CHEMISTRY COMMUNITY

Allen Mayo 1E

Percent protonated is just [final]/[initial] * 100 to account for the percent.

Allen Mayo 1E

K is just the equilibrium constant so Kc and Kp just describe what the equilibrium constant entails for. Kc is for concentration and Kp is for partial pressure.

Allen Mayo 1E

When Delta G is negative then the reaction is spontaneous.

Allen Mayo 1E

If internal energy change is 0 so U=0, U=q+w, so then it would become 0=q+w. This is how it is rearranged to -w=q.

Allen Mayo 1E

Delta G is = 0 because of the fact that the work from both forward and reactions cancels out and work = 0.

Allen Mayo 1E

Concentration is preferred since you are using [x]. Just use mol/L to avoid confusion.

Allen Mayo 1E

Yes it does affect the K. When the temperature changes, the K constant will either start to favor products for reactants depending on if it heated or cooled.

Allen Mayo 1E

If work is done on the system, it should be positive.

Allen Mayo 1E

It is just a different name but all in all, they are the same thing.

Allen Mayo 1E

A closed system has a heat or energy transfer w/ surroundings while an isolated system does not.

Allen Mayo 1E

It should be a rule of thumb to use Kelvin as many constants deal with K such as, R (gas constant). Just add 273 to a Celsius value, this will give you the K conversion of it. Convert to K unless otherwise.

Allen Mayo 1E

A spontaneous reaction is a reaction that can happen without any external aid. It relates to Gibbs FE because it is negative if it is a spontaneous rxn, same with a positive GFE which means it is non spontaneous.

Allen Mayo 1E

There are different gas constants because it relies on the units of the equation so you were right on the dot with assuming that it relates to the units of the given problem. Do pay attention to conversions if the unit is in a different count like KJ or J and such.

Allen Mayo 1E

If ∆U is negative that means U final is less than the initial U. This is b/c ∆U is change in U and it is dependent on the final U subtracted by the initial U. This means the system is losing internal energy. This is vice versa to a positive U

Allen Mayo 1E

What is the relationship between spontaneous processes and GFE? There's an example of what temp Br2 l to Br2 g spontaneous at 1 atm, and it seems that the prof. states at what temp is ∆G negative? I just wanna know this relationship of spontaneous processes and change in GFE.

Allen Mayo 1E

We would use w=-P∆V if the system the equation is being applied in has constant pressure .

Allen Mayo 1E

Entropy's units as I've seen in lectures are usually joules over kelvin and you can find K by adding 273 to the Celsius.

Allen Mayo 1E

Like in another question posted, n is just moles transferred but this can be found by balancing the reaction given to you.

Allen Mayo 1E

In the equation, the n is simply just moles transferred.

Allen Mayo 1E

In a reaction, if the reaction loses electrons then it is oxidized but if it is reduction then the reaction gains electrons

Allen Mayo 1E

The 3rd law of TD, states that a system's temperature reaches 0 Kelvin, the entropy of the system or the S, will reach a constant value, related to the microstates of the system itself.

Allen Mayo 1E

Spontaneous reactions has a net release of energy, which turns it from a higher to a lower state of energy. Further, this can occur by itself without a reaction, hence "spontaneous"

Allen Mayo 1E

Spontaneous reactions does not require any heat for the reaction to go through which in turn has a delta H that is negative. Vice versa with a non-spontaneous reaction because it needs heat and delta H is positive.

Allen Mayo 1E

To clear it up, w is work and work can either be positive or negative so there is no set sign for the value of work. Therefore, it all depends on the value of work.

Allen Mayo 1E

This equation is valid because if one is the system and the other is surroundings, one q will be losing energy and the other will be gaining it. It is transferring heat.

Allen Mayo 1E

Like Jay said earlier, I use a similar strategy by just keeping in mind their prefixes. Exo means outside, and endo means inside. So exo would be external so releasing energy, and endo would be absorbing into.

Allen Mayo 1E

There is an inversely proportional relationship to volume and pressure. When volume increases, pressure decreases and vice-versa.

Allen Mayo 1E

Kw is the equilibrium constant of water and this can be used to find either Ka or Kb based on given values. kw = kb x ka

Allen Mayo 1E

[HA] does not equal to [H3O]. After calculating it will be 0.10-x and H3O = x. Therefore, it will not be the same.

Allen Mayo 1E

Q is a value in thermodynamics that means heat. Or heat transfer because it usually either states whether heat transfers from system or into surroundings hence, qsys=-qsurr.

