Search found 30 matches
- Wed Mar 13, 2019 2:07 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Adding Pt in 6L5 b)
- Replies: 2
- Views: 317
Re: Adding Pt in 6L5 b)
You need Pt in the cell diagram if there is no solid present, so it can act as an electrode to transfer electrons.
- Wed Mar 13, 2019 2:03 pm
- Forum: First Order Reactions
- Topic: 0 Order vs Independent Rate?
- Replies: 1
- Views: 200
Re: 0 Order vs Independent Rate?
I'm not positive, but I think they're the same thing. When a reaction is 0 order, the rate = k[A]^0, and since [A]^0 is equal to 1, the rate will just be rate = k. Since the rate is always the same regardless of the reactant concentration, the rate is independent of the concentration.
- Wed Mar 13, 2019 1:56 pm
- Forum: General Rate Laws
- Topic: Rate equations to know
- Replies: 5
- Views: 514
Re: Rate equations to know
The reference sheet we will be given is posted on ccle, but it looks like we will be given most of the equations and just have to remember which integrated rate law and half life goes with which reaction order
- Tue Mar 05, 2019 3:05 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Ecell
- Replies: 13
- Views: 1423
Re: Ecell
When you do that, make sure you don't negate the E value because that would change it into the standard oxidizing potential. Use the E values exactly as they are given
- Tue Mar 05, 2019 2:54 pm
- Forum: First Order Reactions
- Topic: First order graph
- Replies: 7
- Views: 844
- Tue Mar 05, 2019 2:53 pm
- Forum: First Order Reactions
- Topic: Rate constant and half life
- Replies: 3
- Views: 430
Re: Rate constant and half life
It goes through the reaction faster, so it will go down to half it's original concentration at a faster speed, making it take a shorter amount of time to do so
- Mon Feb 25, 2019 2:13 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Voltage
- Replies: 2
- Views: 300
Re: Voltage
Log base 10 makes it easier when you are working with acid-base solutions because pH is found using log base 10, so it's easier to find the concentration if they're both using the same form of a log
- Mon Feb 25, 2019 2:08 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams
- Replies: 2
- Views: 285
Re: Cell Diagrams
When they are in an aqueous solution together, they are separated by a comma. When one is a solid and the other is in an aqueous solution, they are separated by a line.
- Mon Feb 25, 2019 2:05 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Moles in -nFEcell
- Replies: 5
- Views: 631
Re: Moles in -nFEcell
n represents the number of electrons transferred in a balanced redox reaction
- Wed Feb 20, 2019 4:36 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Test 2
- Replies: 2
- Views: 308
Re: Test 2
My TA said that they will most likely give us values for enthalpy and entropy because making us calculate everything would be a lot of work for only one problem. That being said, it is still possible that they could make us calculate it to be used in the gibbs free energy equation, so I would look o...
- Wed Feb 20, 2019 4:29 pm
- Forum: General Science Questions
- Topic: Midterm
- Replies: 5
- Views: 847
Re: Midterm
Ka2 is the Ka for taking away the second hydrogen in H3PO4
the reaction would be
H2PO4(-) + H2O = HPO4(2-) + H3O(+)
and Ka2 would be
(HPO4(2-)*H3O(+))/H2PO4(-)
the reaction would be
H2PO4(-) + H2O = HPO4(2-) + H3O(+)
and Ka2 would be
(HPO4(2-)*H3O(+))/H2PO4(-)
- Wed Feb 20, 2019 4:20 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Anode v. Cathode
- Replies: 9
- Views: 923
Re: Anode v. Cathode
Dhwani Krishnan 1G wrote:what is the difference between an anode and cathode?
An anode is negatively charged and contains the oxidation part of a reaction, and electrons flow away from it
A cathode is positively charged and contains the reduction part of the reaction, and electrons flow towards it
- Wed Feb 13, 2019 12:32 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: state function
- Replies: 4
- Views: 416
Re: state function
No- both work and heat are not state functions
- Wed Feb 13, 2019 12:30 am
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 3/2R vs 5/2R
- Replies: 8
- Views: 2007
Re: 3/2R vs 5/2R
3/2R is used when it is at constant volume for a monoatomic ideal gas
5/2R is used when it is at constant pressure for a monoatomic ideal gas
5/2R is used when it is at constant pressure for a monoatomic ideal gas
- Wed Feb 13, 2019 12:27 am
- Forum: Administrative Questions and Class Announcements
- Topic: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!
- Replies: 49
- Views: 11461
Re: Lyndon's HOTDOG MIDTERM REVIEW SESSION!! FINALLY!
Help with question 2c)? I understand that since we're cooling it kc will also drop because it's dependent on temperature, but what is the detailed explanation? Since the reaction is breaking bonds, we know that it will be endothermic. When we learned how temperature affects a system, we learned tha...
