I'm looking at the answer for this question and I understand how to do everything, except for the value '-0.59'
I'm assuming that's the lnA from the Arrhenius Equation, however, how do we calculate that with the values given?
Search found 60 matches
- Fri Mar 16, 2018 9:17 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.63
- Replies: 2
- Views: 371
- Fri Mar 16, 2018 9:09 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: 15.61
- Replies: 2
- Views: 345
15.61
How is the equation listed in the answers derived?
I understand it's from lnk=-Ea/RT with changes in T, but my question is where did lnA go?
Thanks!
I understand it's from lnk=-Ea/RT with changes in T, but my question is where did lnA go?
Thanks!
- Fri Mar 16, 2018 7:31 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Homework 11.111
- Replies: 1
- Views: 1184
Re: Homework 11.111
you would use the equation dG = - RT lnK for this question rearrange and substitute in the values you have lnK=- -200000/(8.314*298) = 80.7 solve for K K=1.14*10^35 the question says the equilibrium constant of the first step is 10 times greater than that of the second step so you simply divide the ...
- Fri Mar 16, 2018 7:25 pm
- Forum: Balancing Redox Reactions
- Topic: Ranking elements
- Replies: 2
- Views: 419
Re: Ranking elements
Ti < Mn < Ag
Al < Pb < Cu
Al < Pb < Cu
- Fri Mar 16, 2018 7:17 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: 14.15.a
- Replies: 1
- Views: 281
Re: 14.15.a
Im not sure but maybe it has to do with the states each compound is in. AgBr is a solid while Br- is aqueous. And just to keep things neat the solids (Ag and AgBr) are placed next to each other.
I've seen this done in other examples
I've seen this done in other examples
- Fri Mar 16, 2018 7:06 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: A?
- Replies: 6
- Views: 1051
Re: A?
yes, it is a constant
- Wed Mar 07, 2018 9:27 pm
- Forum: General Rate Laws
- Topic: 15.23 b
- Replies: 3
- Views: 457
15.23 b
Why is the formula used for this question ln[A]0 / [A]t , instead of ln [A]t/[A]0?
- Wed Feb 14, 2018 3:12 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Problem 11.111
- Replies: 2
- Views: 388
Re: Problem 11.111
use the equation deltaG = -RTlnK
and rearrange
lnK= - 200000/(8.314*298)
= 80.7
K = 1.14*10^35
divide by 10 so K2 = 1.14*10^34
and then use it in the same equation as above
deltaG2 = -8.314*298*ln1.14*10^34
= -194 kJ
and rearrange
lnK= - 200000/(8.314*298)
= 80.7
K = 1.14*10^35
divide by 10 so K2 = 1.14*10^34
and then use it in the same equation as above
deltaG2 = -8.314*298*ln1.14*10^34
= -194 kJ
- Tue Feb 13, 2018 10:20 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Heat Transfer with no temperature given
- Replies: 2
- Views: 368
Re: Heat Transfer with no temperature given
I don't know what the question is exactly?
But maybe if the question also suggests/says the change in internal energy is 0, you could use q=-w
and just substitute the given values into w=-PdeltaV ?
Just a thought
But maybe if the question also suggests/says the change in internal energy is 0, you could use q=-w
and just substitute the given values into w=-PdeltaV ?
Just a thought
- Tue Feb 13, 2018 10:15 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Trouton's rule
- Replies: 1
- Views: 321
Re: Trouton's rule
nope. because none of the questions pertaining to trouton's rule are on the syllabus/homework problem list
- Tue Feb 13, 2018 10:12 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Specific Heat capacity
- Replies: 1
- Views: 1388
Re: Specific Heat capacity
so use q=mCdelta T to calculate the heat produced by the water when the temperature is raised from 22.6 to 34.6 so it would look something like this: 115.43*4.184*(34.6-22.6) = 5795.51 J and then for the metal because Qsys=-Qsurr 245.7*C*(34.6-75.2)= -5795.51 J C=-9975.42/-5795.51 =0.580979 =0.581 J...
