Answer:
(6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)
Explanation:
Question; From the given options, the chemical reaction in the question is presented as follows;
C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), given that we have;
[tex]\Delta \text H _f^{\circ}[/tex] for C₆H₁₂O₆ = -1,273.02 kJ/mol
[tex]\Delta \text H _f^{\circ}[/tex] for O₂(g) = 0 kJ/mol
[tex]\Delta \text H _f^{\circ}[/tex] for CO₂(g) = -393.5 kJ/mol
[tex]\Delta \text H _f^{\circ}[/tex] for H₂O(l) = -285.83 kJ/mol
The heat or enthalpy of a reaction, is given as follows;
[tex]\Delta\text H_{rxn}^{\circ} = \sum \text n \cdot \Delta \text H _f^{\circ}(\text {products}) - \sum \text m \cdot \Delta \text H _f^{\circ}(\text {reactants} \text)[/tex]
Therefore, the equation which should be used to calculate [tex]\Delta\text H_{rxn}^{\circ}[/tex], is given as follows;
[tex]\sum \text n \cdot \Delta \text H _f^{\circ}(\text {products})[/tex] = (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol)
[tex]\sum \text m \cdot \Delta \text H _f^{\circ}(\text {reactants} \text)[/tex] = (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)
Therefore;
[tex]\Delta\text H_{rxn}^{\circ}[/tex] = (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)