Search found 31 matches

by Fatima_Iqbal_2E
Tue Mar 13, 2018 11:05 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: arrhenius equation
Replies: 3
Views: 101

Re: arrhenius equation

To add on to this, you can also use the Arrhenius equation when you want to see how the rate constant will be affected by the activation energy and frequency factor. You could also work backwards to find what the temp/Ea/A is at any given k value, provided you have the values for all but one of your...
by Fatima_Iqbal_2E
Tue Mar 13, 2018 11:00 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: enthalpy of sublimation
Replies: 5
Views: 254

Re: enthalpy of sublimation

The numerical value should be given to us, but it is important to know the relationship between , as stated above!
by Fatima_Iqbal_2E
Tue Mar 13, 2018 10:53 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Arrhenius Equation
Replies: 2
Views: 127

Re: Arrhenius Equation

The Arrhenius equation is k=Ae^{-E_{a}/RT} . It allows you to see how the rate constant is related to the frequency factor, temperature, and activation energy. You can use the equation to see how the rate constant changes when one of the three factors affecting it is changed. Additionally, you can a...
by Fatima_Iqbal_2E
Sat Mar 10, 2018 11:15 pm
Forum: Second Order Reactions
Topic: 3rd order?
Replies: 7
Views: 500

Re: 3rd order?

If you want to write the units of K in terms of molarity, it would be , but if you wanted to write it in terms of moles and liters, it would be .
by Fatima_Iqbal_2E
Sat Mar 10, 2018 11:02 pm
Forum: Second Order Reactions
Topic: Units of k for orders above 2nd
Replies: 6
Views: 274

Re: Units of k for orders above 2nd

Ilan Shavolian 1K wrote:isnt there supposed to be an L,L2... at the top??


The M refers to molarity, so the liters would be implied. If m was written, then yes, L would've needed to have been specified.
by Fatima_Iqbal_2E
Sat Mar 10, 2018 11:01 pm
Forum: Second Order Reactions
Topic: Units of k for orders above 2nd
Replies: 6
Views: 274

Re: Units of k for orders above 2nd

A helpful way of finding the units for different orders is to use . The x value would be the order that you are finding your units for. For example, if you are trying to find the units for a 10th order reaction, it would be =. I hope this helps!
by Fatima_Iqbal_2E
Sat Mar 10, 2018 10:56 pm
Forum: Zero Order Reactions
Topic: Half Lives? [ENDORSED]
Replies: 7
Views: 326

Re: Half Lives? [ENDORSED]

Yes, because you would be able to multiply the half-life by the number of times it occurred to find the time it would take to reach the new concentration.
by Fatima_Iqbal_2E
Sun Mar 04, 2018 2:02 pm
Forum: First Order Reactions
Topic: k
Replies: 16
Views: 461

Re: k

I think it would be a good idea to convert time to seconds if the problem doesn't ask otherwise. If needed, after finding k (with seconds as your units of time), you can always use conversion factors to change the seconds into another unit of time.
by Fatima_Iqbal_2E
Sun Mar 04, 2018 1:47 pm
Forum: General Rate Laws
Topic: half-life
Replies: 8
Views: 180

Re: half-life

Adding on to the possible applications of half-life, you can use the half-life to determine how much of a given drug is in a bloodstream x hours after it was initially administered. I think this is actually how the frequency with which you take certain medicines is determined!
by Fatima_Iqbal_2E
Sun Mar 04, 2018 1:43 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: K
Replies: 7
Views: 183

Re: K

Although Dr. Lavelle hasn't gone over it in class yet, the Arrhenius equation
helps determine what factors change the K value. The K increases if the activation energy decreases or if the temperature increases.
by Fatima_Iqbal_2E
Sun Feb 25, 2018 1:22 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Determining N
Replies: 4
Views: 134

Re: Determining N

It doesn't matter, as you would end up with the same n value regardless of how you write the fraction.
by Fatima_Iqbal_2E
Sun Feb 25, 2018 1:07 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: Differential Rate Law
Replies: 3
Views: 96

Re: Differential Rate Law

Yes, as stated before, this kind of reaction would be a zero order reaction, which means that the rate of reaction would depend only on the rate constant, k. I think the rate equation in this scenario would then just be Rate=K, since the exponents on your reactant concentrations would be 0, which wo...
by Fatima_Iqbal_2E
Sun Feb 25, 2018 1:01 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: n
Replies: 8
Views: 178

Re: n

'n' represents the order of the reactants in the rate law and gives insight into the reaction mechanism, you would see it as the exponent on the concentration of the reactants: .
by Fatima_Iqbal_2E
Sat Feb 17, 2018 6:43 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Dissolving
Replies: 2
Views: 87

Re: Dissolving

When Dr. Lavelle showed us an example like this during lecture, we had not yet learned the Nernst equation, so you could use \Delta G^{o} = -nFE^{o} to see if the reaction is spontaneous or not. If the \Delta G^{o} is positive, the reaction would be unfavorable, and the ring would not dissolve. Howe...
by Fatima_Iqbal_2E
Sat Feb 17, 2018 6:39 pm
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Negative E
Replies: 2
Views: 45

Re: Negative E

If E is negative, then is positive, due to the equation , which would tell you that the process you are looking at is not spontaneous.
by Fatima_Iqbal_2E
Sat Feb 17, 2018 6:26 pm
Forum: Balancing Redox Reactions
Topic: Oxidizing vs. Reducing Agent
Replies: 7
Views: 147

