#CH_3OH (g) -> CO (g) + 2H_2 (g)# and #Delta H = 91 "kJ"/"mol"#. What can be inferred about #DeltaS# for the reaction at 600K?
a) It must be positive, since the reaction is thermodynamically unfavorable at 600K.
b) It must be negative, since there are more moles of products than reactants.
c) It must be positive, since delta G is negative and delta H is positive.
d) It must be negative, since delta G is positive and delta H is positive.
a) It must be positive, since the reaction is thermodynamically unfavorable at 600K.
b) It must be negative, since there are more moles of products than reactants.
c) It must be positive, since delta G is negative and delta H is positive.
d) It must be negative, since delta G is positive and delta H is positive.
1 Answer
None of these answers are certainly correct... I would say
#(e)# It should be positive, since there are more mols of products than reactants.
In actuality, the change in standard molar entropy at
#DeltaS_(rxn)^@ = (2 cdot "130.6 J/mol H"_2cdot"K" + 1 cdot "197.7 J/mol CO"cdot"K") - (1cdot "239.7 J/mol CH"_3"OH"cdot"K")#
#=# #"219.1 J/mol"cdot"K"#
And so, at
#DeltaG_(rxn)^@ = "91 kJ/mol" - "298 K" cdot ("219.1 J/mol"cdot"K") xx "1 kJ"/"1000 J"#
#=# #"25.71 kJ/mol"# the reaction is thermodynamically unfeasible...
...But while this reaction is thermodynamically unfeasible at
#color(blue)(DeltaG_(rxn)("600 K")) ~~ "91 kJ/mol" - "600 K" cdot ("219.1 J/mol"cdot"K") xx "1 kJ"/"1000 J"#
#= color(blue)ul(-"40.46 kJ/mol")#
And this makes sense; this is an endothermic process, so by increasing the temperature, the reaction becomes thermodynamically favorable...
Hence,
#a)# is clearly false.It is thermodynamically favorable at
#"600 K"# . We can also say that if it "must be positive because the reaction is thermodynamically unfavorable", it would also have to be true that#"600 K"# is a "low temperature" (for which#DeltaG > 0# when#DeltaH > 0# ), which it is not...
#b)# is clearly false, BECAUSE there are mols of products than reactants.We have already numerically shown that
#DeltaS_(rxn)# is positive here, and it is conceptually sound since the mols of gas increased, increasing the energy dispersal.
#c)# is not well-written, but is the best choice.We do not know ahead of time whether
#DeltaG# is positive or negative, since it is temperature-dependent when#DeltaH > 0# and#DeltaS# is likely positive. So, we cannot say "it must be positive because".
#d)# is clearly false because we already showed that#DeltaS > 0# . This answer neglects the fact that the mols of gas increased.
