Search found 56 matches
- Fri Mar 12, 2021 6:19 pm
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: Intermediates
- Replies: 6
- Views: 313
Re: Intermediates
Yup intermediates aren't included in rate laws but must be substituted after rearranging a fast step.
- Fri Mar 12, 2021 5:18 pm
- Forum: General Rate Laws
- Topic: Intermediate OH- present in overall rate law [ENDORSED]
- Replies: 2
- Views: 240
Re: Intermediate OH- present in overall rate law [ENDORSED]
The [OH-] comes from rewriting the concentration of the intermediate that is a reactant in the slow step. We refer to the fast step that has that intermediate as a product. When you rearrange the equation after setting the reverse and forward reactions equal (which you can do because it is a fast eq...
- Fri Mar 12, 2021 4:59 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Identifying reducing and oxidizing agents
- Replies: 8
- Views: 357
Re: Identifying reducing and oxidizing agents
To add on since you know your overall potential must be positive we can compare the standard potentials to identify which one needs to be flipped to become the oxidation half-reaction.
- Wed Mar 10, 2021 11:49 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Catalysts in the Slow Step
- Replies: 17
- Views: 836
Re: Catalysts in the Slow Step
It wouldn't be in the overall rate law but it is in the reactants of one equation and the products of another. This is different from an intermediate because intermediates are consumed when they react but catalysts are not. Catalysts always show up in the reactants of the first step and the products...
- Wed Mar 10, 2021 11:45 am
- Forum: Experimental Details
- Topic: Fast step at equilibrium [ENDORSED]
- Replies: 4
- Views: 458
Fast step at equilibrium [ENDORSED]
In the lecture, it was explained that for the pre equilibrium approach for the first step we assume the first step is at approximately at equilibrium. Can someone explain why because I'm a bit confused still?
- Sat Mar 06, 2021 12:00 am
- Forum: First Order Reactions
- Topic: How to determine the Order of reactions
- Replies: 8
- Views: 382
Re: How to determine the Order of reactions
Graphing and finding which graph yields the straight line is helpful however you can also solve for rate of reaction algebraically from values determined experimentally. rate=[A] m [B] n then plug in values of A and B and the rates. Use two different trial's data but make sure the concentrations of ...
- Fri Mar 05, 2021 11:54 pm
- Forum: General Rate Laws
- Topic: Gas not involved
- Replies: 3
- Views: 206
Re: Gas not involved
Gas products are favored because they can easily escape the solution preventing the reverse reaction for occur which would interfere with collecting the rate of reaction data. So if you have a closed system the gas would not be able to escape and could react with the products to form the reactants m...
- Fri Mar 05, 2021 11:51 pm
- Forum: Kinetics vs. Thermodynamics Controlling a Reaction
- Topic: what does it mean when kinetics rather than thermodynamics is controlling a reaction
- Replies: 8
- Views: 462
Re: what does it mean when kinetics rather than thermodynamics is controlling a reaction
The occurrence of a reaction may depend more on whether it is more kinetically favored rather than thermodynamically. Usually, thermodynamics states that a reactant becoming a more stable product will always be more favorable however the reaction may not occur if it is not kinetically favorable thus...
- Wed Mar 03, 2021 4:12 pm
- Forum: Second Order Reactions
- Topic: Slope =k
- Replies: 4
- Views: 337
Re: Slope =k
I'm not sure if I'm understanding your question correctly but slope=k because if you look at the equation 1/[A]=kt+1/[A]o it matches the format of y=mx+b. In that equation m is the slope therefore k is the slope in the second-order integrated rate law. I guess I'm asking why only for second-order r...
- Wed Mar 03, 2021 4:04 pm
- Forum: Zero Order Reactions
- Topic: When to use each order
- Replies: 19
- Views: 1020
Re: When to use each order
Order is determined experimentally first. You can only find K once you have determined the order. In the lecture today Professor Lavelle went over how to identify this graphically. You want to look for the graph that will give you a straight line and that slope is K. Do this by graphing [A] vs t, ln...
