English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Without using absolute entropy values, predict the sign of change in S° for each of the following processes.

a.) 2 NO2(g) --> N2O4(g)
b.) 2K (s) + F2 (g) --> 2KF (s)
c.) NaClO3 (s) --> Na+ (aq) + ClO3- (aq)

2006-11-02 06:02:16 · 3 answers · asked by buttrefly007 1 in Science & Mathematics Chemistry

3 answers

Entropy (S) is the measure of disorder in a system. When thinking of problems such as this, it's helpful to think of it in terms of a gas being the highest entropy state, a liquid being kind of in the middle, and then a solid being the lowest entropy state.
A) A gas goes to another gas. There should be no change in S.
B) S is negative. A gas condenses to a solid. You are moving from a higher state of entropy to a lower state.
C) Positive S value. You go from the lowest energy state (solid) to an aqueous solution (liquid), which means the entropy goes up as well.

2006-11-02 06:55:27 · answer #1 · answered by osoloco21 1 · 1 0

In (a), 2 moles of gas go to one mole - negative entropy change.
In (b), a solid and a gas go to a solid. Again negative.
But in (c), a solid goes to two aqueous ions, which is a positive entropy change.

2006-11-06 06:04:39 · answer #2 · answered by Gervald F 7 · 1 0

The first one is probably exothermic, the second one definitely is, and the third one may be endothermic but I can't tell for sure.

2006-11-02 06:06:16 · answer #3 · answered by Anonymous · 0 1

fedest.com, questions and answers