Transition metal carbides, such as TiC, are widely used for the production of cutting and grinding t — Physical Chemistry — Electrochemistry Chemistry Question
Titanium Carbide – A High–Tech Solid
Transition metal carbides, such as TiC, are widely used for the production of cutting and grinding tools, because they are very hard, very corrosion–resistant and have high melting points. Apart from these properties, titanium carbide has a high electric conductivity that is almost independent of temperature, so that it is important in the electronics industry.
What kind of structure is TiC likely to adopt, if the radii are r(Ti4+) = 74.5 and r(C4–) = 141.5 pm?
Model Answer
r(Ti4+)/ r(C4–) = 0.527 NaCl – type
TiC is technically obtained from TiO2 by the reduction with carbon. The enthalpy change of this reaction can directly be measured only with difficulty. However, the heats of combustion of the elements and of TiC can be measured directly. As energy is always conserved and the enthalpy change for a given process does not depend on the reaction pathway (this special application of the First Law of Thermodynamics is often referred to as Hess´s Law), the missing thermodynamic data can be calculated.
Calculate the enthalpy of reaction of the technical production process of TiC:
TiO2 + 3 C TiC + 2 CO
ΔfH(TiO2) = – 944.7 kJ mol–1
ΔfH(CO) = – 110.5 kJ mol–1
ΔrH(TiC + 3/2 O2 TiO2 + CO) = – 870.7 kJ mol–1
Model Answer
(a) TiO2 + CO TiC + 1.5 O2 rH = 870.7 kJ mol–1
(b) C + 0.5 O2 CO rH = – 110.5 kJ mol–1
(a) + 3 (b) :
TiO2 + 3 C TiC + 2 CO
rH = 870.7 + 3(–110.5) kJ mol–1
rH = 539.2 kJ mol–1
In 1919, Born and Haber independently applied the First Law of Thermodynamics to the formation of solids from their elements. In this way, getting exact information about lattice energies for solids was possible for the first time.
Potassium chloride is isotypic to TiC and crystallizes in the NaCl structure.
Use the given data to construct a thermodynamical Born–Haber–cycle of the formation of potassium chloride from its elements and calculate the lattice energy of potassium chloride.
sublimation enthalpy for potassium: K(s) K(g) ΔsubH = 89 kJ mol–1
dissociation energy of chlorine: Cl2(g) 2 Cl ΔdissH = 244 kJ mol–1
electron affinity of chlorine: Cl(g) + e– Cl(g) ΔEAH = – 355 kJ mol–1
ionization energy of potassium: K(g) K+(g) + e– ΔIEH = 425 kJ mol–1
enthalpy of formation for KCl: K(s) + ½ Cl2(g) KCl(s) ΔfH = –438 kJ mol–1
Model Answer
– UL = ΔsubH + ΔIEH + 0.5 ΔdissH + ΔEAH – ΔfH
UL = – (89 + 425 + 122 – 355 + 438 kJ mol–1)
UL = – 719 kJ mol–1
(If the lattice energy is defined in the opposite way the result will be + 719 kJ mol–1)