CHEMISTRY COMMUNITY

Mindy Kim 4C

Adding on to this, this is why we can deduce the rate law directly from the coefficients of the reactants in an elementary reaction. We assume there are no intermediates in elementary reactions and so can derive the rate law just by looking at the reaction.

Mindy Kim 4C

Morris also posted some questions concerning kinetics! You should be able to find it by searching Morris and kinetics.

Mindy Kim 4C

Normally the problem will either tell you which step is the slow, rate determining step, or it will give you the reaction mechanism and the overall rate law for the reaction and you would have to determine from this information which step is the slow step.

Mindy Kim 4C

This is due to activation energy. A reaction proceeds at a faster rate if it has a lower activation energy and vice versa.

Mindy Kim 4C

What is the difference between the differential rate law and the integrated rate law?

Mindy Kim 4C

In the example Dr. Lavelle used in lecture, a reaction can be favorable but be very slow. In this case, kinetics rather than thermodynamics is controlling the reaction.

Mindy Kim 4C

If you look at the reaction: Cu2+ (aq) + Zn(s) --> Zn2+(aq) + Cu(s), you have two aqueous ions and two solid metals. Remember that when setting up the reaction quotient and equilibrium constant, solids and liquids are not included. Therefore, the reaction quotient would simply be the concentrations ...

Mindy Kim 4C

The sentence in the blue box above that section explains it pretty well: "From Ecell=Ecell(standard)-(RT/nF)lnQ rearranged into lnQ= -Ecell/(RTnF)" In this example, Ecell standard is equal to zero because the electrodes are identical in their standard state so there is no voltage in the ce...

Mindy Kim 4C

n represents how many electrons are being transferred in the reaction. The value should be the same whether you are looking at the oxidation or reduction half reactions as the number of electrons should be the same if the half reactions are balanced.

Mindy Kim 4C

The Nernst equation is mainly used for reaction mixtures not at equilibrium. If you are trying to find K using standard potential, the method you described works.

Mindy Kim 4C

Generally, the change in enthalpy or entropy do not change with temperature. The standard entropies and enthalpies may be different for the compounds, but the change in enthalpy or entropy for the reaction does not vary significantly with change in temperature.

Mindy Kim 4C

Yes, galvanic cells include both the cathode and anode.

Mindy Kim 4C

Since the question asks for residual molar entropy, they are asking for the residual entropy for one mole of SOF2. There are 6.022x10^23 molecules (Avogadro's number) in one mole.

Mindy Kim 4C

Yes, you are correct. The bonds broken require energy so they are positive while when bonds are formed energy is released so deltaH is negative.

Mindy Kim 4C

I believe deltaS of the surroundings is zero because even if there is a small change in entropy in the surroundings, your surroundings are so large that such a change would essentially be insignificant and there would be almost no change.

Mindy Kim 4C

To calculate the entropy change for a change in temperature, you want to use the formula deltaS=(moles)(Cv, aka heat capacity at a constant volume)(ln(T2/T1)). You use the heat capacity for a constant volume because you only want the entropy change for a change in temperature so you keep volume cons...

Mindy Kim 4C

In lecture on Monday, Dr. Lavelle used the example of BF2Cl when talking about degeneracy and said the degeneracy of the position of Cl would be 3^10^6. How do you calculate or come to this degeneracy?

Mindy Kim 4C

Yes, the two equations you stated are both correct.
Dr. Lavelle mentioned the third equation briefly, but basically, the internal energy of an ideal gas is dependent on temperature. The 3/2 in the equation comes from the translational motion along the x,y, and z axes.

Mindy Kim 4C

In a vacuum, or free expansion, there is no resistance against the piston. For example, in a non-vacuum system, the external pressure is pushing against the piston. However, in a vacuum, because there is no resistance or external pressure, there is no work being done.

Mindy Kim 4C

To my understanding, all systems fit into one of these three categories. Even if there is some system that does not, I do not think we will be expected to understand how it fits into multiple categories.

