Using Hess's Law to calculate ΔH°rxn

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The discussion focuses on calculating the standard enthalpy change (ΔH°rxn) for a specific chemical reaction using Hess's Law. Participants share their attempts to manipulate given reaction enthalpies to derive the desired ΔH°rxn. One user initially considers multiplying a reaction by 1/2 but is advised against it, leading to a realization about canceling unwanted compounds. Ultimately, the user successfully solves the problem by swapping one reaction and appropriately adjusting another. The conversation highlights the importance of understanding reaction manipulation in thermodynamic calculations.
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Homework Statement


Calculate ΔH°rxn in kJ/mol for:

C3H8(g) + 4 CO2(g) + 4 H2(g) → C7H16(g) + 4 O2(g)

from the following given values of ΔH°rxn:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH°rxn= -2,010 kJ/mol

C7H16(g) + 11 O2(g) → 7 CO2(g) + 8 H2O(g) ΔH°rxn= -4,426 kJ/mol

H2(g) + ½ O2(g) → H2O(g) ΔH°rxn= -241.8 kJ/mol


Homework Equations



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The Attempt at a Solution



I swapped the middle to give +4,426 kJ/mol. Then I tried multiplying the middle by 1/2 and the third one by 4. Am I on the right track? Anything else need to be swapped? I feel like the next step should be obvious; I'm just stuck.
 
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littlebearrrr said:

Homework Statement


Calculate ΔH°rxn in kJ/mol for:

C3H8(g) + 4 CO2(g) + 4 H2(g) → C7H16(g) + 4 O2(g)

from the following given values of ΔH°rxn:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔH°rxn= -2,010 kJ/mol

C7H16(g) + 11 O2(g) → 7 CO2(g) + 8 H2O(g) ΔH°rxn= -4,426 kJ/mol

H2(g) + ½ O2(g) → H2O(g) ΔH°rxn= -241.8 kJ/mol


Homework Equations



None

The Attempt at a Solution



I swapped the middle to give +4,426 kJ/mol. Then I tried multiplying the middle by 1/2 and the third one by 4. Am I on the right track? Anything else need to be swapped? I feel like the next step should be obvious; I'm just stuck.
You almost have it. Just don't multiply the middle one by 1/2.
 
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Ahh, okay thanks! I solved it by swapping the middle and just multiplying the bottom by 4. I figured out what stumped me; it was not knowing how to get the unwanted compounds in the reactions to cancel out. I get it now though.
 
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