Search found 51 matches

by Lena Nguyen 2H
Mon Mar 12, 2018 7:50 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Constant pressure and delta U
Replies: 1
Views: 115

Re: Constant pressure and delta U

Delta U = nCvdeltaT is used to find the heat at constant volume (w = 0), so the use of Cv is appropriate. Although this equation is used in part (b) with constant pressure, finding delta U is broken into two steps, one of which heats at constant volume.
by Lena Nguyen 2H
Mon Mar 12, 2018 7:23 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: problem 15.95
Replies: 1
Views: 78

Re: problem 15.95

The catalyst has no effect on the overall rate of the reaction because it's not increasing the rate of the slow step/rate-determining step, which is the first step. The overall rate of the reaction would increase if a catalyst was applied to the first step.
by Lena Nguyen 2H
Mon Mar 12, 2018 7:10 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: 15.61
Replies: 2
Views: 171

Re: 15.61

It is derived from the Arrhenius equation where one k = Ae^-Ea/RT is divided by k' = Ae^-Ea/RT'. Which k is on top shouldn't matter to get the right answer as long you are consistent with k' matching with T' and k matching with T. By dividing the equations, it gets rid of the A. k/k' = (e^-Ea/RT)/(e...
by Lena Nguyen 2H
Mon Mar 05, 2018 7:30 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: rate constant vs rates
Replies: 2
Views: 122

Re: rate constant vs rates

If the Arrhenius Equation is considered (k = Ae^Ea/RT), the concentration of the reactant doesn't appear and so would not affect the rate constant.
by Lena Nguyen 2H
Mon Mar 05, 2018 5:39 pm
Forum: First Order Reactions
Topic: 15.21 SOS!!!!!
Replies: 2
Views: 152

Re: 15.21 SOS!!!!!

The equation is [A]t / [A]0 = e^-kt. From ln [A]t = -kt + ln [A]0,

ln[A]t - ln [A]0 = -kt
ln [A]t / [A]0 = -kt
[A]t / [A]0 = e^-kt (Exponentiate both sides)
by Lena Nguyen 2H
Mon Mar 05, 2018 5:25 pm
Forum: Second Order Reactions
Topic: 15.19
Replies: 2
Views: 151

Re: 15.19

We know that the reaction in B is second-order because increasing B by a factor (3.02/1.25) increases the rate by a factor (3.02/1.25)^2.
by Lena Nguyen 2H
Mon Feb 26, 2018 10:19 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 14.47
Replies: 3
Views: 162

Re: 14.47

Your answer is correct, and the solutions manuals just rounds down to 10^6 to have one significant figure.
by Lena Nguyen 2H
Mon Feb 26, 2018 8:27 pm
Forum: General Rate Laws
Topic: Homework Problem 15.3
Replies: 4
Views: 267

Re: Homework Problem 15.3

The question asks for the rate of reaction of NO2 rather than the unique average rate of the reaction (which would divide the rate by 2). Therefore, the rate of reaction of NO2 is the change in NO2 over the change in time multiplied by -1.
by Lena Nguyen 2H
Mon Feb 26, 2018 6:39 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Value of k
Replies: 4
Views: 163

Re: Value of k

We haven't learned of a relationship between temperature and k, but k is always positive.
by Lena Nguyen 2H
Fri Feb 23, 2018 12:12 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: homework 14.51
Replies: 1
Views: 138

Re: homework 14.51

is 0 because the electrodes are both silver, but Ecell will be different because the concentrations are not equal and can be calculated using the Nernst Equation.
by Lena Nguyen 2H
Fri Feb 23, 2018 11:27 am
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Homework 14.41
Replies: 2
Views: 121

Re: Homework 14.41

Ecell isn't 0 because of the concentration difference, but Ecell* of the H+ half-reaction equals 0 by definition (also noted in the appendix).
by Lena Nguyen 2H
Thu Feb 22, 2018 8:27 pm
Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
Topic: 14.47
Replies: 1
Views: 128

Re: 14.47

Both the E values can be found by looking up those half-reactions in Appendix 2B.
by Lena Nguyen 2H
Mon Feb 19, 2018 10:33 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: free energy and standard free energy
Replies: 1
Views: 122

Re: free energy and standard free energy

Standard free energy is related to K by the equilibrium composition. If K is over 1, then the standard free energy should be negative as the forward reaction is favored. If K is less than 1, then the standard free energy should be positive as the reverse reaction is favored. Q is for reactions that ...
by Lena Nguyen 2H
Sun Feb 18, 2018 9:44 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.111
Replies: 3
Views: 167

