1. The SDN iPhone App is back and free through November! Get it today and please post a review on the App Store!
    Dismiss Notice
  2. Hey Guest! Check out the 3 MCAT Study Plan Options listed in the 'stickies' area at the top of the forums (BoomBoom, SN2ed, and MCATJelly). Let us know which you like best.

    Also, we now offer a MCAT Test-Prep Exhibitions Forum where you can ask questions directly from the test-prep services.
    Dismiss Notice
  3. Dismiss Notice
Dismiss Notice

Interview Feedback: Visit Interview Feedback to view and submit interview information.

Interviewing Masterclass: Free masterclass on interviewing from SDN and Medical College of Georgia

help me with some aamc practice items!!!

Discussion in 'MCAT: Medical College Admissions Test' started by phatty925, Aug 14, 2002.

  1. phatty925

    phatty925 Senior Member
    7+ Year Member

    Apr 23, 2002
    Likes Received:
    my questions come from passage 17 of the AAMC Physical sciences practice items:

    #112- does Keq never change for a rx since it is a constant? i got confused b/c in #113, it says that a higher Keq would mean a higher yield. but i guess a higher yield does not mean a higher Keq???

    #114- how do you go about figuring this one out? isn't something oxidized if the ratio of oxygens to hydrogens increases?

    #116- how come the answer here is temperature? how come B is wrong?

  2. Note: SDN Members do not see this ad.

  3. dark_scientist

    10+ Year Member

    May 3, 2002
    Likes Received:
    Resident [Any Field]
    I'll take a go at your questions:

    For #113, the question is giving you a hypothetical situation in which you are required to relate two different Keq values. The higher the Keq value, the greater the ratio of products to reactants - thus, a higher yield. For #112 (which I also got wrong) I believe that the Keq value is constant for any particular reaction unless something like the temp is increased.

    For #114, during the oxidation of ethanol, the initial product will be an aldehyde (the oxidation steps are from an alcohol to an aldehyde to a carboxylic acid). The only aldehyde they show is B.

    For #116, B is wrong because if the concentration of reactants changes, the concentration of products will proportionately change to achieve the same equilibrium - the Keq remains constant. However, temperature will effectively alter the equilibrium (I believe according to LeChatlier's principle) to give you a different equilibrium state and thus a different Keq.
  4. Mudd

    Mudd Charlatan & Trouble Maker
    7+ Year Member

    Jun 5, 2002
    Likes Received:
    The passage states that the Keq is 4 at 25C. No matter what ratio of reactants you start with, the Keq is going to still be 4, unless the temperature changes. Keq only changes with temperature. This explains both 112 and 116. The exact values may differ and a reaction can shift to reestablish equilibrium after a perturbation, but if T remains constant, so does Keq.

    For 114, this is an orgo question in the midst of a gen chem passage. As a general rule, primary alcohols oxidize to an aldehyde (lose one H and gain one bond to O), then a carboxylic acid (lose another H and gain another bond to O), and then if it is complete oxidation it eventually forms carbon dioxide, CO2.

    This passage deals with the conversion of a primary alcohol to a carboxylic acid, so an aldehyde is the intermeidate.

Share This Page