Answer:
The change in enthalpy for the formation of ethanol is -276.5 kJ/mol
Explanation:
The given reactions are:
\(C2H5OH(l)+3O2(g)\rightarrow 2CO2(g)+3H2O(l) ----\Delta H_{1}=-1367kJ/mol\)
\(C(s)+O2(g)\rightarrow CO2(g)-----\Delta H_{2}=-393.5 kJ/mol\)
\(H2(g)+ 1/2O2(g)\rightarrow H2O(l)-----\Delta H_{3}=-285.8 kJ/mol\)
The required reaction involves the formation of C2H5OH from C, H2 and O2:
\(3H2(g) + 2C(s) + 1/2O2(g) \rightarrow C2H5OH(l) -----\Delta H_{rxn}=?\)
This can be obtained by reversing reaction (1), multiplying equation (2) by 2, multiplying equation (3) by 3 and then adding the corresponding ΔH values
\(\Delta H_{rxn}= -\Delta H_{1}+ 2(\Delta H_{2})+3(\Delta H_{3})\)
\(\Delta H_{rxn}= 1367 + 2(-393.5)+3(-285.5)=-276.5 kJ/mol\)
