Entropy of Reaction for CO Oxidation at 500 K: 21-Mat-A1 Thermodynamics, December 2018, Q2

Question 2: Entropy of Reaction for CO Oxidation at 500 K (20 marks)

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 CO(g) $+\ \tfrac12$O₂(g) $\rightarrow$ CO₂(g); 298 K entropies and the three temperature-dependent $C_p$ polynomials above.

298 K standard entropies (J K⁻¹mol⁻¹)
Species	$S^{\circ}_{298}$
CO(g)	197.7
CO₂(g)	213.7
O₂(g)	205.1

Find. $\Delta S^{\circ}_R$ at 500 K.

Approach. Get $\Delta S^{\circ}_{298}$ from the tabulated entropies, build $\Delta C_p(T)=C_p(\text{CO}_2)-C_p(\text{CO})-\tfrac12C_p(\text{O}_2)$ from the three polynomials, then add $\int_{298}^{500}\Delta C_p/T\,dT$.

298 K reaction entropy. $$\Delta S^{\circ}_{298}=S^{\circ}(\text{CO}_2)-S^{\circ}(\text{CO})-\tfrac12S^{\circ}(\text{O}_2)=213.7-197.7-102.55=\boxed{-86.55\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
Combined $\Delta C_p(T)$. Collecting the polynomial coefficients term by term ($a+bT+cT^2+dT^3$): $$a=18.9-31.1-\tfrac12(30.8)=-27.6,\qquad b=7.9\times10^{-2}+1.5\times10^{-2}+0.6\times10^{-2}=0.100,$$ $$c=-6.8\times10^{-5}-3.1\times10^{-5}-1.2\times10^{-5}=-1.11\times10^{-4},\qquad d=2.4\times10^{-8}+1.5\times10^{-8}=3.9\times10^{-8}.$$
Integrate $\Delta C_p/T$ from 298 to 500 K. Using $\int(a/T+b+cT+dT^2)\,dT=a\ln(T_2/T_1)+b(T_2-T_1)+\tfrac{c}{2}(T_2^2-T_1^2)+\tfrac{d}{3}(T_2^3-T_1^3)$: $$\int_{298}^{500}\frac{\Delta C_p}{T}\,dT=(-27.6)\ln\!\left(\frac{500}{298}\right)+0.100(202)+\frac{-1.11\times10^{-4}}{2}(500^2-298^2)+\frac{3.9\times10^{-8}}{3}(500^3-298^3)=\boxed{-1.75\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
Total. $$\Delta S^{\circ}_R(500\text{K})=\Delta S^{\circ}_{298}+\int_{298}^{500}\frac{\Delta C_p}{T}\,dT=-86.55-1.75=\boxed{-88.30\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$

Quantity	Value
$\Delta S^{\circ}_{298}$	−86.55 J K⁻¹mol⁻¹
$\int_{298}^{500}\Delta C_p/T\,dT$	−1.75 J K⁻¹mol⁻¹
$\Delta S^{\circ}_R$(500 K)	−88.30 J K⁻¹mol⁻¹

Concept & Theory

Topic. Temperature dependence of a reaction entropy via Kirchhoff-type integration of $\Delta C_p(T)$.

Key relations. $\Delta S^{\circ}(T)=\Delta S^{\circ}_{298}+\int_{298}^{T}\dfrac{\Delta C_p}{T'}\,dT'$, with $\Delta C_p=\sum_{\text{products}}\nu C_p-\sum_{\text{reactants}}\nu C_p$.

Why this works. $S$ is a state function, so the entropy of reaction at any $T$ can be built from the 298 K value plus the entropy picked up (or lost) by heating each species from 298 K to $T$ along a reversible path — the $C_p/T$ integrand is exactly the reversible $dS=\delta q_{rev}/T=C_pdT/T$ for each species, combined with reaction stoichiometry.

Common pitfall. Using $\Delta S^{\circ}_{298}$ unchanged at 500 K — here the correction is small ($-1.75$ J K⁻¹mol⁻¹, about 2% of the total) only because the reactant and product heat capacities partly cancel; for reactions with a large $\Delta C_p$ the correction can dominate.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5–6 (Kirchhoff's law, heat-capacity integration of entropy and enthalpy of reaction).

Question 2 of 7: Entropy of Reaction for CO Oxidation at 500 K

Question 2: Entropy of Reaction for CO Oxidation at 500 K (20 marks)