CHEMISTRY COMMUNITY

Cooper1C

Unimolecular means that the reaction involves one molecule, bimolecular means the reaction involves two molecules, and termolecular means the reaction involves three molecules.

Cooper1C

A catalyst does not change the reactants and products of a reaction - it just changes the pathway. So you should be able to have catalysts and intermediates in the same reaction.

Cooper1C

It's just the notation - it distinguishes k from k'. I don't think there's anything special about it.

Cooper1C

Q: What kind of fish is made out of 2 sodium atoms?
A: 2 Na

Cooper1C

Q: What do you call a wheel made of iron?
A: A ferrous wheel.

Cooper1C

Q: What do you call a clown who's in jail?
A: A silicon.

Cooper1C

k is based on experimental data, so if you're given data you are probably expected to calculate it. If not, I think you can use the table.

Cooper1C

On page 631, the textbook also says that the half-life is not used for second order reactions. Maybe this is because the half-life depends on the initial concentration for zero and second order reactions? This would make it difficult to compare reactions.

Cooper1C

From experiment 2:

17.4 mmol/L*s = k(2.5 mmol/L)(1.25^2 mmol^2/L^2)(1.25^2 mmol^2/L^2)
k = 17.4 mmol/L*s /(1.25^2*1.25^2*2.5) mmol^5/L^5

The mmol and L cancel, so you have

k = 17.4/s /(1.25^2*1.25^2*2.5) L^4/mmol^4

1 mmol = 10^3 mol, so

k = 17.4/(1.25^2*1.25^2*2.5)* 10^12 L^4/mol^4*s

Cooper1C

At standard conditions, you can use the formula $E = E^{\circ} - (\frac{0.05916 V}{n})logQ$ to find the pH. The formula for ph is pH = -log[H+}, so you can find the concentration of hydrogen ions by the Nearst equation.The formula above is already converted from ln to log.

Cooper1C

If you use the formula $E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$ you do not need to change the sign of $E^{\circ}$ for either half reaction because the minus sign does it for you.

Cooper1C

In order to use the formula $\Delta G_{r}^{\circ} = -nFE^{\circ}$ , you can solve for n by writing out the oxidation and reduction half reactions and balancing them.

Cooper1C

Relationship between T and delta U: \Delta U = q (assuming no work is done) If you increase heat added (increase q) then delta U increases. Relationship between T and delta S: delta S \Delta S __{T_{1} \rightarrow T_{2}} = nC(\ln T_{2} - \ln T_{1}) If you look at the natural log graph: If T_...

Cooper1C

The third law of thermodynamics says that the total entropy is always increasing, so delta S total must be positive. The standard enthalpy for a reaction can be positive or negative depending on the reaction. For example, building polymers from monomers has a negative delta S (the complexity is incr...

Cooper1C

The third law of thermodynamics says that the total entropy is always increasing, so delta S total must be positive. The standard enthalpy for a reaction can be positive or negative depending on the reaction. For example, building polymers from monomers has a negative delta S (the complexity is incr...

Cooper1C

Kc is the equilibrium constant in terms of molar concentrations. See section 11.6 in the textbook.

Cooper1C

The question says "what might its residual molar entropy be?" so you would use avogadro's constant.

Cooper1C

Because entropy is a state function and you know the vaporization at 100 degrees C, to find the standard entropy of vaporization at 85 degrees C you can heat the water to 100 degrees C, vaporize it, then cool it to 85 degrees C. This will give you the standard entropy of vaporization at 85 degrees C.

Cooper1C

A negative delta S corresponds to a spontaneous process when the magnitude of T * delta S is less than delta H (which must be negative). This is because delta G must be negative, and delta G = delta H - (T * delta S). A negative delta S would mean that the products have a lower entropy than the reac...

Cooper1C

Reactions are spontaneous when delta G is negative. delta G = delta H - (T * delta S) Releasing heat is favored, and increasing entropy is favored (for example, liquid to gas - especially favored at higher temperatures). You have to consider both cases to make a conclusion about the spontaneity of a...

Cooper1C

Delta S is also a state function.

Cooper1C

I think s->l->g requires energy and g->l->s releases energy.

Cooper1C

You use the enthalpies of formation to calculate delta H, then you have to convert kJ/mol TNT to kJ/L, which is why you need to know the density of TNT.

Cooper1C

The first step is to balance the equation that they describe in the first sentence. CO + H2O -> CO2 + H2 Then you know that the heat absorbed by the calorimeter is equal to the heat released by the reaction. qcal = -qrxn And delta U = q because the reaction occurred in a bomb calorimeter (w = 0 beca...

Cooper1C

Bond enthalpies are averages of many bonds.

