Question

( A=pi i c l_{4}+2 H_{20} rightarrow P_{10} q+4 H C J )
( Delta H_{f} operatorname{Tic}_{4(q)}=-763.2 ) ks/mole
( Delta H_{f} T_{02}=-949.7 mathrm{kJ} / mathrm{mol} e_{1} )
( Delta H_{f} quad t_{20}=-241.8 quad k J / ) mole
( Delta mathrm{H}_{t} quad mathrm{H}_{mathrm{Cl}} )
( =-92.3 mathrm{log} / mathrm{mole} )
Dreaction ( =Delta mathrm{T}_{10} mathrm{O}_{2}+4 mathrm{D}_{mathrm{HC}}-2 mathrm{D}_{mathrm{H}_{2} mathrm{O}}-Delta_{mathrm{TiCl}_{4}} )
( =-949.7+4(-92.3)-2(-241.8)-(-763.2) )
( =-949.7-369.2+483.6+763.2 )
( =-67.1 )

# 3. Use the given standard enthalpies of formation to determine the heat of reaction of the following reaction: TiCl(g) + 2 H2O(g) + TiO2(g) + 4 HCl(g) AH TiC14(g)=-763.2 kJ/mole AH® TiO2(g) = -944.7 kJ/mole AH® H2O(g) = -241.8 kJ/mole AH® HCl(g) = -92.3 kJ/mole (A)-278.1 kJ (B) +369.2 kJ (C) +67.1 kJ (D) -67.1k

Solution