Using Hess's Law to calculate ΔH°rxn

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SUMMARY

The discussion focuses on calculating the standard enthalpy change (ΔH°rxn) for the reaction C3H8(g) + 4 CO2(g) + 4 H2(g) → C7H16(g) + 4 O2(g) using Hess's Law. The given reactions include ΔH°rxn values of -2,010 kJ/mol for C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g), -4,426 kJ/mol for C7H16(g) + 11 O2(g) → 7 CO2(g) + 8 H2O(g), and -241.8 kJ/mol for H2(g) + ½ O2(g) → H2O(g). The correct approach involves swapping the second reaction and multiplying the third reaction by 4 to cancel out unwanted compounds, leading to the final calculation of ΔH°rxn.

PREREQUISITES
  • Understanding of Hess's Law
  • Knowledge of standard enthalpy change (ΔH°rxn)
  • Ability to manipulate chemical equations
  • Familiarity with stoichiometry in chemical reactions
NEXT STEPS
  • Study Hess's Law applications in thermochemistry
  • Learn how to manipulate chemical equations for enthalpy calculations
  • Explore the concept of standard enthalpy of formation
  • Practice calculating ΔH°rxn using various reaction scenarios
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Chemistry students, educators, and professionals involved in thermodynamics and reaction energetics will benefit from this discussion.

littlebearrrr
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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



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.
 
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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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