Cooper1C

I believe that the sign is positive because the definition of enthalpy density is the enthalpy released per liter, so adding a negative sign would mean energy is absorbed.

Cooper1C

I would draw the phase change chart for water first. The first phase change is at 0 degrees and the second is at 100 degrees. If you start at 0 degrees (ice) and go to 20 degrees (liquid), you need to consider both the heat required to change ice to a liquid and the heat required to raise the temper...

Cooper1C

Like balancing chemical reactions, I recommend manipulating the equations that have the least number of the same molecule first, and the should build on it.

Cooper1C

What is the difference between internal energy and enthalpy? H = U + PV. Enthalpy is equal to internal energy plus the product of pressure and volume, so internal energy is included in the enthalpy calculation. Also, internal energy has to do with the kinetic and potential energies of molecules in ...

Cooper1C

: Here is a diagram of a bomb calorimeter from the textbook (page 271)

Cooper1C

Amphiprotic is the ability of a molecule to both accept and donate protons.

Amphoteric is the ability of a molecule to react with both acids and bases.

See page 469 in the textbook.

Cooper1C

I believe the acid/base will not dissociate unless it is in some sort of solvent. In this case, the problem states Ba(OH)2 is in an aqueous solution. Ba(OH)2 is a strong acid, so you do not need to set up an ice box, and you are not given Kb. Because Ba(OH)2 is a strong base, we assume it is 100% io...

Cooper1C

Is there any way to figure out this problem by looking at the Lewis structures?

Cooper1C

If you look at table 4, the full name of the ligand is given and the abbreviation is in parentheses: ethylenediamine (en) enthylenediaminetetraacetato (edta) After the parentheses is a symbol that corresponds to one of the footnotes telling you whether the ligand is bidentrate, tridentate, or hexade...

Cooper1C

The carbons and nitrogens form a long chain, with the hydrogens bonded to each atom. The order in which the atoms are written gives the order of the chain, and because there are two identical segments, the molecule can be drawn as symmetrical. If you count the number of available valence electrons, ...

Cooper1C

This makes sense because in order for it to be a neutral compound, there must be three chlorine atoms. The chromium atom has a +3 charge, so there must be three chlorine atoms, each with a -1 charge. "Trichloride would be redundant."

Cooper1C

The subscript of E is the number of lone pairs around the central atom. You would have to draw the Lewis structure.

Cooper1C

See figure 4.5 in the textbook (page 115): "If the lone pairs had taken the axial positions, they would have been at 90 degree angles from the equatorial positions, resulting in greater repulsion."

Cooper1C

You also have to think about the shape of the molecule. For example, in part (a) (CH2Cl2), the molecule has a tetrahedral shape, so the dipole moments do not cancel, even though they look they could in the Lewis structure -> The molecule is polar.

Cooper1C

Page 118 of the textbook provides good graphics for this concept. For the linear model, the molecule can either have the formula AX2 or AX2E3, so it could have lone pairs around the central atom.

Cooper1C

If you look on page 118 in the textbook, there are really good diagrams that show the shapes. If you think about a molecule with 5 regions of electron density and no lone pairs, the shape is trigonal bipyramidal. Then if you replace the 3 atoms in the plane with 3 lone pairs, the shape is linear.

Cooper1C

It is possible for (b) to have lone pairs if it has 3 lone pairs around the central atom. It will still be a linear shape. See page 118 of the book for a graphic.

Cooper1C

I also believe that organic molecules are usually chains or rings.

Cooper1C

This could be because of resonance. The molecule is an average of the resonance structures, so the bond length is an average as well.

Cooper1C

I looked at the last block the electrons occupy. For example, Bi has 3 electrons in the 6p-block, and because of Hund's Rule and the Pauli Exclusion Principle, each electron will occupy a separate subshell (px, py, pz), so there are three unpaired electrons.

Cooper1C

I believe we are expected to know the s-block, p-block, and first row of the transition metals.

Cooper1C

1 Ångström is 10^-10 meters. Bond lengths are about 1 Ångström.

Cooper1C

If it creates a diffraction pattern (like the electrons did in the slit experiment), it must have wavelike properties.

Or, based on De Broglie's relation, any moving particle with momentum has wavelike properties, but sometimes the wavelength is so small there are no measurable wavelike properties.

Cooper1C

I believe this rule is to eliminate rounding errors in a multi-step problem (50% is rounded up and 50% is rounded down). However, in order to be as precise as possible you should not round your answer until the very end of the problem.

Cooper1C

In case you do not have them on hand, the prefixes are: centi 10^-2 mili 10^-3 micro 10^-6 nano 10^-9 An easy way to remember the prefixes is that they decrease by three powers every "step," with the exclusion of centi (which you can remember by centipede or centimeter, although I do not b...

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