CHEMISTRY COMMUNITY

Jonathan Tangonan 1E

I believe they are the same thing. They are both electrochemical cells.

Jonathan Tangonan 1E

How do we determine what is an oxidizing agent and how do we determine what is a reducing agent?

Jonathan Tangonan 1E

Which has greater residual entropy at 0K: NO or BF3

Why does NO have a greater residual entropy at 0K?

Jonathan Tangonan 1E

k and k' in this case are referring to the rate constants at different temperatures. The k and k' which you are referring to have to deal with forward and reverse reactions. k and k' in this case can also be referred to as k1 and k2 in this instance.

Jonathan Tangonan 1E

The delta U is +158J because in step 2 it returns to its original internal energy. So, the work the system had done which was -158J returns back to its original state by saying delta U of the second step is +158J.

Jonathan Tangonan 1E

This equation is derived from the same reaction at different temperatures where the rate constants are different at separate temperatures so: ln(k 1 )= lnA - Ea/RT 1 ln(k 2 )= lnA - Ea/RT 2 Then we subtract the the equations: lnk 2 -lnk 1 = Ea/R(1/T 1 -1/T 2 ) ln(k 2 /k 1 )= (Ea/R)(1/T 1 -1/T 2 )

Jonathan Tangonan 1E

Well first off you have to balance the equation: AB 4 +2C 2 ->AC2+2B 2 C Then you can set up the rate where: The consumption of AB 4 = deltaAB 4 /deltat= .400M/s Then we can infer that the production of B 2 C= (½) (delta[B 2 4])/delta t which equals .400M/s We plug in the 30M-.15M for delta B 2 4 mu...

Jonathan Tangonan 1E

If you look at the equation delta G= deltaH-TdeltaS, we can determine why we want entropy to be positive for a spontaneous reaction. For a reaction to be spontaneous we want delta G to be a negative value because a negative value indicates spontaneity. There's a table in the textbook Table 9.5 on pa...

Jonathan Tangonan 1E

We can't add the standard potentials together because E is not a state function. But, we can make use of delta G as being a state function through the addition of the delta G values of the half reactions and then solving for the desired half reactions standard potential through the equation delta G=...

Jonathan Tangonan 1E

A double vertical line, || , represents a salt bridge. A single vertical line, | , would represent a phase boundary. Whenever a comma is used that is used to separate substances that are in the same phase. And the phases are included within parentheses And like you said, the reduction goes on the ri...

Jonathan Tangonan 1E

From the reaction profiles that I have seen in the textbook, potential energy was used for the y-axis while the progress of the reaction was used for the x-axis.

Jonathan Tangonan 1E

You would use J/mol. The J*mol^-1 cancel out with the units of the R constant which has JK^-1mol^-1

Jonathan Tangonan 1E

You can calculate this using another form of the Arrhenius equation where:

ln(k₂/k₁) = (E_a/R)(1/T₁-1/T₂)

Where k₁= 1.5 * 10¹⁰ L*mol^-1s^-1, T₁= 298K, T₂=310K, and E_a= 38kJ/mol and then solve for k₂ from there. You get the .59 from the calculations on the right hand side.

Jonathan Tangonan 1E

Enthalpy and entropy are both state functions.

Jonathan Tangonan 1E

It is +158J because it returns back to the original internal energy. Delta U in step 1 is -158J so in order to return to its original internal energy in step 2 it must be +158J.

Jonathan Tangonan 1E

If you were to use the integrated rate law for the constant you could determine it by substituting for the concentration at time t with (.719)(initial concentration) and then isolate k from there. ln[A]= -kt + ln[A initial ] ln(.719[A initial ]) - ln[A] = -kt ln[.719[A initial ]/[A]] = -kt -ln[.719]...

Jonathan Tangonan 1E

https://lavelle.chem.ucla.edu/forum/viewtopic.php?t=3241

This post back from 2013 does a good job explaining what you are asking.

