CHEMISTRY COMMUNITY

EllerySchlingmann1E

Catalysts can be included in the rate law, just intermediates cannot.

EllerySchlingmann1E

If the activation energy for the forward reaction is greater than the reverse reaction then the forward reaction is endothermic because more energy is required in the forward direction. The opposite is true for exothermic reactions: a lower Ea in the forward direction indicates an exothermic reactio...

EllerySchlingmann1E

Ka2 is considered negligible when it is significantly smaller than Ka1. Unfortunately, I don't there is an actual cut off so I don't think you have to worry too much about.

EllerySchlingmann1E

This question is asking you to find the final temperature of the mixture. We want to assume that all of the heat is contained within the system, so q(ice) = -q(water). Now you just want to use the heat of phase change (q=n*deltaH) and the nCdeltaT formulas to find the final temperature, which would ...

EllerySchlingmann1E

In this reaction, nitrogen is being reduced so the reduction half reaction should be:
NO2- + 5H2O + 6e- ---> NH3 + 7OH-

EllerySchlingmann1E

You have to use both. Plug in the calculated (or given) standard potential of the cell into the Nernst equation along with the cell potential measured by the ion-selective electrode (should be given in the problem as well). Then you want to create a Q expression for the reaction and plug that in too...

EllerySchlingmann1E

No, a second order reaction will have a straight line graph and positive slope that is equal to the rate constant only when 1/[A] is graphed vs time. Showing this plot has a straight line is a way of confirming a second order reaction and finding the value of k.

EllerySchlingmann1E

To answer the second part of your question: the decomposition of O3 to O2 can be hypothesized as occurring by different reaction mechanisms that are each made up of different elementary reactions. The decomposition can occur by a one-step reaction mechanism that is composed of just one elementary re...

EllerySchlingmann1E

For this problem, you want to recognize that it is a concentration cell reaction, so standard cell potential is zero and cell potential is equal to -(RT/nF) ln Q. In the 6I.1 table, you can find that Ksp = 6.5×10^−18 for Ni(OH)2 which means [Ni][OH]^2 = 6.5x10^-18. Use your concentration of OH to fi...

EllerySchlingmann1E

Finding the order of a reaction has many applications, including being able to identity the differential and integrated rate laws from the order. Know rate laws allows us to calculate a variety of values such as k and the concentration of a species. Hope that helps!

EllerySchlingmann1E

Because ATP is a reactant, it goes on the bottom of the K and Q expressions. So if the concentration of ATP increases by 4, you would want divide K by 4 to get Q.

EllerySchlingmann1E

We prefer positive voltage values because they are an indicator of spontaneity of the reaction. The equation standard delta G = -nF(standard E) where E is cell potential (or voltage), shows that positive values of E give negative values for standard Gibbs free energy, indicating that the cell reacti...

EllerySchlingmann1E

If there is only one reactant, it will often act as both the reducing and oxidizing agent in different half-reactions.

EllerySchlingmann1E

Change in Gibbs free energy at equilibrium does equal 0; however, standard Gibbs free energy equals -RT ln K. This relates allows for K to be an indicator of spontaneity because at K>1 favors products and produces a negative standard delta G while a K<1 favors reactants and produces a positive stand...

EllerySchlingmann1E

The intention of the van't Hoff equation is to confirm Le Chatelier's principle for the effect of temperature on K not the other way around. I think you want to think of K being dependent on temperature, not vice versa. According to the equation, a change in temperature of a certain amount changes K...

EllerySchlingmann1E

You can also have standard molar Gibbs free energy, which can be used to calculate Gibbs free energy of a reaction. That value is the Gibbs free energy of a substance under standard conditions per mole. Hope that helps!

EllerySchlingmann1E

Gibbs free energy of a substance can be measured under any conditions because it is simply a value; however, change in Gibbs free energy is generally measured at constant temperature and pressure. Hope that helps!

EllerySchlingmann1E

I think any thermodynamics/thermochemistry problems are appropriate because its what we covered last.

EllerySchlingmann1E

for #15, because you were given the standard combustion enthalpies, the question is asking you to write out the chemical equations for the combustions of the reactants and products and use Hess' Law to determine the overall reaction enthalpy. Whereas, for #17, we can use the standard enthalpies of f...

EllerySchlingmann1E

That is what delta U suggests. Especially because internal energy is a state function, it shouldn't matter whether the process is reversible or irreversible, but only for calculating delta U! Reversibility still definitely matters for calculating q and w.

EllerySchlingmann1E

This just touches on the idea that because enthalpy is a state function, the path taken to reach a certain enthalpy doesn't matter so enthalpies can be calculated by adding together the enthalpies of multiple steps of a larger reaction.

