Search found 34 matches
- Thu Mar 14, 2019 8:47 pm
- Forum: General Rate Laws
- Topic: Rate of Formation vs Unique Rate
- Replies: 4
- Views: 436
Rate of Formation vs Unique Rate
is rate of formation the same as unique rate?
- Thu Mar 14, 2019 3:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibruim Assessment
- Replies: 5
- Views: 543
Re: Equilibruim Assessment
Ethan Breaux 2F wrote:don't forget if the x value is very small (i.e. less than 10^-3) you can ignore the x when subtracting it from another value!
I thought if K is less than 10^-3 then we can estimate? Not x?
- Thu Mar 14, 2019 12:03 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibruim Assessment
- Replies: 5
- Views: 543
Re: Equilibruim Assessment
Thank you!!
- Thu Mar 14, 2019 12:02 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium constants with partial pressures and concentrations?
- Replies: 2
- Views: 280
Re: Equilibrium constants with partial pressures and concentrations?
yes! you exclude liquids and solids
- Wed Mar 13, 2019 11:23 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibruim Assessment
- Replies: 5
- Views: 543
Equilibruim Assessment
A mixture initially consisting of 0.250 N2 (g) and 0.500 M H2 (g) reacts to form NH3 (g) which is 0.15 M NH3 (g) at equilibrium. Calculate the concentration of N2 (g) at equilibrium for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) can someone help me find the concentration of N2? Above is everything that i...
- Thu Mar 07, 2019 4:50 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: Unique Average Rate
- Replies: 5
- Views: 555
Re: Unique Average Rate
so the unique average rate of each product/reactant is the same exact thing?
- Thu Mar 07, 2019 4:44 pm
- Forum: General Rate Laws
- Topic: Rate Laws
- Replies: 8
- Views: 654
Rate Laws
Are we expected to know how to get from the differential rate laws to the integrated rate law for the orders, 0, 1, and 2?
- Tue Mar 05, 2019 4:28 pm
- Forum: Second Order Reactions
- Topic: 6th Edition 15.15
- Replies: 2
- Views: 317
6th Edition 15.15
For a reaction, when OH conc. was doubled, the rate doubled. When the CH3OH concentration alone was increased by a factor of 1.2, the rate increased by a factor of 1.2. Write the rate law for the reaction. The answer key says "rate is 1st order in both reactants." But I thought it would be...
- Wed Feb 27, 2019 10:45 am
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Value of n
- Replies: 7
- Views: 910
Re: Value of n
In order to find n (the number of electrons transferred) write down both half reactions and balance them. You always want to make sure that the number of electrons being transferred is the same amount for both half reactions.
- Tue Feb 26, 2019 4:30 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: H+ and OH- in cell diagrams
- Replies: 1
- Views: 244
Re: H+ and OH- in cell diagrams
for this same homework question, how come we don't add H2O in order to balance the OH-?
- Tue Feb 26, 2019 3:58 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell diagram
- Replies: 4
- Views: 385
Cell diagram
How do you identify if a cell diagram needs an inert conductor? Also, which side would you put it on, left or right? Thank you!
- Tue Feb 26, 2019 3:56 pm
- Forum: Balancing Redox Reactions
- Topic: redox homework question
- Replies: 3
- Views: 295
Re: redox homework question
are you sure that oxygen got oxidized? I thought that it remained the same (-2) on both sides.
- Wed Feb 20, 2019 5:34 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Spontaneity in the Gibbs Eqn
- Replies: 4
- Views: 419
Re: Spontaneity in the Gibbs Eqn
When both are negative it would be spontaneous at LOW temps. If both were positive, it would be spontaneous at HIGH temps.
- Wed Feb 20, 2019 5:26 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Deprotonation
- Replies: 3
- Views: 308
Re: Deprotonation
Thank you!
- Wed Feb 20, 2019 5:25 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Electromotive force
- Replies: 2
- Views: 279
Electromotive force
I am still confused on what (emf) is. Can someone please explain and what is its purpose?
- Wed Feb 20, 2019 5:13 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Deprotonation
- Replies: 3
- Views: 308
Deprotonation
What is deprotonation? And how does that relate to Ka2?
- Wed Feb 20, 2019 4:53 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Gibb's free energy Units
- Replies: 4
- Views: 364
Gibb's free energy Units
Do the units of delta G always have to be kj/mol or can we leave them in J/mol?
- Tue Feb 12, 2019 5:40 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Test 1 Question
- Replies: 3
- Views: 374
Re: Test 1 Question
Thank you!!!
- Tue Feb 12, 2019 4:28 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Test 1 Question
- Replies: 3
- Views: 374
Test 1 Question
If a reaction is at equilibrium with deltaH +2.86 kj what would happen if the temperature of the reaction is increased?
- Tue Feb 05, 2019 10:16 pm
- Forum: Phase Changes & Related Calculations
- Topic: isothermal vs reversible
- Replies: 1
- Views: 202
Re: isothermal vs reversible
I know that isothermal means constant temperature, but I need help understanding this more too.
