CHEMISTRY COMMUNITY

Curtis Tam 1J

For the test, will be be given the van't hoff equation and the arrhenius equation in the form ln(k2/k1)=Ea/R(1/t1-1/t2)?

Curtis Tam 1J

They're using the ratios to see how the rate changes with respect to activation energy. I believe you could have plugged in the actual values given in the problem but simply keeping the Ea as a variable makes the calculation easier.

Curtis Tam 1J

The steady state is a much longer approach to finding the correct reaction mechanism. This involves setting the rate of reaction of the intermediate equal to zero and then plugging it into another rate law for another elementary reaction. For preequilibrium the approach is much simpler and simply in...

Curtis Tam 1J

Since this particular reaction is producing product at low temperatures, this means it has a low activation energy, which means that it is under kinetic control as opposed to producing product at high temperatures, which is defined as under thermodynamic control in the problem.

Curtis Tam 1J

Oh wait never mind. If I simplified the balanced equation to stoichiometric coefficients of 1, then the value of n in the Nernst equation changes such that I get the same answer either way. Problem solved :)

Curtis Tam 1J

Determine the potential of each of the following cells: a) Pt(s) / H2(g,1.0 bar) / HCl(aq, 0.075 M) // HCl(aq, 1.0M) / H2(g, 1.0 bar) /Pt(s) In the solutions manual, they give the balanced equation as: 2H+ (aq, 1.0M) + H2(g, 1 atm) ----> 2H+ (aq, 0.075M) + H2 (g, 1 atm) If you go on to calculate E(c...

Curtis Tam 1J

In the book it states that the reaction with a higher activation barrier will be more sensitive to changes in temperature therefore raising its k value. For example, an endothermic reaction has a higher activation energy barrier in the forward reaction than in the reverse. This is why adding heat is...

Curtis Tam 1J

So I noticed that there weren't any homework problems that specifically asked us to calculate out a rate law that matched the overall rate law given by the problem. Most of them had problems that had slow steps whose rate law matched the overall rate law to begin with. For the test, should we expect...

Curtis Tam 1J

On the homework problems, I noticed that for first order half life problems, the solution's manual can count the number of half-lives and multiply that number by the half-life for the reaction. However, for second order problems, it seems like that same operation cannot be carried out. Instead, we m...

Curtis Tam 1J

looking at the reaction***

Curtis Tam 1J

You can't tell if it's first order just by looking at the equation. You can only do so with individual elementary step reactions but not with the total reaction. The order for the net reaction is always experimentally determined which is why it must be given in the problem.

Curtis Tam 1J

For part a you know that the second product is formed under high temps, so that means a large amount of energy is needed to overcome the energy barrier for the product to form. I am also slightly confused on the wording in part b, but I think the rule is that if you have a low Ea, then kinetic contr...

Curtis Tam 1J

In the solutions manual, they were first trying to determine what the concentration of [A]t was before using the first order rate law equation.

Curtis Tam 1J

I believe its because the presence of O2 as a product can have a significant impact on the overall reaction rate. This was probably determined experimentally. I'm pretty sure if we are given a problem where we don't have much background info, we can just use reactants for the rate law :)

Curtis Tam 1J

I'm pretty sure they do exist for products but the point of a rate law is to simplify the reaction so that we are only studying its rate based on the reactants because the concentration of products could affect the rate of reactant consumption.

Curtis Tam 1J

In the book, it simply writes the rate law as rate=k[A][B] for example. But what does the "rate" actually refer to? The reactants? Products? Both? Also, are there going to be any cases where we see a coefficient in the rate law. For example, something like rate=2*k[A][B]. In the book, it i...

Curtis Tam 1J

Basically if the reactant is not affecting the rate of the reaction, it is probably due to the presence of a catalyst. If we have an efficient catalyst, then reactions will go faster. If the catalyst is not very good, the reactions will go slower. Both of these observations would be made without reg...

Curtis Tam 1J

Conceptually I am not totally sure but mathematically, the half life of first order reactions only include the term k, which is the reaction constant. I think if you look at how they plot different concentrations of the reactant versus their respective initial reaction rates, you'll notice that they...

