Given the enthalpy of formation of `CO_(2)(g)` is -94.0 KJ, of CaO(s) is -152 KJ, and the enthalpy of the reaction `CaCO_(3)(s) rarr CaO(s)+CO_(2)(g)` is 42KJ, the enthalpy of formation of `CaCO_(3) (s)` is

Options

A. `-42` KJ
B. `-202` KJ
C. `+202` KJ
D. `-288` KJ

Answer

D

Explanation

`CaCO_(3(s))rarr CaO_((s))+CO_(2(g)) , Delta H=42 kJ`
`Delta H=Sigma (Delta E_(f))_(P)-Sigma (Delta E_(f))_(R )`
`42=[-152)+(-94)]-Delta H_(f)(CaCO_(3))`
`therefore H_(f)(CaCO_(3))=-246-42 =-288 kJ`