EllerySchlingmann1E

For question 3 on the midterm review worksheet, I was wondering why the answer is -312 kJ/mol. I'm sure I'm just making a calculator/stupid error, but I got +312 kJ/mol. Thanks for any help with this!

EllerySchlingmann1E

In Topic 4H in the 7th edition of the textbook, it says that "if the heat capacity is constant throughout the temperature range of interest, then ΔS=nRln(T2/T1)." I was just curious about this equation because I thought for a change in temperature, the appropriate equation to use was ΔS=Cl...

EllerySchlingmann1E

Closed systems can have constant pressure, volume or neither. However, one of the more common examples of a closed system, a beaker with a piston, is usually related to a question in which there is constant pressure.

EllerySchlingmann1E

For this problem, you want to calculate q and w separately and then use the equation delta U = q + w to find the total change in internal energy. q is simply given as 5.5 kJ, but you will need to calculate w using the -P*deltaV equation. Once you have the energy of the work done by the system, you c...

EllerySchlingmann1E

The temperature of a system can remain constant when heat is being supplied if the energy of that heat is being used for the system to do work. For example, if a gas is doing work due to expansion, there may be energy as heat that is driving that expansion and therefore not contributing to an increa...

EllerySchlingmann1E

An isolated system should imply constant volume. Because it cannot exchange energy or matter with its surroundings, a sealed container of constant volume is required for an isolated system.

EllerySchlingmann1E

Because the 22.5 kJ of heat is supplied to the system, it is actually calculated as a negative number, so the solution will turn out positive. Hope that helps!

EllerySchlingmann1E

For 4A7, you want to add together both the heat of the copper and the water to get the total heat of the interaction. I believe it comes to something like: a) q = (300.0g)(4.184 J/°C*g)(80°C) + (400.0g)(0.38 J/°C*g)(80°C) q = 112 kJ b) to find the percentage of the heat that went to raising the temp...

EllerySchlingmann1E

Cv or Cp may be specified, but if not you want to be aware of the other factors acting on the system. Cv is molar heat capacity at constant volume, so if your system is in a sealed container, you would use that one. If your system is at constant pressure, like an open container or with a piston, you...

EllerySchlingmann1E

I know that K is not written with the units mol/L because of the use of concentration in place of activity for species, but I am wondering if we should include mol/L with the concentrations when writing out our K expression on the test and homework. Just not sure what the correct format/practice is ...

EllerySchlingmann1E

In the book, there is a section that details which kinds of cations and anions actually affect the pH of solutions. It's probably a good idea to be familiar with those because they are included in your Ka/Kb calculations while other ions, like group 1 & 2 cations, are not included. Hope that hel...

EllerySchlingmann1E

Not sure if this is a foolproof method, but with those questions, as long as Ka2 is less than Ka1, I think you can ignore it. I did not see any specific parameters either.

EllerySchlingmann1E

For these types of questions, the change in enthalpy (delta H) will probably be given. In that case, positive delta H indicates an endothermic reaction and negative delta H indicates an exothermic reaction. However, a more challenging question may talk about a combustion reaction, which you have to ...

EllerySchlingmann1E

It just depends what is given for gases, but you want all of the species in the reaction to be in either concentration or partial pressure form, not a combination of the two. However, either one will work. Hope that helps!

EllerySchlingmann1E

You have to multiply the concentration by two because of the molar ratio between Ba(OH)2 and hydroxide ions. If you don't recognize the difference in moles, your calculation and pH will be way off. Another reason why it's so important to have a balanced reaction before your start making any assumpti...

EllerySchlingmann1E

For ICE tables, you also want to consider where the reaction is if it not at its initial state in terms of which way the reaction will proceed. For example, if you are given the concentrations of some of the reactants and some of the products in the midst of the reaction, you want to calculate Q and...

EllerySchlingmann1E

When you are looking at changes in the value of K you want to look at whether the reaction is endothermic or exothermic because that will give you a hint as to how K will change if temperature is increased or decreased. For example, if temperature is increased for an endothermic reaction, K increase...

EllerySchlingmann1E

Q and K are not always the same because Q represents [P]/[R] at any time in the reaction while K represents that ratio only when the reaction is at equilibrium. If Q does not equal K, we know that the equation is not at equilibrium.

EllerySchlingmann1E

It just depends on how the partial pressure of the gas is given. If it is given in bar, use that and same for atm. Just make sure to use the correct form of R if you are converting to a concentration using PV=nRT. The difference between the two is numeric but they serve similar purposes for us.

