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Figured I'd start this one out. I'll be registering for the MCAT first thing tomorrow. Who else is taking it on this date? How do you all plan on studying over the school year?
Just got 22 out of 33 on my first 1/3. On Passage 7 I missed almost every single one lol. Oh well
.hey can someone tell Mr what Troll is haha...i see people call people that...I'm guessing its a bad thing lol
In Internet slang, a troll (pron.: /ˈtroʊl/, /ˈtrɒl/) is someone who posts inflammatory,[1] extraneous, or off-topic messages in an online community, such as a forum, chat room, or blog, with the primary intent of provoking readers into an emotional response[2] or of otherwise disrupting normal on-topic discussion.[3] The noun troll may also refer to the provocative message itself, as in: "That was an excellent troll you posted."
While the word troll and its associated verb trolling are associated with Internet discourse, media attention in recent years has made such labels subjective, with trolling describing intentionally provocative actions and harassment outside of an online context. For example, mass media has used troll to describe "a person who defaces Internet tribute sites with the aim of causing grief to families."
Today, TBR bio chap 9&10, entadus's passage, verbal, more PS problems.
I'm taking biochemistry
I think people feel it is comparable to Kaplan in terms of question quality. But I will use it and the kaplan passages since I have them. There seems to be no disagreement that EK101 and the TPR verbal is the best, with people varying between the two as which in particular is the best.
Step right up, folks! Come one, come all! Fun with chemical kinetics, give it a try! 🙂
GOOD LUCK
Passage 2
Researchers are working to elucidate the mechanism of the common reaction:
A + 2C + E ------> F + G ( ΔH = -227 kJ/mol at Standard Temperature and Pressure)
A 3-step mechanism for the reaction has recently been proposed. According to this hypothesis, the reaction involves key intermediates B and D, and includes a rapid initiation step whereby 2 molecules of reactant A reversibly form 1 molecule of activated intermediate B. The proposed mechanism is shown below in Figure 1.
Step 1: 2A <=(forward k1 , backwards k-1)=> B - **fast, reversible**
Step 2: B + 2 C -----------k2---------------------> A + D
Step 3: D + E ------------k3----------------------> F + G
^Figure 1^
A researcher on the team decides to test this mechanism by the method of initial rates. The concentrations of reactants A, C, and E are varied and the rate of formation of product G is determined during the course of the first second of the reaction. The experimental data are provided below (figure 2).
Exp. . . .[A]. . . . . . . [C]. . . . . . . . [E]. . . . . . .initial rate (M/s)
1...........0.1M...........0.1M..........0M................x
2..........0.1M............0.1M........0.05M........1.0*10^-4
3..........0.1M...........0.1M.........0.1M..........1.0*10^-4
4..........0.1M...........0.05M.......0.2M..........2.5*10^-5
5..........0.2M...........0.05M.......0.05M.......5.0*10^-5
^Figure 2^
A second researcher on the team draws an Energy diagram as shown in Figure 3 for an unrelated pair of similar reactions involving Reactant "R", abbreviated as follows:
R -----> P
R -----> P*
Both reactions are known to be exothermic and exergonic at STP.
Figure 3 - see attached image
View attachment 22480
Questions
1. According to the balanced overall reaction in figure 1, what can be inferred about the entropy of reaction and the Gibbs free energy of reaction at temperature T?
A. ΔS < 0 , ΔG < 0
B. ΔS < 0 , ΔG cannot be determined
C. ΔS > 0 , ΔG < 0
D. ΔS > 0 , ΔG > 0
For this question, use ΔG = ΔH - TΔS. Since ΔH is negative (given) and ΔS is negative (4 reactant molecules make 2 product molecules, less disorder), the sign of ΔG will depend on temperature. At high temperatures, the entropic term could overpower the enthalpic term and make ΔG positive. At low temperatures the enthalpic term wins and ΔG is negative.
2. The research team has evidence that step #2 in the proposed mechanism shown in Figure 1 is the slow step. What is the correct and experimentally useful rate law suggested by this mechanism?
A. Rate = k2/k-1 * [A]^2 * [C]^2
B. Rate = k2 * [C]^2
C. Rate = k1k2/k-1 * [A]^2 * [C]^2
D. Rate = k1k2k3 * [A]^2 * [C]^2 * [E]
Note: I had to use (B) instead of in the explanation below, because writing was messing with the bold formatting. Haha.
Since step #2 is the slow step, we start there with the answer choice "B". However, we are not finished because this would include the intermediate B in our rate law, which is nonsense since this is just a consumed intermediate and would not be measurable during the reaction. Therefore we need to substitute out (B), and to do this we use the first step (fast, reversible). We assume an equilibrium is quickly reached, that is, rate forward = rate backward. Therefore k1 * [A]^2 = k-1 * (B). Solving for (B) we get (B) = k1/k-1 * [A]^2. Plugging this in to our rate law from Step 2, we get our final answer (Choice C).
3. What is the experimental rate law the team derived from the data in figure 2?
A. Rate = k * [A][C]
B. Rate = k * [A] * [C]^2
C. Rate = k * [A]^2 * [C]^2
D. Rate = k * [A] * [C] * [E]
Using the data in the table, we quickly find that the rate is independent of [E]. Next, we see that doubling [A] simply doubles the rate, so the reaction is first order in A. Halving [C] we find the rate is one-quarter what it was before, so the reaction is 2nd order in C.
4. In the data in figure 2, experiment 1, what is most likely the measured rate of formation of G (what is the value of "x" )?
