Answer: The standard enthalpy change is -607kJ
Explanation:
The given balanced chemical reaction is,
[tex]3C_2H_2(g)\rightarrow C_6H_6(g)[/tex]
First we have to calculate the enthalpy of reaction [tex](\Delta H^o)[/tex].
[tex]\Delta H^o=H_f_{product}-H_f_{reactant}[/tex]
[tex]\Delta H^o=[n_{C_6H_6}\times \Delta H_f^0_{(C_6H_6)}]-[n_{C_2H_2\times \Delta H_f^0_{(C_2H_2)}][/tex]
where,
We are given:
[tex]\Delta H^o_f_{(C_2H_2)}=230kJ/mol\\\Delta H^o_f_{(C_6H_6)}=83kJ/mol[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(1\times 83)]-[(3\times 230)]=-607kJ[/tex]
The standard enthalpy change is -607kJ