Curtis Tam 1J

I believe it has something to do with simplifying the rate law by assuming that all reactants are in great excess except for one, which will be used to determine the rate of reaction. If something is in great excess, then some time later, it will have minimal change relative to reactants that are in...

Curtis Tam 1J

Not sure if there's a rule for it. Because for example we say glucose is "oxidized" in the combustion equation but technically the carbon I believe is being oxidized.

Curtis Tam 1J

Okay so I get that species being oxidized are the reducing agents and the same logic goes for the species being reduced. So my question is, if something like S2O3 2- is the reducing agent, is the entire molecule said to be oxidized or do we just say the S in the S2O3 2- is oxidized. Or is both accep...

Curtis Tam 1J

Yes it's in the appendix but for some reason, the appendix value is different. Maybe the pH influences the standard E, not very sure. I have the same question as well.

Curtis Tam 1J

When there is no conducting metal available to carry the electrons. In most cases, you will need the pt electrode when there are only ions involved in the redox rxn.

Curtis Tam 1J

Yea in that case we say that the reverse reaction is favored.

Curtis Tam 1J

Problem: 14.33 (a) The standard Gibbs free energy of formation of Tl3+(aq) is +215 kJ.mol-1 at 25°C. Calculate the standard potential of the Tl3+/Tl couple. (b) Will Tl+ disproportionate in aqueous solution? In the book and in the solutions manual, the Standard Gibbs Free Energy of each half reactio...

Curtis Tam 1J

a) Write balanced half-reactions for the redox reaction of an acidified solution of potassium permanganate and iron(II) chloride. b) Write the balanced equation for the cell reaction and devise a galvanic cell to study the reaction (write its cell diagram). I'm pretty sure that Appendix 2B is needed...

Curtis Tam 1J

Is there a reason why OH- is not included in the right hand side of the cell diagram? Because OH- is included in both half reactions.

Curtis Tam 1J

Sometimes the only species are available as ions and are separated by commas on one side of the cell diagram. Does it need to be in any particular order? I feel like there are some inconsistencies in the book problems in regards to how they order the ions

Curtis Tam 1J

Because the surroundings is basically a vacuum. The system is expanding but there is no heat transfer between the system and the surroundings so the surroundings basically remains unchanged.

Curtis Tam 1J

Open system is where you can for example actively take reactant or product out of a system. For example, i think a car gas tank can be an open system. You burn the gasoline and you can manually restore the gasoline.

Curtis Tam 1J

When it says heating at a constant pressure, it refers to the pressure of the system. However, if it is an open system (such as a flask or test tube), then the pressure of the system is the same as the pressure of the surroundings. Yea I think I understand that a bit better now. I looked online and...

Curtis Tam 1J

I'm slightly confused by what the book means when it says heating at constant pressure. Are they referring to the pressure of the system or the pressure of the surroundings (external pressure)? In other words, when you use deltaU=q+w, does q(p) refer to heat transferred when the system is at constan...

Curtis Tam 1J

During the test of an internal combustion engine, 3.00L of nitrogen gas at 18.5C was compressed suddenly (and irreversibly) to 0.500L by driving in a piston. In the process, the temperature of the gas increased to 28.1C. Assume ideal behavior. What is the change in entropy of the gas. The solution w...

Curtis Tam 1J

Are we going to have to know how to derive all the equations for the midterm?

Curtis Tam 1J

Well since this is an endothermic process, I believe some heat would be taken away from the water surrounding the salt. But I'm not sure if this value is even important since I would assume the majority of the change of entropy of the dissolution would be calculating the change in positional entropy...

Curtis Tam 1J

The temperature dependent entropy change equation isn't exactly derived from the reversible isothermal work equation. But, it is similar in a sense: to get that equation, we assume that the change in temperature is infinitely small, that the process is isochoric (volume doesn't change). Then, we su...

Curtis Tam 1J

When we are calculating the entropy change for a reversible process, I understand that you use the equation deltaS=nRln(V2/V1) at constant temperature. For the equation for temperature dependence, deltaS=nCln(T2/T1), would that be used when the temperature is not constant since the final and intial ...

Curtis Tam 1J

I get how to calculate entropy and enthalpy to see how temperature affects gibbs free energy but doesn't the book have to specify which direction it's going in. In problem 63, it uses the same compounds and asks us to calculate based on its decomposition but in this problem it doesn't specify, and t...

