Search found 40 matches
- Fri Mar 16, 2018 10:57 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Reducing Power
- Replies: 5
- Views: 582
Re: Reducing Power
Find the redox equations on the sheet for each of the elements given and the most negative value for E has the most reducing power.
- Fri Mar 16, 2018 10:53 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Test 3 question 5
- Replies: 3
- Views: 567
Re: Test 3 question 5
You have to set the rate equal to k times the concentration of H2O (the moles of H2O (1.8/18.016) times 1/.750L) times the concentration of O2 squared (moles of O2 (1.8/32) times 1/.750L and square that). The equation should look like: 2.7 x 10^2 = k*[(1.8/18.016)*(1/.750)]*[(1.8/32)*(1/.750)] and t...
- Fri Mar 16, 2018 10:48 am
- Forum: Calculating Work of Expansion
- Topic: Irreversible Expansion vs. Reversible Expansion
- Replies: 2
- Views: 567
Re: Irreversible Expansion vs. Reversible Expansion
Since you use the equation w = -P*deltaV for irreversible expansion, you would likely be given the pressure or the change in moles to use for substitution with PV = nRT. For reversible expansion, we use the equation w = -nRT*ln(V2/V1) so you would likely be told it is isothermal and the moles would ...
- Mon Mar 12, 2018 5:43 pm
- Forum: Calculating Work of Expansion
- Topic: different types of work
- Replies: 4
- Views: 433
Re: different types of work
For work you should recognize expansion versus compression but that is as specific as you need to know. The other equations, such as those in the beginning of chapter 8 in the textbook, are not on the equation sheet.
- Mon Mar 12, 2018 5:41 pm
- Forum: *Organic Reaction Mechanisms in General
- Topic: Functional Groups
- Replies: 7
- Views: 2006
Re: Functional Groups
If you are able to draw the Lewis structure of molecules and know what to look for within the structure to determine the functional group, you should be set.
- Mon Mar 12, 2018 5:40 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: slow step
- Replies: 3
- Views: 496
Re: slow step
There can only be one "slow" step because it is considered the slowest of all the steps. Slow is a relative term and describes the speed of one step of reaction in relation to the speed of the other steps.
- Mon Mar 05, 2018 9:25 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: What is Molecularity?
- Replies: 9
- Views: 2427
Re: What is Molecularity?
You can also think about molecularity in regards to the order of a reaction, as first order is unimolecular, second order is bimolecular, and third order is termolecular.
- Mon Mar 05, 2018 9:24 pm
- Forum: First Order Reactions
- Topic: 15.23C
- Replies: 6
- Views: 781
Re: 15.23C
If the question tells you what order the reaction is in, always go with that because it is directly stated in the question.
- Mon Mar 05, 2018 9:22 pm
- Forum: General Rate Laws
- Topic: rate of change of species
- Replies: 2
- Views: 361
Re: rate of change of species
I think this refers to questions like 15.1 on the homework where you are given the equation: N2 (g) + 3H2 (g) -> 2NH3 (g) and are asked what is the relation between the rate of consumption of H2 and the rate of consumption of N2. In this case, the rate of consumption of N2 is 1/3 times the rate of c...
- Mon Mar 05, 2018 9:19 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: derivations
- Replies: 2
- Views: 382
Re: derivations
The derived equations are on the equation sheet. Derivations are helpful for sure so be comfortable with them but you will not need to know how to derive them.
- Tue Feb 27, 2018 12:53 pm
- Forum: General Rate Laws
- Topic: Homework 15.49
- Replies: 1
- Views: 293
Homework 15.49
Could someone please explain how we know that HBr, NO2, and HOBr are first order? All we are given are two reaction equations with no numbers.
- Tue Feb 27, 2018 12:51 pm
- Forum: General Rate Laws
- Topic: initial rate law
- Replies: 5
- Views: 567
Re: initial rate law
Also in a lot of the homework problems, you use the change in the concentration of the product to find the change in the concentration of the reactant.
- Tue Feb 27, 2018 12:49 pm
- Forum: General Rate Laws
- Topic: Writing a rate equation
- Replies: 3
- Views: 409
Re: Writing a rate equation
I don't think you need to ever write it in terms of products because even if you are given the increase or decrease in concentration of products, you can convert it to reactants.
