Because the slow step is the rate-determining step, only changing the slow step will change the rate of the reaction.
Adding a catalyst to one of the fast steps will not affect the rate of reaction because the fast step does not determine the rate.
Search found 86 matches
- Fri Mar 16, 2018 11:18 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Question 15.95
- Replies: 1
- Views: 347
- Fri Mar 16, 2018 11:03 am
- Forum: Reaction Mechanisms, Reaction Profiles
- Topic: sketching reaction profiles
- Replies: 2
- Views: 495
Re: sketching reaction profiles
For a reaction profile, we mainly need to take into account the number of elementary steps, the activation energy of each step, and whether the reaction is exothermic or endothermic. For each elementary step, there is one energy barrier/"hill" in the reaction profile. The slow step will ha...
- Wed Mar 14, 2018 3:06 pm
- Forum: Interesting Applications: Rechargeable Batteries (Cell Phones, Notebooks, Cars), Fuel Cells (Space Shuttle), Photovoltaic Cells (Solar Panels), Electrolysis, Rust
- Topic: Electrolysis
- Replies: 1
- Views: 437
Re: Electrolysis
I think if you're told that it is an electrolytic cell, you should also be able to calculate standard cell potential by putting the reaction with the higher standard reduction potential at the anode rather than cathode. Basically, this would be the reverse of a Galvanic cell, so you could calculate ...
- Wed Mar 14, 2018 10:13 am
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Arrhenius when Temperature Changes
- Replies: 3
- Views: 473
Re: Arrhenius when Temperature Changes
Just make sure that you are consistent with which numbers you use. Keep track of which set of numbers is k1/T1 and which is k2/T2
- Sun Mar 04, 2018 11:48 pm
- Forum: Zero Order Reactions
- Topic: Slopes
- Replies: 3
- Views: 641
Re: Slopes
If you find experimentally certain points on the graphs we discussed in class, you can use the slope to find k. Since k is experimentally determined, this is the way to find k, not the other way around. I wouldn't say you necessarily need to memorize why certain orders correspond with certain graphs...
- Sun Mar 04, 2018 9:54 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Little Product
- Replies: 4
- Views: 668
Re: Little Product
When there is product present, the product will also be reacting to form reactant. So this reverse reaction would occur in addition to reactant forming product, changing the overall rate at which reactant is consumed and product is formed. We therefore use initial rates to avoid having to calculate ...
- Sun Mar 04, 2018 2:49 pm
- Forum: General Rate Laws
- Topic: Rate Law
- Replies: 3
- Views: 503
Re: Rate Law
Rates are always given as positive values. Since the equation for rate is rate= d[]/dt and the concentration of reactants is decreasing, simply using the above equation for the rate of consumption of reactants would yield a negative rate. To correct this and give a positive rate for rate of consumpt...
- Sun Mar 04, 2018 2:43 pm
- Forum: First Order Reactions
- Topic: Significance of the order of a reaction?
- Replies: 1
- Views: 9851
Re: Significance of the order of a reaction?
The order of a reaction tells us how the rate of reaction is affected by the concentration of the reactants. For a zero-order reaction, the rate of reaction is independent of the concentration of reactants, so changing the reactant concentration will have no effect on the reaction rate. For a first-...
- Sun Feb 25, 2018 2:36 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: swapping signs of E values
- Replies: 8
- Views: 3715
Re: swapping signs of E values
I think Dr. Lavelle prefers that we use:
Eocell = Eocathode - Eoanode
so that we don't have to switch the sign of either standard reduction potential
Eocell = Eocathode - Eoanode
so that we don't have to switch the sign of either standard reduction potential
- Sun Feb 25, 2018 12:21 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Galvanic cell set up
- Replies: 8
- Views: 1076
Re: Galvanic cell set up
Since the reactions in Galvanic cells are spontaneous and Eocell = Eocathode - Eoanode, the reaction with the more positive (or less negative) standard reduction potential should be at the cathode so that Eocell > 0.
- Sun Feb 25, 2018 12:06 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Finding Standard Potential Through Compiling Half-Reactions
- Replies: 2
- Views: 378
Re: Finding Standard Potential Through Compiling Half-Reactions
Even though ΔG of a final reaction will be the sum of the ΔGs of its composite reactions, it is important to note that this is not true of standard potential, E o . Since E o is not a state function, you must manipulate the equations to use ΔG, which is a state function. If reaction 1 + reaction 1 =...
- Mon Feb 19, 2018 8:35 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Cell Diagrams [ENDORSED]
- Replies: 4
- Views: 21576
Re: Cell Diagrams [ENDORSED]
inert metal | reactant | product || reactant | product | inert metal The anode is typically on the left and the cathode is typically on the right. The double line represents the salt bridge, and will be replaced with a single line if there is a porous disk instead of a salt bridge. Lines are used to...
