## Search found 53 matches

Sun Mar 04, 2018 5:39 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Arrhenius Behavior
Replies: 3
Views: 437

### Re: Arrhenius Behavior

Hey Ava!!! Exhibiting Arrhenius behavior means that the plot of lnk against 1/T for a reaction gives a straight line (lnk on the y, 1/T on the x). Since the slope of an Arrhenius plot is proportional to the activation energy, the higher the activation energy, the stronger the temperature dependent o...
Sun Mar 04, 2018 5:31 pm
Forum: First Order Reactions
Topic: 15.25
Replies: 4
Views: 301

### Re: 15.25

This comes from the integrated rate law for a first order reaction. The equation is ln[A]=-kt+ln[A]o. If we take everything and raise e to that power, we get e^ln[A]=e^-kt+e^ln[A]o. the terms on the right side can be added (since they have the same base) to become e^(-kt + ln[A]o). e^ln[A] becomes j...
Sun Mar 04, 2018 5:24 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Example 15.2
Replies: 1
Views: 184

### Re: Example 15.2

To find the order given a table of values, you want to find where only the concentration of one reactant is changing. If the other concentrations are constant, then you know that they have not affected the rate, and that the change in concentration of the one reactant is the only influence on the ra...
Sun Mar 04, 2018 5:14 pm
Forum: General Rate Laws
Topic: half life equation
Replies: 1
Views: 139

### Re: half life equation

It depends on if the reaction is zero order, first, or second. The zero order half life equation is [A]o/2k, and depends on the initial amount of the substance you have. For example, it would depend on the initial concentration of mercury you have, and would vary depending on that concentration. For...
Sat Mar 03, 2018 8:07 pm
Forum: Second Order Reactions
Topic: 15.35
Replies: 1
Views: 136

### Re: 15.35

The equation for the half-life of a second order reaction is t1/2=1/k[A]o. This is derived from the integrated rate law for a second order reaction. The equation is 1/[A]=kt+1/[A]o, and we know that at t=t1/2 [A]=1/2[A]o. Therefore, 1/[A]=2/[A]o. When you plug this into the equation, you get 2/[A]o=...
Thu Mar 01, 2018 7:39 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Rate Law [ENDORSED]
Replies: 2
Views: 170

### Re: Rate Law[ENDORSED]

The rate law will be referring to whatever you designate it as. For example, lets say you have the reaction 2NH3 -> N2+ 3H2. If we define the rate as the consumption of NH3, we might have rate=k[NH3]. If we want to describe rate with respect to the other components of the reaction, we use coefficien...
Thu Mar 01, 2018 7:19 pm
Forum: General Rate Laws
Topic: self test 15.2A
Replies: 1
Views: 149

### Re: self test 15.2A

To do this, you need to use the stoichiometric coefficients given in the equation to relate the rates to one another. You know that the rate NH3=rate H2 (2mol NH3/3mol H2), so the rate of NH3=(2/3)rate of H2. You take 1.15 and divide by 2/3, to get an answer of 1.72 mmolH2⋅L-1⋅h-1. To get the unique...
Thu Mar 01, 2018 6:57 pm
Forum: Zero Order Reactions
Topic: Catalysts
Replies: 2
Views: 206

### Re: Catalysts

I think this has to do with enzymes acting as catalysts and their interactions with substrates. Enzymes and substrates have to come together, and eventually the enzyme becomes saturated so increasing the concentration of the reactants with not increase the reaction rate. I think it can also happen i...
Thu Mar 01, 2018 6:55 pm
Forum: Second Order Reactions
Topic: The graph of second order [ENDORSED]
Replies: 5
Views: 447

### Re: The graph of second order[ENDORSED]

Hey! When you look at the integrated rate law for second order, you see that it is 1/[A]=kt + 1/[A]o. If you plug this into the y=mx + b format, your y is 1/[A], the slope is +k, and the y-intercept is 1/[A]o. Since you have positive k, the slope of the line will be positive. (This differs from 1st ...
Thu Mar 01, 2018 6:50 pm
Forum: Balancing Redox Reactions
Topic: One Reactant
Replies: 2
Views: 194

### Re: One Reactant

When there is one reactant, you need to look at the reduction potentials for that reactant and the different products it forms in your equation. The reduction potential for the reactant going to product that is more negative will be the one that is oxidized. The one with a more positive reduction po...
Thu Mar 01, 2018 6:46 pm
Forum: Zero Order Reactions
Topic: Zero Order Reactions and Catalysts
Replies: 2
Views: 207

