1. **Stating the problem:** We are asked to calculate the standard enthalpy change of the reaction ($\Delta H^\circ_{reaksi}$) for the reaction: $$2CO_2(g) + 6H_2(g) \rightarrow C_2H_5OH(l) + 3O_2(g)$$ 2. **Formula used:** The standard enthalpy change of reaction is given by: $$\Delta H^\circ_{reaksi} = \sum \Delta H^\circ_f (produk) - \sum \Delta H^\circ_f (reaktan)$$ 3. **Given data:** $\Delta H^\circ_f CO_2(g) = -393.5$ kJ/mol $\Delta H^\circ_f H_2(g) = -286.0$ kJ/mol $\Delta H^\circ_f C_2H_5OH(l) = -277.6$ kJ/mol $\Delta H^\circ_f O_2(g) = -249.2$ kJ/mol 4. **Calculate enthalpy of products:** $$\Delta H^\circ_{produk} = (-277.6) + 3(-249.2) = -277.6 - 747.6 = -1025.2\text{ kJ}$$ 5. **Calculate enthalpy of reactants:** $$\Delta H^\circ_{reaktan} = 2(-393.5) + 6(-286.0) = -787.0 - 1716.0 = -2503.0\text{ kJ}$$ 6. **Calculate $\Delta H^\circ_{reaksi}$:** $$\Delta H^\circ_{reaksi} = -1025.2 - (-2503.0) = -1025.2 + 2503.0 = 1477.8\text{ kJ}$$ 7. **Answer:** The calculated $\Delta H^\circ_{reaksi}$ is +1477.8 kJ. 8. **Regarding the question if the reaction is not balanced:** The enthalpy change calculation must be based on a balanced chemical equation because the enthalpy values are per mole of each substance. If the reaction is not balanced, the calculation will not be correct. Here, the reaction is balanced as given, so the calculation is valid. Hence, the calculation is correct only if the reaction is balanced as shown.