- Mon Feb 19, 2018 1:07 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Equilibrium constant
- Replies: 3
- Views: 492
Re: Equilibrium constant
Because ΔGo= -nFEo = -RTlnK,
-nFEo = -RTlnK
When standard cell potential is positive, K>1
When standard cell potential is negative, K<1
For positive Eo, larger the standard cell potential of a reaction, the larger the equilibrium constant.
-nFEo = -RTlnK
When standard cell potential is positive, K>1
When standard cell potential is negative, K<1
For positive Eo, larger the standard cell potential of a reaction, the larger the equilibrium constant.
- Thu Feb 15, 2018 11:36 pm
- Forum: Work, Gibbs Free Energy, Cell (Redox) Potentials
- Topic: Delta G=-nFE
- Replies: 5
- Views: 2686
Re: Delta G=-nFE
Moles in the equation ΔG=-nFE is for moles of electrons, so even if the moles of a substance are not explicitly given, you should be able to figure out the moles of electrons through the redox half-reactions.
- Sat Feb 10, 2018 9:26 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.15
- Replies: 3
- Views: 495
Re: 9.15
Heat is always required to melt a substance, so ∆H is positive.
Heat is released when a substance freezes, so ∆H is negative.
∆Hfreezing = ∆Hfusion, and since you are given ∆Hfusion from Table 8.3, you just reverse the sign of ∆Hfusion to get ∆Hfreezing = -6.01 kJ/mol
Heat is released when a substance freezes, so ∆H is negative.
∆Hfreezing = ∆Hfusion, and since you are given ∆Hfusion from Table 8.3, you just reverse the sign of ∆Hfusion to get ∆Hfreezing = -6.01 kJ/mol
- Sat Feb 10, 2018 6:25 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: 9.5
- Replies: 3
- Views: 418
Re: 9.5
Since the problem says that the heat is being transferred from one reservoir to another, you can assume that the heat is leaving the first reservoir (-q) and entering the second reservoir (+q).
- Sat Feb 10, 2018 6:23 pm
- Forum: Calculating Work of Expansion
- Topic: Reversible vs Irreversible Work
- Replies: 1
- Views: 412
Re: Reversible vs Irreversible Work
If the external pressure is higher than the internal pressure, there will be compression rather than expansion of the gas, so you're not comparing it to reversible expansion anymore. However, I would assume that reversible compression would be more work being done on the system than irreversible com...
- Sat Feb 10, 2018 6:14 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Irreversible expansion and entropy [ENDORSED]
- Replies: 1
- Views: 302
Re: Irreversible expansion and entropy [ENDORSED]
ΔS of the system will increase because in free expansion, the volume of the system will still increase, so the particles can occupy more microstates, representing an increase in entropy.
- Sat Feb 10, 2018 6:08 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: 9.13
- Replies: 4
- Views: 506
Re: 9.13
Although it does not specify 1 moleof N2 gas in the book, 9.13 is listed in the solutions manual errors on Lavelle's website. In the corrected answer, Lavelle says to assume both ideal behavior and 1 mol of N2 gas.
- Fri Feb 09, 2018 5:32 pm
- Forum: Concepts & Calculations Using Second Law of Thermodynamics
- Topic: Homework problem 9.47
- Replies: 2
- Views: 463
Re: Homework problem 9.47
Isothermal means that the temperature does not change in the reaction. For all isothermal reactions, there is no change in internal energy, so ΔU=0. In this problem, we are told that ΔU=0 because the reaction is isothermal. Since no work is done in free expansion (w=0) and ΔU = q + w, when ΔU=0 and ...
- Sun Feb 04, 2018 10:40 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Exothermic and Spontaneous
- Replies: 4
- Views: 9977
Re: Exothermic and Spontaneous
Exothermic and endothermic refer to ΔH, or the amount of heat absorbed or released in a reaction. Spontaneous and non-spontaneous refer to ΔG, or the free energy change of the reaction. Because ΔG = ΔH - TΔS, ΔG depends on both ΔH and ΔS. So not all exothermic reactions (ΔH < 0) will be spontaneous ...
- Sun Feb 04, 2018 10:33 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Measuring delta G
- Replies: 4
- Views: 821
Re: Measuring delta G
Since we cannot measure ΔG directly, we use equations with values that we can measure to calculate ΔG. We can measure equilibrium concentrations to calculate the equilibrium constant K, so we can calculate ΔG using K from measured concentrations: ΔG o = -RTlnK If the system is not at equilibrium, we...
