Answer: The value of [tex]\Delta H^o[/tex] for the reaction is, -2512.4 kJ
Explanation:
The chemical equation for the combustion of acetylene follows:
[tex]2C_2H_2(g)+5O_2(g)\rightarrow 4CO_2(g)+2H_2O(g)[/tex]
The equation for the enthalpy change of the above reaction is:
[tex]\Delta H^o_{rxn}=[(3\times \Delta H^o_f_{(CO_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(1\times \Delta H^o_f_{(C_2H_2(g))})+(5\times \Delta H^o_f_{(O_2(g))})][/tex]
We are given:
[tex]\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=227.4kJ[/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{rxn}=[(4\times (-393.5))+(2\times (-241.8))]-[(2\times (227.4)+(5\times (0))]\\\\\Delta H^o_{rxn}=-2512.4kJ[/tex]
Therefore, the value of [tex]\Delta H^o[/tex] for the reaction is, -2512.4 kJ