Search found 32 matches

by Sungyoon_Baek_1A
Mon Mar 12, 2018 3:54 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Oxidation State
Replies: 2
Views: 135

Re: Oxidation State

I'm assuming you meant for the oxidation state for oxygen... Oxygen typically has an oxidation number of -2 when in a molecule. However, when it is by itself and has no overall charge (O2) it is zero.
by Sungyoon_Baek_1A
Mon Mar 12, 2018 1:32 pm
Forum: *Enzyme Kinetics
Topic: Adsorption
Replies: 11
Views: 481

Re: Adsorption

I think the difference between adsorption and absorption is that in absorption, substance goes into another substance else while in adsorption, a substance/molecule sits on top of the other. For heterogeneous catalysts I think the importance of adsorption vs. absorption is that the reactants sit on ...
by Sungyoon_Baek_1A
Mon Mar 12, 2018 1:27 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: 15.79
Replies: 3
Views: 155

Re: 15.79

At high temperatures, the predominant product is CH3CHCHCH2Br. A higher temperature means that there is more energy to overcome a higher activation energy. The high temperatures and there for greater energy allow the reaction to produce CH3CHCHCH2Br. For the product CH3CHBrCHCH2, we do not need a hi...
by Sungyoon_Baek_1A
Mon Mar 05, 2018 2:26 pm
Forum: General Rate Laws
Topic: Average rate of consumption
Replies: 2
Views: 101

Re: Average rate of consumption

Rate will always be positive, rate will never be expressed as a negative number. As you said the equation for R is R = - delta[R]/delta t, but remember that R can also be expressed as R = +delta[P]/delta t. The negative when looking at the forwards rate is there to account for the smaller final reac...
by Sungyoon_Baek_1A
Mon Mar 05, 2018 1:51 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Kinetics [ENDORSED]
Replies: 2
Views: 125

Re: Kinetics [ENDORSED]

Yes, the larger activation energy will be the rate determining step since a larger activation energy means a slower reaction (since it takes more energy for the reaction to proceed forwards). Since the largest activation energy is the slowest reaction and the rate determining step is the slow step o...
by Sungyoon_Baek_1A
Mon Mar 05, 2018 1:48 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Assuming Elementary Rxn [ENDORSED]
Replies: 2
Views: 134

Re: Assuming Elementary Rxn [ENDORSED]

This was a hypothetical reaction in which he told us that the reaction was an elementary and second order reaction. We were only able to assume this because he told us that it was. I don't believe that we would be able to assume this if he had not told us.
by Sungyoon_Baek_1A
Sat Mar 03, 2018 9:40 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: reaction mechanisms reaction rate
Replies: 2
Views: 88

reaction mechanisms reaction rate

On the last lecture, I'm a little confused about the where the rate came from. Do we just know that Rate = k[NO2]^2 because that is the experimentally determined rate law and they gave it to us? And therefore we know that step 1 NO2 +NO2 ---> NO3 + NO is the rate determining step because it goes alo...
by Sungyoon_Baek_1A
Sat Mar 03, 2018 9:31 pm
Forum: Second Order Reactions
Topic: 15.19
Replies: 2
Views: 108

Re: 15.19

By using experiments 1 and 2 we found that the order for A is first order (n = 1). Since we know the order for A, we can use that value to solve for the order of B, so only the C concentration needs to be held constant. The reason we need the other concentrations to be held constant is so that we on...
by Sungyoon_Baek_1A
Sat Mar 03, 2018 9:24 pm
Forum: General Rate Laws
Topic: Rate constant units
Replies: 2
Views: 123

Re: Rate constant units

If you don't want to memorize the units for each order (or just wanna double check) you can derive the units by using the equation Rate = k(concentration^order). The units for rate is concentration per second (I'll just use mol/L for this example). The units for k are unknown/what we are solving for...
by Sungyoon_Baek_1A
Tue Feb 20, 2018 10:23 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Chapter 14 #25
Replies: 1
Views: 76

Re: Chapter 14 #25

You are correct; the stronger reducing agents do have more negative reduction potentials. Cr has a reduction potential of -0.91V while Zn has a reduction potential of -0.76V, therefore, following that logic, Cr is a stronger reducing agent than Zn.
by Sungyoon_Baek_1A
Tue Feb 20, 2018 10:19 pm
Forum: Balancing Redox Reactions
Topic: 14.3
Replies: 2
Views: 105