- Wed Feb 06, 2019 9:43 pm
- Forum: Calculating Work of Expansion
- Topic: Integral Expression for Work
- Replies: 4
- Views: 468
Re: Integral Expression for Work
My TA said that the only equations we will have to use for work are w=-PΔV (for a system with constant pressure) and w=-nRT(ln(V2/V1)) (for a reaction that is isothermal reversible). The integral is more for understanding than actual needing to use the equation to solve it for this class (I think)
- Wed Feb 06, 2019 9:22 pm
- Forum: Phase Changes & Related Calculations
- Topic: Symbol inquiry
- Replies: 4
- Views: 499
Re: Symbol inquiry
Internal energy is represented by "U", and change of internal energy is represented by "ΔU"
the delta symbol (Δ) always represents a change in something
the delta symbol (Δ) always represents a change in something
- Wed Feb 06, 2019 4:18 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: S equation
- Replies: 2
- Views: 250
Re: S equation
-KB is equal to R/Na, where Na is avagadros constant of 6.022 x 10^23, which is how many particles are in 1 mole
-Capital W represents degeneracy, the number of ways of achieving a given energy state
-KB is the boltzman constant, which is equal to 1.381 x 10^-23 J/K
-Capital W represents degeneracy, the number of ways of achieving a given energy state
-KB is the boltzman constant, which is equal to 1.381 x 10^-23 J/K
- Wed Jan 30, 2019 2:36 pm
- Forum: Phase Changes & Related Calculations
- Topic: Week 4 Homework
- Replies: 12
- Views: 1196
Re: Week 4 Homework
I believe you can do any problem from the thermodynamics set, but most of the stuff that we've covered in class is in parts 4D and 4C because he has been going out of the order of the textbook, so those will probably be easier and more helpful because we've learned them already.
- Wed Jan 30, 2019 2:32 pm
- Forum: Calculating Work of Expansion
- Topic: Units for Work
- Replies: 6
- Views: 684
Re: Units for Work
To convert L⋅atm to Joules, you need to use the conversion factor of 101.325 J/1 L*atm, so basically just multiply your answer found using volume and pressure by 101.325 to get the actual answer in joules.
- Wed Jan 30, 2019 2:28 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Hess's Law
- Replies: 4
- Views: 607
Re: Hess's Law
I find that it's easiest to start off with the thing that only exists in one place. For example, in my discussion we did problem 4D.19: find the reaction enthalpy for the reaction [H2 + Br2 -> 2HBr] given: NH3 + HBr -> Nh4Br N2 + 3H2 -> 2NH3 N2 + 4H2 + Br2 -> 2NH4Br We know that we need 2HBr for the...
- Wed Jan 23, 2019 5:43 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Strength of acids based on Ka and pKa
- Replies: 4
- Views: 2040
Re: Strength of acids based on Ka and pKa
pKa and pKb work on the same scale as pH and pOH, respectively. A low pKa indicates a stronger weak acid, whereas a high pKa indicates a weaker acid. In the same way, a low pKb indicates a stronger weak base, whereas a high pKb indicates a weaker base. You can look at the Ka and Kb to compare, but I...
- Wed Jan 23, 2019 5:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating pH of Weak Acids and Bases
- Replies: 2
- Views: 290
Re: Calculating pH of Weak Acids and Bases
Basically, the reason for this is that with such a low Ka, subtracting x won't have that much of an affect on the original concentration. If you were to subtract 1x10^-5 from .1, you would end up with .09999, which is still essentially .1 (especially when you would eventually round for sig figs). La...
- Wed Jan 23, 2019 5:30 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating Equilibrium Constants
- Replies: 2
- Views: 252
Re: Calculating Equilibrium Constants
Yes- in an ice table, the number of x's you put for the change is the same as the amount in the equation. For example, in the equation "N2 + 3H2 = 2NH3", you would subtract 1x from N2, 3x from H2, and add 2x to NH3. Logically, this makes sense because in a reaction, 1 molecule of N2 will c...
- Mon Jan 14, 2019 2:51 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Reaction Quotient Value
- Replies: 5
- Views: 314
Re: Reaction Quotient Value
If Q is equal to K, that means that the concentrations at the time we are evaluating are equal to the concentrations found when the reaction is in equilibrium. This means that the reaction is in equilibrium, so neither reaction is favored because at equilibrium, the forward and reverse reactions occ...
- Mon Jan 14, 2019 2:41 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: 7th edition 5I.35
- Replies: 1
- Views: 215
Re: 7th edition 5I.35
The entire problem is essentially an ice table with the variable "p" substituted in for the change. The First thing you want to do is balance the equation, then find the equilibrium concentrations in terms of "p" and substitute it into the K equation. 2NO <--> N2 + O2 I / 1 bar /...
- Mon Jan 14, 2019 2:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Forward vs. Reverse Reactions
- Replies: 4
- Views: 1622
Re: Forward vs. Reverse Reactions
Kc and Kp both find the equilibrium constant for the reaction, while Qc and Qp can be calculated at any time during a reaction in order to determine which direction the reaction will proceed in. If Q < K, that means that there are more reactants (denominator) than products (numerator) at that time t...
- Thu Jan 10, 2019 1:25 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: how to write the formula for K
- Replies: 6
- Views: 419
Re: how to write the formula for K
When it's written as a subscript of P, that means that it is using partial pressures instead of concentrations. If you are using concentrations, it's okay to write the brackets around it because that indicates that the unit is mol/L. Just double check to make sure which method you are actually using...
- Thu Jan 10, 2019 1:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: The relationship between Kc and K
- Replies: 2
- Views: 248
Re: The relationship between Kc and K
Kc is used to calculate the equilibrium constant using the concentration of the products and reactants at equilibrium. Kc is essentially a more specific way of saying K, because it specifies that you are calculating it using concentration, whereas Kp is calculated using partial pressure. Q, the reac...
- Thu Jan 10, 2019 1:02 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kc and Qc
- Replies: 4
- Views: 43613
Re: Kc and Qc
Kc and Kp both find the equilibrium constant for the reaction, with Kc being calculated using equilibrium concentration and Kp being calculated using equilibrium partial pressures. Qc and Qp can be calculated at any time during a reaction in order to determine which direction the reaction will proce...