- Wed Mar 03, 2021 4:00 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Nernst Equation Units
- Replies: 1
- Views: 139
Re: Nernst Equation Units
Basically, the units aren't a huge deal here because in the problem everything was in standard conditions and both concentration and pressure are in reference to standard conditions, one bar/atm and molarity can be used for K or Q.
- Sat Feb 27, 2021 5:37 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Sapling 7/8 Question 17 [ENDORSED]
- Replies: 4
- Views: 401
Re: Sapling 7/8 Question 17 [ENDORSED]
Yeah, I think because everything is under standard conditions there are no conversions that are necessary. Thus that is why the partial pressures and concentrations are regarded in the same way.
- Fri Feb 26, 2021 10:17 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Half Reactions
- Replies: 2
- Views: 192
Re: Half Reactions
I don't think this would have a half-reaction you can balance but I'm not 100% sure.
- Fri Feb 26, 2021 9:46 pm
- Forum: Balancing Redox Reactions
- Topic: Sapling week 7/8 #5
- Replies: 2
- Views: 219
Re: Sapling week 7/8 #5
Half reactions will be separated by whether it is being oxidized or reduced. When I'm looking for half reactions I separate them by the elements by making sure matching elements are in the products and reactants. So in the case of Cl2O7(g)+H2O2(aq)⟶ClO2^-(aq)+O2(g), one half reaction(not balanced ye...
- Wed Feb 24, 2021 7:07 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Ion-selective electrode
- Replies: 4
- Views: 220
Re: Ion-selective electrode
Yup the ion-selective electrode has the NERNST equation built into it and often tests for the electron potential to solve for a variable in the equation. The other variables are all built in already and gives the concentration of products or reactants for the given electrical potential.
- Wed Feb 24, 2021 6:58 pm
- Forum: Balancing Redox Reactions
- Topic: Oxidizing vs Reducing
- Replies: 55
- Views: 2116
Re: Oxidizing vs Reducing
Oxidizing agents are being reduced and reducing agents are oxidized. Thus if the oxidation number increases it is being oxidized and therefore is the reducing agent because it accepts electrons from the reduction half-reaction. The opposite applies, oxidizing agents are the reactants being reduced b...
- Wed Feb 24, 2021 6:52 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Oxidation Agent
- Replies: 12
- Views: 563
Re: Oxidation Agent
Oxidizing agents are the reactants that are being reduced thus to understand which is the best oxidizer it is asking for the one with the greatest reduction potential the most positive one. A reduction reaction drives the oxidation when the electrons are being released.
- Thu Feb 18, 2021 10:35 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: Orientation Question
- Replies: 4
- Views: 390
Re: Orientation Question
This is to find W but basically, for a molecule with 2 states, it would have the formula 2^n where n is the number of entities. The number of states is the number of possible positions each atom in the molecule can take. So if you have 4 molecules of CO the degeneracy would be 2^4.
- Thu Feb 18, 2021 10:30 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: When to use delta H of formation and when to use delta H
- Replies: 4
- Views: 230
Re: When to use delta H of formation and when to use delta H
dH at standard temperature can be found using the dH of formation using the equation on the sheet. Multiplying dH of formation by the moles gives us the dH of formation for the number of moles that you multiplied it with. This is done usually to find the energy (q) needed for a phase change.
- Thu Feb 18, 2021 10:27 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Textbook Problem 4J.15
- Replies: 1
- Views: 129
Re: Textbook Problem 4J.15
I saw a post earlier that explained you need to see how T affects Gibbs free energy using dG=dH-TdS. We know that this will depend on the -Tds value. As T increases the term will only become more negative if dS is positive. All the compounds have positive entropy except PCl5 so that is the only one ...
- Thu Feb 18, 2021 10:15 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Textbook Question 4C11
- Replies: 5
- Views: 364
Re: Textbook Question 4C11
arisawaters2D wrote:Could you still use the equation q=m*Cs*deltaT?
You would use that equation to find the energy needed to raise the temperature after the ice melted to 20 Celcius.
