Question

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 $\Delta S_{mix} = -Nk_B x \ln x - Nk_B y \ln y$ 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\". 1. 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 $\Delta G_{\alpha \rightarrow \beta}(x)$ you found before (Hints: Recall that entropy contributes to G by the term -TS. Also, make sure all terms have the same units!).

          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
$\Delta S_{mix} = -Nk_B x \ln x - Nk_B y \ln y$
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\".
1. 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 $\Delta G_{\alpha \rightarrow \beta}(x)$ you found before (Hints: Recall that
entropy contributes to G by the term -TS. Also, make sure all terms have the same units!).

$D. We have missed something in this analysis. As more β-D-glucose is converted, a different type of entropy change becomes crucial: the entropy of mixing. You derived an expression for this entropy change in Written Homework 4 (for the mixing of two gases, but the result applies equally here): as the α-D-glucose and the β-D-glucose mix the entropy of the system increases by Δ Smix = -NkB x ln x - NkB y ln y where x is the fraction of β-D-glucose and y the fraction of α-D-glucose in the solution (they add up to 1). Recall that the total number of particles here is N = NA as we are handling quantities p̈er mol.̈ 1. Determine an improved expression for the ẗotalG̈ibbs free energy change Δ Gtotal by adding the contribution due to this change in entropy to the Δ Gα→β(x) you found before (Hints: Recall that entropy contributes to G by the term -TS. Also, make sure all terms have the same units!).$

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