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krek1111 [17]
3 years ago
15

Given the three equations below, what is the heat of reaction for the production of glucose, C6H12O6, as described by this equat

ion? 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
Chemistry
2 answers:
UkoKoshka [18]3 years ago
6 0

Answer:

- 1273.02 kJ.

Explanation:

This problem can be solved using Hess's Law.

Hess's Law states that <em>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.</em>

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

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

  • 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:

<em>6C(s) + 6H₂(g) + 3O₂(g) → C₆H₁₂O₆(s),</em>

<em></em>

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

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

RSB [31]3 years ago
5 0

Answer:

- 1273.02 kJ

Explanation:

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⋅

R

T

P

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