- Tue Feb 13, 2018 10:05 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Ex 9.10 pg. 342
- Replies: 1
- Views: 288
Re: Ex 9.10 pg. 342
they are the same thing you just have to inverse the sign
- Tue Feb 13, 2018 3:59 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: why entropy for aqueous ions are negative
- Replies: 1
- Views: 816
why entropy for aqueous ions are negative
Will we have to know what the entropy for aqueous ions are mostly negative
(like in question 9.89)?
(like in question 9.89)?
- Tue Feb 13, 2018 1:29 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: 9.53
- Replies: 1
- Views: 276
9.53
I understand how we need to use deltaG = deltaH - TdeltaS for this question, and how we need to use the tables in the textbook to find the values for deltaH and deltaS, however, why is the value of delta S listed in the solutions "97.6J/K/mol" ?
Where does that value come from?
Where does that value come from?
- Tue Jan 23, 2018 11:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Mean Bond Enthalpy
- Replies: 2
- Views: 1124
Re: Mean Bond Enthalpy
I think they would give nearly the same result as the average bond enthalpy values are averages of that said bond in a bunch of different molecular structures.
But i'm not 100% sure
But i'm not 100% sure
- Tue Jan 23, 2018 11:46 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.113
- Replies: 3
- Views: 476
Re: 8.113
what does the state (gr) mean?
- Tue Jan 23, 2018 11:44 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Equations and constants sheet 14B
- Replies: 2
- Views: 521
Re: Equations and constants sheet 14B
yep it's decimeter
where 1 decimeter^3 = 1 Liter
where 1 decimeter^3 = 1 Liter
- Mon Jan 22, 2018 11:32 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: HW 8.99
- Replies: 2
- Views: 438
Re: HW 8.99
because we need the amount of mol of product, which would depend on the limiting reagent of the reaction
- Mon Jan 22, 2018 10:50 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heat capacities
- Replies: 5
- Views: 584
Re: Heat capacities
if it's calculating temperature change, then the difference will be the same whether it's in kelvin or celcius
- Mon Jan 22, 2018 10:45 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Average Kinetic Energy
- Replies: 2
- Views: 351
Average Kinetic Energy
Will there be questions involving average kinetic energy (like question 8.103) in the upcoming test?
I don't recall Dr Lavelle going over it in lecture, nor do I see it on his outline 1...
I don't recall Dr Lavelle going over it in lecture, nor do I see it on his outline 1...
- Mon Jan 22, 2018 10:12 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.9 torr conversion
- Replies: 1
- Views: 267
Re: 8.9 torr conversion
torr is another way to express pressure, but thermodynamics answers are usually always in atm so the conversion from torr to atm is on the formula sheet where 1 atm=760 torr!
and for the 1000ml, it's because work is calculated with liters
and for the 1000ml, it's because work is calculated with liters
- Mon Jan 22, 2018 9:55 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 8.61
- Replies: 1
- Views: 296
Re: 8.61
reaction enthalpies may not just be indicating 1 mol of substance so it's only kJ. If it were enthalpy of formation or enthalpy of combustion, then it would be kJ/mol as it's the reaction enthalpy of the formation/combustion of 1 mole of a substance
- Sun Jan 21, 2018 11:56 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 85
- Replies: 1
- Views: 258
Question 85
specifically part b, why is the final step in the answer book shown as the mole (from after calculating it with PV=nRT) being multiplied by 180.6kJ? I would've thought it would be multiplied by 90.3kJ which is reaction enthalpy that produces 1 mole of NO.
- Sun Jan 21, 2018 11:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.25
- Replies: 3
- Views: 520
Re: 8.25
oh yea, I completely forgot about that.
Thanks so much!
Thanks so much!
- Sun Jan 21, 2018 11:48 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Heat and systems
- Replies: 3
- Views: 543
Re: Heat and systems
if you have deltaH/Q, you can look at the sign it has -deltaH means that energy is released by the system (exothermic; the system is losing energy, thus the minus sign) and so energy is being transfered from the system to the surroundings. eg combustion reactions +deltaH means that energy is gained ...
