Another Thermo Question?

PART A

Part A is fairly straightforward, and is just asking you to choose the thermodynamic values to use. Let #Y# be a thermodynamic state function. Then:

If #G# or #H#:

#DeltaY_"rxn"^@ = sum_P nu_P DeltaY_(f,P)^@ - sum_R nu_R DeltaY_(f,R)^@#

If #S#:

#DeltaS_"rxn"^@ = sum_P nu_P S_(P)^@ - sum_R nu_R S_(R)^@#

where:

• #DeltaY_f^@# is some thermodynamic function of formation (such as Gibbs energy of formation or enthalpy of formation) for one pure substance.
• #S^@# is the standard molar entropy for one pure substance, and #P# (or #R#) stands for products (or reactants). This is defined a little differently because the initial state is at #"0 K"#, and #S_("0 K") = 0# from the third law of thermodynamics.
• #nu# is the stoichiometric coefficient of the substance in the balanced chemical reaction.

#"C"_2"H"_2(g) + "H"_2(g) rightleftharpoons "C"_2"H"_4(g)#

We'll choose #G# first (at #"298.15 K"# and #"1 atm"# like usual):

#color(blue)(DeltaG_"rxn"^@) = nu_(C_2H_4(g))DeltaG_(f,C_2H_4(g))^@ - [nu_(C_2H_2(g))DeltaG_(f,C_2H_2(g))^@ + nu_(H_2(g))DeltaG_(f,H_2(g))^@]#

#= (1)("68.11 kJ/mol") - [(1)("209.2 kJ/mol") +(1) ("0 kJ/mol")]#

#= color(blue)(-"141.1 kJ/mol")#

Already we can say that the reaction is spontaneous at room temperature, since #bb(DeltaG < 0)#, which by definition says the reaction is spontaneous, and standard thermodynamic temperature (#"298.15 K"#, or #25^@ "C"#) is room temperature.

Essentially the same math follows for #DeltaH_"rxn"^@#:

#color(blue)(DeltaH_"rxn"^@) = nu_(C_2H_4(g))DeltaH_(f,C_2H_4(g))^@ - [nu_(C_2H_2(g))DeltaH_(f,C_2H_2(g))^@ + nu_(H_2(g))DeltaH_(f,H_2(g))^@]#

#= (1)("52.30 kJ/mol") - [(1)("226.7 kJ/mol") +(1) ("0 kJ/mol")]#

#= color(blue)(-"174.4 kJ/mol")#

which says the reaction is exothermic, i.e. it releases energy into the surroundings.

And for #DeltaS_"rxn"^@#, you have two options. Either follow the same math as above again, which gives you #-"111.98 J/mol"cdot"K"#, or use:

#DeltaG_"rxn"^@ = DeltaH_"rxn"^@ - TDeltaS_"rxn"^@#

and solve for #DeltaS_"rxn"^@# to get:

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

#= color(blue)(-"111.69 J/mol"cdot"K")#
(don't forget to convert from #"kJ"# to #"J"#).

which means the entropy of the system decreases, meaning that the entropy of the surroundings increases.

Either way, there's about a #0.26%# (i.e. negligible) difference between the two methods, so either way works fine when determining #DeltaS_"rxn"^@#.

PART B

I'm surprised you are asked to use the Gibbs-Helmholtz equation... Because this is what I see on Wikipedia:

#((del(DeltaG_("rxn")^@"/"T))/(delT))_P = -(DeltaH_("rxn")^@)/T^2#

If we want the reaction to no longer be spontaneous, then we must have that #DeltaG_"rxn"^@# at some new temperature #T_"nonspont."# be equal to #0#. There's an easy and a hard way to do this...

The easy way

#cancel(DeltaG_"rxn"^@)^"assumed 0" = DeltaH_"rxn" - T_"nonspont."DeltaS_"rxn"^@#

#=> DeltaH_"rxn"^@ = T_"nonspont."DeltaS_"rxn"^@#

#color(blue)(T_"nonspont.") = (DeltaH_"rxn"^@)/(DeltaS_"rxn"^@)#

#= (-"174.4 kJ/mol")/(-"0.11169 kJ/mol")#

#=# #color(blue)("1561.5 K")#

The hard way

From #((del(DeltaG_("rxn")^@"/"T))/(delT))_P = -(DeltaH_("rxn")^@)/T^2#:

Let us multiply both sides by #dT# at constant pressure. Then:

#d(DeltaG_"rxn"^@"/"T) = -(DeltaH_"rxn"^@)/T^2dT#

Integrating both sides from the initial temperature #T^@# to the final temperature #T_"nonspont."#, we get:

#int_(T^@)^(T_"nonspont.") d(DeltaG_"rxn"^@"/"T') = DeltaH_"rxn"^@int_(T^@)^(T_"nonspont.")-1/(T')^2dT'#

where we have assumed that #DeltaH_"rxn"^@# does not change significantly at a new temperature, but we haven't assumed that for #DeltaG_"rxn"^@#. We have also substituted #T' = T# to not confuse the #T# in the #1/T# with the #T# in the integral bounds.

The integral of a derivative cancels out on the left side, and the right side integrates to give #1/T#, so:

#|[(DeltaG_"rxn"^@(T'))/(T')]|_(T^@)^(T_"nonspont.") = DeltaH_"rxn"^@|[(1/(T'))]|_(T^@)^(T_"nonspont.")#

#(DeltaG_"rxn"^@(T_"nonspont."))/T_"nonspont." - (DeltaG_"rxn"^@(T^@))/T^@ = DeltaH_"rxn"^@[1/(T_"nonspont.") - 1/T^@]#

Now, if we set #T^@ = "298.15 K"# and #T_"nonspont."# to be our new temperature at which #DeltaG_"rxn"^@(T_"nonspont.") = 0#, then:

#cancel((DeltaG_"rxn"^@(T_"nonspont."))/T_"nonspont.")^(0) - (stackrel("Known")overbrace(DeltaG_"rxn"^@(T^@)))/T^@ = stackrel("Known")overbrace(DeltaH_"rxn"^@)[1/T_"nonspont." - 1/T^@]#

Now we'd solve for #T_"nonspont."# to find the temperature at which the reaction is no longer spontaneous (i.e. has now crossed over from being spontaneous to being at equilibrium):

#-(DeltaG_"rxn"^@(T^@))/(DeltaH_"rxn"^@T^@) = 1/T_"nonspont." - 1/T^@#

#1/T^@ - (DeltaG_"rxn"^@(T^@))/(DeltaH_"rxn"^@T^@) = 1/T_"nonspont."#

#=> color(blue)(T_"nonspont.") = [1/T^@ - (DeltaG_"rxn"^@(T^@))/(DeltaH_"rxn"^@T^@)]^(-1) = (1/T^@)^(-1)[1 - (DeltaG_"rxn"^@(T^@))/(DeltaH_"rxn"^@)]^(-1)#

#= T^@[1 - (DeltaG_"rxn"^@(T^@))/(DeltaH_"rxn"^@)]^(-1) = ("298.15 K")[1 - (-"141.1 kJ/mol")/(-"174.4 kJ/mol")]^(-1)#

#=# #color(blue)("1561.5 K")#

and you can see the two approaches essentially give the same result (#0%# error if you keep all your decimal places in your calculator).