## Search found 30 matches

Fri Mar 15, 2019 11:20 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Rate of formation, rate in terms of loss
Replies: 3
Views: 82

### Re: Rate of formation, rate in terms of loss

Rate of formation of B is equal to 1/2 rate of decomp of A because A has a coefficient of 2.
Fri Mar 15, 2019 7:08 pm
Forum: Administrative Questions and Class Announcements
Replies: 179
Views: 11033

sonalivij wrote:For worksheet 8 #4d I am getting 144 M/s instead of 140 M/s. Am I doing something wrong?

She rounded to 140 because you need to have two sig figs.
Thu Mar 14, 2019 1:53 pm
Forum: Experimental Details
Topic: Collision Theory
Replies: 8
Views: 285

### Re: Collision Theory

According to the outline, you need to understand how the collision model accounts for the temperature dependence of reactions. So in collision theory, molecules must collide at a minimum kinetic energy in order to break bonds and form new ones. The higher the temperature, the higher the kinetic ener...
Fri Mar 08, 2019 12:52 pm
Forum: Method of Initial Rates (To Determine n and k)
Topic: HW Number 17 (6th ed.)
Replies: 2
Views: 89

### Re: HW Number 17 (6th ed.)

In experiment 1 and 3, the concentration of A also changes so you cannot determine if the change in initial rate is because of the change in [A] or [B]. In experiments 2 and 3, only [B] changes ([C] does not affect rate), so you know that any change in rate is due to the change in [B].
Fri Mar 08, 2019 12:49 pm
Forum: Kinetics vs. Thermodynamics Controlling a Reaction
Topic: Unique Rates
Replies: 3
Views: 122

### Re: Unique Rates

A unique rate is the change in concentration of a reactant or product divided by the stoichiometric coefficient. It is the same for each reactant/product in the reaction but is negative for the reactant and positive for the product.
Fri Mar 08, 2019 12:45 pm
Forum: Experimental Details
Topic: Rate Laws
Replies: 2
Views: 235

### Re: Rate Laws

Rate laws are determined experimentally, so they will be given. You cannot determine rate law from chemical equation unless it is an elementary reaction.
Wed Feb 27, 2019 11:24 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15 a 6th edition
Replies: 1
Views: 70

### Re: 14.15 a 6th edition

The cathode is the half-rxn with a greater Eo. The greater the Eo, the greater its reducing power, so for 14.15a,
AgBr+e- ->Ag+Br- has a lower reducing power in comparison to Ag+ +e- ->Ag, so it is the anode.
Wed Feb 27, 2019 11:17 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Eo an Intensive Property
Replies: 1
Views: 48

### Re: Eo an Intensive Property

Eo is measured under standard conditions and it is a measure of the electron-pulling power of a single electrode.
Wed Feb 27, 2019 11:05 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Reducing power
Replies: 2
Views: 62

### Re: Reducing power

A reducing agent is a compound that acts as an electron donor (is oxidized) and therefore reduces another compound. The reducing power is the ability of a compound to reduce another compound.
Fri Feb 22, 2019 4:37 pm
Forum: Balancing Redox Reactions
Topic: Positive E means Favorable
Replies: 2
Views: 420

### Re: Positive E means Favorable

In the textbook it says that when the standard potential is positive, the reduction half-reaction has a greater electron-pulling power, so the tendency for the half-reaction to occur is stronger, which is what I think he meant by the reactants are favored.
Fri Feb 22, 2019 4:32 pm
Forum: Balancing Redox Reactions
Topic: Standard Oxidation Potentials?
Replies: 1
Views: 56

### Re: Standard Oxidation Potentials?

Oxidation potentials should just be the reverse sign of reduction potentials.
Fri Feb 22, 2019 4:27 pm
Forum: Balancing Redox Reactions
Topic: Test 2 Textbook Sections
Replies: 2
Views: 98

### Test 2 Textbook Sections

Up to what sections of chapter 14 in the textbook (6th ed) should we know for Test 2?
Wed Feb 13, 2019 4:28 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: Delta n
Replies: 7
Views: 408

### Re: Delta n

You can see the change in moles from the balanced chemical equation.
Tue Feb 12, 2019 3:49 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Application of Standard Enthalpy of Formation
Replies: 1
Views: 125

### Re: Application of Standard Enthalpy of Formation

Use the stoichiometric coefficients to find reaction enthalpy. So delta H = 2(-395.72)-2(-296.83). Then to find the enthalpy for 0.030 mol, divide your delta H by two and multiply by 0.030.
Tue Feb 12, 2019 3:40 pm
Forum: Calculating Work of Expansion
Topic: constant external pressure
Replies: 2
Views: 79

### Re: constant external pressure

It should either be given or you could be given enthalpy. The definition of enthalpy is heat at constant pressure, so if you're given enthalpy, you know pressure is constant.
Tue Feb 05, 2019 9:36 pm
Forum: Calculating Work of Expansion
Topic: Units
Replies: 5
Views: 268

