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Consider the equilibrium between the boat forms of an aldohexose (both alpha and beta forms combined)...

Consider the equilibrium between the boat forms of an aldohexose (both alpha and beta forms combined) and the chair forms of the same aldohexose (both alpha and beta forms combined) as follows: boat <--> chair. If the chair form is 1.13 kcal/mol lower in energy than the boat form, the what fraction (expressed as a percent) of the molecules would be in the boat form? Assume a temperature of 25 °C, and assume that all contributions of other forms (e.g., linear) are negligible. Report your answer as a percentage to the nearest ones.

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Answer #1

If chair form is at lower energy,it means gibbs free energy

We know that, ∆G = ∆H -T∆S

∆H is almost same for both boat and chair form

ΔH ≈ΔG

ΔG~ -1.13 kCal/mol

in terms of joules

= -1.13 x 4.18 J/mol = -4.7234 J/mol

-RTlnK = -4.7234 J/mol

K = 1

Lets say boat = A

boat <--> chair

A               0        initially

A-x            x            equilibrium

K = x/(A-x) = 1

x = A -x

A = 2x

% boat form = boat form / total molecules x 100 = A-x / A x 100

x/2x x 100 = 50%

boat = 50 %

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