Question

Someone plz help me here are the photos, I need the last one

Answer 1

(6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)

Step-by-step 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;

`\Delta \text H _f^(\circ)` for C₆H₁₂O₆ = -1,273.02 kJ/mol

`\Delta \text H _f^(\circ)` for O₂(g) = 0 kJ/mol

`\Delta \text H _f^(\circ)` for CO₂(g) = -393.5 kJ/mol

`\Delta \text H _f^(\circ)` for H₂O(l) = -285.83 kJ/mol

The heat or enthalpy of a reaction, is given as follows;

`\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)`

Therefore, the equation which should be used to calculate
`\Delta\text H_(rxn)^(\circ)`, is given as follows;

`\sum \text n \cdot \Delta \text H _f^(\circ)(\text {products})` = (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol)

`\sum \text m \cdot \Delta \text H _f^(\circ)(\text {reactants} \text)` = (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)

Therefore;

`\Delta\text H_(rxn)^(\circ)` = (6 mol)·(-393.5 kJ/mol) + (6 mol)·(-285.83 kJ/mol) - (1 mol)·(-1,273.02 kJ/mol) + (6 mol)·(0 kJ/mol)