Curtis Tam 1J

I believe when it's under constant pressure and temperature. This has something to do with how we derive Gibbs Free Energy to get deltaG=wmax. It's found on page 355.

Curtis Tam 1J

I think if you were to have a free expansion for example, the entropy of the system increases because the volume increases, leading to more possible microstates. However, work is not done on the surroundings because the system is not pushing against any constant pressure. Therefore, the surroundings...

Curtis Tam 1J

Yes! absolutely. If the system expands fast, and the energy lost by work is replaced by heat, you can have an isothermal reversible expansion. Take in mind that, reversible expansion will do more work overall however (the area under the curve is larger). You mean isothermal irreversible expansion r...

Curtis Tam 1J

Is it possible to have an isothermal IRREVERSIBLE expansion? I know that for an isothermal reversible expansion, work done on the surroundings is replaced by heat slowly. However, for an irreversible expansion, I know the expansion is fast but technically can it still be isothermal if enough heat is...

Curtis Tam 1J

I think you don't really need to think in terms of forward or reverse reactions. Just know that in any reaction, it will tend to proceed in a certain way. If the forward reaction doesn't proceed spontaneously, then I believe it should be spontaneous in the reverse direction.

Curtis Tam 1J

Yes, I believe time should be included in equilibrium thermodynamics but now we are going into the realm of differential modeling, which takes into account time delays in dynamical systems. #ls30b

Curtis Tam 1J

I believe the solution manual says H2 is zero right? The enthalpy of formation for Zn2+ is used because ZnCl2 dissociates into Zn2+ and Cl- ions.

Curtis Tam 1J

The expression 3/2RT is derived in the section of the textbook that describes the equipartition theorem. If you use this expression, you will automatically get J/mol.

Curtis Tam 1J

I think it's because at 25 degrees Celsius, H2O would be in the liquid state.

Curtis Tam 1J

Free expansion happens when there is no change in volume or when a reaction, for example, is taking place in a vacuum (no pressure, so P=0).

Curtis Tam 1J

deltaU=q+w

Curtis Tam 1J

I believe it's negative since you have a positive net moles of gas produced. Therefore you are doing work on the surroundings so it's negative :)

Curtis Tam 1J

Here, they give you the enthalpy of combustion, so first you would have to write combustion equations for each of the molecules they give you. But do so for one mole of each combustible. C2H2 + 5/2O2 ----> 2CO2 + H2O C2H6 + 7/2O2 ----> 2CO2 + 3H2O H2 + 1/2O2 ----> 2 H2-=O Rearrange the equations for...

Curtis Tam 1J

Analyze the slopes and heat capacities of the curve. The heat capacity for liquid is double that of solid. Therefore, the flat line for vaporization should be twice as long as the flat line for fusion. Since liquid has a higher heat capacity, it's rising slope should also be less steep than those of...

Curtis Tam 1J

Smh tomato. Check solution manual errors.

Curtis Tam 1J

I think you can just rearrange the equation, q=mc(deltaT) to get c=q/(m*deltaT)

Curtis Tam 1J

I think heat should still be calculated with q=mc(deltaT). The only difference with these ideal gases is that the heat capacities could be any of those found on page 281. I'm not entirely sure if this is what you're referring to though.

Curtis Tam 1J

The enthalpy change given is for every four moles of carbon reacted so that's why you must include that in your stoichiometric calculations :)

Curtis Tam 1J

It's referring to the calorimeter. In this case, internal energy would be negative because heat is given off by the reaction. And because the system is at constant volume, deltaU is equal to q. :)

Curtis Tam 1J

You can tell by looking at the periodic table or by just noticing that certain elements like oxygen are always written as O2(gas) because realistically we never have oxygen (O) by itself in nature since it would be too reactive to stay that way :)

Curtis Tam 1J

Bomb calorimeter is basically a system where a reaction takes place at constant volume. Since volume is constant, then change in internal energy (deltaU) would be equal to just q since deltaV is zero.

Curtis Tam 1J

I believe so!

Curtis Tam 1J

I believe it's delta zero given that you are referring to isothermal expansion/compression of an ideal gas. This means that any work done to move the piston inwards or outwards is restored by heat flow. This makes sense because if deltaU= q + w = 0, then you have q = - w, which basically means the p...