- Sun Feb 04, 2018 10:29 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Gibbs free question
- Replies: 2
- Views: 393
Re: Gibbs free question
At equilibrium, the free energy of the products is equal to the free energy of the reactants.
Since ΔG = Gproducts - Greactants
When Greactants = Gproducts, ΔG = 0
Therefore, at equilibrium, ΔG = 0
Since ΔG = Gproducts - Greactants
When Greactants = Gproducts, ΔG = 0
Therefore, at equilibrium, ΔG = 0
- Sun Feb 04, 2018 10:26 pm
- Forum: Gibbs Free Energy Concepts and Calculations
- Topic: Difference between deltaG and deltaG*
- Replies: 2
- Views: 239
Re: Difference between deltaG and deltaG*
delta G o is for when the reactants and products are all in their standard states. This means that pressure is 1 atm and temperature is 25 o C (298 K). So if any of the given conditions violate these conditions, don't use delta G o . The general rule I use for homework and tests is that if the given...
- Sun Feb 04, 2018 6:22 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Test Number 1
- Replies: 9
- Views: 1085
Re: Test Number 1
There is heat transferred in this situation in order to keep temperature constant.
As a gas expands, it cools. Because the question tells us that temperature is constant, the gas must have absorbed heat from the surroundings to maintain its temperature and counteract the cooling from expansion.
As a gas expands, it cools. Because the question tells us that temperature is constant, the gas must have absorbed heat from the surroundings to maintain its temperature and counteract the cooling from expansion.
- Sun Feb 04, 2018 5:01 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Heat transfer to a liquefied gas.
- Replies: 2
- Views: 440
Re: Heat transfer to a liquefied gas.
When the pressurized liquid escapes as a gas, it undergoes a phase change where it changes from liquid to gas. Enthalpy of vaporization is always positive, meaning that heat is required to change the phase from liquid to gas. So heat is absorbed by the liquid as it changes into the gas phase while i...
- Sat Jan 27, 2018 10:25 pm
- Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
- Topic: Degeneracy (W) [ENDORSED]
- Replies: 8
- Views: 1282
Re: Degeneracy (W) [ENDORSED]
Degeneracy is the number of possible states that a system can have. For example, if one particle can be in one of two sides of a flask (similar to the example in class) the degeneracy is W=2 because the system has two possible states (the particle on either side of the flask). An equation for degene...
- Sat Jan 27, 2018 10:15 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: Enthalpy of Reaction at lower temperatures [ENDORSED]
- Replies: 2
- Views: 1673
Re: Enthalpy of Reaction at lower temperatures [ENDORSED]
At lower temperatures, enthalpy of a reaction (ΔH or q P ) has a bigger influence on the entropy of a reaction because of this equation: \Delta S = \frac{q}{T} When T is lower, ΔS is larger, meaning that q had a larger effect on entropy. For example, if q=10 and T=1, ΔS =10, but if q=10 and T=10, th...
- Sat Jan 27, 2018 10:07 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Ideal Behavior
- Replies: 3
- Views: 411
Re: Ideal Behavior
In general, when a problem says to assume ideal behavior, it is referring to the ideal gas law:
PV=nRT
PV=nRT
- Sat Jan 27, 2018 1:26 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: Calculating degeneracy(W)
- Replies: 3
- Views: 1745
Re: Calculating degeneracy(W)
To modify the equation above, W=2 N for a 2-state system, where N is the number of particles. The example we used was where each particle could be in one of two bulbs of a flask This equation can be used for any number of states, just by changing the base. For a 3-state system: W=3 N The number of s...
- Sat Jan 27, 2018 1:08 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Internal energy as a state function
- Replies: 3
- Views: 509
Re: Internal energy as a state function
Internal energy is a state function because it depends only on the initial and final states of the internal energy. Although work is not a state function, as long as the sum of work done and heat transferred is the same, the change in internal energy will be the same regardless of the pathway taken ...
- Fri Jan 26, 2018 9:19 pm
- Forum: Calculating Work of Expansion
- Topic: Irreversible Pathway Picture
- Replies: 3
- Views: 444
Re: Irreversible Pathway Picture
The reason that no work is done in this picture is because the gas is expanding into a vacuum. There is no external pressure for in the vacuum, so since there is nothing for the gas to push against as it expands, it is not doing any work. The textbook calls this expansion against zero pressure "...
- Sun Jan 21, 2018 2:43 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: enthalpy of reaction
- Replies: 1
- Views: 1475
Re: enthalpy of reaction
If the reaction is conducted at constant pressure , then heat transferred (q p ) will be equal to the enthalpy of the reaction (ΔH). Enthalpy is specifically the heat released or absorbed at constant pressure, so make sure that pressure is constant (reaction is not in a bomb calorimeter) before assu...