- Thu Feb 18, 2021 5:57 pm
- Forum: Calculating Work of Expansion
- Topic: work of expansion and constant pressure
- Replies: 3
- Views: 170
Re: work of expansion and constant pressure
It is likely that if given constant pressure you would use that equation. However, a good clue is to look for what other values are given, if you are given a method to find moles and temperature is given it is likely you have a reversible expansion. If you are only given constant pressure and change...
- Sun Feb 14, 2021 2:39 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Sapling Week 5/6 #18
- Replies: 5
- Views: 306
Re: Sapling Week 5/6 #18
So you got delta G and you set that equal to -RTlnK. Divide delta G by -RT. so you get lnK = deltaG/-RT. Do e^deltaG/-RT to get your answer. If it's still wrong I would double check your delta G value or write your answer in scientific notation if it is a large value of K(i got stuck on that part my...
- Fri Feb 12, 2021 5:16 pm
- Forum: Calculating Work of Expansion
- Topic: Work Equations
- Replies: 2
- Views: 112
Re: Work Equations
In a reversible isothermal reaction, the pressure is not constant so you solve for it using PV=nRT when you plug in P=nrt/V for P when solving for the integral of V1 to V2 you get -nRTlnV2/V1. It was derived in an earlier lecture. It's isothermal because it occurs slowly enough so that there is enou...
- Fri Feb 12, 2021 5:08 pm
- Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
- Topic: Degeneracy vs Entropy
- Replies: 3
- Views: 161
Re: Degeneracy vs Entropy
Everything mentioned above is great! I just wanted to add that degeneracy is usually used to calculate residual entropy which occurs when the temperature of a substance goes to 0K.
- Fri Feb 12, 2021 5:05 pm
- Forum: Van't Hoff Equation
- Topic: Entropy
- Replies: 18
- Views: 839
Re: Entropy
This definitely would just depend on what you are given in a question. If it provides the values needed to solve for entropy I assume it's possible but typically the Van't Hoff equations are used to find K at different temperatures.
- Wed Feb 10, 2021 11:56 am
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs R constant
- Replies: 7
- Views: 258
Re: Gibbs R constant
Yup to add on a lot of the time in Thermodynamics we are dealing with energy and the only R constant that has Joules is 8.314. It is always good to double-check units to make sure the right units cancel out to get the expected units ur answer should have.
- Wed Feb 10, 2021 11:54 am
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: change in S at equilibrium
- Replies: 2
- Views: 73
change in S at equilibrium
This might be a really dumb question but why is S at a max when it deltaS=0? Like the curve prof drew in the lecture was an inverted U shape. I feel like I'm missing something important lol.
- Thu Feb 04, 2021 9:54 pm
- Forum: Calculating Work of Expansion
- Topic: Sapling week 3/4 #15
- Replies: 9
- Views: 492
Re: Sapling week 3/4 #15
yup you want to use the mass to find the volume. First convert the mass to moles and then use that value, R=0.08206, 295K(22C in K) and P=1 to find V using PV=nRT. Once you have that use w=-PdeltaV. Since your pressure is 1 atm w=-delta V. Since you assume you start with no volume of gas your V init...
- Thu Feb 04, 2021 9:47 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Residual Entropy
- Replies: 5
- Views: 160
Re: Residual Entropy
It comes from the positions of an atom can take. At 0K the atoms stop moving but if there is an unknown orientation there is residual entropy. I remember prof saying this is why O2 has no residual energy because there is only one position because you can distinguish between the two molecules since t...
- Thu Feb 04, 2021 9:30 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Sapling #20
- Replies: 4
- Views: 144
Re: Sapling #20
The table gives you different specific heat depending on the property of the gas. So you do need knowledge of VESPR to solve this. Since Xe is just an atom you use 3/2*R. The question gives you your delta T and the mols. The value of the specific heat in the table is the molar specific heat you use ...
- Wed Feb 03, 2021 11:41 am
- 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: W equilibrium
- Replies: 1
- Views: 88
W equilibrium
I'm a bit confused as to why we know W is at a maximum at equilibrium in an isolated situation. I know that means entropy is therefore a max but I'm not sure why this is. Could someone please clarify?