- Tue Feb 05, 2019 10:11 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Bond Enthalpies
- Replies: 2
- Views: 296
Re: Bond Enthalpies
yes, I believe so. When bonds break, energy is required to break the bond so it is positive. However, when bonds are formed, energy is released that is why it is negative. So breaking a bond is endothermic while forming a bond is exothermic.
- Tue Feb 05, 2019 4:26 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 6th edition Hw problem 8.23
- Replies: 1
- Views: 269
6th edition Hw problem 8.23
A calorimeter was calibrated with an electric heater, which supplied 22.5 kJ of energy as heat to the calorimeter and increased the temperature of the calorimeter and its water bath from 22.45 degrees C to 23.97 degrees C. What is the heat capacity of the calorimeter? What equation is used in order ...
- Tue Jan 29, 2019 4:19 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: 6th Edition Problem 8.21
- Replies: 8
- Views: 807
Re: 6th Edition Problem 8.21
heat lost by the copper= -heat gained by the water
so, you use the equation (mass of copper)(Tfinal-TInitial of copper)(C)=-(mass of water)(Tfinal-Tinitial of water)(C). Then solve for Tfinal
so, you use the equation (mass of copper)(Tfinal-TInitial of copper)(C)=-(mass of water)(Tfinal-Tinitial of water)(C). Then solve for Tfinal
- Tue Jan 29, 2019 2:52 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Sig Figs with Enthalpy
- Replies: 4
- Views: 627
Re: Sig Figs with Enthalpy
I think you should always think about, and apply the rules for sig figs. It's a good habit.
- Tue Jan 29, 2019 2:49 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: 6th edition hw problem 8.19
- Replies: 1
- Views: 232
6th edition hw problem 8.19
(a) Calculate the heat that must be supplied to a 500.0-g copper kettle containing 400.0 g of water to raise its temperature from 22.0 degrees C to the boiling point of water, 100.0 degrees C. Why do we need to account for the copper kettle and not just the water in order to find the amount of heat ...
- Tue Jan 22, 2019 4:40 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Finding Ka or Kb
- Replies: 2
- Views: 1327
Re: Finding Ka or Kb
Although some bases or acids are not on the tables, you can look at their conjugate acid or base and then use the equation Ka x Kb = Kw to find the opposite. For example if you have the conjugate base Kb you would manipulate the equation in order to solve for ka and vice-versa. ka=1.00x10^-14/kb
- Tue Jan 22, 2019 3:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Setting up Equilibrium Equations
- Replies: 2
- Views: 220
Setting up Equilibrium Equations
Are we supposed to know how the equilibrium equation looks off the top of our heads when given a solution. For example, in the 6th edition book, problem number 12.69 asks to find the ph for the solution .19 M NH4Cl only. And we obviously need the equilibrium equation to set up the ice table. For hom...
- Tue Jan 22, 2019 2:56 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating the pH - 6D.17 7th edition
- Replies: 2
- Views: 165
Re: Calculating the pH - 6D.17 7th edition
I don't have the 7th edition book but the 6th edition has a similar problem that only gave the initial concentration. What I did was first find ka using the equation ka x kb= kw. Once you find ka you can use the ice table to solve for x. Once you find your x you can plug that into -log(x) which equa...
- Sun Jan 20, 2019 2:51 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sig figs in pH
- Replies: 7
- Views: 751
Re: Sig figs in pH
Dr. Lavelle has 2 links on his website that explain sig figs. It also shows good examples that might be helpful, if you need more clarification.
- Sun Jan 20, 2019 2:48 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Removing products
- Replies: 10
- Views: 3855
Re: Removing products
I think it would affect the ratio definitely, but since K is being manipulated it would no longer be K, it would be Q, so I think Q would be the one affected
- Sun Jan 20, 2019 12:20 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Stoichiometric Coefficients
- Replies: 4
- Views: 393
Re: Stoichiometric Coefficients
I have the same question. I think we would have to factor in the coefficients in the ICE table where we normally account for the change in x. Im not 100% sure though.
- Sun Jan 13, 2019 7:25 pm
- Forum: Student Social/Study Group
- Topic: New to Lavelle
- Replies: 32
- Views: 5118
Re: New to Lavelle
I think doing hw problems is the best way to practice the material. Also, attending office hours is extremely helpful.
- Sun Jan 13, 2019 7:22 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Reaction Quotient Meaning
- Replies: 2
- Views: 208
Re: Reaction Quotient Meaning
I believe you use Q when you are not sure whether or not the reaction is at equilibrium. Or when K is given and you calculate the ratio and it is a different value than K, therefore you would be using Q.
- Sun Jan 13, 2019 7:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating Q
- Replies: 7
- Views: 426
Re: Calculating Q
When calculating both Q or R you put the products over the reactants.