Re: 14.3

To balance the half reaction MnO4- ---> Mn2+ we must first balance all atoms (except oxygen and hydrogen) which is already done since there is one Mn on the reactants side and one Mn on the products side. Next, we have to balance the oxygens by adding H2O. Since there are 4 oxygens on the left side ...
by Sungyoon_Baek_1A
Tue Feb 20, 2018 10:12 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.21
Replies: 1
Views: 67

Re: 14.21

Whenever there is not a solid electrode (if both are aqueous) then, we can use Pt as the electrode. For example in part a, since Cr3+ and Cr2+ are both aqueous, we use Pt as the electrode, making the cell diagram... Pt(s) I Cr3+(aq), Cr2+(aq) II Cu2+(aq) I Cu(s)
by Sungyoon_Baek_1A
Sat Feb 17, 2018 11:32 pm
Forum: Balancing Redox Reactions
Topic: balancing hydrogen
Replies: 1
Views: 96

Re: balancing hydrogen

When balancing for acidic vs basic solutions you can start off by doing everything exactly the same (adding H2O to balance the oxygens and then adding H+ to balance the hydrogens). Since we have H+ in the reaction, this is the equation for an acidic solution. In order to make this a basic solution a...
by Sungyoon_Baek_1A
Sat Feb 17, 2018 11:22 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: Electrolysis to drive non-spontaneous reactions
Replies: 1
Views: 203

Re: Electrolysis to drive non-spontaneous reactions

All non spontaneous does have a negative cell potential because delta G must be positive for non spontaneous reactions and since delta G = -nFE, E must be negative in order for deltaG to be positive. In order to drive any any non spontaneous redox reaction, we must use an elective current to drive t...
by Sungyoon_Baek_1A
Sat Feb 17, 2018 11:08 pm
Forum: Balancing Redox Reactions
Topic: 14.1 Part D
Replies: 4
Views: 145

Re: 14.1 Part D

First we add the two half reactions from part b and c (remember to multiply part b by 3 so that the electrons lost in part b matches the electrons gained in part c). Everything should be balanced. We see that there are 14 H+ on the left and 6H+ on the right so we can cancel all of the H+ on the righ...
by Sungyoon_Baek_1A
Fri Feb 09, 2018 2:29 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: What is a Faraday? [ENDORSED]
Replies: 3
Views: 156

Re: What is a Faraday? [ENDORSED]

A Faraday is the charge of I mole of electrons (the units are coulomb/mol therefore we see that it is charge (coulomb) per mol), but in practicality it is a constant that makes calculations easier to work with and more manageable. The constant value is 1F = 96,485 C/mol
by Sungyoon_Baek_1A
Fri Feb 09, 2018 2:24 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: What makes E an intrinsic property? [ENDORSED]
Replies: 1
Views: 120

Re: What makes E an intrinsic property? [ENDORSED]

We can determine that E is an intrinsic property by using the formula deltaG = -nFE. We see that when the n (moles) value changes, that deltaG will also change (directly related to n). When we rearrange this formula to solve for E, we get E = -deltaG/(nF). since deltaG is directly related to n, chan...
by Sungyoon_Baek_1A
Fri Feb 09, 2018 2:13 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: Concentration and Cell Potential
Replies: 1
Views: 87

Re: Concentration and Cell Potential

A spontaneous process happens when death is negative therefore E must be positive. If we increase the concentration of Mn2+, using Le Chat we know that the reaction will favor the products (to offset the increase of a reactant). To favor the products/favor the forward reaction (spontaneous), E will ...
by Sungyoon_Baek_1A
Thu Feb 01, 2018 11:17 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Reversible vs. Irreversible work amount
Replies: 1
Views: 117

Re: Reversible vs. Irreversible work amount

When we look at the graphs of the reversible and irreversible reactions (volume on the x axis and pressure on the y axis) we know that the area under the curve represents the amount of work done. For an irreversible expansion, the pressure is constant so the work done is only that box under the curv...
by Sungyoon_Baek_1A
Wed Jan 31, 2018 10:01 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.11
Replies: 2
Views: 106

Re: 9.11

yes we are doing P1/P2 since pressure and volume are inversely related. For R we are using 8.314 J/Kmol because in the equation S = n R ln p1/p2, the mol unit of n would cancel out with the mol-1 unit of the R, leaving us with J/K for entropy (which are the correct units). ln p1/p2 has no units sinc...
by Sungyoon_Baek_1A
Tue Jan 30, 2018 4:41 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: 9.25
Replies: 3
Views: 178