- Tue Feb 02, 2021 9:22 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Work Done on Systems
- Replies: 4
- Views: 167
Re: Work Done on Systems
The easiest way to tell is often through which way energy is traveling. If a system is compressed the energy internally increases because the surroundings are doing work on the system. In the same way, we have an endothermic reaction, internal is increasing and work is being done on the system by th...
- Sun Jan 31, 2021 4:49 pm
- Forum: Phase Changes & Related Calculations
- Topic: Sapling Week 3/4 Q10
- Replies: 3
- Views: 250
Re: Sapling Week 3/4 Q10
Yeah so for this question you need to account for the energy from the ice melting. Then use the equation to find the energy changes for the temperatures of the water. There are two values for this the water at 50 degrees (after the ice is melted) and the water at 45 degrees. You will have equations ...
- Fri Jan 29, 2021 5:32 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Making X negligible
- Replies: 34
- Views: 1189
Re: Making X negligible
x is negligible in the denominator of the equilibrium equation when the % ionization is less than 5%. First you can negate the x in the denominator and solve for [H+] or [OH-]. Then take that value to divide it by the initial and multiple by 100. If that number is less than 5% then your value can be...
- Wed Jan 27, 2021 5:16 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Kc vs K
- Replies: 4
- Views: 183
Re: Kc vs K
K is a more generic term for an equilibrium constant. Kc is just a specific form when the equilibrium constant is found from concentrations at equilibrium.
- Wed Jan 27, 2021 11:48 am
- Forum: Calculating Work of Expansion
- Topic: Equilibrium systems
- Replies: 2
- Views: 166
Equilibrium systems
I'm still a bit confused as to why systems at equilibrium have changes in small steps. I would love it if someone could explain it again.
- Mon Jan 25, 2021 7:06 pm
- Forum: Phase Changes & Related Calculations
- Topic: heating curve
- Replies: 8
- Views: 268
Re: heating curve
When steam hits the skin there is a phase change from gas to liquid so the energy needed for the phase change is released. Water at 100 C doesn't release this energy required for phase change since the phase stays the same.
- Fri Jan 22, 2021 8:48 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: amphoteric vs amphiprotic
- Replies: 10
- Views: 251
Re: amphoteric vs amphiprotic
Amphiprotic refers to substances that are both proton acceptors and donors which also means it's both a Bronsted lowry acid and base. However amphoteric substances refer to a substance that is both an acid and base not referring to H+ exchange so a substance that is both a lewis acid and base would ...
- Fri Jan 22, 2021 8:43 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Identifying Strong & Weak Acids/Bases
- Replies: 7
- Views: 232
Re: Identifying Strong & Weak Acids/Bases
Unfortunately these you will have to commit to memory. It helps to make a quizlet or flashcards but naturally the more chem you do these will be easy to remember and identify.
- Thu Jan 21, 2021 10:26 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Ka correlation to strength of an acid
- Replies: 30
- Views: 2047
Re: Ka correlation to strength of an acid
Weak acids have small Ka while strong acids will have larger Ka.
- Thu Jan 21, 2021 10:24 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Weak vs. Strong Acids and Bases
- Replies: 9
- Views: 510
Re: Weak vs. Strong Acids and Bases
You will know when something is a weak base or acid in an equation when the product consists of the ions that form those acids/bases. for example HCl will form H3O+ and Cl-
- Wed Jan 20, 2021 10:58 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: friday week 2 lecture question (approximation of weak acids)
- Replies: 5
- Views: 254
Re: friday week 2 lecture question (approximation of weak acids)
x is approximated to being very small so we assume x barely affects the concentration. If the % ionization is less than 5% the approximation is valid.
- Sun Jan 17, 2021 2:34 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Salt Solution
- Replies: 1
- Views: 92
Re: Salt Solution
Salt solutions give an anion and cation when dissolved in water. When using the equilibrium you work with the equation of the reaction of one of the ions with water. You can treat them the same way when solving as you would for regular equilibriums.
