The free energy change for oxidation of cellulose molecules in a sheet of paper

Question

The free energy change for oxidation of cellulose molecules in a sheet of paper produces CO2 and H2O is large and negative (DGo’ = -677kcal/mol).

C6H10O5+6O26CO2+5H2O

1.      (1pt) Without doing any calculations, would you expect S (entropy change) for this reaction to be positive or negative?

a.       The distribution of carbon, hydrogen, and oxygen atoms is much less random after the reaction than before. As a result, the entropy of the products is smaller than that of the reactants, so the entropy change is positive.

b.      The distribution of carbon, hydrogen, and oxygen atoms is much more random after the reaction than before. As a result, the entropy of the products is greater than that of the reactants, so the entropy change is positive.

c.       The distribution of carbon, hydrogen, and oxygen atoms is much less random after the reaction than before. As a result, the entropy of the products is smaller than that of the reactants, so the entropy change is negative.

d.      The distribution of carbon, hydrogen, and oxygen atoms is much more random after the reaction than before. As a result, the entropy of the products is greater than that of the reactants, so the entropy change is negative.

2.      (2pts) Given this highly exergonic reaction, explain why paper is stable at room temperature in the presence of oxygen (O2).

a.       The distribution of carbon, hydrogen, and oxygen atoms is much less random after the reaction than before. As a result, the entropy of the products is smaller than that of the reactants, so the entropy change is positive.

b.      The distribution of carbon, hydrogen, and oxygen atoms is much more random after the reaction than before. As a result, the entropy of the products is greater than that of the reactants, so the entropy change is positive.

c.       The distribution of carbon, hydrogen, and oxygen atoms is much less random after the reaction than before. As a result, the entropy of the products is smaller than that of the reactants, so the entropy change is negative.

d.      The distribution of carbon, hydrogen, and oxygen atoms is much more random after the reaction than before. As a result, the entropy of the products is greater than that of the reactants, so the entropy change is negative.

Explanation / Answer

1. B. Because now the degree of randomness is increasing. As previously they were bounded and now they are in gaseous form. Number of moleculs is also increasing.

2. Because of high activation energy. The reaction may be exoergic in latter stage. But initially it is stable and has high activation energy

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