- Fri Jan 19, 2018 9:39 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.43
- Replies: 2
- Views: 377
Re: Question 8.43
The reason the answer is c) and not b) is because the slopes of the heating curve for the solid and gas phases must be steeper than the slope of the heating curve for the liquid phase. This is because the heat capacity of the liquid phase is higher than the heat capacities of the solid and gas phase...
- Fri Jan 19, 2018 8:43 pm
- Forum: Phase Changes & Related Calculations
- Topic: Closed Systems [ENDORSED]
- Replies: 8
- Views: 1245
Re: Closed Systems [ENDORSED]
To increase the internal energy of a closed system, you can heat the system and/or perform work on the system.
You can heat the system using an external heat source, and you can perform work on a system by compressing the system (like compressing a bike pump).
You can heat the system using an external heat source, and you can perform work on a system by compressing the system (like compressing a bike pump).
- Fri Jan 19, 2018 8:28 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: Concept Clarification
- Replies: 3
- Views: 294
Re: Concept Clarification
Adding matter to an open system increases the internal energy of that system, and removing matter from an open system decreases the internal energy of the system.
- Fri Jan 19, 2018 8:22 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Question 8.49
- Replies: 2
- Views: 216
Re: Question 8.49
When a reaction produces heat (is exothermic), heat leaves the system. ΔH is negative because this represents heat leaving the system as it is produced.
- Fri Jan 19, 2018 7:58 pm
- Forum: Thermodynamic Systems (Open, Closed, Isolated)
- Topic: Isolated vs. Adiabatic system
- Replies: 8
- Views: 8623
Re: Isolated vs. Adiabatic system
The internal energy of an isolated system will not change over time.
In an adiabatic system, energy is not transferred as heat, but the internal energy can still change if energy is transferred to or from the system as work.
For adiabatic systems, ΔU=w because q=0.
In an adiabatic system, energy is not transferred as heat, but the internal energy can still change if energy is transferred to or from the system as work.
For adiabatic systems, ΔU=w because q=0.
- Sun Jan 14, 2018 10:55 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: Born Habers Cycle [ENDORSED]
- Replies: 4
- Views: 635
Re: Born Habers Cycle [ENDORSED]
Just to note, on the outline for the unit on Thermochemistry and the First Law of Thermodynamics, Dr. Lavelle says to omit section 8.18 concerning the Born-Haber Cycle.
- Sun Jan 14, 2018 8:15 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: 3 Different Methods
- Replies: 2
- Views: 243
Re: 3 Different Methods
Adding onto the previous response, if you are given the standard enthalpies of formation for the reactants and products, you use the equation:
Σ ΔHorxn = Σ ΔHof(products) - Σ ΔHof(reactants)
Σ ΔHorxn = Σ ΔHof(products) - Σ ΔHof(reactants)
- Sun Jan 14, 2018 6:23 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Standard Reaction Enthalpy
- Replies: 1
- Views: 177
Re: Standard Reaction Enthalpy
Some examples of reactants or products not in their standard states is carbon as diamond (instead of graphite) or oxygen as ozone (O3 rather than O2).
So a possible example of a reaction with reactants not in standard states is the conversion of ozone into oxygen:
2O3(g) --> 3O2(g)
So a possible example of a reaction with reactants not in standard states is the conversion of ozone into oxygen:
2O3(g) --> 3O2(g)
- Fri Jan 12, 2018 2:47 pm
- Forum: Phase Changes & Related Calculations
- Topic: Heating Curve
- Replies: 6
- Views: 560
Re: Heating Curve
The length of the lines of the heating curves also represents the amount of heat needed to change the state of the substance. The longer the line, the more heat is required, indicating that the heat capacity is higher.
- Fri Jan 12, 2018 2:24 pm
- Forum: Phase Changes & Related Calculations
- Topic: qv vs qp
- Replies: 4
- Views: 2232
Re: qv vs qp
Dr. Lavelle also specified that Qv is found using a bomb calorimeter, which keeps volume constant, while Qp is found using a coffee cup calorimeter, which allows gases to escape to keep pressure constant.
- Fri Jan 12, 2018 2:21 pm
- Forum: Phase Changes & Related Calculations
- Topic: Standard Reaction Enthalpy
- Replies: 8
- Views: 829
Re: Standard Reaction Enthalpy
Standard reaction enthalpy is specifically the reaction enthalpy for a reaction in which all the products and reactants are in their most stable forms and the pressure is 1 atm. Reaction enthalpy in general could be for a reaction in which ice is used instead of water, or ozone (O 3 ) instead of O 2...
