## Search found 52 matches

Sat Mar 17, 2018 6:24 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: delta G = 0 phase change
Replies: 1
Views: 1205

### Re: delta G = 0 phase change

During a phase change, equilibrium exists between the phases. For example, during the phase change between liquid and gas, the reaction does not "favor" either phase, which means that both liquid and gas molecules will exist. Once you lower the temperature an infinitesimal amount, it will ...
Fri Mar 16, 2018 10:21 pm
Forum: First Order Reactions
Topic: Reaction speed
Replies: 1
Views: 203

### Re: Reaction speed

Yes because rate = k[A]
all other factors being equal, a higher rate constant means a higher speed because rate = mol/L/s
this applies to all of the other rate laws too!
Fri Mar 16, 2018 10:19 pm
Forum: General Science Questions
Topic: Grade
Replies: 8
Views: 844

### Re: Grade

Lavelle sets the grading scale at the end of the quarter, so you can't calculate your letter grade right now based on the points you have in the class. In any case, the scale he sets won't end up hurting you.
Fri Mar 16, 2018 10:16 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: sign of k
Replies: 3
Views: 495

### Re: sign of k

It's all about the wording. For example, the rate of decomposition is positive (even though its concentration is decreasing) because it is the rate at which it is disappearing. This is why we have to take the negative of the d[A]/dt for the rate in A --> B.
Fri Mar 16, 2018 10:13 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: HW #15.85 b
Replies: 1
Views: 125

### Re: HW #15.85 b

Ar is a catalyst; it appears as a reactant and a product, which means it cancels out and it isn't used in the reaction. Since catalysts are involved in the interactions between reactants, they can be included in the rate law, but not in the overall balanced chemical equation.
Sat Mar 10, 2018 2:36 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Activated Complexes [ENDORSED]
Replies: 1
Views: 118

### Re: Activated Complexes[ENDORSED]

Using c as an example: (c) Rate = k[O2][NO] (Products are NO2 and O.) Since you are given the rate law, you know that the reaction depends on the interaction between o2 and NO (you will draw this in your activated complex). You are also given the products, which allows you to visualize which bonds a...
Sat Mar 10, 2018 2:29 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 15.67
Replies: 1
Views: 128

### Re: 15.67

In the problem it says "all other factors being equal," which means that the only thing changing is the rate constant. The concentrations are kept the same. This makes the changes in the rate constant directly proportional to changes in the rate.
Sat Mar 10, 2018 2:27 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Elementary Reaction Rate Law [ENDORSED]
Replies: 1
Views: 89

### Re: Elementary Reaction Rate Law[ENDORSED]

You have to identify the rate determining step, as this is the one that will not have a reverse reaction. Write the rate for the forward reaction for each step, and determine which one looks closest to the experimentally determined rate law. This is usually the rate-determining step. Then, you assum...
Sat Mar 03, 2018 4:46 pm
Forum: General Rate Laws
Topic: 15.13 part a?
Replies: 3
Views: 198

### Re: 15.13 part a?

They're probably referring to the overall order of the reaction. Since there are two reactants, and each is first order, the overall order = 1+1 = 2.
The rate constant's units depend on the overall order.
Sat Mar 03, 2018 4:43 pm
Forum: Zero Order Reactions
Topic: Dependence on concentration
Replies: 2
Views: 168

### Re: Dependence on concentration

If you graph the concentration vs time of a zero-order reaction, you get a straight line. Since Δconcentration/Δtime = rate, an increased concentration has no effect on the rate, since the slope is constant, even though the concentration changes over time. However, if you graph a first order reactio...
Sat Mar 03, 2018 4:35 pm
Forum: General Rate Laws
Topic: 15.17
Replies: 1
Views: 141

### Re: 15.17

In class, Lavelle said that you can use any experiment to plug the values in and find the rate constant. I personally used the last reaction, and got the same answer as the solutions manual.
Thu Mar 01, 2018 12:40 am
Forum: Method of Initial Rates (To Determine n and k)
Topic: 15.19 Answers
Replies: 1
Views: 136

### Re: 15.19 Answers

I am confused about this as well. I am thinking it may have something to do with the mmol^4, as this may mean that you are multiplying by a factor of 10^-3 4 times.
Thu Mar 01, 2018 12:39 am
Forum: General Rate Laws
Topic: Deriving integrated rate laws
Replies: 4
Views: 228

### Re: Deriving integrated rate laws

You need to know the differential rate laws (rate in terms of concentration) and the integrated rate laws (rate in terms of conc+time). The derivations are really helpful to conceptually understand going from the differential to integrated, but both are provided on the constants and equations sheets...
Thu Mar 01, 2018 12:37 am
Forum: First Order Reactions
Topic: First Order Graph
Replies: 11
Views: 497

