The Equilibrium Constant as a Ratio of Probabilities

     Given the balanced chemical equation of a reaction in equilibrium aU + bV <-> cY + dZ, we define as the forward reaction aU + bV -> cY + dZ with reactants U,V and products Y, Z and the reverse reaction as cY + dZ -> aU + bV with reactants Y, Z and products U, V. The coefficients a, b, c, d represent the relative amounts (in moles) of the substances U, V, Y, Z needed for the reaction. That is, a moles of U and b moles of V react to form c moles of Y and d moles of Z. [1 mole is a quantity (6.022 x 10^23) of things just like 1 dozen is a quantity (12) of things]
     By definition, equilibrium is when the rate of the forward reaction equals the rate of the reverse reaction. So when a reaction is in equilibrium, the concentrations of the substances (reactants and products) remains constant even as the reactions continue happening.
     If we add (or subtract) the amount of one (or more) of the substances, the reaction will no longer be in equilibrium. The reactions will continue at different rates until equilibrium is again reached at the same equilibrium concentrations as before.
     For example, if we add substance U to the mixture the increase in the concentration of U will cause an increase in the rate of the forward reaction (more U-molecules reacting with the available V-molecules) tending to increase the product concentrations [Y] and [Z] while decreasing the reactant concentrations [U] and [V]. The product concentrations will increase until the reverse rate of reaction matches the forward rate. Note the forward rate will decrease as the concentrations of [U] and [V] decrease.
      A reaction depends on the right molecules interacting with each other with the right kinetic energy to cause the necessary reaction-causing collisions to effect the redistribution of electrons amongst the constituent molecules. The ambient temperature is a measure of the average kinetic energy of the molecules and the actual molecules have a kinetic energy distribution which we can take as a probability density function about this average. Thus some molecules have higher than average KE and some lower. In particular, the higher the temperature the more higher energy collisions. At the same time, the concentrations of the constituent molecules is also directly related to the rate of reaction-causing collisions. The more molecules we have per volume, the more likely they will collide.
     A mathematical description of the discussion above follows. The rate of a reaction
aU + bV -> cY + dZ is governed by the probability of getting effective collisions between reacting molecules. For example, in the forward reaction we need to get a molecules of U together with b molecules of V in a small enough volume dW and with sufficient kinetic energy to make effective collisions. Assuming for now that the temperature T allows some number of such collisions (the number then being proportional to the temperature) the other factor is the concentrations [U] and [V] of the reactants. The probability of getting a U molecule in a volume dW is directly proportional to [U] and the same for V molecules. Thus the probability of getting the required number of molecules in dW for a reaction to occur is proportional to [U]^a [V]^b. Let P(T) be the probability of a reaction occurring at temperature T given the molecules are within the reaction volume dW. Then the overall probability of a reaction is proportional to the product [U]^a [V]^b x P(T). Since P(T) is fixed for the reaction we have that the probability of an effective collision is directly proportional to [U]^a [V]^b. Thus the rate of the forward reaction is directly proportional to [U]^a [V]^b. Similarly, the rate of the reverse reaction is directly proportional to [Y]^c [Z]^d. Since at equilibrium these rates are equal, we know that the ratio [U]^a [V]^b/[Y]^c [Z]^d must be equal to some constant for a given temperature T. This equilibrium constant is a property of the chemical reaction aU + bV <-> cY + dZ and can be used to predict outcomes of experiments.

The Khan Academy, Pixar and Parabolas

     The Khan Academy is a treasure which I recommend to everyone. Recently they added a short course called Pixar in a Box (link at bottom of entry) in which the first section describes how parabolic arcs are used to make the incredible fields of grass or fur on bears or hairs on heads, etc. It is a fun read and in about a half hour you will get the idea of how these things are done. However, I was left a bit dissatisfied about the explanation of why the construction used really defines a parabolic arc and so I worked out a proof that satisfied myself and perhaps might be of interest to others.
   
     The idea is that if you choose any three distinct points A, B, and C, and divide segments BA and BC into n equal subdivisions each, and connect the dividing points in reverse order, the envelope of the drawn segments will form a polygon and as n increases to infinity, the polygon will approach a parabolic arc. The Khan Academy lesson shows this in pictures which I recommend viewing.

