CHEMISTRY COMMUNITY

Ria Nawathe 1C

Both graphs describe 1st order reactions, the difference is that one expresses the integrated rate law while the other is based on the differential rate law.

Ria Nawathe 1C

You would not include intermediates in the overall rate law. Usually, the intermediate is also in an elementary step that is fast reversible, so you can come up with an expression for the intermediate in terms of other species in the fast reversible elementary step and then replace the intermediate ...

Ria Nawathe 1C

A zero order reaction has a constant rate; it does not depend on concentration. When you plot concentration vs time for a 0 order reaction, the resultant graph should be a line with negative slope, and the slope = -k.

Ria Nawathe 1C

As you stated, the reaction mechanism is the specific pathway or steps through which a reaction occurs. Those steps are typically elementary reactions, and molecularity is a way of classifying these elementary reactions based on the number of species involved in the reaction. The molecularity of an ...

Ria Nawathe 1C

For a reaction mechanism that is proposed to have multiple steps (multiple elementary reactions that comprise the overall reaction), the rate-limiting step is the elementary reaction with the slowest rate. This slowest step is the step that determines the rate of the entire reaction, so the rate law...

Ria Nawathe 1C

Since E(cell) for the given cell is negative, this means that the left half-cell is the cathode and the right half-cell is the anode because of the way the cell potential is reported. This is what the textbook says about this (Focus 6L.3, pg 551): "In summary, the sign of the cell potential rep...

Ria Nawathe 1C

You need to take the square root of the K value because the original reaction you use is 2HF --> 2H+ + 2F-, but the reaction for Ka is HF --> H+ + F-. Dividing the stoichiometric coefficients by 2 is the same as raising the K value to the power of 1/2 (taking the square root), so the result is the K...

Ria Nawathe 1C

a) Because you know that Br- --> BrO3- is the oxidation reaction since the oxidation state of Br changes from -1 to +5, the 2 remaining species must be involved in the reduction reaction. Therefore, O3 --> O2 is the unbalanced reduction half-reaction. b) I think it's just a rule of thumb to always b...

Ria Nawathe 1C

For 6L.7 (a), you essentially have to come up with 2 half-reactions (one oxidation and one reduction) so that when combined the overall reaction gives AgBr(s) <--> Ag+(aq) + Br-(aq). For the oxidation half-reaction: Ag(s) + Br-(aq) --> AgBr(s) + e- For the reduction half-reaction: Ag+(aq) + e- --> A...

Ria Nawathe 1C

Yes, if the E(cell) for the given reaction as written is negative, it means that the reverse reaction is favorable :)

Ria Nawathe 1C

When an electrode has 2 species in the same phase, in what order do we list the species in the cell diagram? For example, for Fe2+ (aq) —> Fe3+ (aq) + e- is the electrode Pt(s)|Fe2+(aq), Fe3+(aq) or the other way around? What is the rule for this?

Ria Nawathe 1C

I believe the convention is to always place the anode on the left in the cell diagram, and the cathode always goes to the right.

Ria Nawathe 1C

You would just look up the standard potentials for the half reactions Cr/Cr3+ and Ag/Ag+ in the table. After you find these values, change the sign of E naught for one of the half-reactions so that adding the 2 values yields a positive voltage. Hope this helps!

Ria Nawathe 1C

The oxidation number of oxygen in O2 and O3 is 0, and in most other compounds it will have an oxidation number of -2. The exception to this rule is peroxides, in which oxygen will have a -1 oxidation number.

Ria Nawathe 1C

If the entropy of the universe increases, then the process is spontaneous/thermodynamically favorable (has a Gibbs free energy value that is negative). If the entropy of the universe does not increase in the course of the reaction/process, then it is not spontaneous/thermodynamically favorable. It h...

Ria Nawathe 1C

Voltage is the same thing as the maximum potential difference but not the same as maximum work. However, voltage/maximum potential difference are related to maximum work through the equation wmax = -nFE. Hope this helps!

Ria Nawathe 1C

To add on, more massive atoms have larger molar entropy because they have a greater number of fundamental particles (electrons, protons, etc.), so they have a greater number of possible states.

