Respuesta :
Answer: The [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 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 that is used to calculate enthalpy change of a reaction is:
[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]
For the given chemical reaction:
[tex]2H_2O(l)\rightarrow 2H_2(g)+O_2(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(H_2)})+(1\times \Delta H_f_{(O_2)})]-[(2\times \Delta H_f_{(H_2O)})][/tex]
We are given:
Enthalpy of substances present in their standard form are taken to be 0.
[tex]\Delta H_f_{(H_2)}=0kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=595.8kJ[/tex]
Putting values in above equation, we get:
[tex]595.8=[(2\times (0))+(1\times (0))]-[2\times (\Delta H_f_{H_2O})]\\\\\Delta H_f_{H_2O}=297.9kJ/mol[/tex]
Hence, the [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.
