Calculate the standard heat of formation of carbon disulphide `(l)`. Given that the standard heats of combusion of carbon `(s)`, sulphur `(s)` and acrbon disulphide `(l)` are `-390, 290.0`, and `-1100.0kJ mol^(-1)`. Respectively.

Explanation

Required equation is
`C(s) +2S(s) rarr CS_(2)(l), Delta_(f)H^(Theta) = ? `
Given
a. `C(s) + O_(2) (g) rarr CO_(2)(g), DeltaH^(Theta) =- 390.0kJ`
b. `S(s) + O_(2) (g) rarr SO_(2)(g), DeltaH^(Theta) =- 290.00kJ`
c. `CS_(2)(l) + 3O_(2)(g) rarr CO_(2)(g) + 2SO_(2)(g), DeltaH^(Theta) =- 1100.00kJ`
Multiply the equation (b) by `2`.
d. `2S(s) +2O_(2)(g) rarr 2SO_(2)(g), DeltaH =- 580.0kJ`
Adding equation (a) and (d) and subtracting (c ),
`[C(s) +2S(s) + 3O_(2)(g) -CS_(2)(l) -3O_(2)(g)`
`rarr CO_(2)(g) + 2SO_(2)(g) - CO_(2)-2SO_(2)]`
`C(s) +2S(s) rarr CS_(2)(l)`
This is the required equation.
Thus `Delta_(f)H^(Theta) =- 390.0- 580.0 + 1100.0 = 130.00 kJ`
Standared heat of formation of `CS_(2)(l) = 130.00 kJ`