Search found 36 matches
- Sun Mar 10, 2019 5:20 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Rate Determining Step
- Replies: 2
- Views: 338
Re: Rate Determining Step
It does not matter which elementary step (first, second, third, etc) is the slow step. It only matters that the slow step in the reaction mechanism determines the rate of the reaction.
- Sun Mar 10, 2019 5:18 pm
- Forum: Second Order Reactions
- Topic: pseudo-first-order reaction
- Replies: 4
- Views: 518
Re: pseudo-first-order reaction
pseudo first order reactions are second order reactions which under some conditions (as described above) behave like first order reactions
- Sun Mar 10, 2019 5:16 pm
- Forum: Second Order Reactions
- Topic: pseudo first order rxns
- Replies: 3
- Views: 369
Re: pseudo first order rxns
At the kinetics review session today, the facilitator said that we likely will not be tested mathematically on pseudo-first order reactions. We just need to know conceptually that pseudo-first order reactions are second order reactions which under some particular conditions (as described above), loo...
- Sun Mar 10, 2019 5:13 pm
- Forum: First Order Reactions
- Topic: Graphs
- Replies: 2
- Views: 389
Re: Graphs
I think you can tell the reaction order of a reaction by looking at the graphs and their axes (as mentioned above). However, I am not sure if you can mathematically compute the reaction order from a graph.
- Sun Mar 10, 2019 5:11 pm
- Forum: Zero Order Reactions
- Topic: how do we tell if a reaction is zero order?
- Replies: 5
- Views: 596
Re: how do we tell if a reaction is zero order?
As my peers have indicated previously, a zero order reaction's rate does not depend on the initial concentration of the reactant. No matter what initial concentration it has, the rate of the reaction will proceed at one certain speed. Also, first and second order reactions have concentration vs. tim...
- Sun Mar 10, 2019 5:09 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: What does unique rate of reaction mean?
- Replies: 6
- Views: 618
Re: What does unique rate of reaction mean?
The unique rate of each reactant and product are the same in a reaction because the changes are with regards to each species' coefficient. The average and instantaneous rates of each reactant and product in a reaction may differ.
- Fri Mar 08, 2019 12:17 am
- Forum: Zero Order Reactions
- Topic: initial concentration
- Replies: 6
- Views: 664
Re: initial concentration
The characteristic of zero order reactions is that they occur at a rate that is independent of the initial concentration. No matter how much of a reactant you have initially, the reaction will proceed at the same rate (that of the rate constant, k).
- Fri Mar 08, 2019 12:16 am
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Units of k
- Replies: 4
- Views: 490
Re: Units of k
To find the units of k depending on the reaction order, simply plug in M/s for Rate and M for the concentrations present. Then solve for the units of k using algebra.
- Fri Mar 08, 2019 12:14 am
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: kinetics vs thermodynamics
- Replies: 3
- Views: 420
Re: kinetics vs thermodynamics
thermodynamics tells us whether a reaction will occur at all (if it has the favorable conditions of being exothermic and spontaneous). kinetics tell us how fast, or the rate of, a given reaction under the condition that it does occur/is thermodynamically favorable.
- Sun Mar 03, 2019 2:13 pm
- Forum: General Rate Laws
- Topic: unique rate
- Replies: 2
- Views: 291
Re: unique rate
The unique rate is the same of products and reactants in that unique reaction.
given aA-->bB+cC,
unique rate=-1/a * [A]=1/b * [B] = 1/c * [C]
Instantaneous rates can be different among reactants and products of a reaction because their coefficients are not considered.
given aA-->bB+cC,
unique rate=-1/a * [A]=1/b * [B] = 1/c * [C]
Instantaneous rates can be different among reactants and products of a reaction because their coefficients are not considered.
- Sun Mar 03, 2019 2:08 pm
- Forum: Balancing Redox Reactions
- Topic: Adding H+ and H2O
- Replies: 13
- Views: 2905
Re: Adding H+ and H2O
To balance H+ in a basic solution, add one molecule of H20 per every H+ you need. Then, add that same number of OH- on the opposite side.
