Answer: +178.3 kJ
Explanation:
The chemical equation  follows:
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(CaO(s))})+(1\times \Delta H^0f_{CO_2}]-[(1\times \Delta H^o_f_{(CaCO_3(s))})][/tex]
We are given:
[tex]\Delta H^o_f_{(CaO(s))}=-635.1kJ/mol\\\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_{rxn}=?[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times (-635.1))+(1\times (-393.5))]-[(1\times (-1206.9))][/tex]
The DH°rxn for the decomposition of calcium carbonate to calcium oxide and carbon dioxide is +178.3 kJ