CHEMISTRY COMMUNITY

Lauren Seidl 1D

My TA told us that steady state will not be on the final because Dr. Lavelle didn't really cover it in class, and that we will only have to use the pre equilibrium approximation approach.

Lauren Seidl 1D

Once you write out the final redox reaction, only the reactants are considered oxidizing or reducing agents. The one that is being oxidized in the reaction is the reducing agent, and the one being oxidized is the reducing agent. Therefore you should write out the final equation before identifying th...

Lauren Seidl 1D

When you balance the equation, you should get AB4(g) + 2 C2(g) ---> AC2(g) + 2 B2C(g). Because the rate of the reaction is constant, we can use the equation rate = (change in concentration)/(change in time). You must multiply the rate by 2 because there are two moles of B2C produced for each mole of...

Lauren Seidl 1D

You use difference of sums to find Gibbs free energy when you are using the Gibbs free energy of formation for the reactants and products. If you are given entropy, enthalpy, and temperature, then you would use deltaG=deltaH - TdeltaS.

Lauren Seidl 1D

Because they divide initial concentration by final concentration, you get a value of 10 in the ln and would therefore set ln(10) = k*t. If you divided final concentration by initial concentration instead, you would get a value of 0.1 in the ln, but you would have to set ln(0.1) = - k*t according to ...

Lauren Seidl 1D

The orders do not have to do with coefficients. If you look at problem 15.17, for example, the reactants all have different coefficients, but that does not affect their orders. You simply look at the concentrations for each when determining order. In 15.23c, however, you do need to use the coefficie...

Lauren Seidl 1D

The answer to part D is in mol*L^-1*s^-1 too, if that makes more sense.

Lauren Seidl 1D

I was confused at first too, and got the same answer as you. I figured that in the solutions manual they converted all the values from mmol to mol, so I did that and got 2.85 x 10^-12. However, in 15.17 they leave the values in mmol and I'm not sure which one they want us to do.

Lauren Seidl 1D

Rate is similar to speed vs. velocity. Though velocity can be positive or negative because it has direction, speed can only be a positive value. The same goes for rate of a reaction, it can only be a positive value because it is simply how fast the reaction is proceeding.

Lauren Seidl 1D

My TA said that it covers up to #39 in the problem set.

Lauren Seidl 1D

I found a great website that tells you exactly how to determine oxidation states, as it would be too long to describe. I would check this out: https://www.chemguide.co.uk/inorganic/redox/oxidnstates.html

Lauren Seidl 1D

There is a typo in the solutions manual. It says on Dr. Lavelle's website that "The correct solution should have a 3 as the subscript for the nickel hydroxide reactant.
2Ni(OH)3 (s) + Cd(s) → Cd(OH)2 (s) + 2Ni(OH)2 (s)"

Lauren Seidl 1D

You can use either of the first two equations, depending on what values are given. If given deltaG directly, then the first equation is easier to use. If only given standard cell potential, then the second equation is easier to use. The third can be used when the reaction is not in equilibrium.

Lauren Seidl 1D

Yes, I also think it was a typo because I'm pretty sure the solutions manual had Fe3+ instead of Fe2+.

Lauren Seidl 1D

Because anode is always written on the left side of the cell diagram, you still have to calculate the Ecell = Ecathode - Eanode, which does give you a negative voltage. This tells you that the reaction is not spontaneous, and that the electrons would flow in the opposite direction.

Lauren Seidl 1D

If you look at the bottom of page 325 under the example, it says to assume that argon is an ideal gas, so I am assuming that is why you would use argon's heat capacity.

Lauren Seidl 1D

w=-p*deltaV, so pressure does in fact affect work.

Lauren Seidl 1D

Reactions at equilibrium are considered reversible, as they go back and forth in order to reach equilibrium. Irreversible reactions, on the other hand, only proceed in one direction.

Lauren Seidl 1D

Because q=n*C*deltaT, and neither moles nor molar heat capacity can be negative, a negative q would mean that delta T is negative. So yes, a negative q means that temperature is decreasing. A negative deltaU does not necessarily mean temperature is decreasing because deltaU=q+w. So even if q was pos...

Lauren Seidl 1D

The easiest way is that deltaS(total) = deltaS(system) + deltaS(surroundings), and deltaS(total) must be 0 for a system at equilibrium, so deltaS(system) = -deltaS(surroundings).

Lauren Seidl 1D

You start with deltaH - TdeltaS = -RTlnK. Then divide both sides by -RT to get (-deltaH/RT) + (deltaS/R) = lnK , or flip it around to get lnK = (-deltaH/RT) + (deltaS/R).

Lauren Seidl 1D

Also, gibbs free energy is equal to enthalpy plus entropy, which are both state functions. Thus, Gibbs free energy must also be a state function.

Lauren Seidl 1D

Also, if delta entropy equals zero then the reaction is at equilibrium.

Lauren Seidl 1D

Delta G combines the values of delta H and delta S to fully determine whether a reaction is favorable or not.

Lauren Seidl 1D

Entropy is the likelihood or probability that that a system will be in a particular state.

Lauren Seidl 1D

Yes, the negative value is indicating that it is lost heat.

