Question

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

Answer 1

- 1273.02 kJ

Step-by-step explanation:

Answer 2

- 1273.02 kJ.

Step-by-step 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.

• We should modify the given 3 equations to obtain the proposed reaction:

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

• We should multiply the first equation by (6) and also multiply its ΔH by (6):

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

• Also, we should multiply the second equation and its ΔH by (6):

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

• Finally, we should reverse the first equation and multiply its ΔH by (- 1):

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

• By summing the three equations, we cam get the proposed reaction:

6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),

• And to get the heat of reaction for the production of glucose, we can sum the values of the three ∆H:

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