1.23 Activity

Chem 152 B Activity 1/24/12 – answers will be posted at chem152B.blogspot.com

1. For a chemical reaction involving only the gas phase, entropy is related to the total number of moles on either side of the equation.  A decrease means lower entropy; an increase means higher entropy.

For the reaction

2H₂(g) + O₂(g) à 2H₂O(g)

do you expect the change in entropy to be positive or negative?  Why?  Do you expect the change in enthalpy to be positive or negative?  Why?  (Think about your background knowledge of this reaction.)  What might this tell you about the spontaneity of the reaction?

You would expect the change in entropy to be negative because there are more moles of gas in the reactants than the products.  You would expect the change in enthalpy be negative because we know this reaction to be exothermic – hydrogen is flammable in air.  DS_univ = DS_sys + DS_surr = DS_sys  - DH/T.  DH is negative so we have a negative number minus a negative number.  When temperature is small, DH will outweigh DS and the reaction will be spontaneous (i.e. DS_univ will be positive).  When temperature is large, DS will outweigh DH and the reaction will not be spontaneous (i.e. DS_univ will be negative).

2. For a chemical reaction involving different phases, the production of gas will, in general, increase the entropy much more than the increase in the number of moles of a liquid or solid.

For the reaction

2HNO₃(aq) + Na₂CO₃(s) à 2NaNO₃(aq) + H₂O(l) + CO₂(g)

do you expect the change in entropy to be positive or negative?  Why?

You would expect the change in entropy to be positive since there are more moles of gas in the products than the reactants.