Now that 8 playing cards have been illustrated, the dealing game will be extended to 48
playing cards. To start the illustration, 24 red cards and 24 black cards will be used to calculate
the multiplicity, the equilibrium constants, and the probability of the red and black cards mixing
together in two card piles; RR, BB, and RB.
Exercise 1: The maximum number of configurations in a 24 red and 24 black shuffle
There are 3224760368100 possible dealings.
Exercise 2:
Some possible outcomes of 48 cards being dealt
A) 12 RR two card piles + 12 BB two card piles + 0 RB two card piles
B) 0 RR two card piles + 12 BB two card piles + 12 RB two card piles
C) 12 RR two card piles + 0 BB two card piles + 12 RB two card piles
D1) 4 RR two card piles + 12 BB two card piles + 8 RB two card piles
E) 6 RR two card piles + 12 BB two card piles + 6 RB two card piles
F) 7 RR two card piles + 6 BB two card piles + 11 RB two card piles
Exercise 3: The calculation of the multiplicity of the above example configurations.
All multiplicities for each pile dealing was calculated using the formula
(2^n)(number of 2 card piles)!/[(RR pile)!(BB pile)!(RB)!] where is the largest number of card
piles in either RR, BB, or RB.
In this 48 total card exercise the 6 RR, 6 BB, and 12 RB dealing led the largest multiplicity of
10234430029824. In the 8 total cards exercise the 1 RR, 1 BB, and 2 RB dealing led the largest
multiplicity of 48. Therefore, the equilibrium of nRR + nBB + 2nRB leads the largest
multiplicity.
Exercise 4: The determination of the equilibrium constants of the card dealings.
one possible outcome:
8 RR, 8 BB, 8 RB
When using the relationship of RR + BB  2RB the equilibrium constant, Keq, is found using
the idea of equilibrium:
Keq= [RB]^2/([RR][BB])
So:
Other possible outcomes
A) 12 RR two card piles + 12 BB two card piles + 0 RB two card piles
B) 0 RR two card piles + 12 BB two card piles + 12 RB two card piles
C) 12 RR two card piles + 0 BB two card piles + 12 RB two card piles
D1) 4 RR two card piles + 12 BB two card piles + 8 RB two card piles
E) 6 RR two card piles + 6 BB two card piles + 12 RB two card piles
F) 7 RR two card piles + 6 BB two card piles + 11 RB two card piles
Error, numeric exception: division by zero
Error, numeric exception: division by zero
As was observed in the 8 total card dealings, some outcomes are impossible when considered
under equilibrium in 48 total card dealings. For an equilibrium to occur in a reaction, some
amount of reactant is needed to make some amount of product. Therefore, RR=0 or BB=0
outcomes cannot occur in equilibrium and RB = 0 cause an equilibrium constant of 0.
Exercise 5:
The possible outcome in exercise 4 above answers the question and shows that the equilibrium
constant is the reaction quotient. Therefore, when the multiplicity is the highest the system is at
equilibrium. So Q=K=4
Most probable outcome:
6 RR, 6 BB, 12 RB lead a reaction quotient of:
The work done below is conducted to determine the probability of the outcome that leads the
largest equilibrium constant of 4.
Exercise 1:
There are 3224760368100 possible dealings.
Exercise 2:
Some possible outcomes of 48 cards
A) 12 RR two card piles + 12 BB two card piles + 0 RB two card piles
B) 0 RR two card piles + 12 BB two card piles + 12 RB two card piles
C) 12 RR two card piles + 0 BB two card piles + 12 RB two card piles
D1) 4 RR two card piles + 12 BB two card piles + 8 RB two card piles
E) 6 RR two card piles + 12 BB two card piles + 6 RB two card piles
F) 7 RR two card piles + 6 BB two card piles + 11 RB two card piles
Exercise 3:
All multiplicities for each pile dealing was calculated using the formula
(2^n)(number of 2 card piles)!/[(RR pile)!(BB pile)!(RB)!] where is the largest number of card
piles in either RR, BB, or RB.
In this 48 total card exercise the 6 RR, 6 BB, and 12 RB dealing led the largest multiplicity of
10234430029824. In the 8 total cards exercise the 1 RR, 1 BB, and 2 RB dealing led the largest
multiplicity of 48. Therefore, the equilibirum of nRR + nBB + 2nRB leads the largest
multiplicity.
Error, missing operator or `;`
Exercise 4:
one possible outcome:
8 RR, 8 BB, 8 RB
Other possible outcomes
A) 12 RR two card piles + 12 BB two card piles + 0 RB two card piles
B) 0 RR two card piles + 12 BB two card piles + 12 RB two card piles
C) 12 RR two card piles + 0 BB two card piles + 12 RB two card piles
D1) 4 RR two card piles + 12 BB two card piles + 8 RB two card piles
E) 6 RR two card piles + 6 BB two card piles + 12 RB two card piles
F) 7 RR two card piles + 6 BB two card piles + 11 RB two card piles
Error, numeric exception: division by zero
Error, numeric exception: division by zero
As was observed in the 8 total card dealings, some outcomes are impossible when considered
under equilibrium in 48 total card dealings. For an equilibrium to occur in a reaction, some
amount of reactant is needed to make some amount of product. Therefore, RR=0 or BB=0
outcomes cannot occur in equilibrium and RB = 0 cause an equilibrium constant of 0.
Exercise 5:
The possible outcome in exercise 4 above answers the question and shows that the equilibrium
constant is the reaction quotient. Therefore, when the multiplicity is the highest the system is at
equilibrium. So Q=K=4
Most probable outcome:
6 RR, 6 BB, 12 RB lead a reaction quotient of:
The work done below is conducted to determine the probability of the outcome that leads the
largest equilibrium constant of 4.
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CardsToT24 is a distribution of 24 red cards and 24 black cards
In this above table, the first column is the card piles of RR, the second column is the card piles of
BB, the third column is the number of card piles of RB, the fourth column is the probability of
getting dealt that combination of card piles, and the fifth column is the equilibrium constant for
that card pile distribution. The probability of getting dealt a distribution that has an equilibrium
constant of 4 is 0.31737025.
The equilibrium constant for the isotope exchange H2+D2  2HD at 25 degrees C and 500
degrees C are 3.3 and 3.8, respectively. The above work shows that playing cards cannot be used
to determine precise equilibrium constant values. However, the playing card illustration does
allow valuable insight into the to determine the probability and approximate equilibrium
constants of distributions.