Answer: The enthalpy of the formation of [tex]H_2O(l)[/tex] is coming out to be -298.415 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. 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]2H_2S(g)+3O_2(g)\rightarrow 2H_2O(l)+2SO_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(H_2O(l))})+(2\times \Delta H^o_f_{(SO_2(g))})]-[(2\times \Delta H^o_f_{(H_2S(g))})+(3\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2S(g))}=-20.63kJ/mol\\\Delta H^o_f_{(O_2)}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.84kJ/mol\\\Delta H^o_{rxn}=-1149.2kJ[/tex]
Putting values in above equation, we get:
[tex]-1149.2=[(2\times \Delta H^o_f_{(H_2O(l))})+(2\times (-296.84))]-[(2\times (-20.63))+(3\times 0)]\\\\\Delta H^o_f_{(H_2O(l))}=-298.415kJ/mol[/tex]
Hence, the enthalpy of the formation of [tex]H_2O(l)[/tex] is coming out to be -298.415 kJ/mol.
