Search found 51 matches
- 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: 1191
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.
- Mon Mar 12, 2018 7:23 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: problem 15.95
- Replies: 1
- Views: 287
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.
- Mon Mar 12, 2018 7:10 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.61
- Replies: 2
- Views: 436
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...
- 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: 378
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.
- Mon Mar 05, 2018 5:39 pm
- Forum: First Order Reactions
- Topic: 15.21 SOS!!!!!
- Replies: 2
- Views: 476
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)
ln[A]t - ln [A]0 = -kt
ln [A]t / [A]0 = -kt
[A]t / [A]0 = e^-kt (Exponentiate both sides)
- Mon Mar 05, 2018 5:25 pm
- Forum: Second Order Reactions
- Topic: 15.19
- Replies: 2
- Views: 396
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.
- 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: 549
Re: 14.47
Your answer is correct, and the solutions manuals just rounds down to 10^6 to have one significant figure.
- Mon Feb 26, 2018 8:27 pm
- Forum: General Rate Laws
- Topic: Homework Problem 15.3
- Replies: 4
- Views: 735
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.
- Mon Feb 26, 2018 6:39 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Value of k
- Replies: 4
- Views: 533
Re: Value of k
We haven't learned of a relationship between temperature and k, but k is always positive.
- 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: 332
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.
- Fri Feb 23, 2018 11:27 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homework 14.41
- Replies: 2
- Views: 426
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).
- 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: 380
Re: 14.47
Both the E values can be found by looking up those half-reactions in Appendix 2B.
- 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: 325
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 ...
- Sun Feb 18, 2018 9:44 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.111
- Replies: 3
- Views: 1669
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.
- Sat Feb 17, 2018 10:20 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.85
- Replies: 1
- Views: 374
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.
- 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: 267
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.
- Thu Feb 08, 2018 5:52 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.15
- Replies: 12
- Views: 3127
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!
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!
- Wed Feb 07, 2018 10:48 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.103
- Replies: 4
- Views: 564
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.
- Wed Feb 07, 2018 8:22 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 11.83
- Replies: 1
- Views: 290
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...
- 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: 254
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...
- Tue Jan 30, 2018 8:38 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.11
- Replies: 4
- Views: 596
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.
- Mon Jan 29, 2018 8:17 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Kb=?
- Replies: 2
- Views: 335
- Sun Jan 28, 2018 12:32 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Increasing temperature to make G negative
- Replies: 3
- Views: 496
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 ...
- Sat Jan 27, 2018 9:55 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Spontineity
- Replies: 5
- Views: 679
Re: Spontineity
Spontaneous reactions are reactions that tend to occur without external influence as in by itself. Dr. Lavelle avoids saying disorder for entropy.
- Sat Jan 27, 2018 8:45 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.13 Entropy Change
- Replies: 2
- Views: 342
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.
- Thu Jan 18, 2018 9:12 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.99
- Replies: 1
- Views: 221
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
Mass = Volume * Density = 800ml * 1g/ml = 800g
- Thu Jan 18, 2018 8:16 pm
- Forum: Phase Changes & Related Calculations
- Topic: Self Test Example 8.1A
- Replies: 1
- Views: 273
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.
- Wed Jan 17, 2018 9:14 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: 8.1 question
- Replies: 8
- Views: 785
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.
- Sat Jan 13, 2018 2:53 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.53
- Replies: 2
- Views: 1960
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.
- Thu Jan 11, 2018 9:22 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Example 8.6
- Replies: 1
- Views: 170
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...
- Tue Jan 09, 2018 9:08 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat vs Energy
- Replies: 2
- Views: 236
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...
- Tue Dec 05, 2017 11:46 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Bond Length
- Replies: 1
- Views: 259
Re: Bond Length
Te is a larger atom, so the H bonded to it would be farther away, creating a weaker bond.
- Tue Dec 05, 2017 12:02 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: 12.55 (d)
- Replies: 1
- Views: 193
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.
Thanks.
- Tue Nov 28, 2017 11:13 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Chemical Reaction Shift
- Replies: 1
- Views: 268
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...
- Tue Nov 28, 2017 9:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Question 11.45 (c)
- Replies: 1
- Views: 239
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.
- Sat Nov 25, 2017 12:44 am
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: 4.75 Part A
- Replies: 6
- Views: 2371
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...
- Sat Nov 25, 2017 12:37 am
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Homework 17.37 Part d
- Replies: 1
- Views: 280
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.
- Tue Nov 14, 2017 11:15 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Chapter 4 #13 Part D
- Replies: 1
- Views: 245
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.
- Mon Nov 13, 2017 10:16 pm
- Forum: Hybridization
- Topic: Ch 4 #95
- Replies: 2
- Views: 296
Re: Ch 4 #95
The first two means the s and p orbitals were from the second energy level (as in n =2).
- Thu Nov 09, 2017 5:24 pm
- Forum: DeBroglie Equation
- Topic: E=hv vs. Work Function?
- Replies: 2
- Views: 2252
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...
- Tue Nov 07, 2017 8:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Resonance Structure
- Replies: 4
- Views: 351
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 ...
- Fri Nov 03, 2017 11:55 pm
- Forum: Ionic & Covalent Bonds
- Topic: Ionic or Covalent?
- Replies: 4
- Views: 553
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...
- Thu Nov 02, 2017 10:46 pm
- Forum: Resonance Structures
- Topic: Lewis Structures: Resonance Hybrid
- Replies: 2
- Views: 438
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.
- Wed Oct 25, 2017 7:43 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: 4s and 3d
- Replies: 2
- Views: 487
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.
- Mon Oct 23, 2017 10:27 pm
- Forum: Trends in The Periodic Table
- Topic: 2.67 part B
- Replies: 4
- Views: 553
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 ...
- Sat Oct 21, 2017 4:32 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Principle Quantum Number [ENDORSED]
- Replies: 3
- Views: 388
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.
- Wed Oct 18, 2017 9:07 pm
- Forum: DeBroglie Equation
- Topic: Difference longest Wavelength and wavelength
- Replies: 2
- Views: 555
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...
- Wed Oct 11, 2017 4:34 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric Effect
- Replies: 8
- Views: 772
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 ...
- Wed Oct 11, 2017 4:18 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Negative Energy
- Replies: 1
- Views: 251
Re: Negative Energy
The negative energy when the electron is descending is the energy it's emitting as a photon.
- Wed Oct 04, 2017 10:36 pm
- Forum: Limiting Reactant Calculations
- Topic: Page F109 from book (Self-Test M.2A)
- Replies: 2
- Views: 662
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 ...
- Tue Oct 03, 2017 5:07 pm
- Forum: Empirical & Molecular Formulas
- Topic: Problem E5
- Replies: 6
- Views: 3423
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.