Question

The second law of thermodynamics states that for any spontaneous process the entropy of the universe must increase. Explain how this statement (increasing entropy) relates to: lambda G_rxn lambda H_rxn lambda S_rxn For this process under standard conditions. H_2O_($) O longleftrightdoublearrow H_2O(f), circle the correct sign you predict for (a), (b), and (c) and explain your reasoning lambda G_rxn + or - lambda H_rxn + or- lambda H_rxn + or-(d) Explain how temperature affects the sign of lambda G in the above process.

Answer 1

1)

ΔG_rxn = ΔH_rxn - TΔS_rxn

The direction of spontaneous change is the direction in which total entropy of the universe increases i.e. the sum of the entropy change of a system and of its surroundings. Here the system is the reaction taking place.

ΔS_univ = ΔS_sys + ΔS_surr

a) For a spontaneous reaction/process, the entropy of the universe, S_univ must always increase, that is,

ΔS_univ > 0 or ΔG_rxn < 0.

b)

At constant T & P,

ΔS_surr = - ΔH_rxn / T

For a spontaneous reaction ΔH can be negative as well as positive.

Endothermic reactions have positive ΔH.

Exothermic reactions have negative ΔH.

c)

For a spontaneous reaction ΔS_rxn can also be negative as well as positive.

Phase changes from solid to liquid or liquid to gas have positive ΔS_rxn, where the degree of freedom of molecules increases.

2)

Under standard conditions H₂O exists in liquid form. Therefore, the reaction is spontaneous.

a) Spontaneous reactions have negative ΔG_rxn.

b) Since the reaction is endothermic, ΔH_rxn is positive.

c) Phase change from solid to liquid has ΔS_rxn > 0.

d) Under standard conditions the reaction is spontaneous.

(i) As we increase the temperature ΔG_rxn will become more negative.

(ii) As we decrease the temperature ΔG_rxn will increase and at equilibrium ΔH_rxn = TΔS_rxn. Further decrease in temperature will make the reaction non spontaneous (ΔG_rxn > 0).