Given the three equations below, what is the heat of reaction for the production of glucose, c6h12o6, as described by this equation? 6c(s) + 6h2(g) + 3o2(g) → c6h12o6(s) c(s) + o2(g) → co2(g), ∆h = –393.51 kj h2(g) + ½ o2(g) → h2o(l), ∆h = –285.83 kj c6h12o6(s) + 6o2(g) → 6co2(g) + h2o(l), ∆h = –2803.02 kj

- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

6C(s) + 6O₂(g) → 6CO₂(g), ∆H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

6H₂(g) + 3O₂(g) → 6H₂O(l), ∆H₂ = (6)(–285.83 kJ) = - 1714.98 kJ.

6CO₂(g) + H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ∆H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

∆Hrxn = ∆H₁ + ∆H₂ + ∆H₃ = (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = - 1273.02 kJ.

thats how its supposed to be