21-Mat-A3 Structure and Characterization of Materials · Dec-10-Met-A3 2018
Question 6 of 7: Heat Balance (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 10-Met-A3, Metal Extraction Processes. Three hours, closed book, one approved calculator (Casio or Sharp). Seven problems of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All seven are solved here, because this set is a study resource rather than an exam script.
Note on the exam title. The printed exam header reads 10-Met-A3, Metal Extraction Processes. The content is extractive metallurgy — mineral processing, mass and heat balances, pyrometallurgical roasting, copper and aluminum production, and electrometallurgy — and is answered as such.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, smelting, mass and heat balances.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — ironmaking, non-ferrous smelting, electrometallurgy.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — pulp density, comminution, froth flotation.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, reaction enthalpies, electrochemistry.
W. G. Davenport et al., Extractive Metallurgy of Copper, 5th ed. — smelting/converting/refining and SX-EW flowsheets.
ASM Handbook, Vol. 2 (Properties and Selection: Nonferrous Alloys); CIM (Canadian Institute of Mining) practice literature for Canadian smelter/refinery practice (e.g. the historic Cominco smelter at Trail, BC, and Rio Tinto's Kitimat, BC aluminum smelter).
Given. Standard enthalpies of formation and molar heat capacities at 25 °C for the reactants and products of the chlorination of titanium dioxide, assumed independent of temperature.
Given data (25 °C)
Species
$\Delta H_f^{\circ}$ (kJ/mol)
$C_p$ (J·K−1·mol−1)
$\mathrm{TiO_2}$ (s)
−945
55.1
$\mathrm{Cl_2}$ (g)
0
33.9
C (graphite)
0
8.5
CO (g)
−110.5
29.1
$\mathrm{TiCl_4}$ (l)
−804
145.2
Find. (a) The balanced equation, (b) $\Delta H^{\circ}$ at 25 °C (298.15 K), and (c) $\Delta H^{\circ}$ at 150 °C (423.15 K).
Approach. Balance the equation from the stated reactants and products, use Hess's law with the tabulated formation enthalpies to get $\Delta H^{\circ}_{298}$, then apply Kirchhoff's law with the reaction's own $\Delta C_p$ (assumed temperature-independent, as stated) to shift the enthalpy to 423.15 K.
Part (a) — balance the equation. Titanium dioxide, graphite and chlorine gas form titanium tetrachloride and carbon monoxide. Balancing Ti, O, C and Cl atom-for-atom:
$$\boxed{\mathrm{TiO_2(s)+2\,C(gr)+2\,Cl_2(g)\longrightarrow TiCl_4(l)+2\,CO(g)}}$$
Check: Ti 1=1; O 2=2 (both O atoms leave in the two CO molecules, none in $\mathrm{TiCl_4}$); C 2=2; Cl 4=4. This is the industrial carbochlorination step of the chloride route to titanium metal and pigment-grade $\mathrm{TiO_2}$.
Part (b) — $\Delta H^{\circ}$ at 25 °C by Hess's law. Sum the formation enthalpies of products less reactants, each weighted by its stoichiometric coefficient:
$$\Delta H^{\circ}_{298}=\big[\Delta H_f^{\circ}(\mathrm{TiCl_4})+2\,\Delta H_f^{\circ}(\mathrm{CO})\big]-\big[\Delta H_f^{\circ}(\mathrm{TiO_2})+2\,\Delta H_f^{\circ}(\mathrm{C})+2\,\Delta H_f^{\circ}(\mathrm{Cl_2})\big]$$
$$\Delta H^{\circ}_{298}=\big[-804+2(-110.5)\big]-\big[-945+0+0\big]=-1025.0-(-945.0)$$
$$\boxed{\Delta H^{\circ}_{298}=-80.0\ \text{kJ/mol}}$$
The reaction is exothermic even before any external heating — a favourable sign for a continuous fluidised-bed chlorinator, which must sustain 900–1000 °C largely from its own reaction heat plus the fuel value of the carbon.
Part (c) — shift to 150 °C by Kirchhoff's law. First find the reaction's own heat-capacity change, using the same products-less-reactants pattern:
$$\Delta C_p=\big[C_p(\mathrm{TiCl_4})+2\,C_p(\mathrm{CO})\big]-\big[C_p(\mathrm{TiO_2})+2\,C_p(\mathrm{C})+2\,C_p(\mathrm{Cl_2})\big]$$
$$\Delta C_p=\big[145.2+2(29.1)\big]-\big[55.1+2(8.5)+2(33.9)\big]=203.4-139.9$$
$$\boxed{\Delta C_p=63.5\ \text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}}$$
Kirchhoff's law then shifts the enthalpy from 298.15 K to 423.15 K, using $\Delta C_p$ as given (temperature-independent):
$$\Delta H^{\circ}_{423}=\Delta H^{\circ}_{298}+\Delta C_p\,(T_2-T_1)=-80.0+\big(63.5\times10^{-3}\big)(423.15-298.15)$$
$$\Delta H^{\circ}_{423}=-80.0+(0.0635\times125.0)=-80.0+7.94$$
$$\boxed{\Delta H^{\circ}_{423}\approx-72.1\ \text{kJ/mol}}$$
Because $\Delta C_p>0$ (the products' heat capacity exceeds the reactants'), raising the reaction temperature makes $\Delta H$ less negative — the reaction gives up slightly less heat per mole at 150 °C than at 25 °C, though it remains comfortably exothermic either way.
