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.

Explanation

The given equations are :
(i) `C(S) + O_(2)(g) to CO_(2)(g) , Delta_(r)H^(@) = -393.3 kJ`
(ii) `S(s) + O_(2)(g) to SO_(2)(g), Delta_(r)H^(@) = -293.72 kJ`
`CS_(2)(l) + 3O_(2)(g) to CO_(2)(g) + 2SO_(2)(g) , Delta_(r)H^(@) = -1108.76 kJ`
The required equation is
`C(s) + 2S(g) to CS_(2)(l), Delta_(r)H^(@) = ? `
Multiply eq. (ii) by 2 and add to eq.(i)
(i) `C(s) + O_(2)(g) to CO_(2)(g), Delta_(r)H^(@) = -393.3 kJ`
`(ii)xx 2` "`2S(s) + 2O_(2)(g) to 2SO_(2)(g) , Delta_(r)H^(@) = -587.44 kJ`
Adding `C(s) + 2S(s) + 3O_(2)(g) to CO_(2)(g) + 2SO_(2)(g) Delta_(r)H^(@) = -980.74 kJ`
Subtract eq.(iii)
(iii) `CS_(2)(l) + 3O_(2)(s) to CO_(2)(g) + 2SO_(2)(g) Delta_(r)H^(@) = -1108.76 kJ`
Subtracting `C(s) + 2S(s) to CS_(2)(l) Delta_(r)H^(@) = + 128.02 kJ`