Standard Gibbs Energy of Ethanol Decomposition at 443 K: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2018, Q5

Question 5: Standard Gibbs Energy of Ethanol Decomposition at 443 K (Part B — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Reaction $CH_3CH_2OH(g)\rightarrow C_2H_4(g)+H_2O(g)$. $\Delta H^\circ_{298}=44.843$ kJ/mol. $\Delta G^\circ_{f,298}$ (kJ/mol): ethanol $-168.16$, ethylene $+68.06$, water $-228.37$. Ideal-gas heat capacities $C_p=A+BT+CT^2+DT^3+ET^4$ (J/mol·K) from Table C-4:

Species	A	B	C	D	E
Ethanol	27.091	1.1055e−1	1.0957e−4	−1.5046e−7	4.6601e−11
Ethylene	32.083	−1.4831e−2	2.4774e−4	−2.3766e−7	6.8274e−11
Water	33.933	−8.4186e−3	2.9906e−5	−1.7825e−8	3.6934e−12

Find. $\Delta G^\circ$ of the reaction at 443 K.

Approach. Get $\Delta G^\circ_{298}$ from formation energies and $\Delta S^\circ_{298}$ from $(\Delta H^\circ-\Delta G^\circ)/T$; propagate $\Delta H^\circ$ and $\Delta S^\circ$ to 443 K with the $\Delta C_p$ integrals, then recombine $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$.

Standard properties at 298 K. From formation Gibbs energies, $$\Delta G^\circ_{298}=68.06+(-228.37)-(-168.16)=+7.85\ \text{kJ/mol},$$ $$\Delta S^\circ_{298}=\frac{\Delta H^\circ_{298}-\Delta G^\circ_{298}}{298.15}=\frac{44{,}843-7{,}850}{298.15}=124.1\ \text{J/mol}\cdot\text{K}.$$
Heat-capacity change of reaction. $\Delta C_p=C_p^{C_2H_4}+C_p^{H_2O}-C_p^{EtOH}$ gives coefficients $\Delta A=38.93$, $\Delta B=-0.1338$, $\Delta C=1.487\times10^{-4}$, $\Delta D=-8.86\times10^{-8}$, $\Delta E=2.53\times10^{-11}$.
Enthalpy and entropy at 443 K (Kirchhoff integrals). $$\Delta H^\circ_{443}=\Delta H^\circ_{298}+\!\int_{298.15}^{443}\!\Delta C_p\,dT=44{,}843+1{,}113=45{,}956\ \text{J/mol},$$ $$\Delta S^\circ_{443}=\Delta S^\circ_{298}+\!\int_{298.15}^{443}\!\frac{\Delta C_p}{T}\,dT=124.08+3.13=127.2\ \text{J/mol}\cdot\text{K}.$$
Gibbs energy at 443 K. $$\Delta G^\circ_{443}=\Delta H^\circ_{443}-T\,\Delta S^\circ_{443}=45{,}956-443(127.2)=\boxed{-10.40\ \text{kJ/mol}.}$$ Although $\Delta G^\circ_{298}=+7.85$ kJ/mol (unfavourable), the large positive reaction entropy (one mole → two moles of gas) turns the reaction favourable on heating: $K=\exp(-\Delta G^\circ/RT)=16.8$ at 443 K.

Quantity	Result
$\Delta G^\circ_{298}$	+7.85 kJ/mol
$\Delta S^\circ_{298}$	124.1 J/mol·K
$\Delta H^\circ_{443}$	45.96 kJ/mol
$\Delta S^\circ_{443}$	127.2 J/mol·K
$\Delta G^\circ_{443}$	−10.40 kJ/mol ($K=16.8$)

Concept & Theory

Topic. Temperature dependence of $\Delta G^\circ$ via Kirchhoff’s law and heat-capacity data.

Key relations. $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ$; Kirchhoff $\Delta H^\circ(T)=\Delta H^\circ_{298}+\int\Delta C_p\,dT$ and $\Delta S^\circ(T)=\Delta S^\circ_{298}+\int(\Delta C_p/T)\,dT$; $\Delta G^\circ=-RT\ln K$.

Why this works. Enthalpy and entropy are state functions, so their values at 443 K are the 298 K anchors plus the heat capacity accumulated between the two temperatures. Recombining them at the target $T$ gives $\Delta G^\circ$ directly — more accurate than assuming $\Delta H^\circ,\Delta S^\circ$ are temperature-independent, because $\Delta C_p$ here is appreciable.

Common pitfall. Using the van’t Hoff shortcut with constant $\Delta H^\circ$ (ignoring $\Delta C_p$), or forgetting to convert $\Delta H^\circ$ to J before dividing by $T$ for $\Delta S^\circ$. Sign of $\Delta G^\circ_{f}$ for ethylene (positive) is easy to mis-enter.

Source. Smith, Van Ness, Abbott & Swihart §4.5–4.6 (heat effects, $\Delta C_p$ integrals) and §13.2 (standard-property changes and $K$ vs. $T$).

Question 5 of 6: Standard Gibbs Energy of Ethanol Decomposition at 443 K

Question 5: Standard Gibbs Energy of Ethanol Decomposition at 443 K (Part B — equal value)