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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) β†’ C6H12O6(s) C(s) + O2(g) β†’ CO2(g), βˆ†H = –393.51 kJ H2(g) + Β½ O2(g) β†’ H2O(l), βˆ†H = –285.83 kJ C6H12O6(s) + 6O2(g) β†’ 6CO2(g) + H2O(l), βˆ†H = –2803.02 kJ

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superman1987

Answer:

- 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.

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

6C(s) + 6Hβ‚‚(g) + 3Oβ‚‚(g) β†’ C₆H₁₂O₆(s),

  • We should multiply the first equation by (6) and also multiply its Ξ”H by (6):

6C(s) + 6Oβ‚‚(g) β†’ 6COβ‚‚(g), βˆ†H₁ = (6)(–393.51 kJ) = - 2361.06 kJ,

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

6Hβ‚‚(g) + 3Oβ‚‚(g) β†’ 6Hβ‚‚O(l), βˆ†Hβ‚‚ = (6)(–285.83 kJ) = - 1714.98 kJ.

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

6COβ‚‚(g) + Hβ‚‚O(l) β†’ C₆H₁₂O₆(s) + 6Oβ‚‚(g), βˆ†H₃ = (-1)(–2803.02 kJ) = 2803.02 kJ.

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

6C(s) + 6Hβ‚‚(g) + 3Oβ‚‚(g) β†’ C₆H₁₂O₆(s),

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

βˆ†Hrxn = βˆ†H₁ + βˆ†Hβ‚‚ + βˆ†H₃ = (- 2361.06 kJ) + (- 1714.98 kJ) + (2803.02 kJ) = - 1273.02 kJ.

Answer:

- 1273.02 kJ

Explanation:

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