CHEMISTRY COMMUNITY

Leah Thomas 2E

When you are calculating work, the negative sign is already taken care of when you do -PΔV=-Δnrt so, therefore, all you have to do is ΔU=q+w and don't have to look at the sign.

Leah Thomas 2E

ΔS(system)=q(system)/T and ΔS(surr)=-q(system)/T so the signs themselves are opposite but also remember that the temperature of the system and the surroundings are not necessarily the same either.

Leah Thomas 2E

Is the degrees of freedom related to how many single and double bonds there are in the molecule?

Leah Thomas 2E

I'm not 100% sure if this logic is correct and I may be thinking backward here but...if a reaction needs a higher temperature to react it must need to overcome a large activation energy because increasing temperature provides energy to overcome that barrier.

Leah Thomas 2E

That might be hard to figure out because functional groups can be integrated within formulas of a bunch of CH2s.

Leah Thomas 2E

I tried just searching it myself and I'm pretty sure it's NO2.

Leah Thomas 2E

Is this the same as using the isolated method of having one reactant in great excess?

Leah Thomas 2E

I would try to at least know how to set up the first part of the derivations because I think there have been past exam questions where the professor will ask you some. If you look at his math assistance on his page it can help you if the math part integrations and stuff are confusing.

Leah Thomas 2E

I'm still confused as to how we know when to use the instantaneous rate and the unique rates?

Leah Thomas 2E

Yes, the reducing power is talking about the element's ions. Keep in mind that an element with greater reducing power can easily be oxidized.

Leah Thomas 2E

I think under different temperature conditions, a reaction can only proceed under a high temperature. The process may be spontaneous but very slow so an increase in temperature will speed up the process. The example given in class was the car to diamond reaction I believe.

Leah Thomas 2E

I'm pretty sure you do the reaction in the anode and then the cathode. Make sure to separate different states with a single line and those with the same states with a comma. A double line goes where the salt bridge is. Do not also forget to add platinum where there is no solid on the anode or cathod...

Leah Thomas 2E

You can also think of it as reactants being used up so it's rate is negative since the amount is declining.

Leah Thomas 2E

K constant just describes how fast a reaction can occur. The only way to speed up a reaction is to change the amount of energy input as in raise its temperature or lower the activation energy needed, I. E. Add a catalyst perhaps.

Leah Thomas 2E

To accurately find n, you must first balance the equation with the correct amount of moles, water, and hydrogen molecules so electrons can cancel out.

Leah Thomas 2E

Additionally if there is no solid in the reactant and product side, you must add it to both.

Leah Thomas 2E

Degeneracy is also a way to measure entropy that is the greater the degeneracy, or amount of ways to rearrange a molecule, the higher the entropy.

Leah Thomas 2E

In general you should know that the gaseous phase of any compound has a higher entropy than the liquid phase because of the greater volume and the vibrational motion the molecules occupy when it is a gas.

Leah Thomas 2E

Also know how to calculate residual entropy using the arrangement of molecules (W). As well as being able to identity when to use total entropy based on the pathways in the problem.

Leah Thomas 2E

When temperature is constant, why do we assume Δ U=0. I understand that Δ U=q+w and q=0 if temperature is constant but what about work? How does that relate to temperature?

Leah Thomas 2E

Can someone explain the difference between ∆G and ΔG° conceptually. Why don't we always need to calculate for ΔG°?

Leah Thomas 2E

I'm pretty sure Dr. Lavelle mentioned that at most times reaction will not have a K=1, or be at equilibrium. However, if K=1 then delta G will be equal to 0 because from the equation delta G=-RTln(K) taking the ln(1)=0 so the overall delta G will also equal 0.

Leah Thomas 2E

Does entropy increase when an ideal gas is compressed isothermally?

Leah Thomas 2E

You must input gasoline to make it run and heat is released when the car is running therefore an engine is an open system.

Leah Thomas 2E

The water is considered the surroundings because that is where the system is happening inside.

Leah Thomas 2E

I would know that at standard states carbon is a solid, N2, O2, F2, Cl2 are gasesm and Br2 is a liquid and I2 is a solid. I helps to know these things especially for standard enthalpies of formation when you are not given an equation or anything.

Leah Thomas 2E

Yes to the above answer. Work can be done by the system or unto the system based on expansion or compression which is a change in volume. Just be wary of open and closed systems.

