CHEMISTRY COMMUNITY

Cris Yuan 2I

Congratulations to Ashley!

Cris Yuan 2I

Here're two pictures from the sides:

Cris Yuan 2I

Here's a fun one about hydrocarbons:

Cris Yuan 2I

If you look at reader p86, it states that the pseudo-equilibrium constant allows us to relate the concentration of the activated complex to the concentrations of the reactants. I think we could use the concentration of the reactants to find out the concentration of the activated complex in problems,...

Cris Yuan 2I

You could see from the diagram that the double bond becomes a single bond, so one of the Cs has received an H+ and the other a Br-. Since the second C received the Br, the H+ most likely goes to the C on the left, which is technically the third carbon because functional group comes first. Br | CH2=C...

Cris Yuan 2I

I think page 82 of the reader explains CH3Cl and other alkyl halides very well:
The C is the electron deficient electrophile site, and because Cl is the nucleophile site, the C-Cl bond could quickly donate an electron to Cl when it attracts a nucleophile.

Cris Yuan 2I

For your second question, since ΔG=ΔH-TΔS, an exothermic reaction means ΔH<0, so unless ΔS<0 AND the reaction is at a very high temperature, the reaction will probably be exergonic as well. When a reaction is exergonic, ΔG<0, so unless ΔS>0 AND the reaction is at a very high temperature, the reactio...

Cris Yuan 2I

The activation complex theory is just the transition state theory that we've learned in class, so you would refer to Reader page 85 about understanding and drawing one. It is a state that has partially formed bonds, while intermediates have fully formed bonds. so since it is a transition state, the ...

Cris Yuan 2I

Since catalysts speed up forward and reverse reactions to the same extent, the addition of a catalyst should not affect equilibrium constant.

Cris Yuan 2I

For catalyst, you can look for something that does not appear in the net equation; in other words, it is not consumed nor produced in the reaction, so they are often used in the first step but produced in the second step. By the way, the intermediates are also absent from the net equation, but you c...

Cris Yuan 2I

After you find the activation energy, plug in the value: ln(k)=-Ea/(R700K)+ln(A) and then use another equation: ln(35)=-Ea/(R350K)+ln(A) then subtract one from the other to get rid of A: ln(k/35)=-Ea/(R700K)+Ea/(R350K) k=35*e^(Ea(1/R350K-1/R700K) and plug in the value of Ea here, and the rest are pr...

Cris Yuan 2I

A. I think it's the same as the problem Professor Lavelle talked about in class today. W=-pΔv=ΔnRT You would balance the equation and use the R for bar and T in Kelvin. B. You would calculate by reaction enthalpy, and the formula is in the reader and the textbook. C. ΔU= qp + w = enthalpy (from B) +...

CHEMISTRY COMMUNITY

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Re: Ashley Sarquiz Award Winter 2017

Re: Excellence in Chemistry Award 2015-16

Re: Post All Chemistry Jokes Here

Re: Pseudo-equilibrium constant

Re: Final Winter 2014 #5B

Re: CH3Cl

Re: exergonic vs exothermic

Re: Ch.15 question 15.85

Re: Catalysts and equilibrium constant

Re: What is the catalyst in this reaction? [ENDORSED]

Re: Quiz prep 2 #11

Re: 8.93 Textbook [ENDORSED]