Lauren Seidl 1D

The kekule structure of benzene suggests alternating single and double bonds, whereas the resonance structure contains six bonds that are between a single and double bond. The resonance structure is more stable, as all six bonds between the carbon molecules are the same.

Lauren Seidl 1D

It is like the hill analogy. The path you take to get to the top does not matter, as you only look at the change in elevation from your starting point to the end point. Therefore, internal energy is not concerned with the "in between", only the beginning and final states.

Lauren Seidl 1D

Specific heat capacity is always constant for each respective substance, regardless of the mass present. Heat capacity on the other hand differs depending on the mass of the substance present.

Lauren Seidl 1D

The work function is also the energy required to remove one electron.

Lauren Seidl 1D

If you look at vaporization for instance, water is not able to increase in temperature past 100 degrees celsius, because after 100 it must be in its gaseous state. So even though more heat is being added, the temperature of the water evaporating remains at 100 degrees celsius and the heat that is be...

Lauren Seidl 1D

You can think of it as the electrons in the p-orbital being more shielded by the electrons in the s-orbital, as it is closer to the nucleus. Therefore, the positive charge of the nucleus reaches the p-orbital electrons much less than s-orbital electrons, making the Zeff on the s-orbital greater than...

Lauren Seidl 1D

You can memorize the VSEPR shape for each VSEPR formula (i.e. AX3E = trigonal pyramidal, or AX6 = octahedral). That way for every structure you draw, you can simply figure out its VSEPR formula and know which shape it denotes. Otherwise, drawing out the VSEPR structure should give you a clear idea a...

Lauren Seidl 1D

Also, you want to minimize the formal charges, so placing the lone electron on C will give you a formal charge of 0 for both atoms.

Lauren Seidl 1D

The bond angles for the atoms will only be slightly less than 90 degrees and slightly less than 180 degrees, not 120 as we do not count the angles between lone electron pairs.

Lauren Seidl 1D

NO2 is a radical, meaning it does not have a full octet on the N atom, and N has one lone electron instead of a lone pair.

Lauren Seidl 1D

You would use the moles of the limiting reagent to find how many moles of your product are being produced, as the amount produced is limited by how much of the limiting reactant is present. This requires the use of molar ratios. So say your equation is 2A+3B->C. If you find A to be the limiting reag...

Lauren Seidl 1D

On Dr. Lavelle's website it says that this review session is from 4:30-6:00 pm in LaKretz 110. Is it at this time or 4:00-6:30? Thank you!

Lauren Seidl 1D

Delocalized charges occur when there is resonance, as the charge occurs in different spots due to the actual structure being a combination of all its resonance structures. It is therefore delocalized.

Lauren Seidl 1D

Delocalized charges occur when there is resonance, as the charge occurs in different spots due to the actual structure being a combination of all its resonance structures. It is therefore delocalized.

Lauren Seidl 1D

The easiest way to spot delocalized pi-bonding is in lewis structures that display resonance structures, as the pi-bond can move around and is therefore "delocalized." Just remember delocalized as not being local to one specific point in the structure.

Lauren Seidl 1D

Yes, it is just these exceptions, as it is more stable to have the d-orbital filled and only half of the s-orbital filled.

Lauren Seidl 1D

I believe that this topic was not heavily emphasized in class and we were not assigned any homework problems on it, therefore it most likely will not be on the final so I wouldn't stress about it too much.

Lauren Seidl 1D

Also, this is not an assigned homework problem so it most likely will not be on the final exam :)

Lauren Seidl 1D

The units used to calculate K and Q must be constant in the equation, so that the units will cancel out and leave K and Q without specific units, as K is a constant and Q is a ratio.

Lauren Seidl 1D

It's easiest to treat ions like other molecules when determining lewis structures, by first adding up all of the valence electrons and either adding or subtracting the electrons (depending on whether it is a cation or anion) to the total number. Then just draw the structure so that the total number ...

Lauren Seidl 1D

When comparing the bond lengths of two different molecules and one has resonance structures, you can determine its relative bond length by taking an average of the bond lengths within the resonance structure, which can then be compared to the bond length of other molecules. This is consistent with t...

Lauren Seidl 1D

Bond length has to do with electrons being shared, and can therefore be determined by the number of bonds. For example, a single bond has the longest bond length, and a triple bond has the shortest bond length.

Lauren Seidl 1D

You can determine the relative bond lengths by looking at the number of single, double, and triple bonds. Triple bonds have the shortest bond length, and single bonds have the longest. Because some structures may have resonance, the bond lengths can be averaged in order to compare it to other struct...

Lauren Seidl 1D

Yes, carbon dioxide and water vapor will always be produced in combustion, but depending on the reactants and due to conservation of mass, other products may be produced. In example H.19, nitrogen is one of the reactants, so it therefore must also be a product. That is why you would get carbon dioxi...

Lauren Seidl 1D

Yes, to approach this problem you must work backwards, using the moles of each product to find the moles of the reactant produced by the equation above it. Use the given moles of NaBr to find moles of Fe3Br8, which is the product of the previous equation above it. Use that to find the moles of FeBr2...

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