CHEMISTRY COMMUNITY

JamesAntonios 1E

So I know for a reaction that produces a net number of moles, w=-(delta n)RT, is this in anyway related to Delta H=\Delta U+\Delta nRT (which is found in the book)? If so, how come we can use this equation under constant volume. I thought that work is zero for constant volume. However, this is the p...

JamesAntonios 1E

I saw this in the book and don't understand. "Potential energy is related to position. Therefore, in an isothermal expansion, why is there no potential energy change as the molecules move further apart?" Is this because temperature is related to the heat energy of the particles, and so the...

JamesAntonios 1E

Is this correct? So I know that when volume is constant, work is zero. However, for a reaction with changing pressure and moles, but constant volume, that is behaving ideally, we can relate deltaPV=delta nRT, in which there would be work, right?

JamesAntonios 1E

A reactant does not show up on the product sides. It is consumed in the reaction. A catalyst, on the other hand, will show up on both sides.

JamesAntonios 1E

I would assume so, as they will aid in finding the y intercepts, and the overall shape of the graph, based on the order.

JamesAntonios 1E

I think so because the rate law is just determined by that one step, so that is the only step that could be the rate determining step.

JamesAntonios 1E

Yes. Discussion sessions meet this week. However, next week, there are no classes.

JamesAntonios 1E

I don't think that molecularity isn't involved in intermediates. Molecularity only applies to the reactants in the rate determining step. If there are intermediates in the slow step, you have to solve for correct pre-equilibrium rate law, I believe.

JamesAntonios 1E

The activation energy determines the rate of reaction. To increase the rate of reaction, the activation energy or energy barrier must be lowered. For example, when rolling a ball up a hill and then letting it roll down, the get up the hill is the activation energy. A catalyst just builds a tunnel th...

JamesAntonios 1E

Yes this trend should continue. Just remember that for zero order, it is actually mol/l*s, not inverted.

JamesAntonios 1E

For a, because the reaction is 2A-->B + C, why isn't there a 2 in the reaction. If we are solving the differential rate law -1/2 d[A]/dt=k[A]^2, we get -1/2 d[A]/[A]^2= kdt. If we multiply the 2 to the other side, we are left with -d[A]/[A]^2=2kdt. If we integrate both sides, we get 1/[A]=2kt + 1/[A...

JamesAntonios 1E

They determined the concentration of H3O+ from steps 2 and 4. In these steps, both BrO3- and Br- are staying the same initial concentration. However, only H3O+ is changing. Therefore, by using these two steps, find the exponent of H3O+. And fc and ln is just another way of finding the exponent. They...

JamesAntonios 1E

No. My TA stated that this won't be necessary. However, Dr. Lavelle has spoken on how it helps to know how to derive each formula, as it demonstrates a deeper understanding of the concepts.

JamesAntonios 1E

We will need to know the overall order, if given experimental data (in order to right a rate law). However, for integrated rate laws, as of right now, we are only are dealing with zero, first, and second orders, and only one reactant.

JamesAntonios 1E

And remember, each side is usually written (anode reactant, anode product) and (cathode reactant, cathode product).

JamesAntonios 1E

E depends on the concentrations of the products and reactants. However, Eo is at standard state, and with respect to a hydrogen electrode.

JamesAntonios 1E

This problem states that "the aluminum air fuel cell is used as a reserve battery in remote location. In this cell , aluminum reacts with the oxygen in air in basic solution. It asks for the oxidation/reduction half reactions. However, for the oxidation, the answer key shows Al-->Al(OH)3. How w...

JamesAntonios 1E

Lucian1F wrote:For a constant volume, not pressure, the work is 0 so U=q=H. The example you stated says U=q at constant volume

According to the example 8.6b, it is at constant pressure.

JamesAntonios 1E

How come at constant pressure, for the first law, delta U=q=delta H? Why is there no work being done? An example of this is on pages 281-282 for example 8.6 where it asks for the change in internal energy at constant pressure.

JamesAntonios 1E

So delta U is 0 when it is an isothermal, reversible expansion. However, what if the expansion is irreversible? Is it still zero? Also, when is q in the first law zero?

JamesAntonios 1E

That could work. However, delta U would have to be given. But we do have an equation for a reversible reaction that is not isothermal for work.

JamesAntonios 1E

Reversible is when there is an infinitesimally small change, the reaction can be "reversed" and is not greatly affected. On the other hand, a irreversible reaction is a large change that is can be specifically pinpointed.

JamesAntonios 1E

Hi! This relates to question 9.73 in the book: "Under what conditions, if any, does the sign of each of the following quantities provide a criterion for assessing the spontaneity of a reaction? (a) G ; (b) H ; (c) S ; (d) delta S total" I thought that if delta S total was negative the rea...

JamesAntonios 1E

Minie 1G wrote:So how would you tell whether the equations, A and B, are spontaneous or not?