Jonathan Tangonan 1E

The derivation for this equation is on page 642 in the textbook but in summary you use the Arrhenius equation at one temperature and at another and subtract the first equation from the second equation getting rid of the lnA. At T1 : lnk1=lnA-Ea/RT1 At T2: lnk2= lnA-Ea/RT2 The subtraction after furth...

Jonathan Tangonan 1E

The .59 comes from the Ea/R(1/T1-1/T2) calculation where Ea=38kJ/mol T1=298K and T2=310K while on the other side of the equation ln(k2/k1) where k1=1.5x10^10.

Jonathan Tangonan 1E

Yes you should. It's on the outline for chapter 15 where we have to "Explain how the collision model and activated complex model account for the temperature dependence of reactions".

Jonathan Tangonan 1E

If we were to look at a reaction that reaches equilibrium like A+B<=>C+D then the equilibrium constant would be the concentration of the products over the reactants where: K= [C][D]/[A][B] If we were to write the forward and reverse rate laws for this reaction, they would be: A+B->C+D where the Rate...

Jonathan Tangonan 1E

https://www.chemguide.co.uk/physical/ba ... rofile.gif

Here's an image that shows the location of the intermediates on a reaction profile.

Jonathan Tangonan 1E

The rate constant and the equilibrium constant of endothermic reactions are increasing with the raising temperature (as temperature goes up k and K). This is because when temperature increases the equilibrium shifts towards the right.

Jonathan Tangonan 1E

I'm pretty sure it is a typo in the book. It should be Fe^3+ + e -> Fe^2+ E=.77

Jonathan Tangonan 1E

I believe this is due to the number of different arrangements either molecule can have. If you were to draw out the cis-MX2Y4 molecule there would be about 12 different orientations that it could possibly have whereas with a trans-MX2Y4 molecule there are 3 different orientations.

Jonathan Tangonan 1E

You would use the equation delta S= nRln(P1/P2) which turns out to be delta S = (1.5mol)(8.3145 J.K^-1.mol^-1)(ln 15atm/.5atm) which equals 42.4 J.K^-1

Jonathan Tangonan 1E

Yes it does indicated that methanol is in a liquid phase, but bond enthalpies refer to gaseous substances. So any substance in a chemical reaction that isn't in a gaseous phase has to be converted from their previous phase into the gaseous phase. So what we are actually looking at is: C(g) + 2H2 + ½...

Jonathan Tangonan 1E

Overall, q and w added together make up the internal energy that happens within a system. Q is the energy transferred as heat where a positive q would indicate an absorption of heat while a negative q would indicate a release of heat from the system. A positive w would indicate that work has been do...

Jonathan Tangonan 1E

I think what's also important to note is that a change that is done reversibly does more work than a change carried out irreversibly if you are asked to identify in that instance which is reversible and which is irreversible. This is provided that the initial and final states are the same.

Jonathan Tangonan 1E

Yes this is true, the change in U= q + w. The change in internal energy is equal to the heat in the case that there are no other processes taking place except that of transferring energy as heat.

Jonathan Tangonan 1E

No what the reaction quotient (Q) is supposed to tell you is where the how the reaction will shift to reach equilibrium. If Q>K then equilibrium will shift towards the left. So when you put it into the ice chart you will put +x on the change for the reactants and will put -x for the change in the pr...

Jonathan Tangonan 1E

cis- and trans- are used to indicate the placements of ligands when there are multiple of them. When cis- is used the similar substances will be arranged on the same side while with trans- this indicates that they will be arranged on opposite sides.

Jonathan Tangonan 1E

03 has a trigonal planar arrangement but has a lone pair so the actual shape is bent. Because of the lone pair occupying one of regions of high electron density, the dipole moment is not cancelled out and is therefore polar.

Jonathan Tangonan 1E

I think what is important to clarify in this question is that there are two several definitions in the textbook which define both acids and bases. Lewis acids and bases involve the giving and accepting of electron pairs while Bronsted Acids and Bases involve the donation and accepting of a proton (H...

