Search found 30 matches
- Sun Mar 10, 2019 2:04 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: negatives in the equation
- Replies: 2
- Views: 150
Re: negatives in the equation
If E is negative, deltaG will be positive and the reaction is unfavorable. If E is positive, delta G will be negative and the reaction is favorable.
- Sun Mar 10, 2019 1:57 pm
- Forum: Van't Hoff Equation
- Topic: Overview
- Replies: 8
- Views: 617
Re: Overview
If you want to find the new K value when the temperature changes, using the Van't Hoff equation will allow you to find the second K value given you have the original K value.
- Sun Mar 10, 2019 1:53 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Molecularity
- Replies: 3
- Views: 215
Re: Molecularity
A unimolecular reaction isn't about collision, but more about a rearrangement of the molecule. For example, decomposition or radioactive decay can occur of a single molecule without any collisions occurring simply because the molecule is breaking apart/rearranging itself into multiple molecules.
- Sun Mar 10, 2019 1:41 pm
- Forum: General Rate Laws
- Topic: Rate Determining Step
- Replies: 5
- Views: 2638
Re: Rate Determining Step
Yes, because the rate determining step is the slowest step in the reaction. That rate determining step determines what the rate law is for the entire reaction.
- Sun Mar 10, 2019 1:28 pm
- Forum: Van't Hoff Equation
- Topic: Van't Hoff use
- Replies: 13
- Views: 1193
Re: Van't Hoff use
The Van't Hoff equation would be used in order to determine the equilibrium constant as the temperature changes. Therefore, when you have two temperatures and two K values, one can use the Van't Hoff equation.
- Sun Mar 10, 2019 1:20 pm
- Forum: General Rate Laws
- Topic: Problem
- Replies: 1
- Views: 107
Re: Problem
The set up is:
d[N2O]/dt = k[N2O]
(0.75 - 0.33 M)/dt = (6.8x10^-3/s)(0.33 M)
Solve for dt.
*I am unsure about which concentration is plugged into [N2O], but I assume it is the final concentration.
Hope this helps.
d[N2O]/dt = k[N2O]
(0.75 - 0.33 M)/dt = (6.8x10^-3/s)(0.33 M)
Solve for dt.
*I am unsure about which concentration is plugged into [N2O], but I assume it is the final concentration.
Hope this helps.
- Sun Mar 10, 2019 1:16 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: k & K
- Replies: 18
- Views: 961
Re: k & K
In kinetics, k is the rate constant, whereas K is the equilibrium constant. Also, K = k/k', which means the the equilibrium constant is equal to the ratio of rate constants of the forward (k) and the reverse (k') elementary reactions.
- Sun Mar 10, 2019 1:14 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Water in a cell diagram
- Replies: 5
- Views: 252
Re: Water in a cell diagram
Only add water in a cell diagram if it is being either oxidized or reduced. If not, there is no point in writing it as it only acts as a solvent and writing aqueous implies that water is present.
- Sun Mar 10, 2019 1:07 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: k' Value
- Replies: 1
- Views: 84
Re: k' Value
Well, one way of finding k' would be to set the reverse rate to the forward rate equal at equilibrium. For example, for the reaction A + B --> C + D, the reaction rates at equilibrium are: k[A][B] = k'[C][D]. You would need to know the concentrations of the products and reactants at equilibrium as w...
- Mon Feb 25, 2019 12:37 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Signs of delta G and E
- Replies: 4
- Views: 199
Re: Signs of delta G and E
As was said previously, in a spontaneous reaction, delta G is negative as free energy is released in the reaction. Also, DeltaG = -nFE, which means that for delta G to be negative, E must be positive. The standard potential is relative to electron-pulling power of the cathode, which means that if th...
- Mon Feb 25, 2019 12:35 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Adding in extra ions
- Replies: 1
- Views: 98
Re: Adding in extra ions
It seems as though the reaction that is shown is a dissolution, which means that AgBr is dissolving into its respective ions, which are Ag+ and Br-. It's similar to the reaction: NaCl (aq) --> Na+ (aq) + Cl- (aq), where the Na is being oxidized and the chlorine is being reduced. Hope this helps!
- Mon Feb 25, 2019 12:26 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Using commas in cell diagram
- Replies: 3
- Views: 175
Re: Using commas in cell diagram
A comma is used when both the oxidized species and the reduced species are in the same phase (usually, both are aqueous) for the same half-reaction. A straight line indicates a phase change between the oxidized species and the reduced species in the same half-reaction.
- Mon Feb 25, 2019 12:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Question 14.25 (Sixth Edition)
- Replies: 3
- Views: 187
Re: Question 14.25 (Sixth Edition)
The more negative the standard potential, the more strongly the reducing agent/element.
- Mon Feb 25, 2019 12:18 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Spontaneous and unstable
- Replies: 4
- Views: 554
Re: Spontaneous and unstable
A thermodynamically stable compound will have a negative standard Gibbs free energy of formation, meaning that it is spontaneous.
A thermodynamically unstable compound will have a positive standard Gibbs free energy of formation, meaning that it is non-spontaneous.
A thermodynamically unstable compound will have a positive standard Gibbs free energy of formation, meaning that it is non-spontaneous.
