Search found 30 matches

by Aarti K Jain 1L
Sun Mar 10, 2019 2:04 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: negatives in the equation
Replies: 2
Views: 58

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:57 pm
Forum: Van't Hoff Equation
Topic: Overview
Replies: 8
Views: 299

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:53 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Molecularity
Replies: 3
Views: 61

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:41 pm
Forum: General Rate Laws
Topic: Rate Determining Step
Replies: 5
Views: 2430

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:28 pm
Forum: Van't Hoff Equation
Topic: Van't Hoff use
Replies: 13
Views: 435

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:20 pm
Forum: General Rate Laws
Topic: Problem
Replies: 1
Views: 41

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:16 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: k & K
Replies: 18
Views: 371

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.
by Aarti K Jain 1L
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: 93

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.
by Aarti K Jain 1L
Sun Mar 10, 2019 1:07 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: k' Value
Replies: 1
Views: 36

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...
by Aarti K Jain 1L
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: 78

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...
by Aarti K Jain 1L
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: 44

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!
by Aarti K Jain 1L
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: 56

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.
by Aarti K Jain 1L
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: 77

Re: Question 14.25 (Sixth Edition)

The more negative the standard potential, the more strongly the reducing agent/element.
by Aarti K Jain 1L
Mon Feb 25, 2019 12:18 pm
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: Spontaneous and unstable
Replies: 4
Views: 88

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.
by Aarti K Jain 1L
Mon Feb 25, 2019 12:16 pm
Forum: Balancing Redox Reactions
Topic: Balancing Charges
Replies: 4
Views: 64

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...
by Aarti K Jain 1L
Mon Feb 25, 2019 12:11 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Negative Delta G
Replies: 2
Views: 53

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...
by Aarti K Jain 1L
Mon Feb 25, 2019 11:55 am
Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
Topic: trends in standard potential
Replies: 2
Views: 52

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...
by Aarti K Jain 1L
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: 93

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 ...
by Aarti K Jain 1L
Sat Feb 02, 2019 12:31 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Work
Replies: 4
Views: 139

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...
by Aarti K Jain 1L
Sat Feb 02, 2019 12:13 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Surroundings
Replies: 11
Views: 150

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.
by Aarti K Jain 1L
Sat Feb 02, 2019 12:11 pm
Forum: Phase Changes & Related Calculations
Topic: PΔV for solids and liquids
Replies: 5
Views: 109

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.
by Aarti K Jain 1L
Sun Jan 27, 2019 1:40 pm
Forum: Phase Changes & Related Calculations
Topic: State Properties
Replies: 3
Views: 81

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)
by Aarti K Jain 1L
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: 72

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.
by Aarti K Jain 1L
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: 81

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...
by Aarti K Jain 1L
Mon Jan 21, 2019 10:09 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Weak acid and its salt
Replies: 3
Views: 54

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.
by Aarti K Jain 1L
Mon Jan 21, 2019 10:03 am
Forum: Non-Equilibrium Conditions & The Reaction Quotient
Topic: pKa and pKb
Replies: 3
Views: 230

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.
by Aarti K Jain 1L
Mon Jan 21, 2019 9:56 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Temperature for Equilibrium
Replies: 4
Views: 74

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.
by Aarti K Jain 1L
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: 101

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...
by Aarti K Jain 1L
Sat Jan 12, 2019 2:23 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Homework 11.79 6th edition
Replies: 2
Views: 59

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...
by Aarti K Jain 1L
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: 455

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...

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