Determining Heat of Formation of n-Butane

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SUMMARY

The standard heat of formation of n-butane (C4H10) cannot be measured directly and is determined using Hess's Law. The combustion reactions provided include the formation of carbon dioxide (CO2) and water (H2O) from carbon (C) and hydrogen (H2). The user attempted to derive the formation equation for n-butane but struggled with balancing the oxygen atoms, ultimately leading to confusion regarding the correct stoichiometry. The discussion also touches on the formation of propanol (C3H8O) and highlights the importance of using standard states in chemical equations.

PREREQUISITES
  • Understanding of Hess's Law and its application in thermodynamics
  • Knowledge of combustion reactions and stoichiometry
  • Familiarity with standard states of substances in thermochemical equations
  • Basic chemical equation balancing techniques
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  • Study Hess's Law in detail and its applications in calculating heats of formation
  • Learn about balancing chemical equations, focusing on combustion reactions
  • Research standard enthalpy of formation and its significance in thermodynamics
  • Explore the formation reactions of other hydrocarbons, such as propanol (C3H8O)
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Chemistry students, educators, and professionals involved in thermodynamics and chemical reaction analysis will benefit from this discussion.

Nellen2222
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Homework Statement



The standard heat of formation of n butane cannot be measured directly but can be determined indirectly from other heats of reaction by applying hess's law. Given the folowing data for the combustion of carbon hydrogen and n butane in oxygen:

C + O2---> CO2 H=-393.5
H2 + 1/2O2 --> H2O
C4H10 + 6.5O2 ---> 4CO2 + 5H2O

Homework Equations





The Attempt at a Solution



Tried several times on scratch paper. The formation of butane is as so : 4C + 5H2-->C4H10

Ive tried the actual formation using CO2 + H2O --> C4H10 But i can't cancel the carbons

Sticking with my other normal equation 4C + 5H2 --> C4H10 (This is what I want, No?)

Tell me if my equation is wrong, cause it proabably is.

I tried multiplying the first step by 4, the second step by 5, and reversing the third step only to get everything to cancel except the oxygen. I get stuck with 6.5 oxygens on the left side and 9.5 Oxygens on the right. Dont know where to g ofrom there. Help please!
 
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4 \times 1 + 5 \times \frac 1 2 = 6.5
 
elaborate...?
 
Here is another question: Write the chemical equation for the standard molar enthalpy of formation of proponol. c3h8O
Is this simply :: 3C + 8H + O2 ----> C3H8O ( I am pretty sure its this one.)

?

Or is it 3CO2 + 4H2O ----> C3H8O + 9/2O2

Thanks
 
Nellen2222 said:
elaborate...?

Where did you get 9.5 molecules of oxygen from?
 
Nellen2222 said:
Is this simply :: 3C + 8H + O2 ----> C3H8O ( I am pretty sure its this one.)

It is not balanced and it uses substances in non standard state. But it is much better than the other one.
 

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