At what temperature is this reaction at equilibrium?

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1 Answer
Truong-Son N.
Mar 13, 2018

Below what temperature does this reaction become spontaneous?

To be specific, you were given standard enthalpies of formation and standard molar entropies... of course, at #25^@ "C"# and #"1 atm"#.

#"CO"_2(g) + "H"_2"O"(g) rightleftharpoons "H"_2(g) + "CO"_2(g)#

Reactions at equilibrium have #DeltaG = 0#, but if we are specifically at #25^@ "C"# as well, then #DeltaG^@ = 0# at reaction equilibrium.

#DeltaG^@ = DeltaH^@ - TDeltaS^@ = 0#

#=> T = (DeltaH^@)/(DeltaS^@)#

Now we need to find the proper #DeltaH_(rxn)^@# and #DeltaS_(rxn)^@# values...

#DeltaH_(rxn)^@ = sum_P n_P DeltaH_(f,P)^@ - sum_R n_R DeltaH_(f,R)^@#

#= [(1)(0) + (1)(-393.8)] "kJ" - [(1)(-110.6) + (1)(-241.9)] "kJ"#

#= -"41.3 kJ"#

where #n# is the mols of something. What do #P# and #R# mean?

#DeltaS_(rxn)^@ = sum_P n_P S_(P)^@ - sum_R n_R S_(R)^@#

#= [(1)(130.7) + (1)(213.8)] "J/K" - [(1)(197.7) + (1)(188.9)] "J/K"#

#=-"42.1 J/K"#

As a result,

#color(blue)(T) = (-"41.3 kJ")/(-"0.0421 kJ/K")#

#=# #color(blue)("981 K")#

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