  
     What I will do here is prove the general statement of this property.

    Statement to be Proved: Given any three distinct points A, B, C, define the sets of points 
Q(t) = (1-t)A + tB, R(t) = (1-t)B + tC, for t in [0,1]. The set of points 
P(t) = (1-t)Q(t) + tR(t) for t in [0,1] define a parabolic arc.

Proof: We know that we have a parabolic arc if we can show that there is a point F and a line L such that for every point P in the set, the distance from P to F equals the distance from P to the line L. 

Define the two-dimensional rectilinear coordinate system {A,B,C} as the system with origin B, a “vertical” v-axis being the line through B and the midpoint M of segment AC where M is on the positive half of the axis, and a “horizontal” u-axis being the perpendicular to line BM through B such that the unit positive u-vector is the cross product of the unit positive v-vector and the unit vector k pointed upward from the uv-plane.

In {A,B,C}, let the points have coordinates A(a₁, a₂), C(c₁, c₂), B(0, 0), 

M( (a₁₊ c₁)/2, (a₂₊ c₂)/2 ) where by definition (a₁₊ c₁)/2 = 0, so c₁ = -a₁ .

Then P(t) = (1-t)Q(t) + tR(t) = (1-t)[(1-t)A + tB] + t[(1-t)B + tC]

= t² (A-2B+C) +2t(B-A) + A, which we can write componentwise as the pairs:

P(t) = [t² (a₁ - 2b₁ + c₁) +2t(b₁ - a₁) + a₁) , t² (a₂ - 2b₂ + c₂) +2t(b₂ - a₂) + a₂) ] 

= [ a₁(1-2t) , t² (a₂ + c₂) + a₂(1-2t) ] for t in [0,1].

Let u = a₁(1-2t) so t = (a₁ - u)/2a₁ and P(u) = [ u , au² + bu + c] for u in [a₁, -a₁] where 

a = (a₂ + c₂)/4a₁² , b = (a₂ - c₂)/2a₁ , c = a a₁² . This shows P(u) forms a parabolic arc (i.e. a quadratic in the {A,B,C} coordinate system).

In particular, we know that for a parabola v = au² + bu + c ,

the vertex V is ( -b/2a , c - b²/4a), the focus F is V+(0, p), and the directrix L is the line v = v₂ - p, where p = 1/(4a). Thus we have proved the statement.

Example: Find an equation for the parabolic arc through A(3,9) and C(2,4) as defined by P(t) above with B(5/2,6).

Let the original rectilinear coordinate sytem which defines A, B, and C for the problem be called the standard xy-coordinate system. In this system the midpoint of segment AC is M(5/2, 13/2). We define the {A,B,C} rectilinear coordinate system as that having origin B(0,0), “vertical” axis as the line through B and M, and the “horizontal” axis as the perpendicular to line BM through B. In this case the “vertical” axis is vertical with respect to the standard coordinates because B and M have the same first coordinate in that system. This tells us that there is no rotation involved in the change of coordinates. Therefore the change of coordinate systems involves only a translation sending point (x,y) in the standard system to the point (x-5/2, y-6) in the new system. 

Thus A(3,9) becomes A(1/2, 3), B(5/2,6) becomes B(0,0), and C(2,4) becomes C(-1/2,-2). Using these translated coordinates, we can write, using the formulas derived in the proof above, P(t)  = [(1/2 - t), t²  - 6t + 3], for t in [0,1] so letting u = (1/2 - t), get P(u) = [ u , u² + 5u + 1/4], for u in [1/2, -1/2]. 

Thus p=1/4, V(-5/2,-6), F(-5/2, -6 +1/4), L is v= -6 - 1/4. These values can be translated back to the standard coordinate system by adding (5/2,6) to get the values for the desired focus and directrix given the original points as F(0, 1/4) and y= -1/4 as well as the vertex (0,0) which of course correspond to the parabola y = x².

More work would be involved in an example where the line through B and M was rotated with respect to the standard y-axis but the theory is the same and of course the proof does not depend on the particulars of the orientation of the {A,B,C} system.

https://www.khanacademy.org/partner-content/pixar/environment-modeling-2/mathematics-of-parabolas2-ver2/a/parabolas-lesson-brief