Given the following bond-dissociation energies, calculate the average bond enthalpy for the Ti-Cl bond.

AH(kJ/mol) TiCl4(g) TiCl;(g) TİC,(g) TICI(g) TiCl3(g) + Cl(g) TİCI,(g) + CI(g) TiCI(g) + CI(g) Ti(g) + CI(g) 335 423 444 519

Solutions

Expert Solution

Average bond energy of Ti-C1 bond:

TiC14(g) → TiC13(g) + C1(g) ΔH = 335kJ

TiCl3(g) → TiCl2(g) + Cl(g) ΔH = 423 kJ

TiCl2(g) → TiCl(g) + Cl(g) ΔH = 444 kJ

TiCl(g) → Ti(g) + Cl(g) ΔH = 519 kJ

ΔH = 1721 kJ

Average Ti-Cl bond = 1721 kJ/4

= 430.25 kJ

Average bond enthalpy of Ti-Cl = 430 kJ.

Average bond energy of Ti-C1 bond:

TiC14(g) → TiC13(g) + C1(g) ΔH = 335kJ

TiCl3(g) → TiCl2(g) + Cl(g) ΔH = 423 kJ

Junaid answered 1 year ago

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