Question

Predict the sign of deltaS for each process:

A) Freezing water

B) sublimatioin of carbon dioxide

c) combustion of gasoline

D) 2H2(g)--> 2H2(g)+O2(g)

E) Zn^2+(aq)+2Br^- -->Zn(Br)2

F)Zn^2+(aq)2Br^- --> Zn(Br)2(aq)

Answer 1

Entropy : The disordernes of molecules are termed as Entropy.

Entropy increases when moles are increases. Entropy increases in the order solid-liquid-gases.

Decrease in entropy is indicated by - ve.

Increase in entropy is indicated bt +ve.

A) In freezing of water, water goes from liquid to solid state therefore entropy decreases = S = -ve.

B) iIn Sublimation, CO2 changes from solid to gases. Therefore entropy increases =  S = +ve.

C) Combustion of gasoline increases moles on product side. Therefore entropy increases =  S = +ve.

D) In this case, both reactants and products are in gaseous state. But product side contains more total 3 moles (2 of H2 and 1 of O2) than reactant side.  Therefore entropy increases =  S = +ve.

E) Total number of moles decreases in product side.  Therefore entropy decreases =  S = -ve.

F) Here also, total number of moles decreases in product side to that of reactant side.  Therefore entropy increases =  S = +ve.