- Sat Dec 09, 2017 8:51 pm
- Forum: Properties & Structures of Inorganic & Organic Acids
- Topic: Strong Bases? [ENDORSED]
- Replies: 2
- Views: 745
Re: Strong Bases? [ENDORSED]
Metal hydroxides are usually strong bases
Examples include NaOH, LiOH, KOH, etc.
Just as strong acids completely dissociate in water, strong bases also completely dissociate in water, so their conjugate acids are EXTREMELY weak, and you assume that none of the strong base remains in solution.
Examples include NaOH, LiOH, KOH, etc.
Just as strong acids completely dissociate in water, strong bases also completely dissociate in water, so their conjugate acids are EXTREMELY weak, and you assume that none of the strong base remains in solution.
- Sat Dec 09, 2017 8:35 pm
- Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
- Topic: Isoelectronic
- Replies: 14
- Views: 2188
Re: Isoelectronic
Isoelectronic means that the two ions have the same number of valence electrons.
F- has valence 8 electrons
Because O- and N2- each have only 7 valence electrons, they are not isoelectronic with F-
F- has valence 8 electrons
Because O- and N2- each have only 7 valence electrons, they are not isoelectronic with F-
- Sat Dec 09, 2017 8:32 pm
- Forum: Acidity & Basicity Constants and The Conjugate Seesaw
- Topic: Question 12.45
- Replies: 3
- Views: 703
Re: Question 12.45
The lower the pKb, the stronger the base.
Because pKa + pKb = 14 (so when pKa increases, pKb decreases), the larger the pKa value for an acid, the stronger the conjugate base.
Because pKa + pKb = 14 (so when pKa increases, pKb decreases), the larger the pKa value for an acid, the stronger the conjugate base.
- Sat Dec 09, 2017 8:21 pm
- Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
- Topic: Pressure and Volume
- Replies: 2
- Views: 487
Re: Pressure and Volume
Increasing pressure by compression (decreasing volume) will cause the reaction to proceed toward the side with fewer moles of gas. Decreasing pressure by increasing volume will cause the reaction to proceed toward the side with more moles of gas. Dr. Lavelle specified in class that just saying that ...
Re: Test 4
Diaquadifluoromercury(III)
- Sat Dec 09, 2017 8:05 pm
- Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
- Topic: Relationship between pH, protanation and molarity
- Replies: 5
- Views: 883
Re: Relationship between pH, protanation and molarity
For an acid, when given the initial molarity and percent deprotonation, you multiply the initial molarity by the percent deprotonation (as a decimal, so over 100) to find the concentration of H 3 O + ions in solution. You then use this concentration to find pH in the normal way (pH=-log[H 3 O + ]) F...
- Sun Dec 03, 2017 6:10 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: ICE table question
- Replies: 5
- Views: 11896
Re: ICE table question
If you are given initial concentrations and all of them are nonzero, calculate the reaction quotient Q. If Q<K, the reaction moves toward the products, so the ICE table would have +x on the products side and -x on the reactants side. If Q>K, the reaction moves toward the reactants, so the ICE table ...
- Sun Dec 03, 2017 6:02 pm
- Forum: Non-Equilibrium Conditions & The Reaction Quotient
- Topic: Notes on 11.37
- Replies: 3
- Views: 633
Re: Notes on 11.37
The rule in general is for a given equation, you raise K to the power of whatever value you multiplied the chemical reaction by.
So if you multiply the equation by 3, the new value of K would be the initial K3
So if you multiply the equation by 3, the new value of K would be the initial K3
- Sat Dec 02, 2017 8:55 pm
- Forum: Hybridization
- Topic: Pi Bonds
- Replies: 2
- Views: 308
Re: Pi Bonds
When a pi bond is delocalized, it is not restricted between two atoms. The electrons are spread out over several atoms in a molecule, giving the affected bonds characteristics (length and strength) between a double and single bond. For example, in benzene (C 6 H 6 ) there are 3 pi bonds delocalized ...
- Sat Dec 02, 2017 8:49 pm
- Forum: Bronsted Acids & Bases
- Topic: Bronsted Acid vs Lewis Acid [ENDORSED]
- Replies: 3
- Views: 638
Re: Bronsted Acid vs Lewis Acid [ENDORSED]
Another difference is that all Bronsted acids are Lewis acids, but not all Lewis acids are Bronsted acids. This is because when a molecule is a proton donor, that proton is an electron pair acceptor, making the molecule both a Bronsted acid(proton donor) and a Lewis acid (electron pair acceptor). So...
- Sun Nov 26, 2017 11:30 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Determining Molecular Shape
- Replies: 3
- Views: 302
Re: Determining Molecular Shape
Since the resonance structures are the same in terms of atom placement and differ only in electron placement, the shapes for different resonance structures will be the same. Shape depends on the number of regions of electron density, so whether there is a single, double, or triple bond has no effect...