If the ΔG is negative for the reaction, it is spontaneous. If it is positive, it is non spontaneous. Another way you could do it is by calculating the total ΔS. If ΔS total is greater than 0, the reaction is spontaneous.

JamesAntonios 1E

Exothermic reactions transfer heat from the system to the surroundings. One of the common formulations of the 2nd Law of Thermodynamics is that the entropy of the universe always increases. This means that exothermic reactions are more likely to be spontaneous. If you consider the relationship ΔG =...

JamesAntonios 1E

I don't understand the explanation of the book. In previous examples, it has said that entropy of dissolving decreases for the surroundings (thermally) and increases for the system (positionally) (positive overall). Therefore, why are some still negative? The answer speaks of errors in reference poi...

JamesAntonios 1E

How do I know whether the entropy has a "+" or a "-"? It depends on the the ratio of final to initial temperature, volume, and/or pressure. You can logically think about by analyzing whether the number of microstates is increasing or decreasing, which can give an indication to w...

JamesAntonios 1E

melissa carey 1f wrote:Also, when do we need to include n, the number of moles, as I see it omitted in some forms of the equations? Thanks!

You pretty much always include them for entropy calculations. Because the units have to cancel out to leave you with J/K.

JamesAntonios 1E

We are not given values for V or T in this problem. Only n and P . Thus, two of the unknowns in the ideal gas equation of state PV = nRT are left unknown. The information that the change occurs isothermally is enough for us to use the formula \Delta S = nRln\frac{V_{f}}{V_{i}} but since we are give...

JamesAntonios 1E

William Lan 2l wrote:If delta P = 0 (constant pressure), why is delta U = q + w? I thought that w = -PdeltaV. If P is 0, there would be no work then right?

That is a variation of the original equation. It all depends on what the environment/surrounding/situation is and whether the reaction is reversible or not.

JamesAntonios 1E

Cristian Carrasco 1F wrote:If a system is expanding would that be the system doing work? and would Delta U be positive or negative

Delta U would depend on both q and w, not just q. If a system is expanding, it is doing work. If it is being compressed, work is being done on the system, and thus is a positive value.

JamesAntonios 1E

Organic chemistry is difficult. Those who study it have alkynes of trouble.

JamesAntonios 1E

so when using delta H, is it wrong for someone to refer to that as heat? delta H is enthalpy right? Heat and enthalpy are only equal when pressure is constant and volume is constant. That is why Dr. Lavelle wrote the equation qp=delta H. If these are not constant, you have to assume that Delta U=De...

JamesAntonios 1E

When is delta H in kJ/mol or just kJ? (because the book uses both units)

JamesAntonios 1E

Yes. That is correct. Isolated let’s nothing in or out.

JamesAntonios 1E

Enthalpy is heat but not heat is enthalpy.

JamesAntonios 1E

Heat capacity doesn’t take into account how much of a substance is there. It doesn’t include grams or moles

JamesAntonios 1E

Yes and no. You are just adding the values together. The values of energy required are always positive and the values of energy released are negative. So just add the positive and negative values.

JamesAntonios 1E

It's basically just about the units. It depends whether you are using grams or moles.

JamesAntonios 1E

They cannot be changed by intrinsically. That allows for them to be added. For example, work cannot be added together because of how "it is prepared." Either force or distance can be changed. But energy "preparation" is the same, and likewise for all state properties, allowing th...

JamesAntonios 1E

The book specifies that adding water will only shift an equilibrium if the other products or reactants are not dilute. However, if they are dilute, it says that we can assume that the concentration is unchanged. Will this be specified, whether the products or reactants are dilute in solution or not,...

JamesAntonios 1E

I've heard many conflicting arguments whether when assigning a radical, formal charge or electronegativity is more important. Some say that the electron is assigned to the atom of the lower electronegativity. However, in the book, there is a homework problem of CL-O where chlorine has a full octet w...

JamesAntonios 1E

If we are given only the structures of two conjugate bases from weak acids and are asked to compare which acid is stronger, how would we tell from the molecular structure? For example, in 12.115A, it asked whether carbonic acid would act as an acid or base in an acetate solution, without looking at ...

JamesAntonios 1E

When we were talking about bond character and polarizability and polarizing power, we said that those with higher polarizing power and higher polarizability express more covalent character than ionic character and becomes less soluble. However, does this only apply to salts? Because, this seems to b...

JamesAntonios 1E

Both your reasons that you stated are equally correct. Remember though than in Le Chatelier's Principle, the concentration of everything can be affected by the addition or subtraction of water.

JamesAntonios 1E

It's okay to use an ICE chart whenever there is a reaction. It just depends what you put for the starting and change values. For a strong base/acid, the change will cause the reactant base/acid to go to 0 while for weak base/acid, there will be some reactant still left over.

