The #K_p# of H2+I2<->2HI is 50.2 at 445°C. #DeltaH_(rxn)^@# for this is 25.84 kj/mol. How can I find the entropy of this reaction?

#DeltaS_(rxn)^@ = "68.54 J/mol"cdot"K"#

Well, recall... for gas-phase reactions,

#DeltaG_(rxn) = DeltaG_(rxn)^@ + RTlnQ_p#

where #DeltaG_(rxn)^@# is defined at an arbitrarily-chosen standard temperature (call it #445^@ "C"#), and #DeltaG_(rxn)# is based on deviation away from standard conditions (#"1 atm"# partial pressures).

So for

#"H"_2(g) + "I"_2(g) rightleftharpoons 2"HI"(g)#,

with #K_p = 50.2# at #445^@ "C"#, we should know that at equilibrium,

• #Q_p = K_p = 50.2#
• #DeltaG_(rxn) = 0#

Therefore:

#DeltaG_(rxn)^@ = -RTlnK_p#

#= -"0.008314472 kJ/mol"cdot"K" cdot (445+"273.15 K") cdot ln(50.2)#

#= -"23.38 kJ/mol"#

At a constant temperature,

#DeltaG = DeltaH - TDeltaS#.

As a result,

#color(blue)(DeltaS_(rxn)^@) = (DeltaH_(rxn)^@ - DeltaG_(rxn)^@)/T#

#= ("25.84 kJ/mol" - (-"23.38 kJ/mol"))/(445 + "273.15 K")#

#= "0.06854 kJ/mol"cdot"K"#

#= color(blue)("68.54 J/mol"cdot"K")#

This makes sense; the molecules descended in symmetry from #D_(ooh)# to #C_(oov)#, so the entropy should increase due to an increased molecular complexity.

This effect dominates since the moles of gas stay the same.