### Re: First Order Graph

If you were to graph just concentration vs time, it would curve downwards, like an exponential decay function. However, if you graph ln of the concentration vs time, then it would be a straight line with a negative slope. This is because taking the log of an exponential is essentially cancelling out...
Sat Feb 17, 2018 2:23 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Periodic Trends
Replies: 1
Views: 100

### Re: Periodic Trends

I think the trend loosely follows electron affinity trends, but there are exceptions. Elements with greater electron affinity seem to generally have a more positive reduction potential because they are more likely to accept electrons.
Sat Feb 17, 2018 2:18 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Electrodes in Cell Diagrams
Replies: 2
Views: 104

### Re: Electrodes in Cell Diagrams

Platinum is the most common, but you can use any inert electrode in this situation (it shouldn't participate in the reaction, it should just conduct).
Sat Feb 17, 2018 2:17 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.13, 14.15
Replies: 1
Views: 102

### Re: 14.13, 14.15

Since #13 already gives you the reaction in the correct direction, you can break it down into half reactions without needing the appendix. But in #15, you have to determine which half reaction to use on the cathode/anode side, so you need to reference the reduction potentials in the appendix. This w...
Sat Feb 10, 2018 4:11 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Problem 9.63
Replies: 1
Views: 110

### Re: Problem 9.63

Yes, it is asking if the Gibbs free energy of its decomposition (breaking up into its elemental parts) is positive. This means that the compound does not spontaneously decompose at room temperature, as spontaneously decomposing would make it unstable. Stable implies that nothing will happen to the c...
Sat Feb 10, 2018 4:07 pm
Forum: Calculating Work of Expansion
Topic: 9.47
Replies: 1
Views: 139

### Re: 9.47

Since it is isothermal, delta H = 0. Since it is free expansion, there is no work being done, as there is no external pressure, which makes P delta V = 0. Since delta U = delta H + P delta V, delta U = 0 as well.
Sat Feb 10, 2018 4:05 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: 8.93
Replies: 3
Views: 154

### Re: 8.93

Water isn't counted into the moles in this situation, since you only plug in the net moles of gas. This is because expansion work is only calculated using moles of gas, since liquids and solids hardly change in volume as a result of expansion.
Sat Feb 03, 2018 3:32 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Entropy Changes Due to Change in Pressure
Replies: 2
Views: 113

### Re: Entropy Changes Due to Change in Pressure

ΔS = nR ln (V1/V2)
using boyle's law for ideal gases:
P1V1 = P2V2
P1/P2 = V2/V1
replacing this in the equation:
ΔS = nR ln (P1/P2)
Sat Feb 03, 2018 3:30 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Universe [ENDORSED]
Replies: 3
Views: 203

### Re: Universe[ENDORSED]

The universe is an isolated system because it cannot exchange heat or matter with its surroundings, as technically it has no "boundary" where you could distinguish external surroundings.
Sat Feb 03, 2018 3:23 pm
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Absolute and statistical entropy
Replies: 2
Views: 180

### Re: Absolute and statistical entropy

I think that absolute entropy is measured relative to temperature in Kelvin, while statistical entropy uses Boltzmann's equations and the existence of microstates. Since entropy based on temperature change is derived using a form of Boltzmann's equation along with the idea of internal energy, I woul...
Sat Jan 27, 2018 11:13 am
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: 125

### Degeneracy related to volume

In lecture, I was wondering why Lavelle replaced
ΔS = kB ln W2/W1 with
ΔS = kB ln V2/V1? I think I may have missed the explanation.
Sat Jan 27, 2018 11:10 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Pressure Internal vs. External
Replies: 2
Views: 396

### Pressure Internal vs. External

Could someone explain why we can replace PΔV = ΔnRT for P(ex)Δv in the equation for work at a constant pressure? For example, say you had a balloon with an ideal gas expanding to twice its volume at an unstated, constant external pressure. The pressure inside the balloon decreases as volume increase...
Sat Jan 27, 2018 11:03 am
Forum: Calculating Work of Expansion
Topic: Irreversible Work Chart
Replies: 5
Views: 217

### Re: Irreversible Work Chart

In lecture, Lavelle said to ignore the vertical brown line. I think it was drawn to compare the irreversible and reversible reactions. But, the external pressure was always constant, it didn't drop down all of a sudden.
Sat Jan 20, 2018 3:57 pm
Forum: Phase Changes & Related Calculations
Topic: Temperature [ENDORSED]
Replies: 2
Views: 124

### Re: Temperature[ENDORSED]

For that problem, since you are using the CHANGE in temperature, your answer will be the same whether you use celsius or kelvin. This is because 0 C = 273.15 K, so its just a transformation of the temperature scale.
Sat Jan 20, 2018 3:52 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.45 question
Replies: 4
Views: 186

