CHEMISTRY COMMUNITY

Brigitte Phung 1F

I believe Lavelle said that we will only work with the Pre-Equilibrium Approach as it is less complex than the Steady-State Approach but still gives us the same result!

Brigitte Phung 1F

The oxidation state is zero for elements in their free (uncombined) state (ex: Zn(s), Al(s)), including those found in nature as diatomic elements (ex: O2, N2, H2).

Brigitte Phung 1F

For 15.61, the ln(k'/k) simply comes from the equation, ln(k'/k) = Ea/R * (1/T1 - 1/T2). This equation is no longer on Lavelle's online equations sheet, however it was on one of his older equation sheets! Here is the derivation: lnk = -Ea/R (1/T1) + lnA lnk' = -Ea/R (1/T2) + lnA lnk' - lnk = [-Ea/R ...

Brigitte Phung 1F

I also believe the problems should always tell you the order of the reaction as 23, 25, 27, and many others also do. But in the case they don't, you can always check the units of k! For your reference: - zero-order: k is in units of mol * L^-1 * s^-1 - first-order: k is in units of s^-1 - second-ord...

Brigitte Phung 1F

Yep! You can find the link titled "Important Final Exam Information" under Exam Information on his website!

Brigitte Phung 1F

In lecture last Friday, Lavelle also said that it would cover all of Kinetics up to (not including) his Reaction Mechanisms slide!

Brigitte Phung 1F

In my discussion, my TA said that we'll only be working with zeroth, first, and second order reactants, however the overall reactions themselves may end up being a higher order (after all the individual orders are added up!). Hope this helps!

Brigitte Phung 1F

I agree. It took me a while to get the correct answer because I kept incorrectly converting from mmol to mol. I'd recommend converting all of the mmols to mols in the beginning and also paying attention to the order with respect to each reactant!

Brigitte Phung 1F

So for part b) of this problem, we need to reuse the first-order integrated rate law again to solve for t. We need to convert from mols of B to mols of A and subtract this value from the initial A concentration to take into account how much A was used up to produce the 0.030 M concentration of B. Th...

Brigitte Phung 1F

I believe you're referring to the single lines in between the I- ions and I2(s) - if so, cell diagram notation involves separating elements/compounds that are in different states (solid/liquid/gas/aq) on both sides!

Brigitte Phung 1F

The double lines represent the salt bridge in galvanic cells. The elements to the left of the double lines represent the anode/oxidation reaction, while the elements to the right of the double lines represent the cathode/reduction reaction.

Brigitte Phung 1F

Here are a few things that are helpful to remember about oxidation numbers
- Oxygen has a oxidation number of -2 in compounds
- Elements in their natural/most stable/free state have oxidation numbers of 0
- Monoatomic ions have oxidation numbers equal to their charge

Brigitte Phung 1F

Delta G is not pathway dependent, so it is only necessary to examine the initial and final states! On the other hand, q, w, and E are not state functions so their pathways are significant in calculations.

Brigitte Phung 1F

I believe that for all tests, we'll be given all the information the homework problems ask us to look for in the appendixes. Hope this helps!

Brigitte Phung 1F

I agree! Lavelle just posted an announcement confirming this :)

Brigitte Phung 1F

I believe that you should work with the balanced equations! For both 9.57 and 9.59, the equations they gave are already balanced so you can go ahead and work with those. For these problems just remember that the ΔHf and ΔGf of O2 are 0. Hope this helps!

Brigitte Phung 1F

For monatomic ideal gasses, at constant volume Cv = (3/2)R and at constant pressure Cp = (5/2)R.

Brigitte Phung 1F

For this problem, you will mainly be using the formulas, ΔS(surr) = - ΔH(system)/T and ΔS(system) = ΔH(system)/T. Note that the negative sign in the ΔS(surr) equation assumes that for exothermic reactions (ΔH of system is negative), the entropy of the surroundings is increased as it accepts heat rel...

Brigitte Phung 1F

When it comes to these problems, it helps to look at the direction of heat transfer from each of the objects/systems involved! When heat is being released by an object, q is negative with respect to that object, whereas when heat is being absorbed by another object, q is positive.

Brigitte Phung 1F

I believe this is referring to the relationship between P and I in the ideal gas law: PV = nRT. If the equations are rearranged as P = nRT/V or V = nRT/P, and n and T are kept constant, you can see that P and V share an inverse relationship. When either P or V goes up, the other one goes up to maint...

Brigitte Phung 1F

The units end up being J/K as kB = 1.381 x 10^-23 J.K^-1 and also ΔS = nRln(V2/V1) cancels out into J.K^-1.

Brigitte Phung 1F

I think it would also help to look at it in terms of the objects themselves. q is positive when heat has been absorbed by the system, whereas q is negative when heat has flowed from the system. Hope this helps!

Brigitte Phung 1F

I agree, while the questions may differ on what they ask for using the final reaction, the problems should give the additional reactions and enthalpies we'll need to complete the final reaction.

Brigitte Phung 1F

This can also be proven with equations. Because ΔH = ΔU + PΔV and ΔU = q + w, you can plug ΔU into the ΔH equation as follows: ΔH = q + w + PΔV. Because w = -PΔV, the last two terms cancel each other out: ΔH = q - PΔV + PΔV, leaving us with ΔH = q at constant pressure. Hope this helps!

Brigitte Phung 1F

I think that because w = -P∆V for expansion problems, the 22 kJ in expansion work becomes equal to w = -P∆V = -(22 kJ). Therefore, when plugging it back into the equation, ∆U = ∆H - P∆V = -15 kJ - (-22 kJ) = - 15 kJ + 22 kJ = +7 kJ. Hope this helps!

