Respuesta :
Answer: The enthalpy of the formation of [tex]CO_2(g)[/tex] is coming out to be -410.8 kJ/mol.Z
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. It is represented as [tex]\Delta H^o[/tex]
The equation used to calculate enthalpy change is of a reaction is:
[tex]\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)][/tex]
For the given chemical reaction:
[tex]2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ[/tex]
Putting values in above equation, we get:
[tex]81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol[/tex]
Hence, the enthalpy of the formation of [tex]CO_2(g)[/tex] is coming out to be -410.8 kJ/mol.
