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

Required equation is
`C(s)+2S(s)toCS_(2)(l),DeltaH_(f)=?` ltbr Given
`C(s)+O_(2)(g)toCO_(2)(g)" "(DeltaH=-393.3kJ)` . . .(i)
`S(s)+O_(2)(g)toSO_(2)(g)" "(DeltaH=-293.72kJ)` . . . (ii)
`CS_(2)(l)+3O_(2)(g)toCO_(2)(g)+2SO_(2)(g)" "(DeltaH=-1108.76kJ)`
First Method: Multiply the eq. (ii) by 2
`2S(s)+2O_(2)(g)to2SO_(2)(g)" "(DeltaH=-587.44kJ)` . . . (iv)
Adding eqs. (i) and (iv) and subtracting eq. (iii),
`[C(s)+2S(s)+3O_(2)(g)-CS_(2)(l)-3O_(2)(g)toCO_(2)(g)+2SO_(2)(g)-CO_(2)-2SO_(2)]`
`C(s)+2S(s)toCS_(2)(l)`
this is the required equation.
Thus, `DeltaH_(f)=-393.3-587.44+1108.76=128.02kJ`
Standard heat of formation of `CS_(2)(l)=128.02kJ`