The second law of thermodynamics states that for any spontaneous process the ent
ID: 1066006 • Letter: T
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.Explanation / Answer
1)
Grxn = Hrxn - TSrxn
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.
Suniv = Ssys + Ssurr
a) For a spontaneous reaction/process, the entropy of the universe, Suniv must always increase, that is,
Suniv > 0 or Grxn < 0.
b)
At constant T & P,
Ssurr = - Hrxn / T
For a spontaneous reaction H can be negative as well as positive.
Endothermic reactions have positive H.
Exothermic reactions have negative H.
c)
For a spontaneous reaction Srxn can also be negative as well as positive.
Phase changes from solid to liquid or liquid to gas have positive Srxn, 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 Grxn.
b) Since the reaction is endothermic, Hrxn is positive.
c) Phase change from solid to liquid has Srxn > 0.
d) Under standard conditions the reaction is spontaneous.
(i) As we increase the temperature Grxn will become more negative.
(ii) As we decrease the temperature Grxn will increase and at equilibrium Hrxn = TSrxn. Further decrease in temperature will make the reaction non spontaneous (Grxn > 0).