Ria Nawathe 1C

Yes, that's the equation you use! Delta U = 982 J = 492 J + w, so w = 982 J - 492 J = 490 J. Hope this helps!

Ria Nawathe 1C

To find K at 150 degrees celsius, you would first need to find delta S naught and delta H naught of the reaction using the standard values in Appendix 2A. Then, you would plug those into delta G naught = delta H naught - T(delta S naught), with T = 150+273 = 423K. Delta G naught = -RTlnK, so K = e^(...

Ria Nawathe 1C

Yes, all of Sapling 5/6 will be on the midterm, plus the questions from Sapling 3/4 not relating to Hess's Law, bond ethalpies, and delta H of formation.

Ria Nawathe 1C

When these problems talk about fundamental particles, they are referring to the number of protons, neutrons, and electrons in the atom. An atom with a greater molar mass and a larger number of electrons, protons, etc. will have a higher standard molar entropy, as there are more possible positions. H...

Ria Nawathe 1C

The equations with T1 and T2 Prof. Lavelle discussed today are for calculating delta S when the temperature is changing. To calculate delta S for isothermal expansions, we would use delta S = qrev/T rather than delta S = nCpln(T2/T1). Hope this helps!

Ria Nawathe 1C

I believe qrev refers to the q value for the reversible isothermal expansion!

Ria Nawathe 1C

I believe reversible reactions are considered more efficient because less energy as heat is lost to the surroundings due to the process being done slowly, and maximum work is done.

Ria Nawathe 1C

At the end of lecture, Prof. Lavelle mentioned that delta Hrxn plays an important role in entropy changes at lower temperatures, but I'm not sure why. Wouldn't delta Hrxn be more significant to entropy changes at higher temperatures?

Ria Nawathe 1C

You would use 8.314 because that value would give you work with units of Joules.

Ria Nawathe 1C

Qp = delta H because by definition, delta H is the change in energy due to heat at a constant pressure.

Ria Nawathe 1C

I don't have an option for CC either :(

Ria Nawathe 1C

Here is the explanation from the textbook: "The work done by the expanding gas is Pexternal * delta V. At this point, you need to adopt the sign convention. When a system expands, it loses energy as work. so, when delta V is positive (an expansion), w is negative. Therefore, w = -Pexternal * de...

Ria Nawathe 1C

External pressure refers to the pressure acting outside the system (on the outside of the piston), and it is the pressure that the system must push against to do work.

Ria Nawathe 1C

We use external pressure!

Ria Nawathe 1C

Can heat capacities be negative? If so, what do negative heat capacities mean?

Ria Nawathe 1C

I think it's a good idea to know the phases of some of the common elements/molecules, but I don't think Prof Lavelle will expect us to know the standard states of some of the more obscure ones.

Ria Nawathe 1C

Typically, the reaction itself is considered the system, and the solvent, typically water, inside the calorimeter is considered the surroundings.

Ria Nawathe 1C

Hi!

In general, breaking bonds is an endothermic process because it requires an input of energy. For an endothermic reaction, the amount of energy required to break the bonds in the reaction will be larger than the energy released form bond formation.

Ria Nawathe 1C

In this case, the system would be the chemical reaction itself, and the surroundings would be the solution inside the calorimeter. Since qp = m*Cp*(delta T) calculates heat with respect to the water/solution surrounding the reaction, it is qsurr. To find qsys (q of the reaction itself), you would ha...

Ria Nawathe 1C

As the previous answer stated, multiplying the bond enthalpy for the O-H bond by 2 just means that you have 2 O-H bonds that are being broken. It helps to draw out the lewis structures of the molecules in question for the bond enthalpy problems :)

Ria Nawathe 1C

As the previous answers stated, an amphoteric substance can behave as both an acid and base in reactions. Amphiprotic refers to a substance that can donate and accept H+ ions (protons). Amphiprotic substances are amphoteric, but amphoteric substances aren't always amphiprotic because some acids and ...

Ria Nawathe 1C

Like the previous answer stated, the enthalpy of vaporization is always greater than the enthalpy of fusion because a phase change from solid to liquid only weakens the intermolecular attractions enough so that the molecules can slide around each other, but liquid to gas requires the intermolecular ...