Re: 11.111

To get rid of e, take the natural log of both sides to get ln 10 = (2.00 x 10^5 + deltaG2)/(8.3145 x 298.15) and solve for deltaG2 from there.
by Lena Nguyen 2H
Sat Feb 17, 2018 10:20 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.85
Replies: 1
Views: 92

Re: 14.85

In Appendix 2B, the half-reactions are listed from most strongly oxidizing to most strongly reducing, so half-reactions at the bottom of the right-hand side are the most strongly reducing. From there, the metals can be ordered by most strongly reducing.
by Lena Nguyen 2H
Sat Feb 17, 2018 9:34 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.5 part a
Replies: 1
Views: 109

Re: 14.5 part a

I don't think the order of ions or molecules matter as long as the anode is on the left side and the cathode is on the right side.
by Lena Nguyen 2H
Thu Feb 08, 2018 5:52 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.15
Replies: 8
Views: 476

Re: 11.15

Start with the equations and
Everything is given except Q, which can be calculated using the partial pressures given. The Gibbs free energy of the reaction can then be calculated.
Hope this helps!
by Lena Nguyen 2H
Wed Feb 07, 2018 10:48 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.103
Replies: 4
Views: 166

Re: 9.103

You wouldn't be able to tell with complete certainty just by looking at it. Calculating the standard Gibbs free energy of the reaction using standard Gibbs free energy of formations would be a way to tell if the reaction is spontaneous.
by Lena Nguyen 2H
Wed Feb 07, 2018 8:22 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 11.83
Replies: 1
Views: 89

Re: 11.83

It's correct that using the standard Gibbs free energy of formations wouldn't work for 150 C because Gibbs free energy is sensitive to temperature changes. Since the standard enthalpy of reaction and standard entropy of reaction are mostly independent of temperature, they are the same at both temper...
by Lena Nguyen 2H
Wed Jan 31, 2018 7:58 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: 9.35
Replies: 1
Views: 116

Re: 9.35

The solutions manual says Container A has higher entropy because it has more particles, though each container has 1.00 mol. I think containers B and C had .5 mol of diatomic molecules because the question said they had 1.00 mol of atoms bound together, so that could explain the solutions manual's an...
by Lena Nguyen 2H
Tue Jan 30, 2018 8:38 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.11
Replies: 4
Views: 170

Re: 9.11

You're correct about the initial and final pressures, but the equation is deltaS = nRln(P1/P2) according to the solutions manual.
by Lena Nguyen 2H
Mon Jan 29, 2018 8:17 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Kb=?
Replies: 2
Views: 115

Re: Kb=?

by Lena Nguyen 2H
Sun Jan 28, 2018 12:32 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Increasing temperature to make G negative
Replies: 3
Views: 195

Re: Increasing temperature to make G negative

The example 9.16 refers to increasing the temperature to make the reaction spontaneous, not exothermic. However, raising the temperature will not always make an endothermic reaction spontaneous. Since \Delta G^{o} = \Delta H^{o} - T\Delta S^{o} , if the endothermic reaction has a positive change in ...
by Lena Nguyen 2H
Sat Jan 27, 2018 9:55 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Spontineity
Replies: 5
Views: 166

Re: Spontineity

Spontaneous reactions are reactions that tend to occur without external influence as in by itself. Dr. Lavelle avoids saying disorder for entropy.
by Lena Nguyen 2H
Sat Jan 27, 2018 8:45 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.13 Entropy Change
Replies: 2
Views: 96

Re: 9.13 Entropy Change

The problem doesn't give us the number of moles, but the solutions manual says to assume n = 1 mol.
by Lena Nguyen 2H
Thu Jan 18, 2018 9:12 pm
Forum: Phase Changes & Related Calculations
Topic: 8.99
Replies: 1
Views: 91

Re: 8.99

The solution uses the given volume and density (same as water so 1g/ml) to find the mass.
Mass = Volume * Density = 800ml * 1g/ml = 800g
by Lena Nguyen 2H
Thu Jan 18, 2018 8:16 pm
Forum: Phase Changes & Related Calculations
Topic: Self Test Example 8.1A
Replies: 1
Views: 111

Re: Self Test Example 8.1A

Given the mass and density, we can use the equation volume = mass/density for both water and ice to figure out the change in volume. From there, work can be calculated.
by Lena Nguyen 2H
Wed Jan 17, 2018 9:14 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: 8.1 question
Replies: 8
Views: 286

