CHEMISTRY COMMUNITY

Gianna Apoderado 1B

A catalyst will be used up in a step in the reaction mechanism but gets produced again in a different step. An intermediate gets produced in a step of a mechanism but gets consumed in another. Also, a catalyst can be included in the overall rate law, while an intermediate cannot.

Gianna Apoderado 1B

I believe the Nernst equation on the equation sheet with log(Q) assumes that the temperature is 298 K (25^oC), while the ln(Q) equation can take into account any temperature.

Gianna Apoderado 1B

Yes, because the concentration of catalysts affect the reaction rate, the higher the concentration, the faster the reaction!

Gianna Apoderado 1B

I believe that when it asks about the reaction, it generally means the forward direction, unless otherwise stated. In this case, the reaction is endothermic because the activation energy of the forward direction is greater than the activation energy of the reverse reaction, which means that the forw...

Gianna Apoderado 1B

Given the species, you can generally predict which reaction will be the reduction and oxidation from a given list of reactions and their standard reduction potentials. The more positive E^o, the stronger the oxidizing agent is, and the more likely that the species will be reduced.

Gianna Apoderado 1B

I was wondering this as well.. I used (3R/2) as C_v for all three gases, and I still got B < C = A, the same result as the one on the "Errors in Solution Manual" page.

Gianna Apoderado 1B

You do end up using the natural log (ln) in this calculation, after you solve for the [A]_t! After you get that value, you plug it in
k = (ln([A]₀/[A]_t)) / t to get k.

Gianna Apoderado 1B

From #45-73 for sure, but there may be integrated exercises (#79 onwards) that might apply to the topics we learned in 15.1-15.6, so i would take a look there just to make sure.

Gianna Apoderado 1B

When you take the integral of the -d[A]/[A]², it gets cancelled out. I attached the step by step integration below:

Gianna Apoderado 1B

Since it was specified that the reactions are second-order, we can use 1/[A] = kt + 1/[A]₀, and just solve for t.

Gianna Apoderado 1B

We may be asked to identify/recognize when we can implement the pseudo-order reaction method, such as when the concentration of one or more reactants are far greater than the concentration of another reactant. Then you would just calculate the rate as if it were a normal ____ order reaction.

Gianna Apoderado 1B

From what I understand from today's lecture, k is the normal rate constant for a reaction, let's say for rate = k[A] N [B] M [C] L . However, if [B] 0 and [C] 0 >> [A] 0 , we can say that rate = k'[A] N , and that the reaction is a pseudo-first order reaction. Comparing the two, we can see differenc...

Gianna Apoderado 1B

I was wondering this as well, since Au³⁺ is supposed to be on the products side. It may have been a solution manual error.

Gianna Apoderado 1B

Since the equation is balanced, you can take a half-reaction from that equation to find the number of electrons, n r . When 2 Ce 4+ --> 2 Ce 3+ , we can see that the overall charge on each side goes from +8 to +6, so, in order to balance that, two electrons need to be added to the left side. This is...

Gianna Apoderado 1B

In the calculation of the standard potential for the left (Metal) reduction, how come the sign of the Cu oxidation potential doesn't become negative? Initially, I had: E o (left) = E o (right) - E o (cell) = -0.34 - 0.689 = -1.029 But when I checked the answer, it was -0.349, meaning the potential o...

Gianna Apoderado 1B

In the book, it says that a spontaneous process has a natural tendency to occur, so it may not even occur at all, depending on the rate of that tendency (which will be covered in Ch. 15)

Gianna Apoderado 1B

The change in free energy (delta G) for a spontaneous process is the energy that is available to do useful work. We were also given an equation in lecture, w_max = delta G, where maximum work is done by a process at constant temperature and pressure. Hope that helps!

Gianna Apoderado 1B

A similar question/post was posted on Chemistry Community a few weeks ago:

viewtopic.php?f=128&t=26506

Someone said while not all isothermal reactions are reversible, reversible reactions are isothermal.

