A scientist measures the standard enthalpy change for this reaction to be 163.2 kJ/mol. CaCO3(s)CaO(s) + CO2(g) Based on this value and the standard formation enthalpies for the other substances, the standard formation enthalpy of CO2(g) is ____ kJ/mol.
CaCO3(s)----->CaO(s)
+ CO2(g)
H0rxn =
[H0f(CaO) +
H0f(CO2)] -
[H0f(CaCO3)]
163.2 = [-635.09 + H0f(CO2)] - [-1206.92]
163.2 = 571.83 + H0f(CO2)
H0f(CO2) = - 408.63 KJ/mol
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