- Mon Feb 25, 2019 12:16 pm
- Forum: Balancing Redox Reactions
- Topic: Balancing Charges
- Replies: 4
- Views: 145
Re: Balancing Charges
In order to find the oxidation number of Mn in MnO4-, first assume, as was said earlier, that the oxidation number of the oxygen is -2. Since there are four oxygens in the molecule, the total charge from the oxygen is -8. However, the charge of the whole molecule is -1. Therefore, the total charge o...
- Mon Feb 25, 2019 12:11 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Negative Delta G
- Replies: 2
- Views: 131
Re: Negative Delta G
From the second law of thermodynamics, we know that the natural progression of a system is to an increase in entropy. Therefore, a reaction in which the total entropy naturally increases is spontaneous. In the equation for Gibbs, deltaG = -T[deltaS(tot)] at constant temperature and pressure. As a re...
- Mon Feb 25, 2019 11:55 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: trends in standard potential
- Replies: 2
- Views: 147
Re: trends in standard potential
The more negative the standard potential, the more strongly reducing the element. Those elements are found on the left of the periodic table. The more positive the standard potential, the more strongly oxidizing the element. Those elements are found toward the upper right. Looking at the list of sta...
- Sun Feb 10, 2019 2:10 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: Degeneracy
- Replies: 4
- Views: 230
Re: Degeneracy
There is more complexity in the SO2F2, whereas FClO3 doesn't have the same complexity. For the FClO3, the position of the Cl changes, so there are four possible places for the Cl to be. For the SO2F2, the oxygens and fluorines can change positions, so there are more orientations/microstates because ...
- Sat Feb 02, 2019 12:31 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Work
- Replies: 4
- Views: 276
Re: Work
In the example of irreversibility (isothermal expansion), the gas isn't doing work on its surroundings as the external pressure is zero. Therefore, P(ΔV) = 0. However, in the reversible example, the gas is doing work on its surroundings. Therefore, the reversible system is always doing more work tha...
- Sat Feb 02, 2019 12:13 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Surroundings
- Replies: 11
- Views: 443
Re: Surroundings
Basically, you get to decide what the system is and what the surroundings are depending on what you are doing and what experiment you are running.
- Sat Feb 02, 2019 12:11 pm
- Forum: Phase Changes & Related Calculations
- Topic: PΔV for solids and liquids
- Replies: 5
- Views: 227
Re: PΔV for solids and liquids
Also, I assume that his focus on gases means that we won't need to focus on how the expression applies to liquids and solids for tests and such.
- Sun Jan 27, 2019 1:40 pm
- Forum: Phase Changes & Related Calculations
- Topic: State Properties
- Replies: 3
- Views: 202
Re: State Properties
There are different formulas for work depending on the type of work done. In this section, I think that we will be working with the equation:
work = -(Pressure)(Change in Volume)
work = -(Pressure)(Change in Volume)
- Sun Jan 27, 2019 1:34 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Heat vs Enthalpy
- Replies: 4
- Views: 154
Re: Heat vs Enthalpy
Heat is the energy that causes changes in a system. Enthalpy is the total state of the system - the heat consumed or released by the system.
- Sun Jan 27, 2019 1:27 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: The 3 methods
- Replies: 5
- Views: 243
Re: The 3 methods
For bond enthalpy, you first have to recognize which bonds are broken and which are formed. Then you add the bond enthalpies for the bonds that are broken and do the same for the bonds that are formed. From there, subtract the bond enthalpy of the formed bonds from the bond enthalpy of the broken bo...
- Mon Jan 21, 2019 10:09 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Weak acid and its salt
- Replies: 3
- Views: 91
Re: Weak acid and its salt
In general, when creating ICE tables, you are only looking at the molecules that affect pH. Therefore, if a cation/anion does not appear in the acid/base reaction, then one does not have to take the cation/anion into account.
- Mon Jan 21, 2019 10:03 am
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: pKa and pKb
- Replies: 3
- Views: 507
Re: pKa and pKb
With regards to Kb, one is looking at the pOH scale instead of the pH scale because OH- ions are produced, not H3O+ during the dissociation/reaction of a base. Therefore, on the pOH scale, having a greater value means that the base is weak.
- Mon Jan 21, 2019 9:56 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Temperature for Equilibrium
- Replies: 4
- Views: 169
Re: Temperature for Equilibrium
Each question should give you the temperature at which the equilibrium value (K) occurs. If no temperature occurs, assume 25 degrees Celsius.
- Sat Jan 12, 2019 2:27 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Adding a liquid
- Replies: 7
- Views: 300
Re: Adding a liquid
Once a reaction has reached equilibrium, it will stay at equilibrium as long as the reaction itself is not affected. Adding a liquid will not change the equilibrium concentrations/pressures of K value so long as the liquid does not react with the gases that have already reacted and reached equilibri...
- Sat Jan 12, 2019 2:23 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework 11.79 6th edition
- Replies: 2
- Views: 122
Re: Homework 11.79 6th edition
There really is no significance for using bar for units as opposed to pascals or atm. The textbook writers probably just want to make sure that the reader can use all the different kinds of units and is consistent when using each one. For this particular problem, I would suggest finding K(p) using t...
- Fri Jan 11, 2019 10:22 am
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: what is lechatliers principle [ENDORSED]
- Replies: 4
- Views: 595
Re: what is lechatliers principle [ENDORSED]
Also, an applied stress can be caused by changes to the environment, such as compression or temperature. As a result, the reaction will favor the side (either reactants or products) that will reduce the stress caused by that change. For example, if a container is compressed, the reaction occurring w...