Curtis Tam 1J

I think what it means is that chemical reactions may produce a net mol of gas which means there is expansion. If there is expansion, then you would have a nonzero value for deltaV and therefore you would also have a nonzero value for work. I'm confused on that as well :(

Curtis Tam 1J

It is the heat capacity of monatomic ideal gases at constant pressure. Constants are found on page 281. I believe you calculate heat with q=(mol of substance)(heat capacity=5/2R)(temp change). :)

Curtis Tam 1J

For benzene without resonance, you would use a bond enthalpy for C single bonded to C and a bond enthalpy for C double bonded to C. However, with resonance, you have a blend of both single and double bond character which means realistically you wouldn't use two distinct enthalpy values for C C bonds...

Curtis Tam 1J

Which chapter?

Curtis Tam 1J

By definition, autoprotolysis generates 10^-7 in concentration for hydronium. When we add such a tiny amount of acid, the amount of hydronium it produces may be so minute as to say perhaps 10^-15. However, when we calculate pH, we get something greater than 7 which cannot be possible because we are ...

Curtis Tam 1J

Most of the time, you just need to calculate Q when there are concentrations on both sides because we wouldn't know for sure which direction to go. However, if only one side of the equation has given concentrations, you can assume the other side is zero, so always shift in that direction.

Curtis Tam 1J

Calculate the difference in energy, deltaE.

Then apply E=hv and solve for v to get the frequency :)

Curtis Tam 1J

Ka would be smaller. The smaller 2+ charge yields less interaction with the oxygen lone pair so the resulting conjugate base would be less stable than that of [Fe(H2O)6]3+

Hope that answers your question :)

Curtis Tam 1J

605011646 wrote:This isn't something to memorize, but Lavelle did mention in class that the acid rain problems in the TB were worth doing too...

Dr. Lavelle didn't really cover it in class so what aspect of acid rain is he talking about. Like the first deprotonation of H2SO4?

Curtis Tam 1J

So this isn't a hw question, but this is more from the lecture notes. Why is HI a really strong acid, and HOI not as strong? I understand that HI has a long bond length and is long, and I understand that having more oxygens stabilizes the resulting anion. However, when combining these concepts, it ...

Curtis Tam 1J

I think this is a topic discussed in 14C. It has to do with the distance between the halogen and hydrogen atom in oxoacids versus binary acids. However, when comparing just HI and HOI alone, I'm not entirely sure how that is determined. Probably we have to look at the pKa's.

Curtis Tam 1J

Aye fam! Nah that wouldn't be necessary cuz you just have to recognize that more oxygens would better stabilize the conjugate base, making the acid stronger. You don't need to do resonance anyways as long as you recognize that the double bond character is spread out evenly over the entire molecule. ...

Curtis Tam 1J

Yes it is one electron short of completing the octet

Curtis Tam 1J

You can think of it in terms of Le Chatelier's Principle. When nitrous acid is dissolved, it produces a weaker conjugate base than that of acetic acid. This means that the conjugate base of acetic acid is better at eating all the hydronium ions (in the solutions manual, it says H+, not sure if they'...

Curtis Tam 1J

I think there's six water molecules attached to the TM because an expanded octet makes the compound more stable. Not entirely sure though.

Curtis Tam 1J

Basically it's used to determine whether or not you can use approximation when solving for x in an ice table. For example, if the percent deprotonation is less than 5, then it is okay to write for example in the denominator that 0.1-x is essentially 0.1.

Curtis Tam 1J

Calculate Q first to see which direction the reaction would go in. Then do an icebox calculation and set your equilibrium concentrations equal to k. Solve for x and plug those values back into the equilibrium concentrations.

Curtis Tam 1J

Part b should be in the answer manual. Basically, when the carbon atom has two regions of electron density, it is sp hybridized. Therefore, its desired bond angle would want to be 180 degrees. However, the bond angle about the carbon has been strained so that it has a bond angle of 120 degrees. It i...

Curtis Tam 1J

A polydentate can bind to multiple sites on the central atom. A chelate is a complex containing a ligand that forms a ring of atoms including the central atom. So a polydentate can be a chelate at the same time. I believe en itself is a ligand so since you have three of them and each is a bidentate,...