- Sun Nov 26, 2017 11:16 pm
- Forum: Hybridization
- Topic: 3 regions of e- density
- Replies: 3
- Views: 555
Re: 3 regions of e- density
The coefficient when describing a hybridized orbital describes the energy level of the orbital being hybridized. So in general, when there are 4 regions of electron density, there will be sp 3 hybridization. Using the coefficient 4 in 4sp 3 means that the element with the hybridized orbitals is in P...
- Sun Nov 26, 2017 10:48 pm
- Forum: Naming
- Topic: Charges on Compounds
- Replies: 1
- Views: 271
Re: Charges on Compounds
To find the charge of a complex, you add the charges of the atoms/ions that make up the complex. In tetracyanonickelate(II): 4 x CN - = 4- 1 x Ni 2+ = 2+ Charge of complex = 2- You know that the nickel in this complex has a 2+ charge from the Roman numerals in the name. Since Table 17.4 is on the ou...
- Sun Nov 26, 2017 10:40 pm
- Forum: Hybridization
- Topic: HMK 4.91
- Replies: 3
- Views: 513
Re: HMK 4.91
The solutions manual also states that because the triple bonded carbons are strained in the ring formation, they may form a diradical in which the triple bond is replaced by a double bond with a single radical electron on each carbon. This would also result in a very reactive molecule because of the...
- Mon Nov 20, 2017 4:21 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Bond Angles
- Replies: 2
- Views: 259
Re: Bond Angles
We need to know the bond angles for the common VSEPR shapes (linear, trigonal planar, tetrahedral, trigonat bipyramidal, etc). However, we do not need to memorize the exact values for experimentally determined bond angles, like those resulting from lone pairs "pushing" the bonded atoms clo...
- Mon Nov 20, 2017 4:13 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Question 4.13 a
- Replies: 3
- Views: 557
Re: Question 4.13 a
The I 3 - molecule has the VSEPR notation AX 2 E 3 . Because it has 5 regions of electron density, begin by looking at the trigonal bipyramidal shape. The bond angles in trigonal bipyramidal are 120 o between the equatorial atoms, and only 90 o for the axial atoms. Because the lone pairs have more r...
- Mon Nov 20, 2017 4:01 pm
- Forum: Properties of Electrons
- Topic: Chem Midterm Question Q3C
- Replies: 1
- Views: 350
Re: Chem Midterm Question Q3C
If the performance of the supercomputer is only based on the speed of the electrons, the number of electrons is unrelated to performance. If you increase the number of electrons, all of the electrons will still have the same speed, there will just be more of them. This is similar to the results of t...
- Mon Nov 20, 2017 3:48 pm
- Forum: Lewis Structures
- Topic: Organic Lewis Structures
- Replies: 2
- Views: 364
Re: Organic Lewis Structures
Generally, with chemical formulas, atoms written next to each other are bonded to each other. So in HOCO, writing the formula this way tells you that the two oxygen atoms are bonded to the carbon, and the hydrogen is bonded to an oxygen. This can be especially helpful in hinting at the common groups...
- Mon Nov 20, 2017 3:41 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lewis structure and Vsper
- Replies: 2
- Views: 296
Re: Lewis structure and Vsper
When drawing Lewis structures, the bonds can be placed at any angle around the central atom. We are not responsible for drawing VSEPR models, but it will help to understand the 3D shape, and you must know the bond angles for the given shapes.
- Mon Nov 20, 2017 3:38 pm
- Forum: Equilibrium Constants & Calculating Concentrations
- Topic: KP versus KC
- Replies: 3
- Views: 509
Re: KP versus KC
Both types of equilibrium constant, whether in regard to molarity or partial pressure, refer to the composition of the equilibrium. Either constant will tell you if the equilibrium sits to the left or to the right.
- Sun Nov 12, 2017 11:32 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Lone Pair Placement
- Replies: 3
- Views: 456
Re: Lone Pair Placement
In general, since lone pairs have the greatest repulsion, put them wherever they will be farthest from each other or from other atoms. This will minimize the repulsion in the molecule and result in the most stable structure.
- Sun Nov 12, 2017 10:34 pm
- Forum: Lewis Structures
- Topic: Radicals
- Replies: 7
- Views: 808
Re: Radicals
As a general rule, yes, an odd number of valence electrons in a molecule means the molecule will be a radical.
- Sun Nov 12, 2017 10:18 pm
- Forum: Trends in The Periodic Table
- Topic: Hydrogen Electronegativity
- Replies: 4
- Views: 5623
Re: Hydrogen Electronegativity
In addition, hydrogen is a very small atom, so its nuclear charge has a strong pull on the electrons in a bond. There is also very little to no shielding because hydrogen only has one electron, so the effective nuclear charge felt by the electron is very high.