- Fri Jan 15, 2021 12:14 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: the different Ks
- Replies: 3
- Views: 777
Re: the different Ks
Adding on to that
Kw can be used to calculate Ka and Kb when given two using the equation Kw=Ka*Kb
Kw can be used to calculate Ka and Kb when given two using the equation Kw=Ka*Kb
- Fri Jan 15, 2021 12:08 pm
- Forum: Ideal Gases
- Topic: Ice method
- Replies: 14
- Views: 669
Re: Ice method
ICE stands for Initial Change and Equilibrium. Usually you start with give K value and one initial concentration and need to find the other concentrations at equilibrium. This is easier to understand with an example problem. 2HI ---> H2 + I2 and [HI] = 0.75 HI H2 I2 I 0.75 0 0 C -2x +x +x E 0.75-2x ...
- Fri Jan 15, 2021 11:52 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Hydronium concentrations less than 10^-7
- Replies: 2
- Views: 69
Re: Hydronium concentrations less than 10^-7
In the lecture, he explains that when hydronium concentrations are less than 10^-7 it hardly affects the pH due to the value being so small so when calculating the new pOH its actually still neutral with a pH of 7 not 10 (which would be the case if you did -log(10^-10) for example). He shows us that...
- Wed Jan 13, 2021 10:57 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Change in volume
- Replies: 4
- Views: 177
Re: Change in volume
If there are more moles on the left that would mean there are more reactants so the equation would be shifted to the right to create more products. When v decreases this causes the concentration to increase more which causes an imbalance on the reactants side so more products will be formed given th...
- Wed Jan 13, 2021 10:52 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Pressure's effect on concentration
- Replies: 4
- Views: 124
Re: Pressure's effect on concentration
I think the key point here is that when you add an inert gas the pressure increases but the volume actually doesn't change and equilibrium is kept. In this case, I believe you assume that you have a rigid container that holds the gas and you are simply adding more gas to it. Therefore adding an iner...
- Sat Jan 09, 2021 11:23 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Sign of x in ICE Box
- Replies: 8
- Views: 380
Re: Sign of x in ICE Box
x is always negative for things on reactants because they are being converted to products and for products, x will be positive. The initial values don't change this.
- Sat Jan 09, 2021 10:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Calculating Q
- Replies: 8
- Views: 281
Re: Calculating Q
Yup Q is also the concentration of products over reactants only these are values of a system that is not at equilibrium. Plus using this value and comparing it to K you can tell which way the reaction will proceed whether it will go to the right or left.
- Thu Jan 07, 2021 8:27 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Finding Kc given K
- Replies: 5
- Views: 204
Re: Finding Kc given K
Maybe provide an example of a question? Kc is the equilibrium constant for an equation using the concentrations at equilibrium.
- Thu Jan 07, 2021 3:39 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: ICE tables
- Replies: 11
- Views: 462
Re: ICE tables
I is the initial concentration, C, is the change, and E is the equilibrium concentration. For example, in A -> B + C, with initial concentration of A being 0.25M and C at equilibrium being 0.15 A B C I 0.25 0 0 C -x +x +x E 0.25-x x x Then you know x = 0.15 and can use it to find K. I just wanted t...
- Wed Jan 06, 2021 3:19 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: solvent in equilibrium constants
- Replies: 8
- Views: 338
Re: solvent in equilibrium constants
solvents aren't included in the equilibrium equation because their concentration barely changes and would effectively cancel out. Aqueous solutions themselves aren't the same as just solvents so they are included.
- Wed Jan 06, 2021 11:50 am
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solvents absence in equilibrium constant eq
- Replies: 3
- Views: 215
Solvents absence in equilibrium constant eq
In the lecture, this was covered but I'm not fully sure about the explanation as to why solvents aren't written in the equilibrium constant equation. Is it that the solvent is in excess and therefore in both the products and reactants so they cancel out? Is there anything else I need to know?
- Mon Jan 04, 2021 3:42 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: Solids Not Having Concentration
- Replies: 7
- Views: 400
Re: Solids Not Having Concentration
It just means that solids and liquids don't change in concentration so they aren't included in the Kc. Concentrations are a measurement of solutes in solvents in a solution which is why a pure substance cannot have a concentration.