: Help on 15.3
Replies: 7
Views: 583

### Re: Help on 15.3

I believe you meant to type in 15.13 instead of 15.3 for this problem. In the solution manual, the mL value of 750. mL is converted to liters which is just .750 L. The problem converts the grams of H2 given into concentration of H2 by dividing by this .750 L and dividing by the molar mass. The same ...
Fri Mar 09, 2018 10:26 pm
Forum: Balancing Redox Reactions
Topic: Acidic vs. Basic Solutions [ENDORSED]
Replies: 4
Views: 306

### Re: Acidic vs. Basic Solutions[ENDORSED]

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&t=28011&sid=fe7b533bb8051e9d3b2b9b0539ec222e
Thu Mar 08, 2018 9:32 pm
Forum: Experimental Details
Topic: 15.19c
Replies: 2
Views: 585

### Re: 15.19c

You could have gotten the answer wrong because of the units. For part c of the question, if you left the units as mmol, the answer would have had mmol^-4 in the units. Since the mmol has an exponent, you would have to multiply the answer by 1000^4 in order to convert it into mol^-4. If you just conv...
Thu Mar 08, 2018 9:21 pm
Forum: General Rate Laws
Topic: Stoichiometric coefficients and order
Replies: 5
Views: 580

### Re: Stoichiometric coefficients and order

As a general rule, I would just not use the stoichiometric coefficients to find the order of a reaction. They line up sometimes but a lot of the times they do not. In many questions where you have to use rate laws, the order will either be given or you can find it without using the coefficients. If ...
Thu Mar 08, 2018 9:17 pm
Forum: General Rate Laws
Topic: Problem 15.3
Replies: 4
Views: 318

### Re: Problem 15.3

This problem is just referring to the average rate which is delta Concentration / delta Time. So the order of the reaction isn't needed to find the rate. It would just be -(320-450)/20.
Thu Mar 08, 2018 9:11 pm
Forum: General Rate Laws
Topic: 15.19
Replies: 2
Views: 229

### Re: 15.19

I don't think it really matters. As long as you write down your units for your final answer, you should be fine. You could always just convert it to both just in case.
Sun Feb 25, 2018 9:05 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: NERNST
Replies: 4
Views: 306

### Re: NERNST

And just to clarify because I know that this confused me initially as well, the lnQ is converted into a logQ using the conversion factor lnx = 2.303 logx. So the 0.0592 value is calculated by (2.303*R*T/F) which would be 2.303*8.314*298/96485.3.
Sun Feb 25, 2018 8:30 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: swapping signs of E values
Replies: 8
Views: 1008

### Re: swapping signs of E values

Yeah I also do what Ridhi does which is flipping the sign of the oxidation reaction since the appendix contains the reduction potentials. Either way works though, it all just depends on what way works best for you.
Sun Feb 25, 2018 7:37 pm
Forum: Zero Order Reactions
Topic: Relation to Reaction Stoichiometry?
Replies: 2
Views: 219

### Re: Relation to Reaction Stoichiometry?

Yeah stoichiometric coefficients are not related to order of reaction im pretty sure. Sometimes they may be the same value but that does not mean they are related. In many cases, the values are not equivalent.
Sun Feb 18, 2018 9:54 pm
Forum: Balancing Redox Reactions
Topic: Balancing Redox Reactions
Replies: 4
Views: 435

### Re: Balancing Redox Reactions

H30+ is basically just H20 and H+ so adding H30+ would just complicate the balancing by including extra molecules that aren't needed when you could just use H+ to balance.
Sun Feb 18, 2018 9:35 pm
Forum: Balancing Redox Reactions
Topic: Basic and Acidic conditions [ENDORSED]
Replies: 11
Views: 3667

### Re: Basic and Acidic conditions[ENDORSED]

If I understand the question correctly, yes H+ is used for acidic conditions and OH- is used for basic. The extra H20 molecules balance out both sides.
Sun Feb 18, 2018 9:21 pm
Forum: Balancing Redox Reactions
Topic: H+ vs H2
Replies: 3
Views: 203

### Re: H+ vs H2

I believe it has to do with the fact that H+ has a charge to it, while H2 does not. This means that you are also adding a positive charge to the reactions which balances out the overall charge. H2 wouldn't give you this charge.
Sun Feb 11, 2018 8:02 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 4 on Midterm Review Worksheet
Replies: 1
Views: 200

### Re: 4 on Midterm Review Worksheet

If you are talking about the Midterm Practice that Lyndon made, the delta S value would be 0 since it is a state property and the final and initial states are identical, meaining there is no net change. He said during the review session that the temperature he wrote in the question, 1234K, should ac...
Sun Feb 11, 2018 7:49 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Delta G
Replies: 2
Views: 192

