Question

The first step in the commercial production of titanium metal is the reaction of rutile (TiO₂) with chlorine and graphite:

TiO_2(s) + 2Cl_2(g) + 2C _(s) $\rightarrow$ TiCl_4(l) + 2CO _(g)

Use the following data to calculate Keq for this reaction at 25° C.

Formula	?H°f (KJ/mol)	S° (J/K mol)
TiO2(s)	-944.0	50.6
Cl2(g)	0	223.0
C (s)	0	5.7
TiCl4(l)	-804.2	252.3
CO (g)	-110.5	197.6

Answer 1

TiO₂(s) + 2Cl₂ (g) + 2C (s) $\rightarrow$ TiCl₄(l) + 2CO (g)

$\Delta$H^o = $\Delta$H^of(products)-$\Delta$H^of(reactants)

$\Delta$H^o =$\Delta$H^of(TiCl₄) + 2 x $\Delta$H^of (CO) - ($\Delta$H^of(TiO₂)+2 x $\Delta$H^of(Cl₂) + 2 x $\Delta$H^of(C))

$\Delta$H^o = -804.2 + 2 x (-110.5) -(-944.0 + 2 x 0 +2 x 0)

$\Delta$H^o = -1025.2 + 944.0 kJ/mol

$\Delta$H^o = -81.2 kJ/mol

$\Delta$S^o = $\Delta$S^of(products)-$\Delta$S^of(reactants)

$\Delta$S^o =$\Delta$S^of(TiCl₄) + 2 x $\Delta$S^of (CO) - ($\Delta$S^of(TiO₂)+2 x $\Delta$S^of(Cl₂) + 2 x $\Delta$S^of(C))

$\Delta$S^o = 252.3 + 2 x (197.6) -(50.6 + 2 x 223.0 +2 x 5.7)

$\Delta$S^o = 647.5 - 508.6 J/K mol

$\Delta$S^o = 138.9 J/K mol

$\Delta$G^o = $\Delta$H^o - T$\Delta$S^o

$\Delta$G^o =  -81.2 kJ/mol - 298 x 138.9 J/K mol

$\Delta$G^o = -122.59 kJ/mol

$\Delta$G^o = -RT ln Keq

-122.59 x 10³ J/mol = - 8.314 J mol^-1 K^-1 x 298 K x ln Keq

ln Keq = 122590/2477.57

ln Keq = 49.47

Keq = e^49.47

Keq = 3.082 x 10²¹