Mindy Kim 4C

Generally, O2, F2, Cl2, N2, and all the noble gases are gases in their standard states. All other elements, except Hg and Br (which are liquids) are solids in their standard states.

Mindy Kim 4C

As Q approaches K, the rate of the forward or reverse reaction might change so that the rates of the forward and reverse reactions are equal.

Mindy Kim 4C

I know that Dr. Lavelle said we can approximate the value of x by assuming concentrations change insignificantly even when x is added. However, we mainly used this method of approximation for weak acids and bases. On the test, should we also use this approximation if Kc<10^-3 or should we mainly use...

Mindy Kim 4C

For most polyprotic acids you do not need to calculate the second dissociation because the Ka2 is typically so small that any H3O+ ions generated from the acid in solution will contribute insignificantly towards the pH. One exception to this is H2SO4, as the acid completely dissociates in the first ...

Mindy Kim 4C

Polyprotic acids can undergo more than one dissociation, or donate more than one proton. Therefore, polyprotic acids will typically have more than one proton (ie. H2CO3, H2SO4, H2S, etc.). Based on this, I believe the question may not specify whether or not the acid is polyprotic.

Mindy Kim 4C

For these polyprotic acids, you can solve them as you would a monoprotic acid because the Ka2 value of these acids is considered so small to the extent that the second reaction contributes insignificantly to the pH. However, for H2SO4, you would take into account both the first and second reactions ...

Mindy Kim 4C

You should write the phases of the molecules just to be safe, and writing the phases will remind you of which molecules should not be included in the equilibrium constant.

Mindy Kim 4C

When the Ka value of the acid is less than 10^-3, we can approximate the value of x by assuming the value of x being subtracted from 0.10 to be insignificant. Even if we subtract the value of x from 0.10, the value will still essentially be 0.10 because of how small x will be.

Mindy Kim 4C

I believe Dr. Lavelle said in lecture that however many sigfigs are in the molarity of H+ or OH- is how many decimal places there should be in your pH or pOH value.

Mindy Kim 4C

For part a in problem 11.7, are the shown flasks chronological? I understand how the answer is the third flask if they were, but I'm unsure how you could get to this answer without knowing the flasks are progressing.

Mindy Kim 4C

Question 1:
If the question gives the amount of gases as partial pressures you can convert to concentration. You can use the ideal gas law (PV=nRT) to convert partial pressures of gases to concentrations if you have both phases in the reaction. Remember that P=(n/V)RT, or concentration=P/RT.

Mindy Kim 4C

Also keep in mind that when you are calculating Q, the reaction composition/mixture given in the problem is not at equilibrium. Therefore, Q will be greater or smaller than K depending on how much more reactants/products there are in the given reaction mixture.

Mindy Kim 4C

Yes, drawing out the Lewis structure will help in determining this. If you look at the Lewis structure for CO3 2-, all oxygens have at least two lone pairs (two oxygen atoms have single bonds with the carbon atom and have 3 lone pairs, one oxygen atom has a double bond with the carbon atom and has 2...

Mindy Kim 4C

Chelating complexes are those that contain one or more ligands that form a ring of atoms that includes the central metal atom. In the case of #35, you can tell that only b will form a chelating complex based on where the NH2 compounds are bonded to the central compound. In molecules a and c, the NH2...

Mindy Kim 4C

We might have to calculate the KA value of a species based on concentrations. For example, if the formula was HA --> H+ + A-, you would multiply the concentrations of [H+] and [A-] and divide this by [HA]. You might also be given the KA values of a compounds and be asked to interpret the meaning of ...

Mindy Kim 4C

If you draw the Lewis structures of both compounds, you will notice that the oxygen atom in diethyl ether (C2H5OC2H5) is in between the C2H5s while butanol (C4H9OH) has an OH- at the end of its molecule. Therefore, while diethyl ether cannot form hydrogen bonds with other diethyl ether molecules, bu...

