Reaction Entropy of Ammonia Synthesis at 700 K: 21-Mat-A1 Thermodynamics, May 2017, Q3

Question 3: Reaction Entropy of Ammonia Synthesis at 700 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 $3H_2(g)+N_2(g)=2NH_3(g)$; $S^\circ_{298}$ and $C_p(T)$ for all three gases as tabulated in the question; target temperature $T=700$ K.

Find. $\Delta S^\circ_R$ at 700 K.

Approach. Compute $\Delta S^\circ_{298}$ from the tabulated standard entropies, form $\Delta C_p(T)=2C_p(NH_3)-3C_p(H_2)-C_p(N_2)$, then integrate $\Delta S^\circ(700)=\Delta S^\circ_{298}+\int_{298}^{700}\Delta C_p/T\,dT$ using Kirchhoff's law for reaction entropy.

Entropy of reaction at 298 K. $$\Delta S^\circ_{298}=2S^\circ(NH_3)-3S^\circ(H_2)-S^\circ(N_2)=2(130.7)-3(197.7)-191.6=\boxed{-523.3\ \text{J}\,\text{K}^{-1}}.$$ The large negative value reflects the loss of moles of gas (4 mol reactant gas → 2 mol product gas).
Heat-capacity difference. Combining the three $C_p(T)$ polynomials term by term, $$\Delta C_p(T)=2C_p(NH_3)-3C_p(H_2)-C_p(N_2)=-40.3-0.098\,T+3.9\times10^{-4}T^2\ \ \text{J}\,\text{K}^{-1}.$$
Integrate Kirchhoff's law. With $\Delta C_p=a+bT+cT^2$ ($a=-40.3$, $b=-0.098$, $c=3.9\times10^{-4}$), $$\int_{298}^{700}\dfrac{\Delta C_p}{T}\,dT=a\ln\!\left(\dfrac{700}{298}\right)+b(700-298)+\dfrac{c}{2}\left(700^2-298^2\right)=4.42\ \text{J}\,\text{K}^{-1}.$$
Reaction entropy at 700 K. $$\Delta S^\circ_R(700\ \text{K})=\Delta S^\circ_{298}+\int_{298}^{700}\dfrac{\Delta C_p}{T}\,dT=-523.3+4.4=\boxed{-518.9\ \text{J}\,\text{K}^{-1}}.$$ The correction is small (≈1 % of $\Delta S^\circ_{298}$) because $\Delta C_p(T)$ stays modest over this range and the $\ln(T_2/T_1)$ and linear/quadratic terms partly offset.

Quantity	Value
$\Delta S^\circ_{298}$	−523.3 J K⁻¹
$\displaystyle\int_{298}^{700}\Delta C_p/T\,dT$	+4.4 J K⁻¹
$\Delta S^\circ_R$ at 700 K	−518.9 J K⁻¹

Concept & Theory

Topic. Temperature dependence of a reaction's standard entropy via Kirchhoff's law.

Key relations. $\Delta S^\circ_{298}=\sum\nu_iS^\circ_i$; Kirchhoff's law $\left(\dfrac{\partial \Delta S^\circ}{\partial T}\right)_P=\dfrac{\Delta C_p}{T}$, integrated as $\Delta S^\circ(T_2)=\Delta S^\circ(T_1)+\int_{T_1}^{T_2}\Delta C_p/T\,dT$.

Why this works. $S^\circ$ tabulated at 298 K is only the reference point; because $S$ is a state function with $dS=C_p\,dT/T$ at constant pressure, the same accounting applies to the reaction as a whole once the products' and reactants' heat capacities are combined into $\Delta C_p(T)$ — integrating $\Delta C_p/T$ (not $\Delta C_p$) is what correctly weights the low-T contribution more heavily.

Common pitfall. Integrating $\Delta C_p\,dT$ (giving $\Delta H$ corrections) instead of $\Delta C_p/T\,dT$ when the target is entropy — the two Kirchhoff integrals look similar but use different integrands.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5–6 (temperature dependence of thermodynamic functions, Kirchhoff's law).

Question 3 of 7: Reaction Entropy of Ammonia Synthesis at 700 K

Question 3: Reaction Entropy of Ammonia Synthesis at 700 K (20 marks)