Temperature Dependence of ΔS° — CO Oxidation at 500 K: 21-Mat-A1 Thermodynamics, May 2015, Q3

Question 3: Temperature Dependence of ΔS° — 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.

Quantity	CO(g)	CO₂(g)	O₂(g)
$S^\circ_{298}$ (J/K·mol)	197.7	213.7	205.1
$C_p(T)$ (J/K·mol)	$31.1-1.5\times10^{-2}T+3.1\times10^{-5}T^2-1.5\times10^{-8}T^3$	$18.9+7.9\times10^{-2}T-6.8\times10^{-5}T^2+2.4\times10^{-8}T^3$	$30.8-1.2\times10^{-2}T+2.4\times10^{-5}T^2$

Find. $\Delta S^\circ_R$ at $T=500$ K for $\text{CO(g)}+\tfrac12\text{O}_2\text{(g)}=\text{CO}_2\text{(g)}$.

Approach. Compute $\Delta S^\circ_{298}$ from the tabulated standard entropies, then apply the Kirchhoff-type correction $\Delta S^\circ(500)=\Delta S^\circ(298)+\int_{298}^{500}\dfrac{\Delta C_p(T)}{T}dT$ using $\Delta C_p(T)=C_p(\text{CO}_2)-C_p(\text{CO})-\tfrac12C_p(\text{O}_2)$.

$\Delta S^\circ_{298}$. $$\Delta S^\circ_{298}=S^\circ(\text{CO}_2)-S^\circ(\text{CO})-\tfrac12S^\circ(\text{O}_2)=213.7-197.7-\tfrac12(205.1)=\boxed{-86.55\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
Assemble $\Delta C_p(T)$. Collecting the constant, $T$, $T^2$ and $T^3$ coefficients of $C_p(\text{CO}_2)-C_p(\text{CO})-\tfrac12C_p(\text{O}_2)$ term by term: $$\Delta C_p(T)=-27.6+0.100\,T-1.11\times10^{-4}T^2+3.9\times10^{-8}T^3\ \ \text{J}\,\text{K}^{-1}\text{mol}^{-1}.$$
Integrate $\Delta C_p/T$ from 298 to 500 K. Term by term, $\int(a/T+b+cT+dT^2)\,dT=a\ln T+bT+\tfrac{c}{2}T^2+\tfrac{d}{3}T^3$, evaluated between the limits: $$\Delta S_{corr}=-27.6\ln\!\frac{500}{298}+0.100(500-298)-5.55\times10^{-5}(500^2-298^2)+1.3\times10^{-8}(500^3-298^3)=\boxed{-1.75\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$
$\Delta S^\circ_{500}$. $$\Delta S^\circ_{500}=\Delta S^\circ_{298}+\Delta S_{corr}=-86.55-1.75=\boxed{-88.3\ \text{J}\,\text{K}^{-1}\text{mol}^{-1}}.$$ The entropy decrease is expected: 1.5 mol of gas (CO + ½O₂) becomes 1 mol of gas (CO₂), so the moles of gas fall — the dominant driver of $\Delta S^\circ<0$ here — and the small extra correction from 298 to 500 K barely changes that.

Quantity	Value
$\Delta S^\circ_{298}$	−86.55 J/K·mol
Kirchhoff correction (298→500 K)	−1.75 J/K·mol
$\Delta S^\circ_{500}$	−88.3 J/K·mol

Concept & Theory

Topic. Temperature dependence of a reaction's standard entropy (Kirchhoff's law applied to $\Delta S^\circ$).

Key relations. $\Delta S^\circ(T_2)=\Delta S^\circ(T_1)+\int_{T_1}^{T_2}\dfrac{\Delta C_p}{T}dT$; $\Delta C_p=\sum n_iC_{p,i}(\text{products})-\sum n_iC_{p,i}(\text{reactants})$.

Why this works. $S$ is a state function, so $(\partial S/\partial T)_P=C_p/T$ can be integrated for products and reactants separately (or, equivalently, for $\Delta C_p$ directly) between any two temperatures — this is the entropy analogue of Kirchhoff's law for $\Delta H^\circ(T)$.

Common pitfall. Integrating $\Delta C_p\,dT$ (no $T$ in the denominator) instead of $\Delta C_p/T\,dT$ — the former is the $\Delta H^\circ(T)$ correction, not $\Delta S^\circ(T)$'s — or dropping the ½ stoichiometric coefficient on $C_p(\text{O}_2)$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5–6 (heat capacity and the temperature dependence of $\Delta H^\circ$, $\Delta S^\circ$).

Question 3 of 7: Temperature Dependence of ΔS° — CO Oxidation at 500 K

Question 3: Temperature Dependence of ΔS° — CO Oxidation at 500 K (20 marks)