### Re: Zero Order Reactions and Catalysts

Hey Audrey!!! :) I think this has to do with enzymes acting as catalysts and their interactions with substrates. Enzymes and substrates have to come together, and eventually the enzyme becomes saturated so increasing the concentration of the reactants with not increase the reaction rate. I think it ...
Thu Feb 22, 2018 2:12 pm
Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
Topic: 14.117
Replies: 2
Views: 420

### Re: 14.117

To get the current, you use the equation I= ΔG/-Et. Plugging into this, we have -10 x 10^J/(-1V)(24h)(3600 s/h)), which gives us an answer that is 115 J/V⋅s. Amps are in units of Coulombs/second, so if we take that answer of 115 J/V⋅s and multiply by 1V⋅C/1J, the Joules and volts cancel, and were le...
Thu Feb 22, 2018 1:56 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.13 d
Replies: 1
Views: 155

### Re: 14.13 d

When you look at the reduction potentials for Au+ and Au3+, you see that Au+ + e- -> Au has E°=1.69 V, and Au3+ + 3e- -> Au has E°=1.40 V. Since 1.40<1.69, this will be the anode, and the reaction must be flipped, so you'll have Au -> Au3+ + 3e-. The cathode reaction will be the same, except it will...
Thu Feb 22, 2018 1:52 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: HW 14. 15 part a
Replies: 1
Views: 5411

### Re: HW 14. 15 part a

For this kind of question, you want to find the reduction potentials of the different components of the reaction you are given. So you need the reduction potential of Ag+ which is Ag+ + e- -> Ag(s) and has an E°=0.80 V, and the reduction potential of AgBr which is AgBr(s) + e- -> Ag + Br- and has E°...
Thu Feb 22, 2018 1:44 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: HW 14.37(C)
Replies: 2
Views: 211

### Re: HW 14.37(C)

You have Cl and H, and the H reaction will be 2H+ +2e- -> H2 with an E°=0V, since H is the standard electrode used to base other potentials off of. Since H is being reduced, the reaction for Cl will be 2Cl- -> Cl2 + 2e-, with an E°= 1.36V. The overall reaction will be 2H+ +2Cl- -> H2 + Cl2, and the ...
Thu Feb 22, 2018 1:32 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.13 d
Replies: 1
Views: 192

### Re: 14.13 d

When you look at the reduction potentials for Au+ and Au3+, you see that Au+ + e- -> Au has E°=1.69 V, and Au3+ + 3e- -> Au has E°=1.40 V. Since 1.40<1.69, this will be the anode, and the reaction must be flipped, so you'll have Au -> Au3+ + 3e-. The cathode reaction will be the same, except it will...
Thu Feb 08, 2018 8:40 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: delta G formation
Replies: 2
Views: 221

### Re: delta G formation

The ΔGf would also be 0 because there is no change between the reactants and products, and H2 is in its most stable form. If you know ΔHf is 0, and ΔGf is 0, then you can also conclude that ΔSf would also be 0 (using ΔG=ΔH-TΔS). Hope this helps!
Thu Feb 08, 2018 8:34 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Circumstances of delta G
Replies: 3
Views: 206

### Re: Circumstances of delta G

ΔG=0 when the system is at equilibrium. For instance, if ice and water in equilibrium then the Gibbs free energy of 1 mol H2O(l) will be the same as the Gibbs free energy of 1 mol H2O(s). ΔS=0 is also equilibrium. If ΔS=0, it means that neither the forward nor the reverse reaction/process is spontan...
Thu Feb 08, 2018 8:24 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 9.65
Replies: 3
Views: 176

### Re: 9.65

I could be wrong on this, but I think since they are only referring to a temperature change, they just want you to look at ΔS. In that case, we can assume ΔH is constant or has no effect on the ΔG. Hope this helps!
Thu Feb 08, 2018 1:49 pm
Forum: Van't Hoff Equation
Topic: van't Hoff Equation
Replies: 2
Views: 370

### Re: van't Hoff Equation

Hey! It doesn't matter if you calculate K1 or K2, as long as you keep it consistent. For instance, if the question gives you two temperatures, 298K and 345K, and you call those T1 and T2, then your K1 would be the value for the 298K and the K2 would be for 345K. One of those K values would be unknow...
Sun Feb 04, 2018 6:05 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Free Energy
Replies: 2
Views: 155