Re: 9.25

Im not sure if there is a general rule since even though two molecules may share the same shape, if they are composed of different atoms they may not share the same number of arrangements (for example CH4 is also a tetrahedral but only has one possible arrangement since all the atoms bonded to the c...
by Sungyoon_Baek_1A
Sat Jan 27, 2018 1:31 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Adiabatic system
Replies: 6
Views: 219

Re: Adiabatic system

I think adiabatic just means that there is no heat flow (insulated) therefore q = 0 and the change in internal energy (delta u) = w
by Sungyoon_Baek_1A
Sat Jan 27, 2018 12:59 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: Degeneracy related to volume
Replies: 1
Views: 97

Re: Degeneracy related to volume

In the example he gave in class, V2/V1 = 2 (since V2 is twice the size of V1). Each gas molecule has two times positions available to it, therefore, W2/W1 also = 2. Since they both equal 2, we are able to substitute out W2/W1 with V2/V1.
by Sungyoon_Baek_1A
Wed Jan 24, 2018 9:59 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond enthalpies vs Standard Enthalpies of Formation
Replies: 2
Views: 106

Re: Bond enthalpies vs Standard Enthalpies of Formation

for bond enthalpies we are calculating the energy of bonds broken minus the energy of the bonds formed (therefore the products are negative and reactants are positive). However, for the standard enthalpies of formation we use the equation of the sum of the moles of each product multiplied by the sta...
by Sungyoon_Baek_1A
Sat Jan 20, 2018 10:10 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Open Beaker [ENDORSED]
Replies: 5
Views: 348

Re: Open Beaker [ENDORSED]

yes. since the open beaker is exposed to the open atmosphere and the atmosphere is very very large in comparison to the small beaker, the pressure is constant.
by Sungyoon_Baek_1A
Sat Jan 20, 2018 5:42 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Pressure's Relationship with Volume
Replies: 2
Views: 111

Re: Pressure's Relationship with Volume

I think he just means that you can't assume that they temperature is going to stay constant. If there was another case in which it did explicitly state that temperature would remain constant you could say that pressure would increase. When the volume decreases and the temperature decreases the molec...
by Sungyoon_Baek_1A
Sat Jan 20, 2018 5:26 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.49
Replies: 2
Views: 159

Re: 8.49

work done by the system(reaction) is always -PdeltaV since whenever work is done it is negative while work done on the system would be positive.
by Sungyoon_Baek_1A
Wed Jan 17, 2018 8:03 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat output of a reaction
Replies: 1
Views: 97

Re: Heat output of a reaction

If the temperature change of the calorimeter is given then you can use qcal = CdeltaT. Or if there is water in the calorimeter and they give you the temperature change of the water then do qh20 = mCdeltaT. then to find the heat of the reaction, we set qrxn = -qcal since the heat lost by the reaction...
by Sungyoon_Baek_1A
Sun Jan 14, 2018 4:44 pm
Forum: Phase Changes & Related Calculations
Topic: Constant temperature
Replies: 3
Views: 128

Re: Constant temperature

The temperature does not change during a phase change because the heat being added or taken away (in the case of solid to liquid; added) is being used to break the forces (hydrogen bond, etc.) between the molecules rather than increasing the average kinetic energy of the molecules (the average kinet...
by Sungyoon_Baek_1A
Fri Jan 12, 2018 4:06 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.67
Replies: 1
Views: 194

Re: 8.67

a) for H20 we know that the formula is 2H2+02=2H20 (couldn't figure out how to type an arrow so I'm using the = to replace it!). We know that H2 has one single bonded, O2 has one double bonded and H20 has two double bonds. So now we have to do the deltaH = bonds broken-bonds formed. You can find the...
by Sungyoon_Baek_1A
Thu Jan 11, 2018 11:56 am
Forum: Phase Changes & Related Calculations
Topic: Compression vs Expansion
Replies: 3
Views: 187

Re: Compression vs Expansion

(from the perceptive of the system) I think compression work will always be positive and expansion work will always be negative since compression is work of the surrounding being done to the system (work is being done on the system) and expansion is always work being done by the system.
by Sungyoon_Baek_1A
Thu Jan 11, 2018 11:52 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy of vaporization
Replies: 1
Views: 84

Re: Enthalpy of vaporization

the enthalpy of vaporization is for benzene is found in table 8.3 just above the self test question. The heat added to this system would be the enthalpy of vaporization multiplied by the moles of sample (so in this case, 39.1g divided by the molar mass of benzene - 78.12g -). By doing this your unit...

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