Ria Nawathe 1C

Yes! Your analysis is definitely correct. For example, something like liquid nitrogen, which only has LDFs, will take significantly less energy/heat to vaporize compared to water as it has very weak intermolecular attractions among its molecules, so its enthalpy of vaporization will be a smaller val...

Ria Nawathe 1C

It's also because your face isn't coming into contact with 1mol (around 18g) of steam at once, just a small amount of steam reaches your face. And as the previous answer stated, the steam has already come into contact with the lower-temperature air, which has cooled it down.

Ria Nawathe 1C

Just a side note: fusion, vaporization, and sublimation aren't really reactions because there aren't any chemical bonds being broken, they're considered phase changes and are physical changes because it's just the intermolecular attractions that are being overcome.

Ria Nawathe 1C

I see a small upwards pointing arrow in the lower right corner of each reply, but I'm not sure if students can click on it.

Ria Nawathe 1C

Like the previous answer stated I think it's an issue with the textbook problem and not your conceptual understanding, because you're approaching it correctly! My textbook also had 2HD(g) <--> H2(g) + D2(g).

Ria Nawathe 1C

Yes, I believe so! Everything that's posted on the Chem 14B class website is what's offered.

Ria Nawathe 1C

I think you're using the wrong K. I used K=160 for H2 + I2 <---> 2HI @ 500K and I got 2.1 * 10^-5 M. Hope this helps!

Ria Nawathe 1C

It is possible to have large Kc values with the Kc expression being cubic. The cubic nature of the expression depends on what information is given in the problem. However, I don't think we'll get problems like this on any exams just because as you said it does get more complicated to solve since we ...

Ria Nawathe 1C

Hi! For this problem, the first thing you should do is create an ICE table based on the graph. The graph tells you the initial concentrations of A (about 28 M), B (0 M), and C (0 M), and it also shows you the equilibrium concentrations of A (about 18 M), B (5 M), and C (10 M). Based on the differenc...

Ria Nawathe 1C

The correct answer is A, equilibrium constants are useful because they are one number that gives us information about the relative concentrations of reactant and product. K is the ratio of the equilibrium concentrations of products to the reactants raised to their respective stoichiometric coefficie...

Ria Nawathe 1C

You're correct, C is false! At equilibrium, the forward and reverse reactions do still occur, just at the same rate.

Ria Nawathe 1C

When will TAs officially be assigned? My discussion section is tomorrow but I don't see who my TA is on myUCLA.

Ria Nawathe 1C

I also think this might be an error in the solutions. Prof Lavelle mentioned in one of the module videos that one of the situations in which there would be no change is if the pressure was increased by adding an inert gas (which doesn't affect the volume and therefore the concentration), but in this...

Ria Nawathe 1C

Hi! Like the previous response stated, a negative delta H indicates an exothermic reaction, as heat is being released/lost from the reaction system. You would consider the heat that is released to be a product of the reaction. Therefore, if temperature is increased, you are essentially increasing a ...

Ria Nawathe 1C

^^ Yeah we only did conceptual stuff, we don't need to know how to calculate the exact probability. However, it might be helpful to know how many nodes (areas with 0 probability of electron density) each subshell in an atom has. The total number of nodes in a subshell = n-1, # of angular nodes = l, ...

Ria Nawathe 1C

Yes, those are correct! The number of orbitals in a subshell is equal to the number of possible ml values for the specific value of l. For 4p, l = 1 so ml = -1, 0, 1; therefore 4p has 3 orbitals.

Ria Nawathe 1C

A conjugate acid is the resulting species when a base accepts a proton, and a conjugate base is the species that results when an acid gives off a proton. For example, the conjugate acid of the base NH3 is NH4+, as NH3 + H+ --> NH4+. The conjugate base of H2SO4 is HSO4-, since H2SO4 donating a proton...

Ria Nawathe 1C

By relative orientations I think they mean the relative arrangement of the hybrid orbitals. For sp3, each of the 4 hybrid orbitals points to one corner of a tetrahedron. For sp2, each hybrid orbital points to a corner of an equilateral triangle. Hope this helps!

Ria Nawathe 1C

I agree with the previous answer, the heme complex + the hystidine protein make up myoglobin.

Ria Nawathe 1C

Yeah I think you have to meet with your TA to go over your midterm!