- Sun Feb 25, 2018 9:46 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Homework 15.19
- Replies: 3
- Views: 517
Homework 15.19
/Is it always necessary to change units to mol from mmol?
- Tue Feb 20, 2018 11:51 am
- Forum: Balancing Redox Reactions
- Topic: 14.5 part a
- Replies: 2
- Views: 370
Re: 14.5 part a
It will become more obvious that oxygen is being reduced once you balance the hydrogens and oxygens in the equation. The half reaction isn't balanced so it is difficult to tell what is happening. The balanced reaction should look like H2O (l) + O3 (g) + 2e- -> O2 (g) + 2OH- (aq). The oxygen is going...
- Tue Feb 20, 2018 11:48 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15 given half reactions?
- Replies: 3
- Views: 519
Re: 14.15 given half reactions?
I also used the appendix. I think we will receive a sheet with half reactions that we will need for the test.
- Tue Feb 20, 2018 11:46 am
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Example 14.3
- Replies: 1
- Views: 266
Re: Example 14.3
In this equation, n stands for the moles of electrons.
- Tue Feb 13, 2018 9:49 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat Capacities of Gases
- Replies: 3
- Views: 667
Re: Heat Capacities of Gases
These are given on the equation sheet.
- Mon Feb 12, 2018 7:44 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.101 (b) [ENDORSED]
- Replies: 4
- Views: 958
Re: 9.101 (b) [ENDORSED]
I had trouble with this one as well. You would think that since the outdoors is so much larger that any temperature change wouldn't affect the entropy as much but the question really just wants you to look at the equation.
- Mon Feb 12, 2018 7:43 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Calorimetry
- Replies: 2
- Views: 372
Re: Calorimetry
Every time there is a temperature change or a phase change you need a separate equation. Usually when an ice cube has a phase change it first has to gain heat or lose heat to reach the right temperature for the phase change. That would be 2 equations. Then, if the ice cube changes temperature after ...
- Mon Feb 12, 2018 7:41 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Delta U = 0
- Replies: 2
- Views: 346
Re: Delta U = 0
Usually when the question says the process is isothermal, delta U will be equal to zero. An isothermal system will also have a delta S(universe) of zero.
- Thu Feb 08, 2018 10:55 am
- Forum: Biological Examples (*DNA Structural Transitions, etc.)
- Topic: problem 9.75
- Replies: 2
- Views: 881
Re: problem 9.75
For cis compounds there are 12 different orientations but for trans compounds there are only 3 different orientations. Therefore, the cis form will have higher residual energy (use Boltzmann entropy calculations).
- Thu Feb 08, 2018 10:52 am
- Forum: Phase Changes & Related Calculations
- Topic: Phase changes with entropy?
- Replies: 6
- Views: 674
Re: Phase changes with entropy?
During a phase change, the temperature does not change but heat must still go into/leave the system in order to cause the phase change. Since heat is entering or leaving the system, the entropy will change and therefore needs to be accounted for.
- Mon Feb 05, 2018 9:53 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Chapter 11 #17
- Replies: 2
- Views: 285
Re: Chapter 11 #17
I also got -2.7 kJ/mol! I think you are right and the answer key has an error in it!
- Mon Feb 05, 2018 9:52 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.63
- Replies: 4
- Views: 454
Re: 9.63
If the value for delta G is negative then energy needs to be added in order to form the compound so therefore it would not be stable, as it doesn't occur spontaneously. However, a negative delta G for a compound means the compound can form without any stimulus or added energy and is therefore a stab...
- Mon Feb 05, 2018 9:47 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.17
- Replies: 10
- Views: 1038
Re: 11.17
I agree. I've put it in several different calculators and have gotten -2.7 kJ/mol every time.
- Wed Jan 31, 2018 2:00 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: A spontaneous reaction
- Replies: 7
- Views: 893
Re: A spontaneous reaction
When you need to go up the hill (low to high), an input of energy is required to do so, so therefore this motion cannot happen on its own and it is not spontaneous. However, if you go down the hill (high to low), the reaction can happen on its own because no energy needs to be put into the system an...