Brigitte Phung 1F

In addition, because the amount of heat absorbed or released at a constant pressure is defined as the enthalpy, we can write that q = ∆H.

Brigitte Phung 1F

In all systems, the matter/substance involved in the system contributes to its energy. Because matter can be exchanged with the surroundings of an open system, the addition or removal of substance from the system can change its energy.

Brigitte Phung 1F

Nope! Lavelle will provide these values on the tests. I believe he just includes the table under the reading because we'll need to reference it for some homework problems and it allows us to compare the heat capacities for different elements/substances. Hope this helps!

Brigitte Phung 1F

I agree. I'm pretty sure we'll be given the values, and all we'll need to do is know how to use them or manipulate them if we need to (i.e. divide reaction enthalpy by the number of moles to find enthalpy for formation of one mole of a substance). Hope this helps!

Brigitte Phung 1F

Heat is path dependent and therefore cannot be determined simply by its current state. When calculating the values of both heat and work, it matters how the end condition is achieved. Therefore, you cannot simply subtract the final condition from the initial condition as you would for a state functi...

Brigitte Phung 1F

I agree with the others. I prefer to think of hybridization as the number of attachments to the central atom, with bonds and lone pairs each acting as one attachment. Boron has three Fluorides attached to it, so it is sp2 hybridized. Hope this helps!

Brigitte Phung 1F

PO43- and PCl^5 do not have dipole moments because they have a symmetrical molecular shape (PO43- is a tetrahedral and PCl^5 is trigonal bipyramidal), which cancels out the dipole moments. Hope this helps.

Brigitte Phung 1F

The shape of (CH3)2Be is only tetrahedral about the carbon atoms as they both have 4 regions of electron density (3 hydrogens about them and are both bonded to Be). However, Be only has two regions of electron density, so the C-Be-C bond is linear with a bond angle of 180 degrees. Hope this helps!

Brigitte Phung 1F

This means that at that time, the ratio of the concentrations of product to reactants is larger than the ratio at equilibrium. This means that at that time, the concentration of the product is higher than it should be at equilibrium, causing the reaction to shift to the left/reactants to reach equil...

Brigitte Phung 1F

In discussion, my TA stated that for the purposes of the test, two of the more important Latin root elements to remember are Copper and Iron. In a coordination compound with a negative charge, Copper is written as Cuprate and Iron is written as Ferrate.

Brigitte Phung 1F

We square our K value when we double the chemical equation because of how doubling the coefficients also doubles the exponents in the formula K = [product]^(product coefficient) / [reactant]^(reactant coefficient). For example, if you had 2A -> B, K = [B]^(1) / [A]^(2). When you double the equation,...

Brigitte Phung 1F

This statement is incorrect in its use of the word always. Higher frequency light only emits electrons with higher kinetic energies if the energy has already surpassed that of the work function; if the light's energy has not yet reached that of the work function, no electrons would be emitted.

Brigitte Phung 1F

Yes, lone pairs count as regions of electron density! An unpaired electron also counts as a single region of electron density.

Brigitte Phung 1F

The DeBroglie Equation is used to find the wavelength of any particle with momentum! Hope this helps.

Brigitte Phung 1F

I agree with Karan! If the molar mass of the compound is given (which I think it was!), you can just multiply it by the percentages to find out how many grams each element would contribute to the total mass of one mole of the compound. You can then directly divide the mass of each element by their r...

Brigitte Phung 1F

I hope these examples of Lewis Acids and Bases also help!
Examples of Lewis Bases: NH3, F-, and OH-
Examples of Lewis Acids: BF3 and H+.

Brigitte Phung 1F

I agree with Ayona. Also,in the powerpoint, it was written that "we apply Pauli Exclusion Principle and Hund's Rule to determine e- configuration," which is what is accomplished through the Aufbau Principle. Hope this helps!

Brigitte Phung 1F

In Nitrogen, all three orbitals in the p subshell are half-filled with one electron each which makes it more stable than Carbon, which has two half-filled and one empty p-orbital. Thus, Carbon has a higher electron affinity than Nitrogen. Hope this helps!

Brigitte Phung 1F

I agree with Kaushal. Also, on his powerpoint, Lavelle wrote that we only need to know the s-block, the p-block, and just the first d-block row, so I don't think we will have to worry much about the other elements!

Brigitte Phung 1F

Yes, the principle quantum number n determines the energy and size, and is also called a shell. The angular momentum quantum number l corresponds to the s/p/d/f-orbitals and describes the shape.

Brigitte Phung 1F

For the most part, I feel that Lavelle will only test us on the topics he covers in lecture. In addition to reviewing the assigned homework and lecture notes, I also like looking at the lecture outlines to see exactly which topics we are expected to understand.

Brigitte Phung 1F

Yes they are! They are just names that are used interchangeably and are both denoted by the symbol Φ.

Brigitte Phung 1F

I also remember Lavelle saying he will give us the conversion factors when the values are in units we do not use as often! But to be safe, it always helps to remember the conversion factors. Here they are for eV and keV: 1 eV = 1.602 x 10^-19 J and 1 keV = 1.602 x 10^-16 J.

Brigitte Phung 1F

After you determine the limiting reactant, which in this case is C6H9Cl3, you would multiply the moles of C6H9Cl3 by its molar ratio with AgCl, which is 1 mol C6H9Cl3 : 3 mol AgCl (from the balanced equation). This gives you the moles of AgCl produced, which you can then multiply with the molar mass...

Brigitte Phung 1F

The problem: Aluminum metal reacts with chlorine gas to produce aluminum chloride. In one preparation, 255g of aluminum is placed in a container holding 535 g of chlorine gas. After reaction ceases, it is found that 300g of aluminum chloride has been produced. (a) Write the balanced equation for the...

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