Chemistry Things
Unit 8...
Unit 7b...
"I Love Lucy" teaches Kinetics
A very important concept in chemical kinetics is the "rate determining step". It is easy to spot in many situations - like here with Lucy and Ethel (although, we chemist prefer "Ethyl"). The overall rate law is governed by the slowest step in any multistep reaction scheme.
Dr. McCord makes a "how to" video
I've been thinking about making more of these "how to" videos like we have in the eBook. However, I was trying to give them a bit of a new look. Have a look and let me know what you think. I am aware of a few minor issues, but overall you can see how to work a first order integrated rate law problem. - Dr. McCord
First Order Kinetics Integrated Rate Law Problem
Unit 7a...
Decay directions on Stability Diagram
The colored arrows on the diagram represent the transitions for beta decay (yellow), positron emission (pink), and alpha decay (green).
The beta decay and positron emission decays always head directly towards the center black boxes which are the stable nuclides. The alpha decays just track back down and will often land on a nuclide that then undergoes positron emission.
The portion of the band of stability is from the diagram in our eBook. Notice how both beta decay and positron emission move 1 diagonal. The alpha decays are all 2 diagonals.
Unit 6...
Protonated vs Deprotonated states
"I'm a protonated state because I HAVE the proton." You might know me best as a plain acid, HA. But remember, I can also be a protonated base, BH+. It doesn't matter which version I am, I still behave as an acid and I donate this proton to a base. My acidic strength is measured by my value for Ka.
"I'm a deprotonated state because I don't have the proton." You probably know me best as a plain base, B. I also exist as a deprotonated acid, A-. I will always behave as a base because I am ready to accept a proton from an acid. My base strength is measured by my value for Kb.
If the only difference between me and my buddy there is that one single proton, then we are a conjugate acid/base pair and have a special relationship which is that: Ka· Kb = Kw
Four Problems Solved the same way...
SO MANY Problems in acid/base theory are answered via the following generic set up.
\[K = {x^2\over C - x}\]
K will either be Ka or Kb and you will typically solve for x and it will be either [H+] (acid problems) or [OH-] (base problems). You then take the log of x and you'll have either pH or pOH.
IF K is small enough (say < 10-4), then the following approximation is valid:
\[K \approx {x^2\over C}\]
and easily solved without using the quadratic formula...
\[x = \sqrt{K\cdot C}\]
Example wordings of questions that ultimately solve like shown above
- What is the pH of a 0.??? M solution of ?????
- What is the percent ionization of ???? in a solution of 0.??? M ?????
- What percentage is protonated in...
- What percentage is deprotonated in...
- Given that a solution of 0.??? M of ???? has a pH of ??.??, what is the value of K?
mass action expression (concentrations)
The mass action expression is an equation that shows the ratio of product concentration states to reactant concentration states. The expression itself is very easy to correctly write as long as you have a balanced chemical equation to reference. So let's consider the following generic reaction in aqueous solution. Each of the lowercase letters represents the coefficient for that species.
\[a\,{\rm A(aq)} + b\,{\rm B(aq)} \rightleftharpoons c\,{\rm C(aq)} + d\,{\rm D(aq)}\]
mass action expression : \(\displaystyle{[{\rm C}]^c[{\rm D}]^d\over[{\rm A}]^a[{\rm B}]^b}\)
Square brackets, [ ], around a species always represents the molar concentration (mol/L) of that species in solution.
This equation is easily written in a formalized way - just put all the product concentrations in the numerator (on top) and all the reactant concentrations in the demoninator (on bottom). The coefficients all become powers to the associated concentration term.
NEWS FLASH!
Pure solids and pure liquids have no meaningful concentration term. Their activities are always exactly 1. This means that they are NEVER part of a mass action expression or equilibrium expression. LEAVE THEM OUT!!
