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.

Options

A. `-128.02 "KJ mole"^(-1)`
B. `+12.802 "KJ mol"^(-1)`
C. `+128.02 "KJ mol"^(-1)`
D. `-12.802 KJ mol"^(-1)`

Answer

C

Explanation

`C_((s))+2S_((s))rarr CS_(2(l)) , Delta H_(f) = ?`
`Delta H_(f)=Sigma (Delta H_("Comb."))_(R )-Sigma (Delta H_("Comb."))_(P)`
`Delta H_(f)=[1xx(-393.3)+2xx(-293.72)]-(-1108.76)`
`Delta H_(f)=+128.02 kJ mol^(-1)`