### Re: Free Energy

Pressure: When there is a certain pressure P, G= G° + RTln(P/P°). If there was no different pressure, then the equation would be G=G°, because the the ln(P/P°) term would be equal to 0 (since P/P° is 1 and ln(1)=0). When you have a different pressure, the RTln(P/P°) term contributes to the value of ...
Sun Feb 04, 2018 5:54 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Max work related to Gibbs Free Energy
Replies: 2
Views: 156

### Re: Max work related to Gibbs Free Energy

It is when there is constant temperature and pressure that ΔG is equal to the max nonexpansion work. Hope this helps!
Sun Feb 04, 2018 5:50 pm
Forum: Van't Hoff Equation
Topic: Van't Hoff Eq.
Replies: 2
Views: 166

### Re: Van't Hoff Eq.

The van't hoff equation is derived from setting the two expressions for ΔG equal to each other. The two expressions we have for ΔG are -RTlnK and ΔH-TΔS. If you set these equal and solve for lnK, you get lnK= -ΔH/RT + ΔS/R. This can be used to calculate the value of K at a different temperature if w...
Sun Jan 28, 2018 4:34 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Week 3 Friday Lecture
Replies: 2
Views: 152

### Re: Week 3 Friday Lecture

It said "therefore ΔHrxn plays an important role at low T" Hope this helps! :)
Sun Jan 28, 2018 4:33 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.11
Replies: 2
Views: 441

### Re: 9.11

The formula is ΔS=nRln(V2/V1). Since we have an ideal gas, the pressure is going to be the inverse of volume, so we can change the equation to be ΔS=nRln(P1/P2). Now, we plug in what we know to find the answer. ΔS=(1.5)(8.314)(ln(15.0/0.500))=42.4 J/K
Hope this helps!
Sun Jan 28, 2018 11:17 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.7
Replies: 2
Views: 169

### Re: 9.7

For part a, you want to use dS=dq/T. Since we know dq=nCdT, we can substitute it in to have dS=(nCdT)/T. When you integrate this, you get ΔS=nCln(T2/T1). From here, you can plug in everything to find ΔS. (for an ideal gas C=(5/2)R) You end up with ΔS= (1)(5/2(8.314))(ln(431/310.8))=6.80 J/K For part...
Sun Jan 14, 2018 5:23 pm
Forum: Phase Changes & Related Calculations
Topic: 8.45 part C
Replies: 2
Views: 212

### Re: 8.45 part C

I think that the ΔH they give you in the question is reaction enthalpy, because absorbing energy would be endothermic, and that is a property that refers to the entire reaction. As far as how you solve it, you want to end up with grams. First, you need to find the number of mol. You need both values...
Sun Jan 14, 2018 5:12 pm
Forum: Phase Changes & Related Calculations
Topic: Question 8.31
Replies: 1
Views: 121

### Re: Question 8.31

Hey! I am not too sure how this correlates to lecture, but I can help you breakdown the question. Since they tell you that krypton behaves as an ideal gas, you can use the heat capacity of ideal gases at constant pressure and temperature that they tell you in the book. The constant pressure one is 5...
Sun Jan 14, 2018 5:02 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Irreversible vs. Reversible [ENDORSED]
Replies: 10
Views: 744

### Re: Irreversible vs. Reversible[ENDORSED]

A reversible process is one that can be reversed by an infinitely small change in a variable. An irreversible process is one where a infinitely small change does not reverse the process of whatever is occurring. Page 265 in the textbook has more information and some good examples. Hope this helps!
Sun Jan 14, 2018 4:57 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond Enthalpies and the use of kJ
Replies: 1
Views: 117

### Re: Bond Enthalpies and the use of kJ

kJ/mol is used for enthalpy of formation, and is for one substance, and you are always considering kJ needed for 1 mol of that substance. kJ/C is used for heat capacity, since that is the energy (in the form of heat) needed to raise the temperature of an object by 1 degree C kJ is used for standard ...
Sun Jan 14, 2018 4:53 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Bond and Standard Formation Enthalpies
Replies: 1
Views: 116