### Re: 8.45 question

Forming a bond creates a more stable molecule than the individual atoms alone, since they have completed their octets. This is why it takes energy to break bonds (you are forcing something from more stable to more unstable) and releases energy to form bonds (you are forming a more stable complex).
Sat Jan 20, 2018 2:08 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.45
Replies: 1
Views: 71

### 8.45

For 8.45 part b, why can we use the standard reaction enthalpy to calculate the heat absorbed when the textbook explicitly states that carbon is not in its pure form? I thought that by definition, standard reaction enthalpy means that this is the enthalpy for when reactants in their standard states ...
Sun Jan 14, 2018 2:42 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Systems
Replies: 3
Views: 242

### Re: Systems

An isolated system is coffee in a sealed thermos because neither energy nor mass (coffee) can leave/enter. A closed system is coffee in a sealed cup, because it can exchange energy in the form of heat with the surroundings, but it still cannot exchange mass.
Sun Jan 14, 2018 2:39 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Standard Reaction Enthalpy
Replies: 1
Views: 107

### Standard Reaction Enthalpy

Does anyone have an example of a chemical equation for which the reaction enthalpy is not the standard reaction enthalpy because one of the reactants/products isn't in standard state? I know of examples in which the phase is not the standard one at that temperature, but what about a molecule?
Sun Jan 14, 2018 2:35 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Irreversible vs. Reversible Processes
Replies: 3
Views: 208

### Re: Irreversible vs. Reversible Processes

I think the pressure is changing for isothermal expansion of an ideal gas. That's only one type of reversible process, so it's not necessarily a rule.
Thu Dec 07, 2017 2:16 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 12.115
Replies: 4
Views: 237

### Re: 12.115

Nitrous acid has a pKa of 3.16, so the pKb of its conjugate base must be 14-3.16 = 10.84 Acetic acid has a pKa of 4.75, so the pKb of its conjugate base must be 14-4.75 = 9.25 Since 10.84 > 9.25, and a large pKb corresponds to a weaker base, the conjugate base of nitrous acid is weaker. This can als...
Thu Dec 07, 2017 1:59 pm
Forum: Bronsted Acids & Bases
Topic: Determining the Stronger Acid
Replies: 2
Views: 233

### Re: Determining the Stronger Acid

An oxoacid will more readily lose H+ if the resulting anion is stabilized. This occurs when electronegative atoms that are part of the molecule delocalize the negative charge on one atom, as this makes the anion more stable overall. For this problem, since Cl is more electronegative than Br, it has ...
Fri Dec 01, 2017 6:17 pm
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Resonance and Polydentates
Replies: 1
Views: 166

### Re: Resonance and Polydentates

I think polydentate depends on lone pairs, and whether the molecule is large enough for it to bind at two sites.
Fri Dec 01, 2017 6:15 pm
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 11.89 part b
Replies: 2
Views: 231

### Re: 11.89 part b

You have to convert from kPa to bar, and 100kPa = 1 bar, so each value has to be divided by 100.
Wed Nov 29, 2017 2:38 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 11.89 Part b
Replies: 2
Views: 260

### Re: 11.89 Part b

Kp is calculated using the unit "bar", so you have to convert kPa to bar.
100 kPa = 1 bar, so you have to divide each one by 100.
Wed Nov 29, 2017 2:35 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Mole Ratios in ICE Tables
Replies: 2
Views: 847

### Re: Mole Ratios in ICE Tables

In the "change" row of your ICE chart, you have to use the stoichiometric coefficients of the chemical equation to represent the change. For example, if your equation was 3H2 + N2 --> 2NH3 and you had the following initial values, your table would look like this: I 0.4 0.4 0 C -3x -x +2x E...
Mon Nov 20, 2017 9:46 pm
Forum: Hybridization
Topic: lone pair effect on hybridization
Replies: 3
Views: 200

### Re: lone pair effect on hybridization

Usually, a lone pair means that one of the hybrid orbitals is filled with a pair of paired electrons, while the others have unpaired electrons.
Mon Nov 20, 2017 9:43 pm
Forum: Hybridization
Topic: Homework question #43
Replies: 2
Views: 202

### Re: Homework question #43

You would expect the bond angle to increase. This is because of the trend in bond angles in these hybrids: sp - 180 sp2 - 120 sp3 - 109.5 As the number of p orbitals increases, the "p-character" of the hybrids increases, and therefore the "s-character" decreases. So as the s char...
Fri Nov 10, 2017 10:41 pm
Forum: Dipole Moments
Topic: Drawing dipole moments [ENDORSED]
Replies: 3
Views: 356

