Question 5 of 7: Equilibrium Constant and Reaction Thermodynamics for NO₂ Dissociation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, H₂, H₂O(g) and CO₂ from the NIST-JANAF Thermochemical Tables.
Question 5: Equilibrium Constant and Reaction Thermodynamics for NO₂ Dissociation (20 marks)
Given. 1 mol NO₂ dissociates as $\text{NO}_2=\text{NO}+\tfrac12\text{O}_2$ at a total pressure equal to the standard-state pressure $P^\circ=10^5$ Pa; the analyzed ratio $P_{NO}/P_{NO_2}=0.6$ at 500 K and $3.6$ at 1000 K.
T (K)
PNO/PNO₂
500
0.6
1000
3.6
Find. $K_p$ at 500 K and 1000 K, then $\Delta H^\circ$ (assumed T-independent) and $\Delta G^\circ$ at 300 K.
Approach. Let $x$ be the extent of dissociation per mole of NO₂ fed; express every species' mole fraction in terms of $x$, use the given ratio to solve for $x$ at each $T$, then build $K_p=y_{NO}\sqrt{y_{O_2}}/y_{NO_2}$ (valid directly in mole fractions because the total pressure equals $P^\circ$). Van't Hoff's equation between the two $K_p$ values gives $\Delta H^\circ$; $\Delta G^\circ=-RT\ln K_p$ at either point gives $\Delta S^\circ$, and both together extrapolate $\Delta G^\circ$ to 300 K.
Extent of reaction from the ratio. With 1 mol NO₂ feed: $\text{NO}_2=1-x$, $\text{NO}=x$, $\text{O}_2=x/2$ mol, total $=1+x/2$. Mole fractions are proportional to moles at fixed total $P$, so $P_{NO}/P_{NO_2}=x/(1-x)$. At 500 K, $x/(1-x)=0.6\Rightarrow x=\boxed{0.375}$; at 1000 K, $x/(1-x)=3.6\Rightarrow x=\boxed{0.7826}$.
(a) $K_p$ at 500 K. Mole fractions: $y_{NO_2}=0.5263$, $y_{NO}=0.3158$, $y_{O_2}=0.1579$. Since $P_{tot}=P^\circ$,
$$K_p=\frac{y_{NO}\sqrt{y_{O_2}}}{y_{NO_2}}=\frac{(0.3158)\sqrt{0.1579}}{0.5263}=\boxed{0.238}.$$
(b) $K_p$ at 1000 K. Mole fractions: $y_{NO_2}=0.1563$, $y_{NO}=0.5625$, $y_{O_2}=0.2813$, giving
$$K_p=\frac{(0.5625)\sqrt{0.2813}}{0.1563}=\boxed{1.909}.$$
(c) $\Delta H^\circ$ from van't Hoff. $\ln(K_2/K_1)=-\dfrac{\Delta H^\circ}{R}\left(\dfrac1{T_2}-\dfrac1{T_1}\right)$ with $T_1=500$, $T_2=1000$ K:
$$\Delta H^\circ=\frac{-R\ln(K_2/K_1)}{1/T_2-1/T_1}=\frac{-(8.314)\ln(1.909/0.238)}{1/1000-1/500}=\boxed{+17.3\ \text{kJ/mol}}$$
(endothermic, consistent with dissociation increasing strongly with $T$).
(d) $\Delta G^\circ$ at 300 K. First evaluate $\Delta G^\circ=-RT\ln K_p$ at a data point, e.g. 500 K: $\Delta G^\circ_{500}=-(8.314)(500)\ln(0.238)=+5.96$ kJ/mol. With $\Delta H^\circ$ assumed constant, $\Delta S^\circ=(\Delta H^\circ-\Delta G^\circ_{500})/500=(17{,}300-5960)/500=22.7$ J/mol·K. Extrapolating linearly to 300 K:
$$\Delta G^\circ_{300}=\Delta H^\circ-T\Delta S^\circ=17{,}300-(300)(22.7)=\boxed{+10.5\ \text{kJ/mol}}.$$
Quantity
Result
(a) Kp (500 K)
0.238
(b) Kp (1000 K)
1.909
(c) ΔH°
+17.3 kJ/mol
(d) ΔG° (300 K)
+10.5 kJ/mol
Check
The container is described only by its initial loading pressure; because the amount of gas changes as NO₂ dissociates, the working assumption above is that the mixture's total pressure at equilibrium remains at the reference pressure $P^\circ=10^5$ Pa, which is what lets $K_p$ be built directly from mole fractions. This is the standard simplification for this class of textbook problem.