### Re: Bond and Standard Formation Enthalpies

For standard formation enthalpies, the standard formation enthalpy of a substance is always per mol, so you have a value that is kJ/mol. To find standard reaction enthalpy, you need an answer that is in kJ. Thus, you multiply the standard formation enthalpy by n (the stoichiometric coefficient in th...
Sun Jan 14, 2018 4:45 pm
Forum: Phase Changes & Related Calculations
Topic: Constant temperature
Replies: 3
Views: 167

### Re: Constant temperature

Let's say we are talking about the process of melting. Once we hit the melting point of the substance, the molecules gain enough energy to move past one another rather than being locked in a solid form. All the added energy at this temperature is being used to overcome the attractive forces between ...
Thu Nov 30, 2017 1:15 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Dien
Replies: 1
Views: 185

### Re: Dien

If you draw out the structure, you see that the 3 nitrogens all have one lone pair. Since they have this lone pair, it makes them capable of bonding, making the overall ligand have 3 bonding sites. This makes the entire thing qualified as tridentate. The easiest way to determine this is to draw out ...
Thu Nov 30, 2017 1:08 pm
Forum: Lewis Structures
Topic: 4.7A Homework
Replies: 3
Views: 694

### Re: 4.7A Homework

For part b, the OSCL angles are all the same because the long pair repels them equally. For part c, since the shape is a trigonal pyramidal, the angles will be slightly less than 109.5 (106-107).
Thu Nov 30, 2017 1:03 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Changes in K
Replies: 4
Views: 234

### Re: Changes in K

The equilibrium constant is also unaffected by changes in pressure. As a rule, changes in concentration and pressure do not affect K, and changes in temperature do affect K. If the reaction is endothermic, production formation is favored, and if the reaction is exothermic, reactant formation is favo...
Thu Nov 30, 2017 12:57 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Exothermic vs. Endothermic?!
Replies: 5
Views: 2159

### Re: Exothermic vs. Endothermic?!

Endothermic reactions will have a positive ΔH, because heat is being added, and exothermic reactions will have a negative ΔH, since heat is being released. If you see a positive ΔH value associated with the reaction and the temperature is increased, this will favor the formation of products. If you ...
Thu Nov 30, 2017 12:47 pm
Forum: Sigma & Pi Bonds
Topic: Localization
Replies: 3
Views: 387

### Re: Localization

It is delocalized if there are other bonds near that pi bond because then the electrons will be spread out over all of the bonds and not specifically localized to that pi bond.
Thu Nov 30, 2017 12:36 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: Effects with Water
Replies: 2
Views: 210

### Re: Effects with Water

If the solution is in water (aka aqueous), then adding/taking away water doesn't affect it since water is technically on both sides of the reaction. Water will affect the reaction when it is in a gas phase, or is not present as the solvent. Also, make sure you check whether pressure or volume are ch...
Thu Nov 30, 2017 12:31 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.13
Replies: 3
Views: 192

### Re: 11.13

You can use the concentrations if you want to, but parts a and c have all reactants and products in the gas phase, so its easier to use pressure and not convert to concentration. For part b, since the components that are counted in Q are aqueous, it makes more sense and is faster to use concentratio...
Tue Nov 14, 2017 10:38 am
Forum: Determining Molecular Shape (VSEPR)
Topic: 4.43
Replies: 2
Views: 206

### Re: 4.43

The question is asking what will happen as s character is increased in the hybrid. The s character refers to the contribution of the sigma bond in the hybridization. As you go from sp3 to sp2 and so on, the s character increases. For example, in sp3, the makeup is 75% p character and 25% s character...
Tue Nov 14, 2017 10:31 am
Forum: Determining Molecular Shape (VSEPR)
Topic: 109.5 degrees
Replies: 3
Views: 309

### Re: 109.5 degrees

The bond angles have been experimentally determined, and you can use the VSEPR theory/model to coordinate the shape/geometry with the bond angle that has already been determined for that shape. Hope this helps!
Tue Nov 14, 2017 10:27 am
Forum: Dipole Moments
Topic: 4.25 part a
Replies: 2
Views: 224

### Re: 4.25 part a

The C-Cl bond is polar, and the C-H is almost nonpolar(it has small polarity, the importance is that it is different in polarity from C-Cl). Even if the arrangement is symmetrical, the different polarities of the C-H bond and the C-Cl bond will have an effect on the entire molecule and make the mole...
Tue Nov 14, 2017 10:21 am
Forum: Hybridization
Topic: Ch4 #43
Replies: 1
Views: 157