Re: Oxidizing vs. Reducing Agent

For me, it helps to remember OIL (oxidation is losing e-) RIG (reduction is gaining e-). If a species is losing electrons when going from reactants to products, the oxidation number increases, so the species is being oxidized and is a reducing agent. However, if a species is gaining electrons when g...
by Fatima_Iqbal_2E
Sat Feb 10, 2018 2:34 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: Units Cancel
Replies: 1
Views: 75

Re: Units Cancel

When you take the ln of something, if you are converting from one unit to another by multiplying by a certain constant (like you do when converting from Kpa to atm), the ln woud stay the same, since your p1/p2 would remain the same. However, for temperature, this wouldn't be the case, because you ar...
by Fatima_Iqbal_2E
Sat Feb 10, 2018 2:23 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: negative entropy [ENDORSED]
Replies: 3
Views: 128

Re: negative entropy [ENDORSED]

A negative change in entropy would mean that the number of microstates is decreasing, and you'd have a better idea of where specific molecules are. I don't think it's possible to have a negative entropy, as you can never have a degeneracy less than 1, so the formula S=k_{b}lnW would never lead to an...
by Fatima_Iqbal_2E
Sat Feb 10, 2018 2:19 pm
Forum: Balancing Redox Reactions
Topic: Problem 14.1
Replies: 2
Views: 62

Problem 14.1

The problem asks to identify the elements undergoing changes in oxidation state and indicate the initial/final oxidation numbers. The solution manual says that Cr is reduced from 6+ to 3+, but I'm kind of confused as to where they got the 6+ from. Is that just because they multiplied the 2 Cr by 3+,...
by Fatima_Iqbal_2E
Sun Feb 04, 2018 12:48 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: negative entropy [ENDORSED]
Replies: 10
Views: 252

Re: negative entropy [ENDORSED]

The formula used to determine entropy, S= lnW requires that W be a positive number, since you can't have a negative number of microstates, and thus, can't have a negative degeneracy. Because of this, I don't think there is any way that you'd be able to have a system with a negative entropy.
by Fatima_Iqbal_2E
Sun Feb 04, 2018 12:07 pm
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: 9.13
Replies: 5
Views: 149

Re: 9.13

zanekoch1A wrote:So for future reference if they do not specify number of moles and say assume ideal gas conditions this automatically means 1mol of substance?


I would assume so.
by Fatima_Iqbal_2E
Sun Feb 04, 2018 11:36 am
Forum: Van't Hoff Equation
Topic: Change in Entropy Remains Constant
Replies: 3
Views: 131

Re: Change in Entropy Remains Constant

I was also confused about the same thing. Would we always assume that the change in entropy remains constant when using this equation?
by Fatima_Iqbal_2E
Sat Jan 27, 2018 11:04 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: Third Law of Thermodynamics
Replies: 4
Views: 134

Third Law of Thermodynamics

Could someone please explain what exactly the Third Law of Thermodynamics is? I remember going over it in lecture, but I don’t quite understand it.
by Fatima_Iqbal_2E
Sat Jan 27, 2018 10:50 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Degeneracy (W) [ENDORSED]
Replies: 8
Views: 375

Re: Degeneracy (W) [ENDORSED]

In lecture, Dr. Lavelle describes degeneracy as the number of ways that a given energy state could be achieved. You could find this number by using W=2^N if you had a 2-State system, 3^N if you had a 3-State system, etc.
by Fatima_Iqbal_2E
Sat Jan 27, 2018 3:53 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: Finding W
Replies: 3
Views: 127

Re: Finding W

Depending on the number of states there are in a given system (2-state, 3-state, etc), you could use the formula x^N to find the degeneracy, where X is the number of systems in the state, and N is the number of atoms in the system.
by Fatima_Iqbal_2E
Thu Jan 18, 2018 11:09 am
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: enthalpy
Replies: 3
Views: 116

Re: enthalpy

Not necessarily. Delta Hf is the enthalpy of formation, while Delta Hrxn is the enthalpy of reaction. Generally, problems will ask you to find the enthalpy of reaction using the enthalpies of formation of the reactants and products. Additionally, the enthalpies of formation are for one molecule, whi...
by Fatima_Iqbal_2E
Thu Jan 18, 2018 10:22 am
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: the sign of q and w [ENDORSED]
Replies: 6
Views: 201

Re: the sign of q and w [ENDORSED]

Because that is the general formula, both q + w would be positive, as you'd be adding their contributions together. If there was a -q or a -w, you would then add it into the equation as a negative number.
by Fatima_Iqbal_2E
Thu Jan 18, 2018 10:01 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 8.55
Replies: 2
Views: 94

Re: 8.55

I was getting stuck at the same part, I think there's a typo in the problem.
by Fatima_Iqbal_2E
Sat Jan 13, 2018 2:00 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat Capacity
Replies: 3
Views: 127

Re: Heat Capacity

Yes, you are correct in saying that the heat capacity would always be the heat needed to raise the temp by 1C.
by Fatima_Iqbal_2E
Sat Jan 13, 2018 1:26 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Calorimeters
Replies: 3
Views: 100

Re: Calorimeters

I was confused about the same thing, and I found this website helpful:
https://sciencing.com/a-calorimeter-work-4925148.html
There isn't much emphasis on coffee cup vs. bomb calorimeters, but I think it helps with understanding generally the use of calorimeters.
by Fatima_Iqbal_2E
Thu Jan 11, 2018 2:04 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Enthalpy of sublimation
Replies: 3
Views: 101

Re: Enthalpy of sublimation

Because the enthalpy of vaporization is H(vapor)-H(liquid), and the enthalpy of fusion is H(liquid)-H(solid), when you add the two enthalpies together, it becomes H(vapor)-H(liquid)+H(liquid)-H(solid). The H(liquid) would cancel out, leaving you with H(vapor)-H(solid), which would give you the entha...

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