Ria Nawathe 1C

The textbook has a set of rules in Focus 9C.1 (Toolbox 9C.1) for writing the formulas and naming the compounds! For writing the formula, the textbook says to write the chemical symbol of the central metal atom first and then write the symbols of the ligands in alphabetical order. For different ligan...

Ria Nawathe 1C

You would place a lone pair on the equatorial vs axial plane based on which position would limit the number of 90 degree interactions with the other atoms in the molecule, since these interactions are unfavorable.

Ria Nawathe 1C

Just a heads up, the textbook uses the -ido convention but the Sapling HW uses o!

Ria Nawathe 1C

Hi! I had trouble with this one too, but the book has a helpful list of rules in Focus 9C.1 for naming coordination compounds/complexes. First, everything enclosed by the brackets is part of the complex ion. The complex ion has an overall charge of +1, as Cl is -1 and the whole coordination compound...

Ria Nawathe 1C

So I have no clue, so I tried to research what it might be, and I'm getting conflicting answers, one of which says that it's simply perpendicular to each other (normal to each other), and the other of which says they are standardized to one another. The first definition makes the most sense to me, ...

Ria Nawathe 1C

What does it mean for hybrid orbitals to be normalized vs unnormalized?

Ria Nawathe 1C

I don’t think you can assume that any molecule with a central atom that has lone pairs is polar. For example, XeF2 is non polar, and it has 3 lone pairs on the central atom. The lone pairs are arranged at 120 degrees to each other and the F-Xe-F bond angle is 180 degrees, so there is a 0 net dipole ...

Ria Nawathe 1C

I don't think there will be discussions on Thursday or Friday, but I'm not sure if students in those discussions will have to make it up.

Ria Nawathe 1C

Just to add on, an unpaired electron on a radical like NO2 is also considered to be a region of electron density.

Ria Nawathe 1C

Dr. Lavelle mentioned that the energy of the hybrid orbitals is the average of the energies of the atomic orbitals used.

Ria Nawathe 1C

FionaHunter21 wrote:Are there dipole moments between all atoms with a difference in electronegativity or just the polar bonds?

I believe all bonded atoms with a difference in electronegativity exhibit a dipole, but the magnitude/strength of the dipole is given by mu = |q|*d (charge * distance between the atoms).

Ria Nawathe 1C

I agree with the previous answer, Dr. Lavelle mentioned in one of his lectures that the unpaired electron in radicals like NO2 is considered a region of electron density.

Ria Nawathe 1C

The effective nuclear charge refers to the protons/nucleus of an atom, not the electrons. In terms of periodic trends, the effective nuclear charge increases as you go across a period because the atomic number increases and electrons are being added to the same energy level. It decreases down a grou...

Ria Nawathe 1C

Cations, molecules with electron deficient central atoms (like BF3), and molecules with polar double bonds (like CO2) tend to be lewis acids since they are more likely to accept an electron pair. Anions and molecules with lone pairs (like H2O) tend to be lewis bases since they have electrons to dona...

Ria Nawathe 1C

A dipole moment is a partial positive charge and partial negative charge within a molecule, though the molecule is as a whole electrically neutral. For example, the hydrogens in H2O have a partial positive charge and the oxygen has a partial negative charge due to the electronegativity difference be...

Ria Nawathe 1C

To add on, in terms of periodic trends polarizability tends to increase as you go down a group because the atomic radius and the number of electrons increase.

Ria Nawathe 1C

For anions, polarizability increases as you go down a group since the atomic radius and number of electrons increases and the effective nuclear charge decreases. In general, the larger the radius and the more electrons an atom has, the more polarizable. For polarizing power, the smaller the atomic r...

Ria Nawathe 1C

London dispersion forces are induced dipole-induced-dipole interactions. Basically, when non-polar molecules collide their electron densities are shifted, forming a temporary delta negative and delta positive side. This induces the neighboring molecule to form a temporary dipole as well and produces...

Ria Nawathe 1C

I believe Professor Lavelle mentioned it's the energy per mole of the given interaction. For example, if there was 1 mole of Na+ and Cl- interactions, the amount of energy released due to the ion-ion interactions would be 250 kJ.

