Question

Consider each of the following reactions and decide whether the standard entropy change for the reaction will be, (a) Positive; or (b) Very close to Zero; or (c) Negative.

Matching pairs

N2 (g) + 3 H2 (g) = 2 NH3 (g)
Mg (s) + CO2 (g) = MgO (s) + CO (g)
C6H12O6 (s) + 6 O2 (g) = 6 CO2 (g) + 6 H2O (l)
2 NO2 (g) = N2O4 (g)
BaSO4 (s) + 2 H2O (g) = BaSO4.2H2O (s)
PCl5 (s) = PCl3 (s) + Cl2 (g)

My answers are:
very close to zero
negative ???
positive
negative
negative
positive

i am not sure of the second one ?? u see the entropy of CO is less than CO2 due to the different arrangements however it can be also close to zero what do u guys think ???

Answer 1

You should get a second opinion on these answers but here is my reasoning. Check my thinking.
a. Four moles goes to two moles. I think S decreases.
b. Both Mg and MgO are crystalline solids. About the same. CO and CO2 both gases. About the same at the same T. So S is about zero.
c. solid and gas goes to gas and liquid. More disorder on the right. S positive.
d. decrease in moles again.
e. gas on left and none on the right. S is negative.
f. gas on the right. none on the left. S is positive.

Answer 2

so the first one is wrong too hmmm ! yeah ur reasoning makes sense ... i will ask my chem teacher too

Answer 3

hmmm interesting