Using the data given below, pick the correct statement about the reaction:
`TiO _(2) (s) + 2Cl _(2) g) to TiCl _(4) (l) + O _(2) (g)` [Given that `Delta _(f) H ^(@)` for `TiO _(2 ) (s), TiCl _(4) (l), Cl _(2) (g) and O _(2) (g)` are ` -944.7, - 804.22, 0.0, 0.0 kJ mol ^(-1)` respectively. Also `Delta S ^(@)` for ` TiO _(2) (s), TICl _(4) (l), Cl _(2) (g) and O _(2) (g) ` are `50.3,252.3, 233.0 , 205.1 J mol ^(-1) K ^(-1),` respectively.]

Options

A. The reaction is exothermic at standard conditions.
B. The standard entropy is favourable for the reaction in forward direction.
C. The reaction is not spontaneous at standard condition at `25 ^(@)C`
D. Reaction becomes spontaneous at higher temperature.

Explanation

`Delta H ^(@) = [ H _( TiCl _(4)(l)) ^(@) + H _(O _(2) (s)) ^(@) ] - [ H _(TiO _(2)(s)) ^(@) + H _(Cl _(2(g))) ^(@)] = (- 804. 2 + 0) - ( - 944.7 + 0) = 140 . 5 kJ mol ^(-1)`
`Delta S ^(@) = [ S _( TiCl_(4)(l)) ^(@) + S _(O _(2 (g))) ^(@) ] - [ S _( TiO _(2) (s)) ^(@) + S _(Cl _(2) (s))^(@) ]= (+ 252. 3 + 205. 1 ) - ( + 50 . 3 + 2 xx 233)`
`=- 58. 9 J mol ^(-1) K ^(-1) =- 0. 0 5 89 kJ mol ^(-1) K ^(-1) , Delta G ^(@) = Delta H ^(@) - T Delta S ^(@) = 140. 5 - 298 xx (-0.0589)`
`= - 58.06 kJ mol ^(-1),` Since charge in free energy is positive, the reaction is nto spontaneous.