- Sun Mar 03, 2019 2:06 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell diagrams and solids
- Replies: 9
- Views: 1020
Re: Cell diagrams and solids
If there is a solid or liquid (like Hg (mercury)), an inert solid like Pt(s) is not needed
- Sat Feb 23, 2019 3:31 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: 14.9 6th Edition
- Replies: 2
- Views: 281
Re: 14.9 6th Edition
I am also not sure about how I is being oxidized since its oxidation number is -1 on both sides of the equation. As for the number of moles of electrons transferred, I believe we have to refer to the electrochemical series. The reduction equation for Ce4+ + e- --> Ce3+ in the table in Appendix 2B in...
- Sat Feb 23, 2019 3:27 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Salt Bridge
- Replies: 3
- Views: 339
Re: Salt Bridge
In addition, the salt bridge neutralizes the two solutions as they become charged due to electron transfer between the metal electrodes.
- Sat Feb 23, 2019 3:25 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Test 2 Help
- Replies: 7
- Views: 758
Re: Test 2 Help
I asked him about that equation after class on Friday and he says it is fair game because we can derive it ourselves from deltaG=-RTlnK and deltaG=-nFE but setting them equal to each other and dividing both sides by -RT.
- Mon Feb 18, 2019 10:19 am
- Forum: Phase Changes & Related Calculations
- Topic: Negative and Positive values of delta H and w
- Replies: 6
- Views: 3852
Re: Negative and Positive values of delta H and w
w>0 means work is being done on the system
w<0 means work is being done by the system
H>0 means the system requires heat (heat flows in)
H<0 means the system release heat (heat flows out)
w<0 means work is being done by the system
H>0 means the system requires heat (heat flows in)
H<0 means the system release heat (heat flows out)
- Mon Feb 18, 2019 10:16 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Battery
- Replies: 5
- Views: 524
Re: Battery
I would also say a battery is a closed system because although it cannot transfer matter into its surroundings, it can certainly transfer electrical energy.
- Mon Feb 18, 2019 10:13 am
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Endothermic and Exothermic Graphs
- Replies: 3
- Views: 1075
Re: Endothermic and Exothermic Graphs
We often discuss endothermic and exothermic reactions as absorbing heat and releasing heat, respectively. Heat is a form of energy, which is why energy is on the y axis.
- Mon Feb 11, 2019 9:48 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: General entropy question
- Replies: 9
- Views: 854
Re: General entropy question
Adding heat, increasing temperature, making the substance a gas, increasing the volume of the substance all increase entropy
- Mon Feb 11, 2019 9:42 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: why is H negative when calculating S?
- Replies: 4
- Views: 460
Re: why is H negative when calculating S?
When you go from a liquid to a solid, delta S is negative because the molecules are more ordered (less disordered).
- Mon Feb 11, 2019 9:22 am
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 24229
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
I got +312kJ initially too, but I think the question is more complicated in that you have to write out the combustion equations of the three equations given and then use Hess' Law. When you do that, you get -312kJ.
- Tue Feb 05, 2019 10:54 pm
- Forum: Calculating Work of Expansion
- Topic: +/- work values
- Replies: 5
- Views: 521
Re: +/- work values
An abstract way to think about work's + vs - values is that when work is being done ON the system, w is positive because there is "work added into the system." Conversely, when work is done BY the system, w is negative because "the system is losing work."
- Tue Feb 05, 2019 10:50 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Bomb Calorimeter
- Replies: 4
- Views: 557
Re: Bomb Calorimeter
Like above comments have mentioned, there is no change in volume in a bomb calorimeter. With this in mind while calculating change in internal energy (delta U) where delta U=q+w and w=-PdeltaV, w would be equal to 0 since delta V is 0. Therefore, the internal energy change (delta U) of a bomb calori...
- Tue Feb 05, 2019 10:48 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Reversible vs Irreversible
- Replies: 2
- Views: 316
Re: Reversible vs Irreversible
Isothermal, reversible expansions generate the maximum work. Since work is the area under a curve (of P on the x-axis and V on the y-axis), imagine the area shaded under the curve of the reversible expansion to be almost touching the curve itself since the changes in volume are so small. Alternative...
- Sun Feb 03, 2019 10:12 am
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: What does "reversible isothermal" mean?
- Replies: 1
- Views: 170
What does "reversible isothermal" mean?
The textbook references "reversible isothermal" expansions and concludes that these expansions have an internal energy change of 0. Does this correlate to Dr. Lavelle's "isolated system" (reaction in a bomb calorimeter) example from lecture 11?
