A scientist measures the standard enthalpy change for the following reaction to be 149.2 kJ :

Question

CaCO3(s)CaO(s) + CO2(g)

 

Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CaO(s) is kJ/mol.

Answer 1

There is missing information necessary to answer this question. you need the standard enthalpies of formation of the other substances. Also I am assuming the equation is CaCO3(s)--> CaO(s) + CO2(g)

 

In any case, you need to use the formula for enthalpy of reaction:

 

ΔH reaction= (ΣΔproducts)-(ΣΔreactants)

 

In a chemistry book, I found that enthalpy of the other molecules

 

ΔH for CaCO3(s)= −1206.9 kj/mole

ΔH for CO2(g)= −393.509 kj/mol

 

Set up the equation and plug the values

 

ΔH reaction= [(ΔHCO2) + (ΔHCaO)] - [(ΔHCaCO30]

 

149.2 kj= [(-393.509) + (ΔHCaO)] -[(-1206.90]

 

ΔHCaO= 149.2 - 1206.90 + 393.509= -664.191 kj/mol