Gianna Apoderado 1B

Residual entropy is the entropy of a sample at T = 0, that comes from positional disorder surviving at that temperature. In other words, it is the disorder retained by some solids at T = 0.

Gianna Apoderado 1B

I also got the same answer as Kelly! Make sure to double check your signs, you may have misplaced a negative sign or positive sign.

Gianna Apoderado 1B

I would stick with using C(p,m) in that deltaS equation, which as said by Joyce, is (5/2)R at constant pressure.

Gianna Apoderado 1B

By the way, the table for bond dissociation energies is Table 3.3, which is page 93 in the textbook!

Gianna Apoderado 1B

Yes, like Mitch said above, delta U has two components, as seen in the equation deltaU = q + w (the first law of thermodynamics). Even if change in temperature is 0, and q is 0, work can still be done and affect internal energy.

Gianna Apoderado 1B

From what I’ve seen, delta S is given in both J/K and J/K•mol-1 depending on the context of what you need it for! (Ive always assumed that the “per mole” is implied)

Gianna Apoderado 1B

We can also view entropy as representative of the amount of energy that is unavailable to do mechanical work in a closed system!

Gianna Apoderado 1B

Like Michelle, I assumed the r subscript stood for reaction, but when I looked it up, it said r stood for rotational?

Gianna Apoderado 1B

Yes, what Ammar said above is true! In addition, as you work through the problem you can see that you'll have to get rid of N₂, which isn't in the net equation, and flipping the second equation is the only way to get rid of it.

Gianna Apoderado 1B

Hi! For test 1, Lavelle said to look through Outline 1 on his website for the concepts we will be tested on, except for isothermal problems. Since the equipartition theorem is not on there, I would say that we don't need to study it.

Gianna Apoderado 1B

You can also look at it as finding the q of the reaction, or q_rxn. So, using the relationship of q_rxn = -q_cal, your calculated value of the specific heat of the calorimeter and the change in temperature, you can find the solution (much like how Joyce did!)

Gianna Apoderado 1B

I would stick with atm for pressure and liters (L) for volume when calculating work using w=-P*deltaV.

Gianna Apoderado 1B

Going by the sig figs of the given values, I agree that it should be 3 sig figs. (Maybe the book isn't as strict with sig figs, I'm not sure)

Gianna Apoderado 1B

Yes! To add on to what Lisa said, intensive properties only depend on the type of the substance, so no matter the size or amount of substance you have, the property does not change. For example, the color of copper does not change whether you have 2 g of it or 50 g.

Gianna Apoderado 1B

I don't really get why you need to add the heat of the kettle and the heat of the water together. I thought it was just asking for the heat supplied to the kettle to heat it up to 100 degrees Celsius. Hi, I assume you mean it was just asking for the heat needed for the water to boil. We need to tak...

Gianna Apoderado 1B

Gas has two molar heat capacities because there is one for constant volume (Cv) and one for contant pressure (Cp) !

Gianna Apoderado 1B

Since Dr. Lavelle didn't cover it in lecture or included it in the topic outlines on his website, I believe we don't need to know it for the final!

Gianna Apoderado 1B

The conjugate seesaw is a term describing the relative strength of an acid/base to its conjugate base/acid. The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate base.

Gianna Apoderado 1B

If I recall correctly, the only table we were meant to be quite familiar with was Table 17.4. The other tables in the chapters we've discussed may be helpful in understanding concepts, but I don't believe we have to have them memorized.

Gianna Apoderado 1B

Hi! Oxidation numbers are not always positive! For example, for a sulfur ion S^2-, its oxidation number is -2.

Gianna Apoderado 1B

Well, I can see why NH3 will come before the Cl, because it is called ammine, while the chloride is called chlorido. I'm not too sure why NO2, or nitro, comes before chlorido though.

Gianna Apoderado 1B

That might have just been a book error, or maybe a rare exception to the rule, because according to Toolbox 17.1 we do name ligands in alphabetical order!

Gianna Apoderado 1B

Yes! We take both the geometry of the molecule as well as the electronegativity of each atom in that molecule to determine polarity. We can use the periodic table and its trends to determine which atom is more electronegative than the other (but take note of the exceptions to this rule - such as the...