Leah Thomas 2E

Dr. Lavelle additionally may ask the "why does steam cause severe burns?" question in relation to the heating curve of water.

Leah Thomas 2E

Professor Lavelle mentioned that Δ H=q at constant pressure but how could that be true if Δ H is a state property while q (heat) is not a state property?

Leah Thomas 2E

Also, ΔH of vaporization is the change of enthalpy so to explain the equation you have to take the enthalpy of vapor subtracted from the enthalpy of liquid because it references again final-initial. In this case, vapor is the final product.

Leah Thomas 2E

In problem 8.57 it states that the delta H for H2 is -286 kJ/mol but I though Professor Lavelle mentioned that at hydrogen's most stable form, delta H is equal to zero. Someone please explain.

Leah Thomas 2E

All I was told about nodes is that angular nodes equal n-l-1 and radial nodes equal n-l. Is this right and there is more I need to know about them other than there is 0 electron density in that region?

Leah Thomas 2E

A TA actually went over this in one of the review sessions. I think he explained that bismuth is a very poor conductor and even though it is a metal, it has more nonmetal properties because it such a bad conductor and therefore is labelled as amphoteric.

Leah Thomas 2E

Sometimes adding or taking away water has an effect on the shift of the reaction and sometimes is doesn't because it is considered a liquid. How do you know when that occurs/ when is water considered a medium or part of the reaction affected? This was on the modules and homework problems.

Leah Thomas 2E

Also I think for a molecule to be polydentates can also be chelates that are a ring structure created between atoms that must include the central atoms so be careful of those type of problems as well.

Leah Thomas 2E

So when Q is greater than K that means the products are favored but Professor Lavelle stated that the reaction will never overshoot equilibrium so how can you have more products than reactants?

Leah Thomas 2E

You only need to change gases to concentration from partial pressure if an aqueous solution is present in the equation meaning it is not homogenous.

Leah Thomas 2E

Can someone explain the different between a polar/nonpolar bond vs a polar/nonpolar molecule because I know that a molecule may be polar but the bond is nonpolar but how?

Leah Thomas 2E

Ok wait so the question is not assuming that the energy required to do work has already been reached?

Leah Thomas 2E

You might have also been confused because usually the positive spin of the electron is drawn first and then the negative spin when filling the orbital but the 3d10 electron can still have either spin since you don't know specifically the electron that is filling that last d orbital.

Leah Thomas 2E

Can someone explain how the electron affinity of noble gases work relative to its column and even those of the column above and below it?

Leah Thomas 2E

When atoms are alone, they tend to be reactive to create that noble gas configuration (octet rule) which is why bonding makes atoms have lower energy overall. Also why noble gases are less reactive, they have the octet and want to stay that way.

Leah Thomas 2E

Can someone explain why there are more exceptions when filling electrons for the lanthanides? Like there are more energy levels so they become closer and...
Just for context.

Leah Thomas 2E

Also you can think of it that if an atom can attract an electron easily, then the ionization energy will be big because it want to hold on to that electron for as long as it can.

Leah Thomas 2E

Can someone please explain the what the quantum number ml stand for? I how you get the number but what does it represent and how do you know based on which number is given to you?

Leah Thomas 2E

Also by having the initial energy be at n=5, you get a negative energy in your calculations which is what you want since the energy is being emitted (electron is losing that energy). Just look at the context of the question though.

Leah Thomas 2E

Dr. Lavelle mentioned this briefly but if an electron is somehow excited to the second energy level and then is able to jump to the third energy level, when it emits energy does it go back to the second energy level then the first or just drops straight back to the first?

Leah Thomas 2E

Need a quick answer. So the speed of light is constant and is what we use in our equation. However, when dealing with the photoelectric effect, energy is emitted unto a metal at a different a certain speed and is it emitted off at a slower speed/energy since there is some sort of threshold energy (w...

Leah Thomas 2E

I was confused on this matter too The only way I was able to answer the question was process of elimination because I knew all the other ones were false. No sure why this one is true though. Can anyone please explain?

Leah Thomas 2E

In my high school chemistry class my teacher was pretty lenient about not writing the states of matter, but now I regret it because it makes me lazy when writing out chemical equations. You should get used to writing out the entirety of the equation as it will assist you in solving word problems wit...

Leah Thomas 2E

I think that the solutions manual made a mistake. I am sure that your are supposed to divide by the initial molar mass of the compound to find the percent change in mass. Either way the answer does remain the same.

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