EllerySchlingmann1E

If the gas is in a reaction with other aqueous solutions whose equilibrium concentrations are given as concentrations, you may need to convert the partial pressure of your gas to a concentration using the ideal gas equation. In a reaction, you want to use all pressures or all concentrations, not a m...

EllerySchlingmann1E

Inductive effect deals with the creation of a permanent dipole in a bond due to electron fluctuations throughout a molecule, it was never discussed in lecture, so I wouldn't worry about it.

EllerySchlingmann1E

Hi, Sorry, this doesn't answer your question but I have a similar question so I figured I would add it to this thread. In the textbook, the naming of the compound says to drop the ending of ions and add "-ato" for -ate ions, "-ido" for -ide ions and "-ito" for -ite ions...

EllerySchlingmann1E

Chelating ligands can bind cations tightly because they can bind in multiple locations instead of just one. The multiple binding sites hold the cation more tightly.

EllerySchlingmann1E

The way Professor Lavelle explained it in class is that you can calculate the oxidation state/number of a transition metal by looking at the charges of the atoms or ions bonded to in a coordinate complex as well as the overall charge of the complex. The oxidation state/number is the charge of the co...

EllerySchlingmann1E

Larger atoms have stronger intermolecular forces because they are more polarizable (their electron clouds are more easily distorted). This is because they have more electrons surrounding their nucleus and those electrons are not held as tightly by the nucleus due to large energy shells. When we are ...

EllerySchlingmann1E

To find the formal charge for C2O4 2- you want to draw the Lewis structure as the book instructs, so with only a single bond between the carbons. This allows for each of the carbons to double bond with one oxygen and single bond with the other. The formal charge of oxygen when it is double bonded is...

EllerySchlingmann1E

For salt, the water completely dissociates the Na and Cl ions because its polarity; the water surrounds all of the separated ions. A similar thing happens with NiCl but here the water actually bonds to the Ni to create a coordinate compound instead of just being electrostatically attracted like with...

EllerySchlingmann1E

It definitely depends on which kinds of elements and ions you want to talk about when discussing polarizing power. As mentioned above, polarizing power describes a cation's ability to create a dipole moment in an anion by disturbing its electron cloud. Because polarizing power follows the same trend...

EllerySchlingmann1E

That's right, the textbook is simply saying that both SOCl angles are the same because they have symmetrical shape within the molecule. You are also correct in thinking about the lone pair creating a smaller bond angle for the bound atoms, it's just that that was not what the textbook was asking.

EllerySchlingmann1E

The shape of the molecule is angular/bent because one of the regions of electron density around the Cl atom is a lone pair, which is not counted when deciding the name of the shape. However, because there are three regions of electron density around the Cl atom, the electron arrangement is trigonal ...

EllerySchlingmann1E

Drawing the most stable Lewis structure helped me a lot in getting my answer, because the reasoning has to do with the number of bonds between the designated atoms in each of the species.

EllerySchlingmann1E

The A in the VSEPR model has an implied subscript 1 because it is the central atom. I think the purpose is to denote that central atom, but you don't need the superscript because it is implied.

EllerySchlingmann1E

Polarizing power and polarizability are two halves of the same coin. Polarizing power means the power an atom, ion, or molecule has to polarize, or create a dipole in another atom, ion, or molecule. On the other hand, polarizability denotes how easily a species can be polarized. We tend to talk abou...

EllerySchlingmann1E

Fluctuating electron density definitely results in temporary dipoles (van der Waals forces), but you also might want to think of the fact that these densities are constantly shifting and setting off chain reactions of fluctuating electron densities within the atoms or molecules around it. The consta...

EllerySchlingmann1E

It is a necessary step for the purposes of drawing a correct Lewis structure. If the ion is not bracketed with its charge then there can be significant confusion about the charge of the ion as well as the formal charges of the atoms in the ion. It is best to always put the brackets and denote charge.

EllerySchlingmann1E

The trends for ionization energy, electron affinity, and electronegativity are all the same. Electronegativity and electron affinity are very similar, but one major distinction is the fact that electron affinity of an atom doesn't change while its electronegativity can depending on how it is bonding...

EllerySchlingmann1E

It is correct to think of the 3d^5 4s^1 and 3d^10 4s^1 electron configurations being more stable and lower energy than if the 4s orbital had electrons with paired spins, but as Dr. Lavelle mentioned in lecture, this is because the 3d subshell wants symmetric electron distribution, which is achieved ...

EllerySchlingmann1E

I'm pretty sure Dr. Lavelle said we only needed to know/be able to figure out electron configurations through the 4th period, so I would not worry about it. To answer your question about Cerium, there are a ton of exceptions to the rules that govern the first few periods in the later periods because...

