Calculate the standard heat of formation of carbon disulphide (l). Given that the standard heats of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are -393.3, -293.72 and -1108.76 kJ `mol^(-1)` respectively.

Answer

`-879.1 kJ mol^(-1)`

Explanation

Aim `: C(s) + 2S(s) rarr CS_(2)(l), DeltaH = ?`
Given `: (i) CS_(2)(l) + 3O_(2)(g) rarr CO_(2)(g) + 2SO_(2)(g), Delta H - 110.2 kJ mol^(-1)`
(ii) `S(s) + O_(2)(g) rarr SO_(2)(g) , Delta H= - 297.4 kJ mol^(-1)`
(iii) `C(s) +O_(2)(g) rarr CO_(2)(g), Delta H =- 394.5 kJ mol^(-1)`
Eqn. (iii) ` +2 xx ` Eqn. (ii) - Eqn. (i) gives the required result.