- Sun Jan 21, 2018 11:44 pm
- Forum: Calculating Work of Expansion
- Topic: When to use equation [ENDORSED]
- Replies: 3
- Views: 413
Re: When to use equation [ENDORSED]
for the first equation, usually the question says "isothermic" (as in the assigned text book questions)
but it's not going to be on this coming test! :)
but it's not going to be on this coming test! :)
- Sun Jan 21, 2018 11:43 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: System Types
- Replies: 7
- Views: 840
Re: System Types
the example Dr lavelle gave in class was just to think of 3 test tubes - Open system: both matter and energy can travel out, which would be an open test tube with no insulation - Closed system: only energy can travel out, not matter, which would be a closed test tube - Isolated system: neither matte...
- Sun Jan 21, 2018 10:21 pm
- Forum: Calculating Work of Expansion
- Topic: Types of Work [ENDORSED]
- Replies: 3
- Views: 449
Re: Types of Work [ENDORSED]
Im pretty sure this is the only type of work because a) Dr Lavelle didn't talk about the rest in lecture and b) the practice questions he set aren't about any other types of work (to the extent of my knowledge)
- Sun Jan 21, 2018 5:38 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 8.25
- Replies: 3
- Views: 520
8.25
for question 8.25, I understand everything up until the point where the answer key says Qreaction + Qcalorimeter =0, which then leads to the final answer. Can someone explain why the heat from the reaction + the heat of the calorimeter equals to 0?
Thanks :)
Thanks :)
- Sun Jan 21, 2018 4:48 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Standard Enthalpies of Formation
- Replies: 3
- Views: 283
Re: Standard Enthalpies of Formation
the two are different but that's how you would do a calculation of the enthalpy of reaction through bond enthalpies
- Sat Dec 09, 2017 4:05 pm
- Forum: Naming
- Topic: Ligand Prefixes: Di- vs. Bis-
- Replies: 1
- Views: 352
Re: Ligand Prefixes: Di- vs. Bis-
the most common example would be diethylenetriammine where if it was (dien)2 then it would be didiethylenetriammine, so instead of having "didi-" you switch the first 'di' out and replace it with 'bis' so it's clearer
--> bisdiethylenetriammine
--> bisdiethylenetriammine
- Sat Dec 09, 2017 12:27 am
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Percentage deprotonation
- Replies: 2
- Views: 486
Re: Percentage deprotonation
because a representative equation for acid reactions is HA --> H3O+ + A- where Kc=[H3O+][A-]/[HA] if you were to use an ice table to calculate molarities, Kc would become: Kc= x^2 / [HA]-x as both have the same amount of moles and because the concentration of [H3O+] and [A-] are the same, it can be ...
- Sat Dec 09, 2017 12:20 am
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 3
- Views: 467
Re: Bond Angles
I think as far as what was taught in class, yes. molecules might have angles of <109.5 if there are lone pairs present (eg in bent/angular shapes) as they have greater repulsion than normal bonds. molecules with angles of >109.5.. I would assume there's only molecules with 120 degrees (as far as wha...
- Thu Dec 07, 2017 9:29 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 11.79
- Replies: 1
- Views: 326
Re: 11.79
you add the change in N2 to the original value so now PN2=4.68 and the rest of the initial values are the values they give you.
and from here you just solve for x as you would any other ICE table.
and from here you just solve for x as you would any other ICE table.
- Thu Dec 07, 2017 9:25 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Reaction shift
- Replies: 3
- Views: 222
Re: Reaction shift
the reaction will favor the reactants and thus shift left as K must remain constant
and K=[P]/[R] so if P increases, R will increase to keep K remain the same
and K=[P]/[R] so if P increases, R will increase to keep K remain the same
- Tue Nov 28, 2017 12:11 am
- Forum: Naming
- Topic: Di, Tri, Tetra vs. Bis, Tris, Tetrakis
- Replies: 1
- Views: 2550
Re: Di, Tri, Tetra vs. Bis, Tris, Tetrakis
so you would use di, tri, tetra on ligands that don't have said prefixes in their names already, eg dichlorido while if you were to write (dien)2, instead of it being didiethlynetriamine which would be confusing, it's bisdiethylenetriamine, same with (en)2 for example, it would be bisethylenediamine...