Ria Nawathe 1C

An example of a molecule with a permanent dipole is H2O; the difference in electronegativity between the O and Hs causes the electrons to be pulled towards O, creating a permanent delta negative and delta positive side. Temporary dipoles occur when molecules that are otherwise non-polar, like N2, co...

Ria Nawathe 1C

I think the electrons on the anion are referred to as polarizable and the cation as polarizing and not the other way around because the protons in the nucleus of the cation don't "move" due to the attractive force they experience from the electrons of the anion. It is the electrons that ar...

Ria Nawathe 1C

In terms of thermodynamics, if you think about the molecule/compound as a system, anything that increases the energy of the system, for example putting in energy to break bonds, makes delta E positive. When energy is released, delta E is negative. The intermolecular/interionic interactions have nega...

Ria Nawathe 1C

For non-resonance structures, electrons are localized.

Ria Nawathe 1C

I believe angular is 2 atoms bonded to the central atom with 1 lone pair, and bent is 2 atoms bonded to the central atom with 2 lone pairs.

Ria Nawathe 1C

Will the final also be multiple choice?

Ria Nawathe 1C

Here is the textbook's explanation for why it increases stability (it's in Focus 2B): "As well as delocalizing electrons over the atoms, resonance also lowers the energy below that of any single contributing structure and helps to stabilize the molecule. This lowering of energy occurs for quant...

Ria Nawathe 1C

The angular momentum quantum number l, which tells us about the subshells in each shell, is determined by the principle quantum number (n=3 in this case). For n=3, l can be 0...n-1 which is 0, 1, and 2, corresponding to s, p, and d.

Ria Nawathe 1C

Delocalized electrons are electrons that do not "belong" to a particular atom or are not associated with a specific bond in the molecule. Here is the textbook's explanation for why it increases stability (it's in Focus 2B): "As well as delocalizing electrons over the atoms, resonance ...

Ria Nawathe 1C

Within a particular energy level (n), all electrons with the same angular momentum quantum number (l) are in the same sub-shell, and all electrons with the same n, l, and ml (magnetic quantum number) are in the same orbital.

Hope this helps!

Ria Nawathe 1C

What is the difference between radial and angular nodes, and how many do each of the orbitals (s, p, d, f) have?

Ria Nawathe 1C

He drew 3 up-spin electrons first because according to Hund's Rule, electrons occupy empty orbitals in the same sub-shell before pairing up. As for why the up-spin electrons are drawn first, I think it just has to do with convention.

Ria Nawathe 1C

If you think about EM radiation as a wave of oscillating electric and magnetic fields that are perpendicular to each other, the frequency is related to how the electric field changes, or oscillates, with time. The greater the frequency, the faster the field changes/oscillates. The smaller the freque...

Ria Nawathe 1C

Within a particular energy level (n), all electrons with the same angular momentum quantum number (l) are in the same sub-shell, and all electrons with the same n, l, and ml (magnetic quantum number) are in the same orbital. Here's an explanation with a helpful graphic: https://chemistry.stackexchan...

Ria Nawathe 1C

That's what I heard too, some of the TAs said it was all multiple choice.

Ria Nawathe 1C

Delta v should be 10 m/s. I believe this is an error in the book's solution. Here's a full list of errors in the solution manual:

https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf

Ria Nawathe 1C

As the previous answer stated, the equation is E = hv. The photoelectric effect demonstrated that light exhibits properties of particles and also demonstrated the relationship between energy and frequency of photons.

Ria Nawathe 1C

^^ Thanks!

Ria Nawathe 1C

The circular standing wave model is just a math model that was developed to reconcile electrons having wave-like properties and having quantized energy states. In this model, each stable circular standing wave (with whole number wavelengths) around the nucleus represents an energy level (n=1, 2, 3.....

Ria Nawathe 1C

Does anyone know if we'll also be given the wavelengths corresponding to the different regions of the EM spectrum, or will we need to just know it?

Ria Nawathe 1C

They're not mandatory but they are helpful, and the videos are pretty short so it won't take too much time to check them out!

Ria Nawathe 1C

How do you identify which energy level is n1 and which is n2 for frequency = R(1/(n1)^2 - 1/(n2)^2)? Is n1 always the initial energy level and n2 always the final energy level?

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