Re: 8.1 question

Thermos bottles are supposed to be insulated to prevent heat from escaping. The book mentions that thermos bottles are isolated systems on p. 260.
by Lena Nguyen 2H
Sat Jan 13, 2018 2:53 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: 8.53
Replies: 2
Views: 104

Re: 8.53

The solution involves using the heat capacity of the calorimeter, so we would use , which doesn't require the mass of the calorimeter.
by Lena Nguyen 2H
Thu Jan 11, 2018 9:22 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Example 8.6
Replies: 1
Views: 108

Re: Example 8.6

For part (a), the first formula is the equation q = n\Delta TC_{V, m} from the textbook but modified so that \Delta T is on its own side. From \Delta T = T_{f} - T_{i} , the final temperature is calculated. \Delta U = q at constant volume because no expansion work is done (assuming no nonexpansion w...
by Lena Nguyen 2H
Tue Jan 09, 2018 9:08 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat vs Energy
Replies: 2
Views: 141

Re: Heat vs Energy

They are similar, but I don't think they can be used interchangeably. Heat is a specific type of energy, one that can be transferred by a temperature difference, which is why they have the same units. A change in energy can also come from work being done, but work is not heat. When using heat capaci...
by Lena Nguyen 2H
Tue Dec 05, 2017 11:46 pm
Forum: Acidity & Basicity Constants and The Conjugate Seesaw
Topic: Bond Length
Replies: 1
Views: 120

Re: Bond Length

Te is a larger atom, so the H bonded to it would be farther away, creating a weaker bond.
by Lena Nguyen 2H
Tue Dec 05, 2017 12:02 pm
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 12.55 (d)
Replies: 1
Views: 102

12.55 (d)

I didn't understand the solutions manual explanation for HCOOH and its electron withdrawal ability for part d on 12.55. Can someone explain it?

Thanks.
by Lena Nguyen 2H
Tue Nov 28, 2017 11:13 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Reaction Shift
Replies: 1
Views: 111

Re: Chemical Reaction Shift

The shifting to the right or left still refers to the equilibrium constant. The reaction shifts to the right if K is greater than 1 x 10^3 and the reaction shifts to the left if K is less than 1 x 10^-3. A large K means that there are more products than reactants at equilibrium, which is why the rea...
by Lena Nguyen 2H
Tue Nov 28, 2017 9:24 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Question 11.45 (c)
Replies: 1
Views: 136

Re: Question 11.45 (c)

F2 having a larger equilibrium constant means that more product is formed compared to reactant in F2 than Cl2, showing that the dissociation of F2 is easier than the dissociation of Cl2. Since Cl2 dissociates less frequently than F2, it is more stable.
by Lena Nguyen 2H
Sat Nov 25, 2017 12:44 am
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: 4.75 Part A
Replies: 1
Views: 250

Re: 4.75 Part A

When you multiplied the molar mass of 32.04 g/mol by the percentage of carbon, hydrogen, and oxygen, the numbers you get are the amount in grams of each element. To get the number of moles, you have to divide the grams by the molar masses of each element. Then you should get the correct molecular fo...
by Lena Nguyen 2H
Sat Nov 25, 2017 12:37 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Homework 17.37 Part d
Replies: 1
Views: 115

Re: Homework 17.37 Part d

EDTA is said to be one of few ligands that are hexadentate on pg. 744 of the textbook, but it doesn't mention an explanation. I'm not sure if EDTA as hexadentate is just something we have to know.
by Lena Nguyen 2H
Tue Nov 14, 2017 11:15 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Chapter 4 #13 Part D
Replies: 1
Views: 132

Re: Chapter 4 #13 Part D

One of the two possible answers does have a triple bond between both nitrogen atoms. I think the other one is included because if the formal charges are calculated, both structures have an atom with a +1 charge and an atom with a -1 charge, making them similar in lowest energy.
by Lena Nguyen 2H
Mon Nov 13, 2017 10:16 pm
Forum: Hybridization
Topic: Ch 4 #95
Replies: 2
Views: 132

Re: Ch 4 #95

The first two means the s and p orbitals were from the second energy level (as in n =2).
by Lena Nguyen 2H
Thu Nov 09, 2017 5:24 pm
Forum: DeBroglie Equation
Topic: E=hv vs. Work Function?
Replies: 2
Views: 651

Re: E=hv vs. Work Function?