- Tue Nov 28, 2017 12:08 am
- Forum: Naming
- Topic: Question 17.31c and d
- Replies: 1
- Views: 309
Re: Question 17.31c and d
I think either one is ok, the (OH2) just makes the fact that it is O that's bonded to the metal clearer
- Tue Nov 28, 2017 12:07 am
- Forum: Ideal Gases
- Topic: Partial pressures
- Replies: 4
- Views: 736
Re: Partial pressures
use the equation PV=nRT and substitute the values in
- Mon Nov 27, 2017 9:54 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Polydentate Ligands(17.33)
- Replies: 1
- Views: 216
- Mon Nov 27, 2017 9:52 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Compressing a Reaction Mixture
- Replies: 1
- Views: 667
Re: Compressing a Reaction Mixture
adding inert gas doesn't change the concentration of the reactants/products as the moles and the volume of R and P remain constant while increasing the pressure, and thus decreasing the volume does - C=n/V so because the number of moles of the reactants/products remain the same but the volume decrea...
- Mon Nov 27, 2017 9:48 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework 11.1d
- Replies: 1
- Views: 259
Re: Homework 11.1d
change in concentration doesn't affect K, and the only way this is possible is that if the concentration of reactants are increased, the concentration of products will increase as well to keep the P/R ratio the same
- Mon Nov 27, 2017 9:45 pm
- Forum: Shape, Structure, Coordination Number, Ligands
- Topic: Diethylenediamine
- Replies: 2
- Views: 500
Re: Diethylenediamine
If you're talking about (en)2, it would be called bisethylenediamine and yes; while diethylenetriamine (dien) would have 3 bonding sites each
- Mon Nov 27, 2017 9:44 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium constants
- Replies: 1
- Views: 160
Re: Equilibrium constants
K can either indicate Kc (equilibrium constant with concentrations) or Kp (equilibrium constant using partial pressure)
- Sun Nov 26, 2017 10:24 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 11.7
- Replies: 5
- Views: 566
Re: 11.7
use the method of how you would find the equilibrium constant
K=[X]^2 / [X2]
=[(12/17)*0.1]^2 / [(5/17)*0.1]
=0.17
the 12/17 and 5/17 are the number of molecules that have/ haven't dissociated, as given in the diagrams of the 4 containers
K=[X]^2 / [X2]
=[(12/17)*0.1]^2 / [(5/17)*0.1]
=0.17
the 12/17 and 5/17 are the number of molecules that have/ haven't dissociated, as given in the diagrams of the 4 containers
- Sun Nov 26, 2017 10:21 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone Pairs [ENDORSED]
- Replies: 3
- Views: 459
- Sun Nov 26, 2017 10:21 pm
- Forum: Hybridization
- Topic: 4.41
- Replies: 4
- Views: 1940
Re: 4.41
So first, draw out the Lewis Structure - use valence electrons of each atom to determine the bonding (eg every carbon makes 4 bonds), then once you have the structure drawn out it'll be more straightforward to determine the hybridization - basically the number of electron densities of each carbon, a...
- Sun Nov 05, 2017 8:07 pm
- Forum: Formal Charge and Oxidation Numbers
- Topic: Formal Charge
- Replies: 8
- Views: 1071
Re: Formal Charge
when the formal charge = 0
- Sun Oct 29, 2017 10:14 pm
- Forum: Ionic & Covalent Bonds
- Topic: Homework 3.5 part c [ENDORSED]
- Replies: 3
- Views: 525
Re: Homework 3.5 part c [ENDORSED]
[Ar] 3d10
basically Ga Z=31
but it's Ga3+ so 31-3=28
basically Ga Z=31
but it's Ga3+ so 31-3=28
- Thu Oct 26, 2017 9:09 pm
- Forum: Trends in The Periodic Table
- Topic: Exceptions to Ionization Energy Trend
- Replies: 1
- Views: 349
Re: Exceptions to Ionization Energy Trend
Maybe the ones where we're able to work out from the periodic table? Like the electron configuration of nitrogen is 1s2 2s2 2p3 (stable, half filled) while Oxygen has an electron configuration of 1s2 2s2 2p4 (one extra electron from being stably half filled), so taking an electron from the oxygen is...
- Thu Oct 26, 2017 9:02 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Periodic Trends
- Replies: 2
- Views: 198
Re: Periodic Trends
Ionisation Energy for atoms would be different when they're in the gaseous state than in other states - which is the easiest way to compare the energy required between different atoms as there is very little attraction with other atoms. In liquids/solids, the intermolecular forces will distort the v...