### Re: Delta G

I know you can assume Delta G is 0 when a solution/reaction is at equilibrium. This is how the equation standard delta G = -RTlnK is derived.
Sun Feb 11, 2018 6:32 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: Finding entropy of vaporization for water at room temp
Replies: 2
Views: 268

### Re: Finding entropy of vaporization for water at room temp

You can calculate the entropy of vaporization at a temperature like 25 degrees by adding up the entropies required to heat the liquid water to 100 degrees, the entropy of vaporization at 100 degrees, and the entropy required to cool the vapor to 25 degrees. Adding these entropies together would give...
Sun Feb 04, 2018 11:21 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Test 1 Question
Replies: 4
Views: 357

### Re: Test 1 Question

An ideal cooler is not a closed system because coolers keep drinks cold. This means that heat is not exchanged with its surroundings. Isolated systems do not exchange heat with their surroundings while closed systems do. If you thought it was closed because coolers are not always 100% effective and ...
Sun Feb 04, 2018 11:03 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Gibbs free question
Replies: 2
Views: 173

### Re: Gibbs free question

I believe you would also set ΔG = 0 when you want to find the range of temperatures that a reaction needs to be spontaneous. Since when ΔG < 0 the reaction is spontaneous, solving for T in the equation ΔG = ΔH - TΔS = 0 would give you the temperature needed for the reaction to be in equilibrium. Dep...
Sun Feb 04, 2018 10:44 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: 9.23
Replies: 1
Views: 204

### Re: 9.23

You would expect COF2 to have a higher molar entropy since it has more possible positions/microstates than BF3. Since BF3 is a Boron atom surrounded by 3 Fluorine atoms, there is only one state that it can be in. COF2, on the other hand, is a Carbon atom surrounded by an Oxygen atom and 2 Fluorine a...
Sat Jan 27, 2018 3:07 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Chapter 9, Question 7
Replies: 2
Views: 213

### Re: Chapter 9, Question 7

When a monatomic ideal gas is at constant pressure, the R constant is multiplied by 5/2. When a monatomic ideal gas is at constant volume, it is multiplied by 3/2. I do not believe we need to know where these numbers came from just that we need to use them in these situations of constant pressure or...
Wed Jan 24, 2018 10:36 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Molar Kinetic Energy
Replies: 2
Views: 220

### Re: Molar Kinetic Energy

Danah Albaaj 1I wrote:Also a follow-up, can someone explain why the answer to part c is 12.5 kJ/mol instead of 12.5 J/mol?
Thank you!

The answer is 12.5 J/mol. You may have a more outdated version of the solution manual or something if it says kJ/mol.
Wed Jan 24, 2018 7:09 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.73
Replies: 2
Views: 191

### Re: 8.73

For part a, the C-H bonds are not included because when C2H2 becomes C6H6 there are still C-H bonds. These C-H bonds aren't destroyed, they are instead just moved into a different formation. The only bonds that are actually destroyed and then made again are the C C triple bonds in C2H2 and the benze...
Sun Jan 21, 2018 10:55 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Homework Problem
Replies: 2
Views: 142

### Re: Homework Problem

Since NO2 has more atoms than NO, it has more bond vibrations that can absorb energy. With more energy, there is a bigger molar heat capacity. It makes sense because with more atoms there's more potential to have high energy than if you have fewer atoms. So, the more complex a molecule is, the bigge...
Sun Jan 21, 2018 10:48 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.49
Replies: 5
Views: 351

### Re: 8.49

Whenever no temperature is given but a temperature is needed to solve, you assume the temperature is 25.0 degrees Celcius which is 298 K.
Sun Jan 21, 2018 10:39 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.101E
Replies: 1
Views: 120

### Re: 8.101E

Yes I believe so, since the enthalpy of formation of 02 is 0, it does not need to be included in calculating the enthalpy of the reaction since multiplying anything by an enthalpy of formation of 0 will result in just 0.
Mon Jan 15, 2018 4:01 pm
Forum: Phase Changes & Related Calculations
Topic: Most stable phases for halogens
Replies: 3
Views: 230

### Re: Most stable phases for halogens

He also mentioned how O2, N2, and H2 are all gases in their purest form. So I believe the only ones we need to know are C, H2, O2, N2, and the four main halogens.
Mon Jan 15, 2018 3:51 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Intensive vs extensive property
Replies: 2
Views: 206

### Re: Intensive vs extensive property

Yeah David is right. Extensive properties depend on mass so like heat capacity is the amount of energy needed to change the temperature of some substance by 1 degree C which varies with the amount of the substance. Specific heat capacity is an intensive property. It is the energy needed to change th...
Mon Jan 15, 2018 3:37 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Units
Replies: 2
Views: 172

### Re: Units

I believe that when you are calculating something like ΔH° f that is a standard reaction enthalpy of something (like the standard enthalpy of combustion or formation, etc) you use the units kJ/mol but when it is just ΔH you only use kJ. Basically, it seems that in the textbook when there is a degree...