Answer:
[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\ \ \ ;\ \ \ \Delta H=2651.2kJ[/tex]
Explanation:
Hello,
In this case, in order to answer to the requirement, we first should invert the given reaction since it is the formation of barium carbonate and we need its decomposition:
[tex]BaCO_3(s)\rightarrow BaO(s)+CO_2(g)[/tex]
Thereby, the enthalpy of reaction is inverted, to positive since it is the contrary reaction:
[tex]\Delta H=662.8kJ[/tex]
Nevertheless, we need to specify it for the formation of 4 moles of carbon dioxide it means:
[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\\\Delta H=662.8kJ*4[/tex]
Which finally results in the following thermochemical expression:
[tex]4BaCO_3(s)\rightarrow 4BaO(s)+4CO_2(g)\ \ \ ;\ \ \ \Delta H=2651.2kJ[/tex]
Regards.
