Chapter 20, Problem 20.53QA

Interpretation Introduction

To calculate:

The change in the free energy of the hydrolysis reaction of the maltose to glucose.

Answer to Problem 20.53QA

Solution:

The change in free energy of the hydrolysis reaction of maltose to glucose is $-16.4 kJ$.

Explanation of Solution

1) Concept:

The change in the free energy of the reaction is calculated from the standard molar free energy of products and reactant as

${∆G}_{rxn}^o=\sum n_{product}{∆G}_{f, product}^o-\sum n_{reactant}{∆G}_{f, reactant}^o$

2) Formula:

${∆G}_{rxn}^o=\sum n_{product}{∆G}_{f, product}^o-\sum n_{reactant}{∆G}_{f, reactant}^o$

3) Given:

i) ${∆G}_{f, \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{t}\mathrm{o}\mathrm{s}\mathrm{e}}^o=-2246.6 kJ/mol$

ii) ${∆G}_{f, \mathrm{g}\mathrm{l}\mathrm{u}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}}^o=-1274.4 kJ/mol$

iii) ${∆G}_{f, water}^o=-285.8 kJ/mol$

iv) $Maltose+H_{2}O\to 2 Glucose$

4) Calculation:

Plug in the values and calculate the free energy change for the hydrolysis of maltose to glucose,

${∆G}_{rxn}^o=\sum n_{product}{∆G}_{f, product}^o-\sum n_{reactant}{∆G}_{f, reactant}^o$

${∆G}_{rxn}^o=[n_{glucose}\times {∆G}_{f, \mathrm{g}\mathrm{l}\mathrm{u}\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}}^o]-\left[n_{maltose}\times {∆G}_{f, \mathrm{m}\mathrm{a}\mathrm{l}\mathrm{t}\mathrm{o}\mathrm{s}\mathrm{e}}^o+ n_{water}\times {∆G}_{f, \mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}^o\right]$

${∆G}_{rxn}^o=\left[2 mol \left(\frac{-1274.4 kJ}{mol}\right)\right]-\left[1 mol \left(\frac{-2246.6 kJ}{mol}\right)+1 mol \left(\frac{-285.8 kJ}{mol}\right)\right]$

${∆G}_{rxn}^o=-2548.8 kJ+2532.4 kJ$

${∆G}_{rxn}^o=-16.4 kJ$

The change in the free energy of the hydrolysis reaction of maltose to glucose is $-16.4 kJ$.

Conclusion:

We used the free energy change formula related to the standard molar free energy for the products and reactants.