### Re: Units

You have to convert L atm into joules. 1 L atm = 101.325 J
Tue Feb 05, 2019 9:30 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: Determining the Number of Orientations/Degeneracy & Microstates
Replies: 3
Views: 134

### Re: Determining the Number of Orientations/Degeneracy & Microstates

Degeneracy is how many microstates there are. By the number of states for the base of the degeneracy equation, it means the number of ways the molecule can be arranged. For example, for carbon monoxide, it can either be arranged C-O or O-C, which is why the base is 2 and w=2^N
Tue Feb 05, 2019 9:25 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Cv vs Cp
Replies: 3
Views: 113

### Re: Cv vs Cp

The heat capacity at constant pressure (Cp) is greater than Cv because when heat is added at constant pressure, the gas is allowed to expand and some of the heat is used as work. At constant volume, the gas cannot expand so no work is done and all the heat goes into raising the temperature of the sy...
Wed Jan 30, 2019 7:40 pm
Forum: Phase Changes & Related Calculations
Topic: HW problem 4A.5
Replies: 1
Views: 96

### Re: HW problem 4A.5

A reversible process occurs when a system is at equilibrium (P internal = P external). It can be reversed by an infinitely small change in a variable. For example, if the pressure inside a cylinder is about equal to the pressure outside of it, the piston can move in either direction if there is a sm...
Wed Jan 30, 2019 7:25 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 6th edition 8.41
Replies: 1
Views: 51

### Re: 6th edition 8.41

You need to set q(ice)= -q(water)
q(ice) consists of both the heat required to melt the ice at 0 degrees as well as the heat required to raise the ice from 0 degrees to the final temperature.
So q(ice)=mCs(Tf-0.0) + n x enthalpy of fusion
q(water is just mCs(Tf-45)
Mon Jan 28, 2019 12:09 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat capacity
Replies: 1
Views: 45

### Re: Heat capacity

Heat capacity is the ratio of the heat supplied to the rise in temperature, so heat capacity=q/delta T
Wed Jan 23, 2019 11:26 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Chemical Eq Part #3 Post-Module Assessment
Replies: 1
Views: 63

### Re: Chemical Eq Part #3 Post-Module Assessment

After setting up an ICE table you should get the expression 1.80=x^2/(3-x). Simplifying gives you the quadratic equation x^2+1.80x-5.4=0. Solving for x you should get x=1.59. Since the equilibrium concentration for PCl5 is 3-x, [PCl5]=1.41, [PCl3]=[Cl2]=1.59
Wed Jan 23, 2019 11:19 am
Forum: Ideal Gases
Topic: atm vs. bar?
Replies: 23
Views: 534

### Re: atm vs. bar?

Bar is the standard unit of pressure but 1 atm is equal to approximately 1.01 bar so when calculating equilibrium constant they are about the same.
Wed Jan 23, 2019 11:13 am
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Ch.11 #25
Replies: 1
Views: 41

### Re: Ch.11 #25

We used the synthesis reaction because that is what is given in table 11.2 and the problem tells you to refer to the table for the value of Kc. If you solved for the decomposition of HI you would get 1/Kc since it is the reverse reaction.
Tue Jan 15, 2019 2:17 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE Tables [ENDORSED]
Replies: 11
Views: 331

### Re: ICE Tables[ENDORSED]

You can use an ICE table when you are given initial concentrations and you need to find the equilibrium concentrations. The products will not always start at zero but the values will be given.
Mon Jan 14, 2019 10:40 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Pressure [ENDORSED]
Replies: 2
Views: 80

### Re: Pressure[ENDORSED]

Adding an inert gas has no effect on the equilibrium concentration so it does not change the constant.
Mon Jan 14, 2019 12:44 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Catalysts [ENDORSED]
Replies: 3
Views: 90

### Re: Catalysts[ENDORSED]

Catalysts do not change the equilibrium constant and has no effect on the equilibrium composition. It speeds up both the forward and reverse reactions to the same extent, so dynamic equilibrium is not affected.
Wed Jan 09, 2019 3:08 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: Equilibrium table
Replies: 1
Views: 39

### Re: Equilibrium table

You can use the approximation where x is negligible when you have a small K value.
Wed Jan 09, 2019 3:05 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: When can "-X" be disregarded?
Replies: 2
Views: 61

### Re: When can "-X" be disregarded?

According to the textbook, x can be disregarded when it is less than 5% of the initial value. You can do this when you have a small K value.
Mon Jan 07, 2019 8:52 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: 6th Edition Ch11 Question 1d
Replies: 3
Views: 60

### Re: 6th Edition Ch11 Question 1d

If you increase the concentration of reactants, it will form more product, however this doesn't mean that the concentration of product will be greater than the concentration of reactants at equilibrium. Partial pressure is the pressure exerted by a gas in a mixture and is expressed in atm, bar, Torr...