Respuesta :
The question is incomplete, here is the complete question:
Calculate the enthalpy of formation (kJ/mol) of MgO(s).The enthalpy of reaction for the equation as written is -100.8 kJ/mol. If the answer is negative, enter the sign and then the magnitude.
[tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]
Given:
,
Answer: The enthalpy of formation of MgO(s) is -601.7 kJ/mol
Explanation:
Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}][/tex]
For the given chemical reaction:
[tex]MgO(s)+CO_2(g)\rightarrow MgCO_3(s)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(1\times \Delta H_f_{(MgCO_3(s))})]-[(1\times \Delta H_f_{(MgO(s))})+(1\times \Delta H_f_{(CO_2(g))})][/tex]
We are given:
[tex]\Delta H_f_{(MgCO_3(s))}=-1096kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_{rxn}=-100.8kJ/mol[/tex]
Putting values in above equation, we get:
[tex]-100.8=[(1\times (-1096))]-[(1\times \Delta H_f_{(MgO(s))})+(1\times (-393.5))]\\\\\Delta H_f_{(MgO(s))}=-601.7kJ/mol[/tex]
Hence, the enthalpy of formation of MgO(s) is -601.7 kJ/mol