### Re: Ch4 #43

The s character refers to the contribution of the sigma bond in the hybridization. As you go from sp3 to sp2 and so on, the s character increases. For example, in sp3, the makeup is 75% p character and 25% s character, but in sp2 the makeup is only 60% p character and 33% s character. The more s cha...
Thu Nov 02, 2017 7:43 pm
Forum: Lewis Structures
Topic: Expanded rules
Replies: 4
Views: 194

### Re: Expanded rules

The reason why it is depends on the d orbital is because the d orbital being in the valence shell allows the atom to accommodate extra electrons that it otherwise would not be able to if it has a lower subshell. Hope this helps!
Thu Nov 02, 2017 7:41 pm
Forum: Lewis Structures
Topic: Expanded Octet
Replies: 3
Views: 194

### Re: Expanded Octet

Anything in period 3 or higher has the potential to have an expanded octet because the d orbitals can accommodate the extra electrons. The best way to determine if something has an expanded octet or not is by first looking at the period it is in and whether or not it has a d orbital, then counting t...
Thu Nov 02, 2017 7:37 pm
Forum: Octet Exceptions
Topic: F orbitals
Replies: 4
Views: 392

### Re: F orbitals

f orbitals come into play once you get to lanthanides and actinides, but I don't think you will need to know anything about how the f orbital works in lewis structures or bonding for chem 14A. However, they typically follow the same rules as s, p, and d block elements.
Thu Nov 02, 2017 7:29 pm
Forum: Ionic & Covalent Bonds
Topic: #25 on Chapter 3
Replies: 6
Views: 495

### Re: #25 on Chapter 3

You can use the oxidation numbers (charges) on the elements to find the subscript for the element it is paired with. For part a), Magnesium has an oxidation number of +2 and Arsenic has an oxidation number of -3, so you use the "criss cross" rule. The way this works is you have Mg^+2 and A...
Thu Oct 19, 2017 12:13 pm
Forum: Properties of Electrons
Topic: Rydberg's formula
Replies: 6
Views: 725

### Re: Rydberg's formula

Rydberg's equation is used when you want to find the wavelength, frequency, or energy of an electron moving from one level to another. Most of the time you will be given the starting and final energy levels (n1 and n2), but sometimes the question might ask you to solve for n1 or n2. In this case, ma...
Thu Oct 19, 2017 12:02 pm
Forum: Properties of Light
Topic: EM spectrum
Replies: 6
Views: 485

### Re: EM spectrum

You don't need to know specific wavelengths, but I think it is helpful to know the relative wavelengths of the whole spectrum. For example, gamma-rays or x-rays have shorter wavelengths than microwaves and etc.
Sat Oct 14, 2017 10:41 pm
Forum: Properties of Light
Topic: 1.15 [ENDORSED]
Replies: 2
Views: 236

### Re: 1.15[ENDORSED]

When you look at a digram of values, since the wavelength is 102.6 nm, it corresponds to the Lyman series. Anything in the Lyman series is going to have n=1 as the beginning energy level. Hope this helps!
Sat Oct 14, 2017 10:32 pm
Forum: Properties of Light
Topic: 1.3
Replies: 5
Views: 436

### 1.3

I understand most of this question, but I am a little confused on choice c which says "the extent of the change in the electrical field at a given point decreases". The answer key says the electrical field corresponds to amplitude, but how exactly are they related?
Thu Oct 05, 2017 9:32 pm
Forum: Significant Figures
Topic: Integers and Exact Numbers- Sig Figs [ENDORSED]
Replies: 4
Views: 495

### Re: Intergers and Exact Numbers- Sig Figs[ENDORSED]

I think this means that you don't use the integer value as your lowest number of sig figs. For example, if you are multiplying 3 by 5.5, you would look at 5.5 which has 2 sig figs, and not the 3 which only has one. Hope this helps!
Thu Oct 05, 2017 9:18 pm
Forum: Limiting Reactant Calculations
Topic: M. 15 Fundamentals
Replies: 4
Views: 510

### Re: M. 15 Fundamentals

To derive the formula for aluminum chloride, you can use the oxidation numbers for the elements involved. Aluminum has an oxidation number of 3 so you would write it as Al and then +3 as an exponent. Cl has an oxidation number of -1 so you would write it as Cl with -1 as the exponent. Now you have A...