A. 0 M/s
B. 1.0*10^-4 M/s
C. 2.5*10^-5 M/s
D. Cannot be determined
This is a tricky question since at first glance we are convinced the reaction rate is independent of [E]. Therefore you would be tempted to pick answer choice B. However, realize that without any reactant "E", it would be impossible to form product "G". Since we are measuring not the disappearance of "A" and "E", but rather the rate of appearance of "G", we would not measure any "G" being formed.
5. What is the correct value of k in the experimental rate law from Figure 2?
A. 10^-2 M/s
B. 10^-2 M^-2
C. 10^-1 M/s
D. 10^-1 M^-2
Sorry I forgot to put seconds^-1 in the correct answer choice (my bad). This question requires you to have the rate law correct in question 3. The rate is in units of (M/s), and [A] in units of (M), and (B)^2 would therefore be in units (M)^2. To make the units cancel properly, the rate constant must have units M^-2 * s^-1. Sorry I messed that one up, we need seconds on the bottom somewhere in there 😛
6. In Figure 3, what could be a method for favoring the formation of P* over P?
A. The use of a catalyst to lower the activation energy
B. Changing the energy of the reactants
C. Increasing the temperature of the reaction
D. Decreasing the temperature of the reaction
A catalyst would not work since a catalyst will lower the energy of the transition state, but not change which product we get. This question is about thermodynamic vs kinetic control.
Since P is the more stable product, we consider it to be the thermodynamic product. The high activation energy barrier to get to P would mean a high temperature would favor formation of P.
Since P* is the kinetic product (lower activation energy barrier), it would form faster. At low temperatures, reactant would be less likely to climb the high barrier to P, and comparatively more likely to form P*.
7. What is true about the reactions indicated in Figure 3?
A. P* is thermodynamically favored and Δ(ΔG) is negative.
B. P is thermodynamically favored and Δ(ΔG) is positive.
C. P is thermodynamically favored and the sign of Δ(ΔG) has not been defined.
D. P* is thermodynamically favored and Δ(ΔG) is positive.
P is lower in energy than P*, and so it is thermodynamically favored. Δ(ΔG) on the energy diagram is marked as the difference in energy between P and P*, however it does not tell us which order to subtract them. The sign is not important, what is important is to realize P* is higher than P.
Not sure what I can tell you regarding your courseload and mcat, just do what you can to have the right time to prep. I think if physics is your worst section and you havent worked any problems in a while, then it is ok to rework some of your older material just to have the concepts and equations fresh again. it isnt like verbal where you MUST practice new material to make progress. speaking of verbal, my suggestion is to make sure you are FRESH AND NOT TIRED when doing these, otherwise you might not make the gains you are hoping for. It really does require solid attention and concentration to improve, just my opinion.I'm taking biochemistry
I haven't seen any Kaplan material, but know a lot of people don't like BR. I have EK 101 which I haven't started yet and have been alternating between PR and BR. My BR verbal has been getting progressively worse, but PR hasn't been too bad so I'm wondering if I should be worried or not.
NED I've been wondering. What's your study plan since you're not doing SN2? I need to reconfigure mines so i wanted to see what others are doing that are non SN2 plans
That's awesome! Small victories quite often build up into bigger ones. I've grasped a number of Chemistry and some Physics passages that I couldn't for last year's exam although my bio is struggling which is odd. I can't wait to finish content review and see how my exam scores go. I really hope it doesn't follow the same pattern as last year.
Nutty,^ that was directed towards you. Apparently my phone forgot how to quote.
How far does everyone have until Content Review is done?
How far does everyone have until Content Review is done?
Done with content review 🙂 Just have to brush up on whatever weak points are revealed in passages and full length practice exams. Also, I still have lots of verbal practice needed.How far does everyone have until Content Review is done?
Uh.... no! There are 7 weeks to go! 😛#8weekstogo
Question of the day!! haha jk but anyway...
What is the difference between a coulombic charge and potential? I cant seem to grasp this important concept. Plus I have a test on wednesday lol
Thanks guys
For those of you doing the Sn2ed schedule, do you guys literally read over all the chapters on that review day?
How far does everyone have until Content Review is done?
just out of curiosity, but for those who are taking classes + studying for the MCAT this semester, how many hours do you put in per day? Or maybe it's better to ask, how do you fit in your mcat studying during the school days?
I'm also struggling with finding time to devote to mcat study while taking classes. By the time I get all the crap done for my other classes I'm pretty beat and have a hard time focusing. Anyone have any other tips for this?
kind of in the same boat. i haven't been very good about studying but i find after work studying is best done with the coffee.
haha I'm the say way, but with Mtn Dew. I'm pretty sure the stuff's flowing through my arteries/veins at this point
that's no good. among other problems that might result the carbonation in the mountain dew will cause an increased level of blood co2 leading to decreased blood ph from carbonic anhydrase activity, which will result in a decreased hb affinity for oxygen. symptoms may include lethargy with prolonged exposure. also bubbles in the blood might cause some issues. furthermore, intravenous high fructose corn syrup will result in severe metabolic issues.
tldr if you need to take soda intravenously make sure it's flat and uses real sugar. may i suggest Mexican Coca cola, or perhaps some Texas Dr Pepper?
For those of you who outline each chapter in great detail and use BR Biology, how did you outline the metabolic pathways chapter? I can't imagine we're expected to memorize each step in these pathways.
I take notes in great detail but I use EK Bio supplemented with selected portions of TBR Bio. The Metabolic Pathways chapter is insane. Know the purpose of the major pathways and the starting substrates and products.