EllerySchlingmann1E

Yes, the experimentally observed bond lengths for benzene are known/given quantities. In benzene, the fact that the bond lengths are the same matters because it makes the structure symmetrical and therefore lower energy, but you don't have to just know the bond lengths of resonant structures, they w...

EllerySchlingmann1E

The fact that the potassium ions just float around the phosphorus indicate that the bonds between the atoms are considered ionic, therefore the phosphorus atoms possesses all of the atoms in the model. The way the potassium atoms are floating around the phosphorus I think is pretty arbitrary, they c...

EllerySchlingmann1E

Wave functions and orbitals are interchangeable. Orbitals are simply a way to visually represent the wave functions of electrons in certain energy states. So orbitals are the way math functions model electrons. With Dr. Lavelle's mention of experimental data, I think he was just trying to reiterate ...

EllerySchlingmann1E

E in that equation actually equals the energy of an electron in that energy shell, not the change in the energy. But yes, you can connect the E=-hR/n^2 to the E=hv equation.

EllerySchlingmann1E

Hi,

Just wondering if anyone can explain the difference between angular and radial nodes in terms of nodal planes for orbitals. They are mentioned in the textbook, I don't know how to differentiate them. Thanks so much!

EllerySchlingmann1E

I think you want to know that 4 of the 5 d-orbitals have 4 lobes and that the 5th has 2 lobes and the donut, but like Dr. Lavelle said in class, we won't be asked to draw the boundary surfaces necessarily. It just helps with visualization.

EllerySchlingmann1E

Hi, Energy and energy level are related through the formula: E(n)=-hR/n^2, where E is the energy of the electron, n is the energy level, h is Planck's constant, and R is the Rydberg constant. You're right, in the equation energy and energy level are inversely proportional, but this is because we hav...

EllerySchlingmann1E

Hi, delta E is negative because we are talking about the energy of the electron. In class, Dr. Lavelle mentioned how the energy of electron get more negative as they get closer to the nucleus, therefore, when the electron moves from the 4th to the 2nd energy state, it loses (emits) energy in the for...

EllerySchlingmann1E

Hi,

Yeah, that's totally right. If you have the measured moles of a reactant and want to find the mass of a product, you just need to multiply by the molar ratio of the compounds and then by the molar mass of the product.

EllerySchlingmann1E

Hi, In order to solve this equation without using Rydberg's equation, you need to find the energy difference of the energy levels with E=hv, which you seem to have. The trick is then to use the negative of that answer when plugging into the deltaE=E(final)-E(initial) equation. Remember, the energy t...

EllerySchlingmann1E

Hi, keV stands for kilo electronvolts, so you want to convert that to just electronvolts before using the conversion factor for Joules which is 1 eV/ 1.60218 * 10^-19 J. I'm not sure how often we will use keV but I'm pretty sure the conversion is given on our equations and constants sheet for tests....

EllerySchlingmann1E

I think doing all of the homework problems as well as the audio-visual modules is the way to go, because the syllabus said that homework would sometimes be used for tests. Furthermore, the questions in the assessments for the modules were developed by Dr. Lavelle so they may give some sense of what ...

EllerySchlingmann1E

You do need to take molar ratio into consideration. While amount of moles definitely matters, if the molar amounts are similar for the reactants, comparing their molar ratios with the product you are trying to find will give you a definite limiting reactant. In order to incorporate the molar ratio i...

EllerySchlingmann1E

Hi, I am currently reading Appendix 1C in our textbook about scientific notation. It says that when rounding off numbers ending in 5, the number should be rounded to the "nearest even number." For example, 2.65 would be rounded to 2.6. I feel very much like I have been lied to my entire li...

EllerySchlingmann1E

Hi, Molecules and formula units are not interchangeable, though they do refer to very similar things. Molecules refer to compounds in which the atoms are bonded covalently, while formula units refer to compounds with atoms that are ionically bonded. People tend to use "molecule" as a blank...

EllerySchlingmann1E

Hi, From what I understand, there is no difference in the calculations for atoms, molecules, and formula units. Because a mole is just 6.022X10^23 things, it doesn't matter whether those things are atoms, molecules, formula units or bicycles. You want to use Avogadro's number as your conversion fact...

EllerySchlingmann1E

Hi, This question definitely confused me as well. However, I figured out that while the number of moles and molecules stay the same in the solution as a whole, the molarity, or the number of moles/molecules (once you convert with Avogadro's number) in that 10. mL after 90 dilutions is almost nonexis...

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