Concentrations only make sense for dissolved species (think "aqueous" phase) or for gases (think pressure, which is really a concentration term for gases).
mass action expression (pressures)
The mass action expression can also be expressed with pressures instead of concentrations. This is of course the usual way of doing things for gaseous reactants and products. Consider once again the generic reaction like before except this time all the species are gases.
\[a\,{\rm A(g)} + b\,{\rm B(g)} \rightleftharpoons c\,{\rm C(g)} + d\,{\rm D(g)}\]
mass action expression : \(\displaystyle{P_{\rm C}^{\,c}P_{\rm D}^{\,d}\over P_{\rm A}^{\,a}P_{\rm B}^{\,b}}\)
Same idea as with the concentration version except now all the terms are partial pressures of each of the species.
mass action expression (mixed terms)
The mass action expression can have mixed terms - both concentrations (solution species) and pressures (gas species). Here is a real example from electrochemistry where hydrogen ions are reduced to hydrogen gas. This is a 1/2 reaction, which is why electrons are included in the reaction but not the mass action expression.
\[2{\rm H^+(aq)} + 2e^- \rightleftharpoons {\rm H_2(g)}\]
mass action expression : \(\displaystyle{P_{\rm H_2}\over [{\rm H^+}]^2}\)
Notice that partial pressure is used for the gas and concentration is used for the solution species.
mass action expression (activities)
OK, We are REALLY Supposed to be using ACTIVITIES!
OK, so everything previous to this was really just an approximation of the real mass action expression. The true thermodynamic mass action expression is based on activities, and not really "plain" concentrations and pressures.
\[a\,{\rm A(aq)} + b\,{\rm B(aq)} \rightleftharpoons c\,{\rm C(aq)} + d\,{\rm D(aq)}\]
true mass action expression : \( \displaystyle{ {a_{\rm C}^c \cdot a_{\rm D}^d} \over {a_{\rm A}^a \cdot a_{\rm B}^b} }\)
Of course activities are really just the values of the concentrations and the pressures ratioed to a standard concentration and pressure which happen to be 1 M for concentration and 1 atm for pressure.
reaction quotient, Q
The reaction quotient is symbolized with the capital letter Q. Q is simply the numerical value for the mass action expression under any stated set of conditions (concentrations and/or pressures). So Q can essentially be any number between \(0\) and \(\infty\). There is no such thing as a negative concentration or pressure which is why you cannot have negative values of Q.
Take all the concentrations and pressures of all the species in the system and put them into the mass action expression. Then calculate the number - that is Q! It will range from infinitely small (zero, or approaching zero) OR infinitely large (approaching \(\infty\)). Students should THINK of the pivot point within this range as a 1 (one). Generally, when the mass action express calculates to greater than one, then this is a situation where the system is more product favored. Values smaller than one generally mean that the system is more reactant favored.
The value of Q is what is important because you will use that value to determine in which way the system will respond so as to reach the equilibrium state.
IMPORTANT! Q can be all possible values
Q can be every possible value between zero and infinity because you can pick any value you want to go into the mass action expression. You will usually pick the real concentrations and/or pressures that are given in a problem though.
Q is a single number that represents
the current state of the system.
the equilibrium constant, K
The equilibrium constant is symbolized with the capital letter K. Unlike Q, K is a very specific value for a specific reaction under specific conditions. K is a recasting of ΔG° for the reaction.
K is also equal to the mass action expression just like Q, except that all the concentrations and pressures must be their equilibrium values! So think of K as a subset of Q at which equilibrium occurs. There will be lots and lots of possible equilibrium values for a given reaction - just like there are lots of solutions to calculating various points that lie on a straight line. If you know the formula for the straight line (\(y=mx+b\)), then you can generate an infinite number of matched (x,y) coordinates that fall on that line.
A given reaction under specific conditions (lets say standard conditions for reference purposes) will have a very specific value for \(\Delta G^\circ_{\rm rxn}\). The magnitude and the sign of that standard free energy change tells us about the reaction being either strongly or weakly (big kJ or little kJ) spontaneous or non-spontaneous (positive or negative) as written. The equilibrium constant is a recasting of standard free energy. And by recasting I mean that there is an equation that relates the two values. What is that equation?
\[\Delta G^\circ = -RT\ln K\]
now flip that equation
\[K = e^{-\Delta G^\circ\over RT}\]
So the secret is out... K is just a mathematical conversion of ΔG°.
Unit 5...
- no extra content -