Answer
$$CH_3CH_2OH(l) \longrightarrow CH_3OCH_3(l) $$
$\Delta H^o_{rxn} = 74.3 \ kJ/mol$
Work Step by Step
1. Reverse the second equation:
$$2 CO_2(g) + 3 H_2O (g) \longrightarrow CH_3OCH_3 (l) + 3O_2(g)$$
$\Delta H^o_{rxn} = 1309.1 \ kJ/mol$
2. Sum the equations:
$$2 CO_2(g) + 3 H_2O(g) + CH_3CH_2OH(l) + 3O_2(g) \longrightarrow CH_3OCH_3(l) + 3O_2(g) + 2 CO_2(g) + 3 H_2O (g)$$
Removing the repeated ones:
$$CH_3CH_2OH(l) \longrightarrow CH_3OCH_3(l) $$
$\Delta H^o_{rxn} = 1309.1 \ kJ/mol - 1234.8 \ kJ/mol = 74.3 \ kJ/mol$
