=-1.15K(J)/(m)ol

\ C. The quoted value of

\\\\Delta G_(\\\\alpha ->\\\\beta )=G_(\\\\beta )-G_(\\\\alpha )=-1.43k(J)/(m)ol

corresponds to

\\\\alpha

-D-glucose ending in a state of

100%\\\\beta

-D-glucose, meaning

G_(initial )=G_(\\\\alpha )

while the

G_(final )=G_(\\\\beta )

. If instead only a fraction

x

ends up as

\\\\beta

-D-glucose while the rest

(1-x)

ends as

\\\\alpha

-D-glucose then

G_(final )=xG_(\\\\beta )+(1-x)G_(\\\\alpha )

.\ Use this to find

\\\\Delta G

if, starting with

100%\\\\alpha -D

-glucose only

(1)/(5)

of it ends up as the

\\\\beta -D

form.\ 2\ Let's analyze the possibility of equilibrium in this process. For the general case where only a fraction

x

of the

\\\\beta

-D-glucose ends in solution, determine the change

\\\\Delta G_(\\\\alpha ->\\\\beta )(x)

in Gibbs free energy. Pick a few

x

values and plot the function.\ According to your plot, for what fraction

x

would the Gibbs free energy reach its smallest value?\ D. We have missed something in this analysis. As more

\\\\beta

-D-glucose is converted, a different type of entropy change becomes crucial: the entropy of mixing. You derived an expression for this entropy change in Written Homework 4 (for the mixing of two gases, but the result applies equally here): as the

\\\\alpha

-D-glucose and the

\\\\beta

-D-glucose mix the entropy of the system increases by\

\\\\Delta S_(mix )=-Nk_(B)xlnx-Nk_(B)ylny

\ where

x

is the fraction of

\\\\beta

-D-glucose and

y

the fraction of

\\\\alpha

-D-glucose in the solution (they add up to 1). Recall that the total number of particles here is

N=N_(A)

as we are handling quantities "per mol".\ Determine an improved expression for the "total" Gibbs free energy change

\\\\Delta G_(total )

by adding the contribution due to this change in entropy to the

\\\\Delta G_(\\\\alpha ->\\\\beta )(x)

you found before (Hints: Recall that entropy contributes to

G

by the term - TS. Also, make sure all terms have the same units!).

=1.15KJ/mol C. The quoted value of G=GG=1.43kJ/mol corresponds to -D-glucose ending in a state of 100%-D-glucose, meaning Ginitial=G while the Gfinal=G. If instead only a fraction x ends up as -D-glucose while the rest (1x) ends as -D-glucose then Gfinal=xG+(1x)G. 1. Use this to find G if, starting with 100%-D-glucose only 51 of it ends up as the -D form. 2 2. Let's analyze the possibility of equilibrium in this process. For the general case where only a fraction x of the -D-glucose ends in solution, determine the change G(x) in Gibbs free energy. Pick a few x values and plot the function. 3. According to your plot, for what fraction x would the Gibbs free energy reach its smallest D. We have missed something in this analysis. As more -D-glucose is converted, a different type of entropy change becomes crucial: the entropy of mixing. You derived an expression for this entropy change in Written Homework 4 (for the mixing of two gases, but the result applies equally here): as the -D-glucose and the -D-glucose mix the entropy of the system increases by Smix=NkBxlnxNkBylny where x is the fraction of -D-glucose and y the fraction of -D-glucose in the solution (they add up to 1). Recall that the total number of particles here is N=NA as we are handling quantities "per mol". 1. Determine an improved expression for the "total" Gibbs free energy change Gtotal by adding the contribution due to this change in entropy to the G(x) you found before (Hints: Recall that entropy contributes to G by the term TS. Also, make sure all terms have the same units!)