Mindy Kim 4C

The problems in the books that asked you to identify which molecules have the shortest and longest bond lengths gave molecules with different orders of bonds. Therefore, to determine which molecules have shorter and longer bond lengths, I would first draw the Lewis structures for the molecules and t...

Mindy Kim 4C

I believe both the O-S-Cl and Cl-S-Cl bonds have the same bond angle (less than 109.5 degrees). The reason the solutions manual says there is only one bond angle for part b is because both these bonds have the same bond angle, so there is not more than one bond angle present in the molecule. Hope th...

Mindy Kim 4C

Higher bond orders are shorter than single bonds. Therefore, this difference in bond length might have a small effect on how electrons are pulled by atoms. However, I don't think this significantly impacts polarity of the molecule and I don't think we will need to know this for any test.

Mindy Kim 4C

Shape is important when determining whether or not structures are polar or nonpolar. For example, if you look at the structure of H2O (bent/angular), the molecule would be nonpolar if the structure was linear (the two dipole moments cancel each other out). However, due to its bent structure, the dip...

Mindy Kim 4C

The difference in pi and sigma bonds lies in how the orbitals in a bond are oriented. In a sigma bond, the orbitals of the atoms involved in the bond overlap end to end while in a pi bond, the orbitals overlap in two places side by side. Therefore, pi bonds cannot be rotated without breaking the bon...

Mindy Kim 4C

Double and triple bonds are shorter than single bonds so they might slightly affect bond angles in the molecular structure. However, I do not think we have to know specific angles for double/triple bonds.

Mindy Kim 4C

An example of this is a molecule with three atoms and two bonding pairs and one lone electron pair. The electron geometry for this molecule would be trigonal planar because there are 3 different electron densities. However, the molecular shape for this molecule would be bent because only the atoms a...

Mindy Kim 4C

I believe only lone pairs around the central atom affect molecular shape. The molecular shape is given for each central atom in the molecule and lone pairs on atoms that are not the central atom do not affect the bond angles considered in molecular shape.

Mindy Kim 4C

Yes, I got phosphorus as well. The total number of valence electrons counted from the Lewis structure is 32 electrons. Subtracting 21 electrons for 3 chlorine atoms and 6 electrons for the oxygen, you obtain 5 valence electrons. The element in period 3 with 5 valence electrons is phosphorus.

Mindy Kim 4C

Polarity has to do with where the electrons tend to lie, based on the dipole moments of the molecules. I think polarity would be impacted by the formal charge which should be considered when drawing resonance structures. Therefore, a resonance structure could be polar and have a dipole moments while...

Mindy Kim 4C

Why is distance inversely proportional to the interaction potential energy? Shouldn't it be that the closer two molecules are, the higher the potential energy?

Mindy Kim 4C

Why are hydrogen bonds stronger than ion-dipole interactions? I would have thought an interaction involving an ion would be much stronger than a hydrogen bond.

Mindy Kim 4C

Valence electrons are those after the preceding noble gas. Therefore, if you write the electron configuration for the element, whichever electrons come after the preceding noble gas are all valence electrons. For example, for manganese, the electron configuration is [Ar]3d54s2. Therefore, manganese ...

Mindy Kim 4C

Photons do not have any mass to them but they do have momentum. This momentum can be transferred to objects (those with a small mass and high velocity). This transfer of momentum allows photons to exert force on objects. However, keep in mind that this phenomenon is only observed with objects of sma...

Mindy Kim 4C

Degeneracy is basically the concept of orbitals being in the same energy level. For example, the 2px, 2py, and 2pz orbitals are all degenerate with each other because they all have the same energy level of n=2.

