Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2014, Q7

Question 7 — Heat Balance (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. One mole of solid TiO₂ reacts with graphite and chlorine gas to give liquid TiCl₄ and gaseous CO, with standard formation enthalpies and constant (temperature-independent) heat capacities supplied for every species at 25 °C.

Given data (per mole of each species as written in the balanced equation)
Species	ΔH°_f (kJ/mol)	C_p (J K⁻¹ mol⁻¹)
TiO₂(s)	−945	55.06
Cl₂(g)	0 (element)	33.91
C (graphite)	0 (element)	8.53
CO(g)	−110.5	29.12
TiCl₄(l)	−804	145.2

Find. (a) The balanced equation; (b) ΔH° of reaction at 298.15 K (25 °C); (c) ΔH° of reaction at 403.15 K (130 °C).

Approach. Balance the chlorination/reduction reaction on Ti, O, Cl and C; get ΔH°(298 K) from the standard formation enthalpies (elements taken as zero); then use Kirchhoff's law with the reaction's ΔC_p (assumed constant) to shift that enthalpy to 403 K.

Balance the equation. One TiO₂ supplies the two oxygen atoms needed to form two CO, which fixes two C and, via the four Cl needed for TiCl₄, two Cl₂: $$\boxed{\mathrm{TiO_2(s) + 2\,C(gr) + 2\,Cl_2(g) \longrightarrow TiCl_4(l) + 2\,CO(g)}}$$ Checking every element: Ti 1=1; O 2=2 (2×CO); C 2=2; Cl 4=4 (2×Cl₂ = 2×2 in TiCl₄). This is the carbochlorination step of the chloride-route Kroll process for titanium metal.
Compute ΔH° at 298 K from formation enthalpies. With elements (Cl₂, C) at zero, $$\Delta H^\circ_{298}=\left[\Delta H^\circ_f(\mathrm{TiCl_4})+2\,\Delta H^\circ_f(\mathrm{CO})\right]-\left[\Delta H^\circ_f(\mathrm{TiO_2})\right]$$ $$\Delta H^\circ_{298}=\left[(-804)+2(-110.5)\right]-\left[-945\right]=(-1025)-(-945)$$ $$\boxed{\Delta H^\circ_{298}=-80.0\ \text{kJ per mole TiO}_2\ \text{reacted}}$$ The reaction is exothermic, consistent with chlorination reactions of this type running with little or no external heating once started.
Compute the reaction's ΔC_p. Summing products minus reactants, each weighted by its stoichiometric coefficient, $$\Delta C_p=\left[145.2+2(29.12)\right]-\left[55.06+2(33.91)+2(8.53)\right]$$ $$\Delta C_p=203.44-139.94$$ $$\Delta C_p=+63.5\ \text{J K}^{-1}$$ ΔC_p is positive, so the reaction becomes less exothermic as temperature rises.
Apply Kirchhoff's law to shift ΔH° from 298 K to 403 K. With ΔC_p assumed constant over the interval, $$\Delta H^\circ_{T_2}=\Delta H^\circ_{T_1}+\Delta C_p\,(T_2-T_1)$$ Using $T_1=298.15$ K and $T_2=403.15$ K, the 25 °C-to-130 °C span converts to exactly $\Delta T=105$ K regardless of the zero-point offset: $$\Delta H^\circ_{403}=-80\,000\ \text{J} + (63.5\ \text{J K}^{-1})(105\ \text{K})=-80\,000+6667.5$$ $$\boxed{\Delta H^\circ_{403}=-73\,332.5\ \text{J} = -73.33\ \text{kJ per mole TiO}_2\ \text{reacted}}$$
Sanity-check the direction of the shift. Because ΔC_p is positive, the products' sensible-heat requirement grows faster with temperature than the reactants', so less net heat is released at the higher temperature — the reaction becomes 6.7 kJ less exothermic per 105 K, which is exactly what the small positive ΔC_p of 63.5 J/K predicts for a 105 K rise ($63.5\times105/1000=6.67$ kJ), confirming the arithmetic.

Final results — Question 7
Part	Quantity	Result
(a)	Balanced equation	TiO₂(s) + 2C(gr) + 2Cl₂(g) → TiCl₄(l) + 2CO(g)
(b)	ΔH° at 25 °C (298 K)	−80.0 kJ
(c)	ΔH° at 130 °C (403 K)	−73.33 kJ
—	Supporting: ΔC_p of reaction	+63.5 J/K

Topic. Reaction enthalpy from formation data and its temperature dependence (Kirchhoff's law).

Key relations. $\Delta H^\circ_{rxn}=\sum \nu_i\,\Delta H^\circ_{f,i}(\text{products})-\sum \nu_i\,\Delta H^\circ_{f,i}(\text{reactants})$; Kirchhoff's law $\Delta H^\circ_{T_2}=\Delta H^\circ_{T_1}+\Delta C_p\,(T_2-T_1)$ for constant $\Delta C_p$.

Why this works. Enthalpy is a state function, so ΔH° at any temperature can always be built from a reference value plus the net sensible-heat difference between products and reactants as the system is carried from the reference temperature to the one of interest — there is no need to know any enthalpy value directly at 130 °C, only the heat capacities. This is exactly the calculation that sizes the cooling or heating duty of a chlorinator: knowing how ΔH° itself shifts with operating temperature is what lets a plant predict whether a reactor needs external heat input as it scales up or changes throughput.

Common pitfall. The two recurring errors are forgetting that elemental Cl₂ and graphite have zero standard formation enthalpy by convention (so they must not be omitted from ΔC_p, only from the ΔH°_f sum), and using the Celsius temperatures directly in Kirchhoff's law instead of converting to kelvin — here it happens to make no difference because $\Delta T$ is the same in either scale, but only because both temperatures are consistently shifted by the same 273.15 K offset; the safer habit is to convert first every time.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (heat capacity, Kirchhoff's law) and Ch. 6 (Hess's law, standard enthalpies of formation); Habashi, Textbook of Pyrometallurgy, Ch. 8 (chlorination metallurgy, the Kroll process).

Question 7 of 7: Heat Balance (20 marks)

Question 7 — Heat Balance (20 marks)