Dismiss Notice
Visit Interview Feedback to view and submit interview information.

Solutions and exothermic heats of solution

Discussion in 'MCAT Study Question Q&A' started by rskhan29, Jun 18, 2008.

  1. rskhan29

    rskhan29 Junior Member
    7+ Year Member

    Joined:
    Jul 6, 2006
    Messages:
    42
    Likes Received:
    0
    Status:
    Medical Student
    Hey guys,
    Does an exothermic heat of solution raise the temperate of the solution? I don't understand why it would-the system is the solvent and solute, so it should lose heat.
    And does an exothermic heat of solution result in increased velocity of the particles in the fluid?

    According to an examkrackers question, the answer to both my questions is yes. Apparently temperature and the velocity of molecules increases in a solution with an exothermic heat of solution. Does anyone know why?
     
  2. Note: SDN Members do not see this ad.

  3. Kaustikos

    Kaustikos Archerize It
    7+ Year Member

    Joined:
    Jan 18, 2008
    Messages:
    12,216
    Likes Received:
    4,155
    Okay, here's the way I understood this concept. If you add solute to a solvent and they mix together and the end-product is in a more stable, lower energy state; then the product has released energy (exothermic) this energy causes an increase in temperature of the solution. This "lost heat" is what causes the temperature to increase.

    For the second part; my only reasoning is that since you're increasing the temperature, you HAVE to be increasing the velocity of the molecules. Higher temperature = higher KE = higher velocity.

    Now, take for example an endothermic reaction. This is one where the product is formed but requires energy absorption in order to happen. The resulting product would have absorbed energy to get to its stable state and caused an overall decrease in temperature from the original temperature of the solute/solvent independently. The energy of the solution, so to speak, went into making this product.

    If the summation of energy from the reaction (energy of bond breaking/making) results in a positive value, it means the solution required a lot more energy to break the original bonds and than was released from the bonds formed between the solute and solvent.
    If the summation of energy from the reaction results in a negative value, then you know that the bonds formed released more energy than the energy required to break the original bonds.

    Hope that helps, though I don't know for sure if this is what you were talking about.
     
  4. engineeredout

    10+ Year Member

    Joined:
    May 11, 2008
    Messages:
    3,417
    Likes Received:
    570
    Status:
    Fellow [Any Field]
    Maybe you're getting hung up on the fact that the delta-Q is negative for an exothermic reaction even though heat is being released to the surroundings?

    Energy is required (not released) when bonds are broken. Energy is released when bonds are formed. If you're forming bonds that are stronger then the original bonds, then the system is going to release the heat of the original bonds and then some on top of it. This is an exothermic reaction. Opposite for endothermic.

    And yeah like the guy above said, increase the temperature of a system, increase the kinetic energy. Temperature is a measure of the average kinetic energy of a system.
     
  5. rskhan29

    rskhan29 Junior Member
    7+ Year Member

    Joined:
    Jul 6, 2006
    Messages:
    42
    Likes Received:
    0
    Status:
    Medical Student
    Thanks for the replies,
    But I do understand what you've explained. What I don't understand is why a release of energy, via heat during an exothermic reaction, would increase the kinetic energy and and thus temperature of the solution. Shouldn't the temperature of the surroundings increase, but not the solution itself?
     
  6. Kaustikos

    Kaustikos Archerize It
    7+ Year Member

    Joined:
    Jan 18, 2008
    Messages:
    12,216
    Likes Received:
    4,155
    The surrounding is the solution. If you're asking why the heat doesn't leave the solution and go into the air, that's because the energy released is contained in the solution. The molecules are releasing the energy into their immediate surrounding, which is the liquid. In the long run, though, yes you could have the container absorb the energy and the solution lower in temperature. But, the solution will still increase in temperature initially.
    If you want to test this out/find an example - try taking sample of concentrated sulfuric acid (5N) and diluting it with millieq water.
     
  7. unsung

    10+ Year Member

    Joined:
    Mar 12, 2007
    Messages:
    1,356
    Likes Received:
    13
    Status:
    Resident [Any Field]
    Remember, the heat of the solution refers to the sum of the change in enthalpies of these 3 steps:

    1. break solute-solute intermolecular bonds (endothermic)
    2. break solvent-solvent intermolecular bonds (endothermic)
    3. form solute-solvent intermolecular bonds
    (exothermic)

    So an overall exothermic heat of solution means the exothermic step #3 outweighs the endothermic first 2 steps. I.e. the bonds that form are stronger than the bonds that you start out with.

    As for why the solution itself experiences an increase in temperature, the answer is really that the solution is like the "surroundings" that the reaction occurs in. So if the reaction is exothermic, it releases the heat into the surroundings (the solution), so the solution increases in temperature.

    This is more clear if you think about a coffee cup calorimeter, where the whole reason that setup works to measure the heat of solution of the reaction, is that whatever heat released by the reaction goes into changing the temperature of the solution. The heat doesn't just get out into the air (ideally) somehow and bypass increasing the temperature of the solution. Similarly, if the reaction is endothermic, where is it going to remove the heat from? It's going to grab the heat out of the solution that it's in. Thus, the temperature of the solution will be found to have decreased. The solution is the surroundings of the reaction.

    Does that help?
     
  8. rskhan29

    rskhan29 Junior Member
    7+ Year Member

    Joined:
    Jul 6, 2006
    Messages:
    42
    Likes Received:
    0
    Status:
    Medical Student
    Okay, thanks guys. That was helpful.
     

Share This Page