The equation is kinetic energy = photon energy + work function. If you're finding the energy of the photon given frequency or wavelength, you would use E = hv. For electrons given velocity or kinetic energy, you would use DeBroglie's equation. The work function is the threshold energy, which you wou...
by Lena Nguyen 2H
Tue Nov 07, 2017 8:30 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Resonance Structure
Replies: 4
Views: 191

Re: Resonance Structure

Resonance structures are when the location of double bonds or triple bonds can be drawn in different locations while still being the same molecule. An example would be (NO3)- can have its double bond on any of the oxygen. This is just for Lewis structures, though. In reality, the molecule will be a ...
by Lena Nguyen 2H
Fri Nov 03, 2017 11:55 pm
Forum: Ionic & Covalent Bonds
Topic: Ionic or Covalent?
Replies: 4
Views: 213

Re: Ionic or Covalent?

Ionic or covalent bonds are decided by the differences in electronegativity. Bonds have partial ionic and covalent character, but if the the difference in electronegativity is greater than 2, it is considered ionic. If the difference is less than 1.4, it is considered covalent. If the charges betwee...
by Lena Nguyen 2H
Thu Nov 02, 2017 10:46 pm
Forum: Resonance Structures
Topic: Lewis Structures: Resonance Hybrid
Replies: 2
Views: 202

Re: Lewis Structures: Resonance Hybrid

Yes, the formal charges are used. If you calculate the formal charges for the atoms in part (a), the first structure of Xe and F is preferred because they are both 0.
by Lena Nguyen 2H
Wed Oct 25, 2017 7:43 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: 4s and 3d
Replies: 2
Views: 152

Re: 4s and 3d

4s is only lower than 3d when 3d doesn't have electrons. Once electrons enter the 3d orbitals, 3d is lower in energy than 4s. So an electron would be removed from 4s before 3d.
by Lena Nguyen 2H
Mon Oct 23, 2017 10:27 pm
Forum: Trends in The Periodic Table
Topic: 2.67 part B
Replies: 4
Views: 233

Re: 2.67 part B

For carbon and nitrogen, we would need to consider their electron configurations. Adding an electron to carbon would be adding it to an empty p-orbital, but for nitrogen, adding an electron means adding it to a p-orbital with an electron already inside it. This is more difficult because of electron ...
by Lena Nguyen 2H
Sat Oct 21, 2017 4:32 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: Principle Quantum Number [ENDORSED]
Replies: 3
Views: 197

Re: Principle Quantum Number [ENDORSED]

The principle quantum number is the energy level of the electron, denoted by n (which is the same n as in the Rydberg Equation). If the energy level goes up, n also goes up.
by Lena Nguyen 2H
Wed Oct 18, 2017 9:07 pm
Forum: DeBroglie Equation
Topic: Difference longest Wavelength and wavelength
Replies: 2
Views: 216

Re: Difference longest Wavelength and wavelength

When you find the longest possible wavelength, you will use 0 for the kinetic energy and then figure out the wavelength normally. For finding a wavelength, it's similar but there is usually other information, such as the velocity of the ejected electron, which would give you the kinetic energy. So i...
by Lena Nguyen 2H
Wed Oct 11, 2017 4:34 pm
Forum: Photoelectric Effect
Topic: Photoelectric Effect
Replies: 8
Views: 355

Re: Photoelectric Effect

In the photoelectric experiment, the electron is completely removed or ejected from the metal. In this case, the energy the electron is hit with only has to be greater than the threshold energy and not an specific amount. This experiment wasn't about moving electrons to higher energy levels, it was ...
by Lena Nguyen 2H
Wed Oct 11, 2017 4:18 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Negative Energy
Replies: 1
Views: 137

Re: Negative Energy

The negative energy when the electron is descending is the energy it's emitting as a photon.
by Lena Nguyen 2H
Wed Oct 04, 2017 10:36 pm
Forum: Limiting Reactant Calculations
Topic: Page F109 from book (Self-Test M.2A)
Replies: 2
Views: 181

Re: Page F109 from book (Self-Test M.2A)

To identify the limiting reactant, you need to use the amounts of each reactant in moles. So converting each reactant to moles, (5.52g Na)(1 mol/22.99g Na) = .24 mol Na (5.1g Al2O3)(1 mol/101.957g Al2O3) = .05 mol Al2O3 For every 1 mol of Al2O3, you need 6 mol of Na to complete the reaction because ...
by Lena Nguyen 2H
Tue Oct 03, 2017 5:07 pm
Forum: Empirical & Molecular Formulas
Topic: Problem E5
Replies: 6
Views: 1119

Re: Problem E5

Your answers are correct, and you can check the answers to any odd problems in the back of the book or in the solutions manual.

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