# Another Thermo Question?

## Consider the reaction C2H2(g) + H2(g) <->C2H4(g). Use the data in the table above for the following: A: Estimate Delta G, H and S for the reaction above at 25C, and use your results to show that the reaction is spontaneous at room temperature. B: As the temperature increases, the reaction ceases to be spontaneous. Use the Gibbs-Helmholtz equation to predict the temperature at which the reaction is no longer spontaneous for 1 atm pressure in each gas.

Nov 25, 2016

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 $H$:

$\Delta {Y}_{\text{rxn}}^{\circ} = {\sum}_{P} {\nu}_{P} \Delta {Y}_{f , P}^{\circ} - {\sum}_{R} {\nu}_{R} \Delta {Y}_{f , R}^{\circ}$

If $S$:

$\Delta {S}_{\text{rxn}}^{\circ} = {\sum}_{P} {\nu}_{P} {S}_{P}^{\circ} - {\sum}_{R} {\nu}_{R} {S}_{R}^{\circ}$

where:

• $\Delta {Y}_{f}^{\circ}$ is some thermodynamic function of formation (such as Gibbs energy of formation or enthalpy of formation) for one pure substance.
• ${S}^{\circ}$ 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 $\text{0 K}$, and ${S}_{\text{0 K}} = 0$ from the third law of thermodynamics.
• $\nu$ is the stoichiometric coefficient of the substance in the balanced chemical reaction.

${\text{C"_2"H"_2(g) + "H"_2(g) rightleftharpoons "C"_2"H}}_{4} \left(g\right)$

We'll choose $G$ first (at $\text{298.15 K}$ and $\text{1 atm}$ like usual):

$\textcolor{b l u e}{\Delta {G}_{\text{rxn}}^{\circ}} = {\nu}_{{C}_{2} {H}_{4} \left(g\right)} \Delta {G}_{f , {C}_{2} {H}_{4} \left(g\right)}^{\circ} - \left[{\nu}_{{C}_{2} {H}_{2} \left(g\right)} \Delta {G}_{f , {C}_{2} {H}_{2} \left(g\right)}^{\circ} + {\nu}_{{H}_{2} \left(g\right)} \Delta {G}_{f , {H}_{2} \left(g\right)}^{\circ}\right]$

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

$= \textcolor{b l u e}{- \text{141.1 kJ/mol}}$

Already we can say that the reaction is spontaneous at room temperature, since $\boldsymbol{\Delta G < 0}$, which by definition says the reaction is spontaneous, and standard thermodynamic temperature ($\text{298.15 K}$, or ${25}^{\circ} \text{C}$) is room temperature.

Essentially the same math follows for $\Delta {H}_{\text{rxn}}^{\circ}$:

$\textcolor{b l u e}{\Delta {H}_{\text{rxn}}^{\circ}} = {\nu}_{{C}_{2} {H}_{4} \left(g\right)} \Delta {H}_{f , {C}_{2} {H}_{4} \left(g\right)}^{\circ} - \left[{\nu}_{{C}_{2} {H}_{2} \left(g\right)} \Delta {H}_{f , {C}_{2} {H}_{2} \left(g\right)}^{\circ} + {\nu}_{{H}_{2} \left(g\right)} \Delta {H}_{f , {H}_{2} \left(g\right)}^{\circ}\right]$

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

$= \textcolor{b l u e}{- \text{174.4 kJ/mol}}$

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

And for $\Delta {S}_{\text{rxn}}^{\circ}$, you have two options. Either follow the same math as above again, which gives you $- \text{111.98 J/mol"cdot"K}$, or use:

$\Delta {G}_{\text{rxn"^@ = DeltaH_"rxn"^@ - TDeltaS_"rxn}}^{\circ}$

and solve for $\Delta {S}_{\text{rxn}}^{\circ}$ to get:

$\frac{\textcolor{b l u e}{\Delta {S}_{\text{rxn"^@) = (DeltaH_"rxn"^@ - DeltaG_"rxn}}^{\circ}}}{T}$

$= \textcolor{b l u e}{- \text{111.69 J/mol"cdot"K}}$
(don't forget to convert from $\text{kJ}$ to $\text{J}$).

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 $\Delta {S}_{\text{rxn}}^{\circ}$.

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 $\Delta {G}_{\text{rxn}}^{\circ}$ at some new temperature ${T}_{\text{nonspont.}}$ 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"^@

$\implies \Delta {H}_{\text{rxn"^@ = T_"nonspont."DeltaS_"rxn}}^{\circ}$

$\textcolor{b l u e}{{T}_{\text{nonspont.") = (DeltaH_"rxn"^@)/(DeltaS_"rxn}}^{\circ}}$

$= \left(- \text{174.4 kJ/mol")/(-"0.11169 kJ/mol}\right)$

$=$ $\textcolor{b l u e}{\text{1561.5 K}}$

The hard way

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

Let us multiply both sides by $\mathrm{dT}$ at constant pressure. Then:

$d \frac{\Delta {G}_{\text{rxn"^@"/"T) = -(DeltaH_"rxn}}^{\circ}}{T} ^ 2 \mathrm{dT}$

Integrating both sides from the initial temperature ${T}^{\circ}$ to the final temperature ${T}_{\text{nonspont.}}$, we get:

${\int}_{{T}^{\circ}}^{{T}_{\text{nonspont.") d(DeltaG_"rxn"^@"/"T') = DeltaH_"rxn"^@int_(T^@)^(T_"nonspont.}}} - \frac{1}{T '} ^ 2 \mathrm{dT} '$

where we have assumed that $\Delta {H}_{\text{rxn}}^{\circ}$ does not change significantly at a new temperature, but we haven't assumed that for $\Delta {G}_{\text{rxn}}^{\circ}$. We have also substituted $T ' = T$ to not confuse the $T$ in the $\frac{1}{T}$ 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 $\frac{1}{T}$, 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 ${T}^{\circ} = \text{298.15 K}$ and ${T}_{\text{nonspont.}}$ 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 ${T}_{\text{nonspont.}}$ 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)

$= {T}^{\circ} {\left[1 - \left(\Delta {G}_{\text{rxn"^@(T^@))/(DeltaH_"rxn"^@)]^(-1) = ("298.15 K")[1 - (-"141.1 kJ/mol")/(-"174.4 kJ/mol}}\right)\right]}^{- 1}$

$=$ $\textcolor{b l u e}{\text{1561.5 K}}$

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