Calcination of Sodium Bicarbonate — Reaction Equilibrium: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2016, Q4

Question 4: Calcination of Sodium Bicarbonate — Reaction Equilibrium (Part B — 25 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. Solid–solid–gas decomposition with two gaseous products (CO₂, H₂O) and pure-solid reactant/product (activities = 1). Started from vacuum, so the two gases are produced in equimolar amounts and each partial pressure is half the total: $p_{CO_2}=p_{H_2O}=P_{tot}/2$. Data: $P_{tot}=0.826$ kPa at 303.15 K and 166.97 kPa at 383.15 K. Reference $P^\circ=100$ kPa (1 bar), $R=8.314$ J/mol·K.

Find. (a) $\Delta H_{rxn}$; (b) $\ln K(T)$; (c) the temperature at which $p_{CO_2}=1$ bar.

Figure 3 — van’t Hoff plot: $\ln K$ is linear in $1/T$ with slope $-\Delta H/R$. The line reaches $\ln K=0$ (i.e. $K=1$, $p_{CO_2}=p_{H_2O}=1$ bar) at 113.4 °C.

Approach. Write $K$ in terms of the two partial pressures, evaluate it at both temperatures, then use the integrated van’t Hoff equation (constant $\Delta H$) to get $\Delta H$ and the $\ln K$–$1/T$ line, and solve that line for the temperature where $K=1$.

Equilibrium constant at each temperature. With solids at unit activity, $K=(p_{CO_2}/P^\circ)(p_{H_2O}/P^\circ)=(P_{tot}/2P^\circ)^2$: $$K_1=\left(\frac{0.826}{2(100)}\right)^2=1.71\times10^{-5},\qquad K_2=\left(\frac{166.97}{2(100)}\right)^2=0.697.$$
(a) Heat of reaction (integrated van’t Hoff). $\ln\dfrac{K_2}{K_1}=-\dfrac{\Delta H}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$, so $$\Delta H=-\frac{R\ln(K_2/K_1)}{\left(1/T_2-1/T_1\right)}=-\frac{8.314\,\ln\!\left(\tfrac{0.697}{1.71\times10^{-5}}\right)}{\left(\tfrac{1}{383.15}-\tfrac{1}{303.15}\right)}=\boxed{+128\ \text{kJ/mol.}}$$ The large positive value confirms a strongly endothermic calcination.
(b) Equilibrium constant as a function of T. With $\Delta H$ constant, $\ln K=-\dfrac{\Delta H}{RT}+C$. The slope is $-\Delta H/R=-15417$ K, and matching at $T_1$ gives $C=\ln K_1+15417/T_1=39.88$: $$\boxed{\ln K=39.88-\frac{15417}{T}\ \ (T\text{ in K}).}$$
(c) Temperature for $p_{CO_2}=1$ bar. If $p_{CO_2}=1$ bar then, being equimolar, $p_{H_2O}=1$ bar too, so $K=(1)(1)=1$ and $\ln K=0$: $$0=39.88-\frac{15417}{T}\;\Rightarrow\;T=\frac{15417}{39.88}=386.6\ \text{K}=\boxed{113.4\ ^\circ\text{C.}}$$

The result is intuitive: 1 bar CO₂ corresponds to a total decomposition pressure of 2 bar, which the bicarbonate reaches just above 110 °C — consistent with the measured 166.97 kPa (≈ 0.83 bar CO₂) already at 110 °C.

Quantity	Value
$K$ at 30 °C / 110 °C	$1.71\times10^{-5}$ / 0.697
(a) Heat of reaction $\Delta H_{rxn}$	+128 kJ/mol (endothermic)
(b) $\ln K(T)$	$39.88-15417/T$
(c) $T$ for $p_{CO_2}=1$ bar	386.6 K = 113.4 °C

Concept & Theory

Topic. Heterogeneous reaction equilibrium (decomposition pressure) and the van’t Hoff temperature dependence of $K$.

Key relations. $K=\prod(p_i/P^\circ)^{\nu_i}$ with pure-solid activities = 1; van’t Hoff $\dfrac{d\ln K}{dT}=\dfrac{\Delta H}{RT^2}$, integrated form $\ln(K_2/K_1)=-\dfrac{\Delta H}{R}(1/T_2-1/T_1)$; linear model $\ln K=C-\Delta H/(RT)$.

Why this works. Because the reactant and product solids are pure phases, only the gaseous species appear in $K$; starting from vacuum forces $p_{CO_2}=p_{H_2O}$, so a single measured total pressure fully determines $K$. Two $(T,P)$ points then fix the two constants in the van’t Hoff line, from which any other equilibrium condition follows.

Common pitfall. Using the total pressure directly in $K$ instead of the individual partial pressures (each is $P_{tot}/2$), or forgetting to divide by the reference pressure so that $K$ carries spurious units.

Source. Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics, 8th ed., §13.4–13.5 (equilibrium constant, effect of temperature); heterogeneous equilibria §13.9.

Question 4 of 6: Calcination of Sodium Bicarbonate — Reaction Equilibrium

Question 4: Calcination of Sodium Bicarbonate — Reaction Equilibrium (Part B — 25 marks)