Wednesday, January 24, 2007
Wednesday, Day 2
Honors: we discussed the physical properties of vapor pressure and boiling point as related to intermolecular forces of attraction. The STRONGER the intermolecular forces of attraction (IMFA), the lower the fraction of molecules that can overcome their intermolecular attractions at a given temperature (avg. KE of molecules in sample), therefore the lower the vapor pressure. Also, the STRONGER the IMFA, the higher the temperature required to get the substance to have a vapor pressure equal to that of the opposing atmospheric pressure, therefore, the higher the boiling point (temperature) of the substance.
Tomorrow, we will finish our discussion of IMFA as related to melting points and solubility; then we will briefly discuss/review the types of solids and attractions OR bonds within the solid lattice of particles. We then MUST begin the math of chem unit, so I will at least give you a brief introduction to how to relate various quantities in a chemical reaction/equation.
Regents: we focused on intermolecular forces of attraction (IMFA) and gave examples of substances (and, for some, mixtures) that exhibit a given type of IMFA. It is important to be able to know, just by drawing out the Lewis structure of a given molecule, what type of IMFA that the molecule has and also how relatively strong that IMFA is. We will further discuss how the IMFA's relate to the various physical properties of substances.
AP: we finished deriving the time-dependent rate law and the the half-life equation for a second order reaction. We then did the same for a zeroth order reaction. We then did a few examples involving the second order rate equation.
Note: the raw data of concentration vs. time will not yield constant "half-life" time intervals for a second order reaction; only first order reactions half constant half-lives that are independent of reactant concentration.
It is important to relate the equations to the expected graphs, to know what the slopes indicate, and to be able to calculate the rate constant, WITH UNITS, from the graphs.
We talked a bit about collision theory, which will lead to the Arrhenius equation that relates a rate constant to the activation energy requirement for a given reaction.
NOTE: I did misspeak at the very end of class; I SHOULD HAVE SAID that the Arrhenius equation shows that THE LOWER the activation energy, THE HIGHER (faster) the rate constant (that is accounted for by the negative sign in the equation:
k= A e^(-Ea/RT)... also, from the last test, I noticed that not everyone uses a small (lowercase) k for the rate constant; that is VERY BAD because BIG K is RESERVED for the "equilibrium constant" ONLY. Be careful with that.
Thanks.
p.s. I don't think you all know how good you are at descriptive chem; on the last test, all but one of you got almost ALL of them right!
With a little more experience, question 4 may net you the easiest 15 points on the AP exam in the shortest amount of time. Consider that the national AVERAGE score on that question is 5 out of 15 !!!
Tomorrow, we will finish our discussion of IMFA as related to melting points and solubility; then we will briefly discuss/review the types of solids and attractions OR bonds within the solid lattice of particles. We then MUST begin the math of chem unit, so I will at least give you a brief introduction to how to relate various quantities in a chemical reaction/equation.
Regents: we focused on intermolecular forces of attraction (IMFA) and gave examples of substances (and, for some, mixtures) that exhibit a given type of IMFA. It is important to be able to know, just by drawing out the Lewis structure of a given molecule, what type of IMFA that the molecule has and also how relatively strong that IMFA is. We will further discuss how the IMFA's relate to the various physical properties of substances.
AP: we finished deriving the time-dependent rate law and the the half-life equation for a second order reaction. We then did the same for a zeroth order reaction. We then did a few examples involving the second order rate equation.
Note: the raw data of concentration vs. time will not yield constant "half-life" time intervals for a second order reaction; only first order reactions half constant half-lives that are independent of reactant concentration.
It is important to relate the equations to the expected graphs, to know what the slopes indicate, and to be able to calculate the rate constant, WITH UNITS, from the graphs.
We talked a bit about collision theory, which will lead to the Arrhenius equation that relates a rate constant to the activation energy requirement for a given reaction.
NOTE: I did misspeak at the very end of class; I SHOULD HAVE SAID that the Arrhenius equation shows that THE LOWER the activation energy, THE HIGHER (faster) the rate constant (that is accounted for by the negative sign in the equation:
k= A e^(-Ea/RT)... also, from the last test, I noticed that not everyone uses a small (lowercase) k for the rate constant; that is VERY BAD because BIG K is RESERVED for the "equilibrium constant" ONLY. Be careful with that.
Thanks.
p.s. I don't think you all know how good you are at descriptive chem; on the last test, all but one of you got almost ALL of them right!
With a little more experience, question 4 may net you the easiest 15 points on the AP exam in the shortest amount of time. Consider that the national AVERAGE score on that question is 5 out of 15 !!!
# posted by C @ 5:20 PM 