### Re: Drawing dipole moments[ENDORSED]

In lecture, Dr. Lavelle said that we should draw the dipole moment pointing towards the negative charge, but the book says the opposite. Which one is the proper convention?
Fri Nov 10, 2017 10:36 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Drawing lewis structures
Replies: 5
Views: 340

### Re: Drawing lewis structures

Usually, if the best structure is still unable to make all the atoms have a formal charge of 0, you draw the "next best" possible Lewis Structures, since these will contribute to the structure. Most of the time, if all atoms have a formal charge of 0, you only need to draw that one. There ...
Fri Nov 03, 2017 5:35 pm
Forum: Ionic & Covalent Bonds
Topic: Question 3.27
Replies: 2
Views: 214

### Re: Question 3.27

Just to clarify, the reason you cross the charges is because in order to obtain the chemical formula for the neutral compound, the charges have to cancel. So for c for example, it takes 3 Li + ions to cancel the 3- charge of one Nitrogen atom. The chemical formula is always the lowest amount of atom...
Fri Nov 03, 2017 5:13 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Molecular Shape and Structure
Replies: 1
Views: 148

### Re: Molecular Shape and Structure

Since it takes less energy to remove electrons from the atom with the lower ionization energy, it has a weaker hold on its electrons than the other atoms. Thus, it is more "willing" to share its electrons. Since the central atom forms the most bonds in a molecule, it makes sense for the at...
Wed Oct 25, 2017 11:04 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: What exactly is an excited state? [ENDORSED]
Replies: 3
Views: 300

### Re: What exactly is an excited state?[ENDORSED]

The excited state of an atom is a state in which all the electrons are NOT in their lowest possible energy levels. Most commonly, this means that the electron has jumped from its ground state to a higher energy level. But it can also include differing spins, or different placement of the electrons. ...
Wed Oct 25, 2017 10:49 pm
Forum: Trends in The Periodic Table
Topic: Electron Affinity and Ionization Energy
Replies: 4
Views: 452

### Re: Electron Affinity and Ionization Energy

Trends in ionization energy: 1. increases down a period (as the atomic number increases, the effective nuclear charge increases on each electron, making them more difficult to remove) 2. decreases down a group (as n increases, the valence electrons get farther away from the nucleus, making the elect...
Thu Oct 19, 2017 12:07 pm
Forum: *Shrodinger Equation
Topic: Atomic Orbitals and Energy [ENDORSED]
Replies: 3
Views: 319

### Re: Atomic Orbitals and Energy[ENDORSED]

As you move away from the nucleus, the energy level increases, so the electrons in orbitals farther away from the nucleus have greater energy. However, I'm not able to understand what "farther away" really means, since all of the diagrams of the orbitals show that they are centered around ...
Thu Oct 19, 2017 11:47 am
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Probability Density
Replies: 1
Views: 210

### Probability Density

The book mentions that the probability of finding an electron is greatest where the undulations of the wave are the greatest. I understand this mathematically (wave function squared in this regions has the highest value), but conceptually, can someone explain why this occurs?
Fri Oct 13, 2017 4:46 pm
Forum: DeBroglie Equation
Topic: Deriving the function [ENDORSED]
Replies: 5
Views: 480

### Re: Deriving the function[ENDORSED]

Yes, De Broglie's equation applies to everything that has momentum (so your particle must have mass and velocity). I think it was derived by relating quantum mechanics to classical mechanics. So, E = mc^2 and E = hf E = mc^2 = hf **set equations equal to one another** E = mc^2 = h(c/λ) **substitute ...
Fri Oct 13, 2017 4:29 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Spectroscopy - Balmer Series
Replies: 1
Views: 188

### Spectroscopy - Balmer Series

Lavelle mentioned that the Balmer Series is for changes in energy levels from or to n = 2, and that the change between n = 2 and n = 1 emits/absorbs UV light (since this is Lyman Series and requires more energy). Does this mean that if you were to take a Hydrogen atom in its ground state and subject...
Thu Oct 05, 2017 4:17 pm
Forum: Limiting Reactant Calculations
Topic: Problem M11 part a
Replies: 5
Views: 445

### Re: Problem M11 part a

I think the problem with combining both equations is that you end up deciding between P4 or O2 as the limiting reactants for the production of P4O10, when really you should be deciding between P4O6 and O2. Also, you assume that all of the given amount of oxygen is used in the production of P4O10, so...
Thu Oct 05, 2017 3:46 pm
Forum: SI Units, Unit Conversions
Topic: Sig Figs [ENDORSED]
Replies: 16
Views: 2073

### Re: Sig Figs[ENDORSED]

I think that the problems are computed using whichever values are in the given periodic table. So if your periodic table says 14.007, the answer key probably uses that value to find their final answer. This means that if you were to round, you might end up with a slightly different value for the las...

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