Question

Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3H2(g) + 2C(s) + ½O2(g) → C2H5OH(l) C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l), ΔH = –1367 kJ/mol C(s)+O2(g)→CO2(g), ΔH = –393.5 kJ H2(g)+½O2(g)→H2O(l), ΔH = –285.8 kJ

Answer 1

It would be -277.6 KJ/mol

Answer 2

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

$\Delta H_{rxn}= -\Delta H_{1}+ 2(\Delta H_{2})+3(\Delta H_{3})$

$\Delta H_{rxn}= 1367 + 2(-393.5)+3(-285.5)=-276.5 kJ/mol$