Question

1. Calculate ∆Sº for the reaction below:

N2(g)+O2(g)⇄2NO(g)

where ∆Sº for N2(g), O2(g), & NO(g), respectively, is 191.5, 205.0, & 210.6 J/mol-K

a. -24.7 J/K
b. 24.7 J/K
c. 185.9 J/K ***
d. -185.9 J/K

2. For which of these is there a decrease in entropy?
a. Cl2(g)→2Cl(g)
b. O2(g)→O2(aq)
c. C(s)+O2(g)→CO2(g)
d. NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l) ***it created a liquid

Answer 1

1. dSo rxn = (n*dSo products) - (n*dSo reactants)
2. You want the process in which movement of the molecules is decreased.

Answer 2

1. 205.0 + 191.5 = 396.5
210.6 - 396.5= -185.9 j/k

Answer 3

You didn't follow instructions.
"where ∆Sº for N2(g), O2(g), & NO(g), respectively, is 191.5, 205.0, & 210.6"
so N2 = 191.5
O2 = 205.0
NO = 210.6

From my previous post:
N2(g)+O2(g)⇄2NO(g)
dSo rxn = (n*dSo products) - (n*dSo reactants)
dSo rxn = (2*dSo NO) - (1*dSo N2 + 1*dSo O2 =
2*210.6 - 191.5 - 205.0 = ?

Answer 4

I change my answer to 24.7...forgot the 2