Jonathan Tangonan 1E

I believe when the molecular formula isn't fully condensed as it is not in the conjugate acid of CH3NH2. The way that it is written would be the best representation of the actual formal structure. In this case, CH3NH3+ is most likely the best representation of this structure where Carbon has the 3 H...

Jonathan Tangonan 1E

A homonuclear diatomic molecule is a 2 atom molecule made up of the same element. A heteronuclear diatomic molecule is a molecule composed of two atoms of two different elements.

Example of a heteronuclear diatomic molecule: HF
Example of a homonuclear diatomic molecule: O2

Jonathan Tangonan 1E

There are multiple electron geometries that can produce a linear shape depending on the number of lone pairs a centra atom has. For instance, in the case of a molecule whose central atom has 5 regions of high electron density and and 3 lone pairs it will produce a linear shape. This is the same for ...

Jonathan Tangonan 1E

I believe at one point in Dr. Lavelle's lectures he said that anything smaller than 10^-15m would be out of the range of an object being measurable.

Jonathan Tangonan 1E

It changes to ferrate because the "ate" indicates that the overall charge of the complex is negative. Fe originally comes from the latin ferrum and that's where the ate is added onto.

Jonathan Tangonan 1E

Yes so each bond line that you see would count as one electron and each lone pair would have two electrons and you would just subtract that from the usual amount of valence electrons that surround the atom.

Jonathan Tangonan 1E

Yes the difference is going to be the difference between the arrangement of the electrons and the regions of the high electron density and the number of lone pairs around the central atom. With a molecule with an arrangement of AX3E2 where there are five regions of high electron density and 2 lone p...

Jonathan Tangonan 1E

For "a" on 3.11 if you look at the ground electron state configuration of Co it is [Ar] 3d7 4s2. As we are looking for a 3+ ion, when 3 electrons are taken from a Co atom they are first removed from the s subshell and then they are lost through the d subshell resulting in the [Ar] 3d6 conf...

Jonathan Tangonan 1E

Yes lone pairs are considered as one region of high electron density when counting the regions around the central atom.

Jonathan Tangonan 1E

As you go down a group of atoms their radii increases. Thus, there are more levels/ shells shielding the outermost electrons from the pull of the nucleus making it easier for valence electrons to be removed. As you go across a period more electrons are filling the valence shell as that happens the r...

Jonathan Tangonan 1E

A) When you're are given the n and l to find the total number of possible electrons you determine the possible ml values and and multiply the number of possible ml values by 2. ml is l, l-1,...-l. n=2; l=1 so the subshell is p. This means that ml can be -1,0,1 and 2 electrons can fill each of those ...

Jonathan Tangonan 1E

Just to clarify I believe that the l quantum number is going to be positive. The ml or the magnetic quantum number is when you will start getting negatives where for example where l=1, ml can be -1, 0, or 1.

Jonathan Tangonan 1E

An atomic orbital is basically the most probable location an electron is going to be located which is based off a wave function. The orbitals are mainly classified by three different quantum numbers the n, l, and ml. The n determines the size and energy of the orbitals this is the defining factor as...

Jonathan Tangonan 1E

To add on, I think something important to note is the line " No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50*10^16 Hz." This information gives us what we would call the threshold frequency or the minimum frequency required to eject e...

Jonathan Tangonan 1E

(a) The chemical equation before being balanced would be Al + Cl2 -> AlCl3. After it is balanced it would look like 2Al + 3Cl2 -> 2AlCl3 (b) For this one you have to determine the limiting reactant so you need to convert the masses of the reactants to moles and then compare the ratios of the stoichi...

Jonathan Tangonan 1E

Both ways are correct they only teach it that way so the concept is easier to grasp. Personally, I convert from grams to moles and just divide the smallest from there and you end up with the same thing anyways so its only a matter of preference.

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