Gianna Apoderado 1B

Also, we know that CH2 has a larger bond angle than CH4 because CH2 has less electron regions, only 3, so when we apply the electron repulsion theory, they can be further apart from each other than a molecule with 4 electron regions. Hope that helps!

Gianna Apoderado 1B

Yes, CH2 would have a bond angle less than 120 due to the lone pair! The so-called electron "cloud" takes up a large volume of space, and electron repulsion from that lone pair pushes down on the bond angles.

Gianna Apoderado 1B

Here is a diagram clarifying what Ashley said above! You can see that the 3 lone pairs and their electron "clouds" have repulsion, such that the bonds between the central atom and the two other atoms are manipulated to a linear shape. https://image.slidesharecdn.com/tang04-vsepr-1209261830...

Gianna Apoderado 1B

Yes, just to reiterate, every single bond, as well as the first bond (so only 1!) in a double or triple bond, counts as a sigma bond.

Gianna Apoderado 1B

For me, how I identify polar vs nonpolar molecules is by looking at the lewis structures of the molecules, and determining whether on not the shape allows for dipoles (if there are any) to cancel. Take two common examples: CO2 and H20. When you draw the lewis structure for H20, you'll see that the s...

Gianna Apoderado 1B

From what I understand, hydrogen has relatively high electronegativity, especially compared to the other elements in its group, due to only having one electron in its 1s shell. Atoms are more stable and "happiest" with a full shell, so hydrogen has a higher tendency to attract an electron ...

Gianna Apoderado 1B

For test 3 question 2b, we had to place Cl, Na, and Br- in increasing atomic or ionic radius. I had them in the order of Cl < Br- < Na, but the correct answer was Cl < Na < Br-. I understand that Br- is larger than the neutral atom Br due to the extra electron, but how is it larger than Na, when the...

Gianna Apoderado 1B

31. Using the mole ratio provided in the chemical equation, 1 mole of A means that you'll use 0.5 moles of B. (You could convert this into grams if you know the grams per mole of B). 33. Since the limiting reagent is A, the max amount of product you could produce has to be calculated from the amount...

Gianna Apoderado 1B

Like Nina said above, periodic trends are not always 100%! There are some exceptions to the "rules" that apply generally to the elements.

Gianna Apoderado 1B

Electron affinity, the change in energy when an electron is added to a neutral atom to form a negative ion, affects elements/electrons in the sense that the higher the electron affinity, the greater the attraction for a electron/the easier it is for that element to gain electrons.

Gianna Apoderado 1B

I believe the answer is D! A, B, and C all describe the same equation, which has to do with the photoelectric effect. The equation for D was not derived as a result from the photoelectric experiment!

Gianna Apoderado 1B

I believe the answer is UV! If I'm not mistaken, ultraviolet radiation is used because it has a short wavelength (high frequency), which means it has a higher energy per photon (E=hv). Even with low intensity light, it can eject electrons from the surface.

Gianna Apoderado 1B

Hi! Since we are referring to the Balmer series, after the excited electron moves from a higher energy level back down to n=2, it does not move down to n=1, because then we'd be referring to the Lyman series. I've included a link to an image similar to the one shown in Friday's lecture, which hopefu...

Gianna Apoderado 1B

From what I understood from the reading, the Balmer series is made up of lines in the visible spectrum. I'm assuming you mean the Lyman series, not the Rydberg series (because there was only mention of a Rydberg constant in the book). The Lyman series is made up of lines in the ultraviolet spectrum....

Gianna Apoderado 1B

It said on the website that the test will cover Fundamentals, so I'm assuming we will have to know some of the concepts in the textbook in that section that weren't covered in class. I'm not entirely positive though!

Gianna Apoderado 1B

Hello Rita, From what I recall, there's a system of writing chemical formulas called the Hill system. Whenever a chemical formula contains carbon atoms, carbon is listed first, then hydrogen (if it contains hydrogen), then the rest of the elements in alphabetical order. If there are no carbon atoms,...

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