Chem 2 ACS final study questions help

[Kaxa2000]

I'm studying for the ACS final for chem 2. I tried asking these ?'s on yahoo answers with no success so hopefully someone here can help. Just answering one of these 4 ?'s will be helpful.

What is the relationship between the equilibrium constant (Kc) of a reaction and the rate constants for the forward (kf) and reverse (kr) reactions?

A. Kc = kfkr
B. Kc = kf/kr
C. Kc = 1/(kfkr)
D. Kc = kf - kr

Please explain your answer.

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Two reactions w/ different activation energies have the same rate at room temp. Which statement correctly describes the rates of these two reactions at the same, higher temp?

A. The reaction w/ the larger activation energy will be faster
B. The reaction w/ the smaller activation energy will be faster
C. The two reactions will continue to occur at the same rates
D. A prediction cannot be made w/o additional info

Please explain your answer.
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Also, I read on here from prev posts(http://forums.studentdoctor.net/showthread.php?t=429214) that the ACS final for chem 2 is cumulative from all the way back to Chem 1? If that's true then I'm screwed....I don't think my prof mentioned that. The front cover of the ACS study guide says

"If you are taking a first-term exam, only the first five topic groups pertain. For a second-term exam, the final five topic groups are most important; although the second-term topics do build upon knowledge acquired in the first term."

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[tharper14]

Well I'll answer the easy one, I'm not in the mood to do calculations or look up stuff in my chem book.

For question 2, the reaction with the smaller activation energy will be faster, because well, it has a smaller activation energy and they are both at the same temp. It can't be the one with the larger activation energy, because it requires more energy for the reaction to take place. Unless more energy (higher temp) is applied to the system or a catalyst is provided, the reaction with the smaller activation energy will go faster. Gee, I really suck at explaining 😎

I think the answer to question 1 is B, no explaination... it just is.

 

[Kaxa2000]

Thanks for answering. I thought that it was the smaller activation energy answer too, but on the answer sheet the answer is A. You're right...the answer to the first one is b, but I don't understand why.

 

[mrblah]

Thanks for answering. I thought that it was the smaller activation energy answer too, but on the answer sheet the answer is A. You're right...the answer to the first one is b, but I don't understand why.

because its a law (rate constant law or something). I honestly don't remember the name, but I had to memorize it also.

 

[tootoo30]

For problems 9 and 10, you just have to do division to find the order. For number 9, Divide number 2 to by number 1 to find the 👎 and then divide number 3 by number 2 to find (m) and those will be your rate order. For number 10, divide number 2 by number 1 to find 👎 and then divide number 4 by number 3 to find (m). I just substituted the chemical name with 👎 and (m).

For question one, if I'm not mistaken, the equillibrium = products/reactants

HTH