- Tue Jan 29, 2019 11:35 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Delta U vs Delta H
- Replies: 6
- Views: 1603
Re: Delta U vs Delta H
It is also noteworthy that the equation deltaH=deltaU+(P)deltaV is only applicable in the case that the heat is supplied to the system at a constant pressure. deltaU is also equal to q + w, where q describes molecules moving in random direction and w describes molecules moving in a definite directio...
- Tue Jan 29, 2019 11:29 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Standard enthalpy of formation vs. enthalpy of formation
- Replies: 3
- Views: 341
Standard enthalpy of formation vs. enthalpy of formation
Is the difference between "standard enthalpy of formation" and "enthalpy of formation" just that the reactants and products of the "standard enthalpy of formation" reaction are in their standard states and those in a reaction asking for the "enthalpy of formation&q...
- Tue Jan 29, 2019 11:21 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Lecture
- Replies: 5
- Views: 495
Re: Lecture
Moreover, the heating curve showed that water vapor burns more than liquid water even though they are at the same temperature (100˚C, shown on Y axis), because water vapor has more heat than liquid water (shown by the fact that it is further right on the X axis, which represents heat).
- Mon Jan 21, 2019 10:43 am
- Forum: Administrative Questions and Class Announcements
- Topic: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
- Replies: 179
- Views: 24229
Re: DOWNLOAD SESSION WORKSHEETS HERE - Sun 7-9PM (Karen)
Did anyone happen to get around 0.077g for #2 on the acid-base review worksheet? The key says 0.10g, so I am unsure whether Karen rounded or I did the question incorrectly.
- Wed Jan 16, 2019 2:45 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Exothermic vs. Endothermic Reaction
- Replies: 9
- Views: 3243
Re: Exothermic vs. Endothermic Reaction
It helps to see an energy graph to think about why exothermic has a negative change in H and endothermic has a positive change in H. In exothermic reactions, heat is leaving the system, so the graph will end up at a lower delta H value in the products compared to the reactants. Therefore, the change...
- Wed Jan 16, 2019 2:41 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: 11.69 (6th Edition)
- Replies: 3
- Views: 258
Re: 11.69 (6th Edition)
The equilibrium constant K is a constant that describes the ratio of concentrations of products/concentrations of reactants at equilibrium (when the rate of the forward reaction is equal to the rate of the reverse reaction). Therefore, it is only "relevant" when the reaction is at equilibr...
- Wed Jan 16, 2019 2:34 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Number 115 Chapter 11 HW
- Replies: 2
- Views: 226
Re: Number 115 Chapter 11 HW
I believe the "quick way" Dr. Lavelle described in class to determine which way the reaction shifts when the pressure increases only applies for moles of gas, not gas and aqueous solutions. The quick way is to see which side has fewer moles of gas and the reaction will proceed in that dire...
- Wed Jan 09, 2019 7:39 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Homework problem 11.89 6th edition
- Replies: 1
- Views: 189
Homework problem 11.89 6th edition
Could anyone please explain why the total pressure is 100 for 11.89 in the 6th edition? 11.89 The following plot shows how the partial pressures of reactant and products vary with time for the decomposition of compound A into compounds B and C. All three compounds are gases. Use this plot to do the ...
- Tue Jan 08, 2019 10:04 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Equilibrium constants Kp notation
- Replies: 2
- Views: 267
Re: Equilibrium constants Kp notation
For Kp, the solution manual shows the values of partial pressures in parentheses. Just putting CO2 in brackets like [CO2] is the concentration of CO2, which is correct for a Kc expression but incorrect for Kp. Instead, for pressures, I believe we should use parentheses like so: (P subscript CO2).
- Tue Jan 08, 2019 9:59 pm
- Forum: Ideal Gases
- Topic: Activity
- Replies: 3
- Views: 290
Re: Activity
Instead of using activity, we use molarities (concentration) in our calculations of the equilibrium constant value which is a close approximation to activities (more accurate but harder to measure).
- Tue Jan 08, 2019 9:53 pm
- Forum: Ideal Gases
- Topic: Conversion
- Replies: 1
- Views: 90
Re: Conversion
Dr. Lavelle used PV=nRT to convert between pressure and concentration. Concentration is in molarity, which is defined by moles over L. In the Ideal Gas Law, moles is expressed by "n" and L is expressed by "V," the volume. If we divide both sides of the Ideal Gas Law by V, we will...