Allen Mayo 1E

Endothermic reactions absorbs energy and exothermic reactions release it. Like it was said in another post in this forum, if net energy in the reactions leans toward forming bonds, then the reaction is endothermic because it uses more energy to form bonds than to break them. This is vice-versa.

Allen Mayo 1E

Solids and liquids pressure is negligible because changes in its pressure does not affect, there is no effect in volume. So when gas pressure changes, there is more of an effect to the substance.

Allen Mayo 1E

The prof. said in lecture that the Qp is equal ∆H, which is enthalpy change. So, when solving, usually Qp is the ∆H.

Allen Mayo 1E

I believe each method is useful based on what is the given for the problem. There is not a "preferred" method.

Allen Mayo 1E

Molar heat capacity is intensive because its not based on the amount of the substance placed in the system. It is fixed.

Allen Mayo 1E

Specific heat capacity depends on the amount of the substance so heat required to raise 1 gram of the substance by 1 degree Celsius. Heat capacity is the heat required to raise a substance by 1 degree C.

Allen Mayo 1E

I believe that there isn't a term for it rather a definition. It is just a path function, the professor talks about this briefly in the same lecture he talks about state properties and he explains it in that path functions represent properties like heat and work which require to find the path it too...

Allen Mayo 1E

Hess's Law is just the the previous idea mentioned in the previous lecture that enthalpy are state properties that can be added or subtracted. Therefore, it is why you see Prof. Lavelle add the multi-step reaction of the nitrogen oxide to form nitrogen dioxide example. He simply adds the two as soon...

Allen Mayo 1E

I am sure that vapor and gas mean the same thing but is used in the class interchangeably.

Allen Mayo 1E

I am sure that vapor and gas mean the same thing but is used in the class interchangeably.

Allen Mayo 1E

Heat is not a state property because as he said in the lecture it entirely depends on the fact that it depends on the path taken to get to the value of heat. Enthalpy is a state property because it does not depend on path taken and can be quantified from final to initial.

Allen Mayo 1E

These equations are the same. The P=conc(rT) is a rearranged version of PV=nRT. When dividing the V from P to the other side, it then can equal to P = n/v *rT. And we know that n/v is can equal to molar conc., because moles over liters is mol L-1, which is concentration! Also, this is just used when...

Allen Mayo 1E

Prof. Lavelle showed us an example between weak acids and bases and it ended up being CH3COOH + H2O <-> H3O+ + CH3COO- and it then set up in the equilibrium table as 0.10 - x x x. I know that it turns out to be that the Ka is able to be approximated but I wanted to try the quadratic equation just to...

Allen Mayo 1E

Q is only equal to K when the reaction is at equilibrium. However, the difference is that when the reaction is not at a given equilibrium, the Q can use the given reactants and products at a certain point to find the direction in which the reaction will happen.

Allen Mayo 1E

Solids and liquids are not included in an equilibrium reaction because of the fact that they are pure substances. Additionally, this means there is no concentration for solids and liquids are likely used to be solvents in a reaction. This means that gas and aqueous solutions are able to be used beca...

Allen Mayo 1E

It's pretty much based on preference. Mol L -1 is essentially the same as M but in lectures the professor might use mol L-1 just to show the conversions if need be and to clarify that a certain unit did not just come out of nowhere.

Allen Mayo 1E

In Lec. 2 around 44:50, Prof. Lavelle approximates because the constant is very very small, almost negligible. I understand the idea of approximation because it is an almost negligible value but I would like to know if there is a rule of thumb based on if a certain value will be subject to approxima...

Allen Mayo 1E

R in the PV=nRT, ideal gas law is just a constant. It is actually a gas constant that equals to R = 8.314 J*(K/mol). This is just the value I know from my time in Physics but I imagine there are different based on the conditions that the units are given in.

Allen Mayo 1E

K is just the equilibrium constant. Prof. Lavelle does go into detail with this but all in all, Kc is just K with specifics. This means it is the equilibrium constant with respect to the molar concentration hence, c as c = molar concentration. This is also the same with partial pressure when dealing...

Allen Mayo 1E

K is just a ratio for the reaction. It is simply a constant for the reaction. Also, when dividing the concentrations, their units cancel out, which then ends up at the same conclusion, it is unitless!

Allen Mayo 1E

Prof. Lavelle stated this on Lecture 1, it's around the midpoint, maybe a little less than the midpoint of the lecture. He explained that the reverse of the reactions eq constant should just be the inverse of K so 1/K. He shows that it is pretty much flipping the structure of the equilibrium constan...

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