JamesAntonios 1E

When determining acid strength comparisons, which is more important: bond length or polarity of the bond? It seems to differ between oxoacids/carboxylic acids and acids like HF, HBr, HI, and HCl.

JamesAntonios 1E

I'm out of chemistry jokes. I should zinc of a new one.

JamesAntonios 1E

In this situation, liquid water is a part of the reaction. According to Le Chatelier's Principle, if more reactant is added then more CO2 will react and more product will be made. If the reaction did not include water however, then the addition of water will not affect the amount of product that is...

JamesAntonios 1E

P can have an expanded octet. You want to achieve the lowest normal charge possible. That is when O has two bonds and two lone pairs. This gives it a formal charge of 0. If it only had one bond, it would have a formal charge of -1.

JamesAntonios 1E

If a compound is a tridentate for example, will it only act as a tridentate, or could it be a bidentate or monodentate?

JamesAntonios 1E

The bars are another way of writing molarity or concentration. []=molarity

JamesAntonios 1E

Look at their hybridization and how many orbitals are half filled as well as their location in the periodic table.

JamesAntonios 1E

Remember when it asks for sub shell designation, it means both n and l as well (e.g. 4s, 2p, etc...).

JamesAntonios 1E

Usually with bond angles and molecular/electron shape, it is pure memorization and nothing more. However, most of the bond angles are relatively easy to figure out by using logic and working through the lewis structure.

JamesAntonios 1E

If a molecule has an unpaired electron, where does this electron go? Should the Lewis structure be made so that the lowest formal charge is achieved and that determines the placement of this electron?

JamesAntonios 1E

Remember, atoms want to be in the lowest energy and most stable state. This is either when all the orbitals are filled or none are filled. For group 13, they lose 3 electrons so that they reach noble gas configuration. Noble gasses are in the most stable state. They neither accept nor donate electro...

JamesAntonios 1E

A resonance structure means that the molecule is the same but the bonds can be interchanged. The bonds blur together in (dotted line). In isomers, this is referring to only organic compounds. The bonds do not blur with one another. Resonance structures can refer to both organic and inorganic molecul...

JamesAntonios 1E

Chemists never die. They just reach equilibrium.

JamesAntonios 1E

When in ground state and having lost or gained no electrons, atomic radius increases from right to left and down. However, when ions come into play, because those on the right of the periodic table gain electrons, they have an expanded shell and thus have a larger radius than cations. But, in all, a...

JamesAntonios 1E

Which takes priority? Should you have a lower formal charge with an expanded octet or higher formal charge with the 8 valence e-? And should you try to the central atom to have the lowest formal charge in all?
Thanks!

JamesAntonios 1E

Why exactly do electrons fill up the 4s orbital before filling up the 3d? And why when writing out the configuration must we write 3d before 4s? It all depends on the where the atom is on the periodic table. If d is 4/5, or 9/10, then d fills up first (goes to the higher number). If not these scena...

JamesAntonios 1E

Rydberg's Equation describes the change in energy. Change is always final-initial, except in very few specific cases.

JamesAntonios 1E

It's easier to think about in terms of E=-hR/n^2. Then, calculated the change of final-initial. Or use energy to get wavelength or frequency. Hope this helps!

JamesAntonios 1E

Set up the equation like this. The change in momentum times the change in position has to be greater than or equal to h/4pi. Momentum is mv of an electron. v is the unknown but you do know the position. So, plug in these numbers (9.110x10^-31 kg)(x m/s)(350.x10^-12 m)>/(6.626x10^-34 Js)/4pi. Solve t...

JamesAntonios 1E

EM radiation can be seen through the particle lens. If the frequency is decreasing, the energy of the field/wave is decreasing as well. With relation to an electric field, if the frequency, hence the energy decreases.

JamesAntonios 1E

Laura Riccardelli wrote:How often will be asked to apply Rydberg's equation in this class?

Usually, you can assume to apply the Rydberg equation whenever there is a change in energy level, say from n=2 to n=1, etc... Other than that scenario, the Rydberg equation is not that useful.

JamesAntonios 1E

joshuaxian wrote:During a combustion reaction of an organic compound, is water vapor usually produced, or are there situations in which liquid water is produced instead?

usually water vapor unless otherwise stated.

JamesAntonios 1E

As a follow-up question for Problem E35, I was wondering how you know the conversion of 1 mol Ca5((PO4)3)OH is exactly 1 mol Ca5((PO4)3)F ? Is it assumed since the problem doesn't state otherwise? That was the whole point of the percent increase problem. If you have a certain number of grams of Ca5...

JamesAntonios 1E

Do you know if the problems will specifically tell us how many significant figures to use? No. With the numbers that are given, you should know how many sig figs to use. But when using molecular mass, the best number is 3-4 sig figs. When in doubt, use 3 sig figs. Other than that, always apply the ...

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