- Sun Oct 15, 2017 9:19 pm
- Forum: Properties of Light
- Topic: Lyman vs Balmer Series
- Replies: 4
- Views: 579
Re: Lyman vs Balmer Series
electrons in the Lyman series emit light in the UV region, while electrons in the Balmer series emit light in the visible light region.
So you would look at the wavelength of the light emitted and deduce it from that.
(Im not sure if I answered what you were asking, but yeah)
So you would look at the wavelength of the light emitted and deduce it from that.
(Im not sure if I answered what you were asking, but yeah)
- Sun Oct 15, 2017 9:16 pm
- Forum: Properties of Electrons
- Topic: Atomic Spectra [ENDORSED]
- Replies: 4
- Views: 518
Re: Atomic Spectra [ENDORSED]
I think in the case of ~Atomic Spectra~, it's not so much the attraction of electrons to protons, but instead the absorption and emission of energy.
n=1 is the ground state/ energy level for electrons, which is the furthest "way down" an electron can go after emitting energy.
n=1 is the ground state/ energy level for electrons, which is the furthest "way down" an electron can go after emitting energy.
- Sun Oct 15, 2017 9:13 pm
- Forum: Properties of Electrons
- Topic: Hydrogen series
- Replies: 5
- Views: 601
Re: Hydrogen series
there is also a "Brackett series" which is n=4 but i'm 99.9% sure we don't need to know this
- Sun Oct 15, 2017 9:11 pm
- Forum: Photoelectric Effect
- Topic: Photoelectric effect [ENDORSED]
- Replies: 4
- Views: 604
Re: Photoelectric effect [ENDORSED]
An experiment showed that if a light beam had high enough frequency it could make a surface eject an electron. The emphasis is on frequency here as the intensity of the light had no effect on the electron, which showed that the light was not acting like a classic wave. --> this boils down to the ide...
- Sun Oct 15, 2017 9:00 pm
- Forum: Properties of Light
- Topic: Planck's Constant
- Replies: 10
- Views: 1236
Re: Planck's Constant
whenever you need to use an equation with the constant "h" in it.
eg E=hv
good thing is, the value is on the formula sheet so we don't have to memorize it :)
eg E=hv
good thing is, the value is on the formula sheet so we don't have to memorize it :)
- Sun Oct 15, 2017 8:59 pm
- Forum: Properties of Light
- Topic: Question 1.15
- Replies: 2
- Views: 510
Re: Question 1.15
1. convert the wavelength given to m 102.6x10^-9m 2. figure out the frequency by rearranging c=wavelength*frequency v=c/wavelength = 2.998*10^8 / 102.6*10^-9 = 2.922x10^15 3. as the question says "in the ultraviolet spectrum", one can concur that the final energy level is n=1 4. put known ...
- Sun Oct 08, 2017 8:25 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Chapter 1 Test this week [ENDORSED]
- Replies: 2
- Views: 331
Chapter 1 Test this week [ENDORSED]
So it's posted on the website that there's going to be a test on Chapter 1 up to 1.5 in the textbook this week during the discussion section you're enrolled in. I'm in discussion 1B on Tuesday mornings so I was just wondering if the test was going to be on this week's discussion section (October 10)...
- Tue Oct 03, 2017 10:32 pm
- Forum: Limiting Reactant Calculations
- Topic: m11
- Replies: 7
- Views: 997
Re: m11
my bad i only did half the question, but yes what @Ivy Lu 1C said
- Tue Oct 03, 2017 10:30 pm
- Forum: Student Social/Study Group
- Topic: Fundamental M.25
- Replies: 3
- Views: 1534
Re: Fundamental M.25
@Hyein Cha 1K
I do it the same way
I do it the same way
- Tue Oct 03, 2017 10:21 pm
- Forum: Limiting Reactant Calculations
- Topic: m11
- Replies: 7
- Views: 997
Re: m11
I'm assuming this is the white phosphorus to phosphorus oxide question. P4 + 3O2 --> P4O6 masses for both the reactants = 5.77g get the molar mass of the elements from a periodic table and divide the mass with it to get the number of moles of each particle n (P4) = 5.77/(30.97*4) = 0.0466 n (O2) = 5...