- Sun Nov 12, 2017 9:47 pm
- Forum: Octet Exceptions
- Topic: Expanded Octet
- Replies: 6
- Views: 2380
Re: Expanded Octet
The elements that typically have less than an octet are hydrogen, helium, lithium, and beryllium. Boron also may form molecules where it only has 6 electrons. For expanded octets, any element in period 3 or later with a d-orbital can have an expanded octet. This includes sulfur, phosphorus and chlor...
- Sun Nov 12, 2017 9:32 pm
- Forum: Determining Molecular Shape (VSEPR)
- Topic: Trigonal Bipyramidal with 1 lone pair and 4 bonds
- Replies: 3
- Views: 630
Re: Trigonal Bipyramidal with 1 lone pair and 4 bonds
In the equatorial plane, there are 3 regions of electron density, which can be separated by approximately 120 o bond angles. This is favorable compared to axial atoms above and below the equatorial plane, which have only approximately 90 o bond angles with equatorial atoms. Since lone pairs have mor...
- Sun Nov 05, 2017 9:21 pm
- Forum: Lewis Structures
- Topic: Lewis Structure Format
- Replies: 2
- Views: 437
Re: Lewis Structure Format
To clarify, when there is a bracket with a charge around a Lewis structure with multiple atoms, that means that the molecule shown is a polyatomic ion. The bonds between the atoms within the ion itself are covalent, but the ion will form ionic bonds with other ions. When there are bonds shown as lin...
- Sun Nov 05, 2017 1:33 pm
- Forum: Resonance Structures
- Topic: Homework problem 3.57
- Replies: 1
- Views: 405
Re: Homework problem 3.57
The atom in the center in the molecules here is always the one with the lowest ionization energy, so these do follow the guideline laid out in class. The guideline for ionization energies is that ionization energy decreases as you go down a group and increases right to left across a period. In the s...
- Sun Nov 05, 2017 1:18 pm
- Forum: Lewis Structures
- Topic: Lewis Acid/Base vs. Regular Acid/Base
- Replies: 1
- Views: 238
Re: Lewis Acid/Base vs. Regular Acid/Base
Lewis acids and bases are defined in terms of electrons, while regular (Bronsted) acids and bases are defined in terms of protons. A Lewis acid accepts an electron pair and a Lewis base donates an electron pair. A Bronsted acid donates a proton and a Bronsted base accepts a proton. If something is a...
- Sun Nov 05, 2017 1:06 pm
- Forum: Trends in The Periodic Table
- Topic: why carbon has a higher electron affinity than nitrogen.
- Replies: 5
- Views: 7674
Re: why carbon has a higher electron affinity than nitrogen.
Nitrogen has a half-filled 2p subshell, so that there is one electron in each orbital. This creates an unusually stable atom because of half-shell stability. Because nitrogen is relatively stable on its own, it has a relatively low electron affinity. It is similar to the reason that chromium has an ...
- Mon Oct 30, 2017 10:42 am
- Forum: Formal Charge and Oxidation Numbers
- Topic: Ionization energy vs. electron affinity [ENDORSED]
- Replies: 5
- Views: 1141
Re: Ionization energy vs. electron affinity [ENDORSED]
Another way to think about this is that a low ionization energy corresponds to likelihood to form a cation (because it is easier to remove the electron) and a high electron affinity corresponds to likelihood to form an anion (because more energy is released when an electron is added). This is helpfu...
- Mon Oct 30, 2017 10:35 am
- Forum: Trends in The Periodic Table
- Topic: Exceptions to trends [ENDORSED]
- Replies: 6
- Views: 4906
Re: Exceptions to trends [ENDORSED]
The ionization energies of the Group 15 elements is higher than that of the Group 16 elements because in Group 15, each p-orbital has one electron, which is relatively stable because it is a half-filled subshell. Group 16 elements have one set of paired electrons in the p-orbital, and the repulsion ...
- Sun Oct 22, 2017 3:46 pm
- Forum: Electron Configurations for Multi-Electron Atoms
- Topic: nodal planes [ENDORSED]
- Replies: 3
- Views: 785
Re: nodal planes [ENDORSED]
Nodal planes are planes on which there is no chance that an electron will be found. The s-orbitals have no nodal planes, but the p- and d- orbitals do. They are mainly of interest when we are trying to figure out the location of an electron in a given orbital. The electron will never be at the nucle...
- Sun Oct 22, 2017 1:31 pm
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: Magnetic Quantum numbers vs. Spin Magnetic
- Replies: 2
- Views: 567
Re: Magnetic Quantum numbers vs. Spin Magnetic
The magnetic quantum number of an electron tells us precisely which orbital within a subshell that the electron occupies. The spin magnetic quantum number has values of either +1/2 (spin up) or -1/2(spin down). This tells us if the electron spins counterclockwise (spin up) or clockwise (spin down). ...