Mindy Kim 4C

To solve this problem using the Rydberg formula: Remember that the formula calculates frequency, not energy, based on the energy levels of the electron. Therefore, the formula is frequency=R(1/ninitial^2-1/nfinal^2) Plugging everything in, you get: 1.14x10^14 Hz=3.29x10^15 Hz(1/ninitial^2-1/16) nini...

Mindy Kim 4C

We discussed:
writing electron structures (configurations) of atoms: Pauli Exclusion Principle, Hund's Rule, Aufbau/Building-Up Principle
order of orbitals (based on which state has the lower energy)
exceptions to electron configuration rules

I would recommend reading P. 44 to P.48.

Mindy Kim 4C

I don't think we need to know how to solve Schrodinger's wave function equation or do any calculations with it. However, we do need to know how to identify orbitals and the relationship of them to the different quantum numbers.

Mindy Kim 4C

For part b, you can find the number of sodium atoms in 5.00 mg of sodium by doing the following: (5.00 mg Na)(1 g Na/1000 mg Na)(1 mol Na/22.99 g Na)(6.022x10^23 atoms/1 mol Na) = 1.3097 x 10^20 atoms Na (1.3097x10^20 atoms)(3.37x10^-19 J/atom)=44.1 J The reason why you multiply the energy from part...

Mindy Kim 4C

In the Atomic Spectra Module, there is a question that states, "An excited hydrogen atom emits light with a frequency of 1.14x10^14 Hz to reach the energy level n=4. In what principle quantum level did the electron begin?" I understand how to solve this problem with Rydberg's equation, but...

Mindy Kim 4C

Either Hz or s^-1 is accepted in your final answer. However, when using units in calculations, I find it easier to use s^-1 because you can see units being canceled or multiplied. Usually, other units will include s^-1 instead of Hz (such as J=kgm^2s^-2).

Mindy Kim 4C

First, you want to find the amount of Joules emitted in 2.0 seconds: (32 J/s)(2.0 s) = 64 J Next, you can find the energy of one photon with just the wavelength by using the equation E=hc/(wavelength): E=((6.626x10^-34 Js)(3.0x10^8 m/s))/(4.20x10^-7 m) E=4.733x10^-19 J To find how many photons were ...

Mindy Kim 4C

The light source is typically from the UV region because light from the UV region has shorter wavelengths and higher frequency, hence higher energy.

Mindy Kim 4C

When finding the wavelength of radiation emitted by a hydrogen atom in a transition from one energy level to another, the initial or starting energy level is n1 and the final energy level is n2. Therefore, n1=4 and n2=2. Your change in energy will end up being negative because when an electron moves...

Mindy Kim 4C

First, you want to convert the 5.00 mg of sodium atoms to how many atoms of Na there are. You can do this through the following: (5.00 mg Na)(1 g Na/1000 mg Na)(1 mol Na/22.99 g Na)(6.022x10^23 atoms/1 mol Na) which equals 1.3097 x 10^20 atoms of Na. Then, to find the energy emitted by 5.00 mg of so...

Mindy Kim 4C

And in regards to naming acids, compounds with a suffix of -ate have the suffix of "-ic" and "acid." Compounds with a suffix of -ite have the suffix of "-ous" and "acid." Therefore, since the compound is nitric acid, the compound is nitrate, which is NO3-. Add...

Mindy Kim 4C

Because this problem has multiple elements that are present in multiple compounds, I think it will take longer to solve. However, I balanced this equation by starting with nitrogen to get 2 C10H15N + O2 --> CO2 + H2O + CH4N2O. Then, I balanced the carbon in the carbon dioxide molecule to get 2 C10H1...

Mindy Kim 4C

I believe the same applies for molar masses given in the problem. Since these molar masses are still conversion factors, they would also not apply to significant figures in your final answer.

Mindy Kim 4C

No, in problems only the significant figures given in the problem matter for the significant figures in your final answer. For example, if you use the conversion factor 35.45 g/mol, this number would not change the number of significant figures in your final answer. And if the problem gave 3.45 g as...

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