- Wed Jan 31, 2018 1:53 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Why should deltaG=0?
- Replies: 4
- Views: 812
Re: Why should deltaG=0?
You can also think of it like delta G goes from negative to positive with 0 as the "threshold" so to find at what temperature this threshold is at, you set deltaG equal to 0.
- Wed Jan 31, 2018 1:52 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Dependence of free energy on pressure
- Replies: 2
- Views: 286
Re: Dependence of free energy on pressure
I was also confused by this in class but in the equation G = G^o + RTln(P/P^o), I believe that the P is the atmospheric pressure and the P^o is always 1.00 atm because it's standard. Therefore, when P is more than 1.00 atm, (P/P^o) is positive and when P is less than 1.00 atm, (P/P^o) is negative.
- Wed Jan 31, 2018 1:19 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: Homework 9.25
- Replies: 1
- Views: 323
Homework 9.25
When I plug (1.38x10^-23)*ln(6^6.02x10^23) into my calculator, I get a result that says "overflow" but the answer is 14.9 J/K. How is this possible?
- Thu Jan 25, 2018 12:04 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Confusion about entropy formula
- Replies: 6
- Views: 578
Re: Confusion about entropy formula
If there is a heat reservoir in the surroundings, the temperature will be constant and therefore heat will be reversible. If a system is doing work, it is losing energy but heat from the surroundings can be added to the system to replace the energy lost. For instance, during a phase change, work is ...
- Thu Jan 25, 2018 12:57 am
- Forum: Phase Changes & Related Calculations
- Topic: Chapter 8 Homework #117
- Replies: 1
- Views: 187
Chapter 8 Homework #117
Why is the net production 2/3 mole of gas? There are 2 moles of reactants and 4 moles of products so how is the change 2/3?
- Wed Jan 24, 2018 10:29 pm
- Forum: Phase Changes & Related Calculations
- Topic: Chapter 8 Homework #99
- Replies: 2
- Views: 499
Chapter 8 Homework #99
How do you figure out the enthalpies of formation? The enthalpy for ZnCL2 isn't in the appendix of the book.
- Thu Jan 18, 2018 11:43 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.29 heat capacity
- Replies: 3
- Views: 353
Re: 8.29 heat capacity
From what I understand, there are three forms of movement that affect the internal energy of a molecule, and therefore the heat capacity. These are: translational, linear rotational, and nonlinear rotational. For more complex molecules, all three motions are occurring and therefore the molecule has ...
- Thu Jan 18, 2018 11:40 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Q 8.13
- Replies: 9
- Views: 958
Re: Q 8.13
In this question, the part we are trying to figure out is the work that can be done by the system, or on the system. In general, if the system does work, such as expansion, the work value will be positive and if the system has work done on it, such as compression, the work value will be negative.
- Thu Jan 18, 2018 11:31 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 5
- Views: 451
Re: Hess's Law
Since enthalpy is a state function, it doesn't matter what path is taken to the final point, the only values that matter are final and initial. That is why you can "skip" intermediate steps by adding equations.
- Thu Jan 18, 2018 11:26 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: enthalpy
- Replies: 3
- Views: 435
Re: enthalpy
When you subtract the enthalpies of formation of the reactants from the enthalpies of formation of the products, the result is the enthalpy of the reaction.
- Mon Jan 15, 2018 1:12 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Homework Problem 8.31
- Replies: 2
- Views: 180
Re: Homework Problem 8.31
For monatomic gases you can use the formula C[p][/M] = (5/2)*R as well as C[V][/M] = (3/2)*R
- Thu Jan 11, 2018 11:53 am
- Forum: Phase Changes & Related Calculations
- Topic: Compression vs Expansion
- Replies: 3
- Views: 458
Compression vs Expansion
Is the work done during compression always positive and the work done during expansion always negative?
- Thu Jan 11, 2018 11:51 am
- Forum: Phase Changes & Related Calculations
- Topic: Open, Closed, or Isolated Systems
- Replies: 6
- Views: 10437
Re: Open, Closed, or Isolated Systems
It would be a closed system because while the mercury cannot leave the thermometer, heat can be transferred to or from the thermometer, therefore not making it isolated.