Curtis Tam 1J

I think this is actually related to the le chatelier's principle which is beyond 11.8. When you have a greater concentration of reactant, for example, more product will be produced to "minimize the effect of the stress" to maintain the same Kc value.

Curtis Tam 1J

Calculate Q :)

Curtis Tam 1J

I did that problem too and got confused. I think it's cuz you want to focus on the ratio of O2 raised to its nonzero coefficient to O3 raised to its nonzero coefficient. This ratio would never change because this produces the value Kc but that doesn't necessarily mean that the ratio of O2 to O3 woul...

Curtis Tam 1J

A sigma bond is a direct overlap, whereas pi bonds overlap side by side on both sides of an internuclear axis. I don't believe there is a way for two atoms to directly overlap their orbitals two times. Pi bonds can because you can have two sets of pi bonds perpendicular to each other.

Curtis Tam 1J

Just look at the electron arrangement for the atom and assume that it will be slightly less than the bond angles for that particular arrangement. For example, water has a bent shape but it's electron arrangement is tetrahedral. Therefore, we would assume that it's slightly less than 109.5.

Curtis Tam 1J

Are you talking about hybridization. If so, I believe TM can since they can hybridize to have sp3d2 orbitals since many of them can form coordination complexes.

Curtis Tam 1J

Hey,

Just wondering if anyone knows if a single bonded oxygen is hybridized. My TA said it's not but I just wanted to confirm. Thx!

Curtis Tam 1J

Yes, it just wanted to emphasize that even if you changed the single bond between the oxygen and the Sb to a double bond, the shape would not change.

Curtis Tam 1J

I believe you approximate when you know the Kc value is really small, meaning any change "x" would be negligible, though it wouldn't be equal to zero. Is this what you are talking about?

Curtis Tam 1J

The shape for ch2cl2 is tetrahedral. The lewis structure drawing is really misleading because it makes you think that the chlorine atoms are directly opposite from each other on a 1D surface. In reality though, the angles are 109.5 degrees so wherever you place the chlorine atoms, they will always b...

Curtis Tam 1J

I believe so since you don't have any concentration of solids in the first place.

Curtis Tam 1J

Lol got the same question wrong on the midterm if that's what you're asking about. The bonds themselves are polar. The molecular overall however is nonpolar. Since the test was covering chapters 1 through 3, you shouldn't have used the idea that dipoles can cancel each other out. That was taught in ...

Curtis Tam 1J

Yes you must take the lone pairs into consideration. When there are more than 5 bonding regions, let's just say 6, then you would have a hybridization of sp3d2. I don't think you will run into many situations where there are 7 bonding regions but if there are im guessing you would increase the numbe...

Curtis Tam 1J

Yes lowest energy is when the formal charges are closest to zero. In the midterm i believe that the structures all had formal charges of zero for every element, including the radical for the carbon. But in any case, it's too late to change anything anyways so welp

Curtis Tam 1J

n denotes the energy of the shell so it would correlate to the period of the periodic table. d and f would be l values which which is not the same thing as the "nth" period. In the case of d, electrons will fill the orbitals lagging one period behind the s orbitals. For f, electrons will f...

Curtis Tam 1J

If you were to calculate the formal charge of the sulfur in this Lewis structure, you would get a formal charge of 0, which is favorable.

Curtis Tam 1J

When getting the +2 oxidation state, you would want to remove the outermost electrons which would be in the 4s orbitals. Hope that helps!

Curtis Tam 1J

The f block starts filling 2 blocks behind. It also goes in an irregular order since the energy differences between the higher principal quantum numbers get smaller and smaller. Therefore, there is a greater variation in electronic configuration with negligible deviations in energy from configuratio...

Curtis Tam 1J

The magnetic quantum number refers to the different orbitals of a sub-shell (for example, the p x , p y , and p z orbitals). It is different from the angular momentum quantum number (l) that only describes whether it's s, p, d, or f. ml can take on the positive and negative values of l. For example...

Curtis Tam 1J

I think it could be used but if the "velocity" of the work function is not given (I haven't seen a problem that gives velocity as part of a work function) then maybe it's not valid. I also know that photons have no rest mass so I don't know what you would put for "m".

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