- Sun Oct 22, 2017 1:11 am
- Forum: Wave Functions and s-, p-, d-, f- Orbitals
- Topic: electron configurations [ENDORSED]
- Replies: 2
- Views: 357
Re: electron configurations [ENDORSED]
To clarify, the 2p electron subshell has 3 orbitals, 2p x , 2p y , and 2p z , each of which can hold 2 electrons. According to Hund's Rule, if electrons in the same subshell can occupy different orbitals, then they will do so. So when an atom has 2 electrons in the 2p subshell, they will occupy the ...
- Sun Oct 15, 2017 1:20 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: H-Atom Equation
- Replies: 3
- Views: 481
Re: H-Atom Equation
To elaborate on the last answer: When you calculate ΔE, you do final-initial. In this case, that means E n (final) - E n (initial) When the electron is moving from a higher energy state to a lower energy state, ΔE will be negative. When the electron is moving from a lower energy stat...
- Sat Oct 14, 2017 11:32 pm
- Forum: Properties of Electrons
- Topic: Speed/Energy of Electrons
- Replies: 4
- Views: 669
Re: Speed/Energy of Electrons
I think your confusion is mainly mixing up speed, energy, and frequency. The speed of light is constant (3.00 x 10 8 m/s), and is equal to the frequency times wavelength of light. The frequency(v) of light is different for different types of light, and the energy(E) of light is related to the freque...
- Sat Oct 14, 2017 10:40 pm
- Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
- Topic: Question about Atomic Spectroscopy
- Replies: 4
- Views: 592
Re: Question about Atomic Spectroscopy
Because each element has a different number of protons, each element's nucleus has a different positive charge. This variation in positive charge for each element causes the electrons in each element to be bound to the atom by a different amount of energy. The slight differences in configuration cau...
- Sat Oct 14, 2017 8:22 pm
- Forum: Einstein Equation
- Topic: Planck's Constant
- Replies: 3
- Views: 478
Re: Planck's Constant
Constants usually have the units that they do because of the units of the quantities that they relate.
So in the equation E=hv, Planck's constant relates frequency(in seconds-1) and energy (Joules). In order for this equation to make sense, Planck's constant must have the units of Joule-seconds.
So in the equation E=hv, Planck's constant relates frequency(in seconds-1) and energy (Joules). In order for this equation to make sense, Planck's constant must have the units of Joule-seconds.
- Fri Oct 06, 2017 6:03 pm
- Forum: Balancing Chemical Reactions
- Topic: Determining States of Matter in Reactions
- Replies: 7
- Views: 1541
Re: Determining States of Matter in Reactions
For the purposes of chemical equations, aqueous solutions and liquids are different, and it is important to make this distinction when writing out the equation.
An aqueous solution is indicated by the symbol (aq), while a liquid is indicated by the symbol (l).
An aqueous solution is indicated by the symbol (aq), while a liquid is indicated by the symbol (l).
- Fri Oct 06, 2017 1:32 pm
- Forum: SI Units, Unit Conversions
- Topic: Prefixes [ENDORSED]
- Replies: 6
- Views: 1010
Re: Prefixes [ENDORSED]
As far as I know, the common prefixes need to be memorized.
The ones we went over in lecture are
giga: 109
mega: 106
kilo: 103
deci: 10-1
centi: 10-2
milli: 10-3
micro: 10-6
nano: 10-9
pico: 10-12
The ones we went over in lecture are
giga: 109
mega: 106
kilo: 103
deci: 10-1
centi: 10-2
milli: 10-3
micro: 10-6
nano: 10-9
pico: 10-12
- Thu Oct 05, 2017 10:50 pm
- Forum: Accuracy, Precision, Mole, Other Definitions
- Topic: Formula Units
- Replies: 2
- Views: 507
Re: Formula Units
Discussing the formula unit of an ionic compound is equivalent to discussing a molecule of a covalently bonded compounds. It is technically incorrect to call a unit of NaCl a molecule, although that may be what you have heard in past chemistry classes. Call it a formula unit instead. Fundamentals Se...
- Thu Oct 05, 2017 10:10 pm
- Forum: Balancing Chemical Reactions
- Topic: Writing chemical formulas [ENDORSED]
- Replies: 5
- Views: 1149
Re: Writing chemical formulas [ENDORSED]
When we are given the name of the compound in a problem and we are supposed to find the chemical formula for an equation with no other information, the compound is usually an ionic compound and we are supposed to be able to figure out the formula using the oxidation numbers of its elements. For exam...