# Calculate the Distribution of Butane Conformations at a Particular Temperature

Calculate the Distribution of Butane Conformations at a Particular Temperature

Assume that we have an equilibrium mixture of the gauche and anti conformations at 25°C (298 K).

gauche⇋anti

K= [anti ] [ gauche]

To calculate the value of K, we will use the following equation.

K=e −G RT

∆G will be calculated using ∆G=∆H-T∆S

∆H is the difference in potential energy between the anti and gauche conformations. At 25˚C, this value is -3.8 kJ/mol.

While it might appear that there is no entropy change in this process, ∆S is not equal to zero. There are two possible gauche conformations with the same potential energy, while there is only one possible anti conformation. Thus there exists more possible microstates for the gauche conformation than for the anti. To calculate the value of ∆S, we will use the following equation.

∆S=RlnWanti-RlnWgauche

R is the gas constant (8.31 J/mol K) W is the number of possible microstates for each conformation

Thus for butane: ∆S=Rln1 – Rln2

=0-(8.31 J/mol K x ln2) =5.76 J/mol K

Now back to ∆G

∆G = -3.8×103 J/mol K – (298K x -5.76 J/mol K) =-2100 J/mol

Returning to the calculation of K

K=e 2100 J /mol

8.31J /mol K×298K

K=2.3 To calculate the percent anti conformation present at 298K, we recall that the value of K is this process is essentially the ratio of anti:gauche

Thus, since the ratio is 2.3:1.0, the percentage can be calculated as follows.

2.3 2.31.0

×100

= 70% (anti)

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