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# The combustion of pentane, c5h12, occurs via the reaction C5H12

### Resolved Question:

The combustion of pentane, c5h12, occurs via the reaction C5H12 + 8O2 -> 5CO2 + 6H2O with the heat of formation values given: c5H12: -35.1kJ/mol; CO2: 393.4kJ/mol; H2): -241.8kJ/mol. Calculate the enthalpy for the combustion of penthane in kJ.
Submitted: 6 years ago.
Category: Homework
Expert:  David Scrafton replied 6 years ago.

Hi Honi Mai, I'd be happy to answer this question for you,

Would you be able to check something for me however, as the enthalpy of formation of CO2 should be a very negative value. Should it be -393.4 kJ/mol?

Can you check the values you have been given and I'll calculate the answer for you if you would like.

David.

Customer: replied 6 years ago.
sorry it's -393.5kJ/mol

thank you
Expert:  David Scrafton replied 6 years ago.

Ok, give me a few minutes and I'll calculate the answer.

Thanks,

David.

Expert:  David Scrafton replied 6 years ago.

Hi Honi Mai,

I've finished the answer, I've written it out as I think it is easier than trying to put all the formulae into typed format.

I can attach the image and send it to you on Just Answer, is this ok?

Thanks,

David.

Customer: replied 6 years ago.
ya that's fine. thanks so much
Expert:  David Scrafton replied 6 years ago.
You can view this answer by clicking here to Register or Login and paying \$3.
Category: Homework
Satisfied Customers: 111
Experience: BSc(Hons) - 1st class honours in Chemistry, MPhil (Chemistry), MBBS (Medicine)
Expert:  David Scrafton replied 6 years ago.

Thanks for accepting, look out for the decimal points as they're a little faint, but you should still be able to see them. Also, just wondering, is the enthalpy of formation for C5H12 +35.1 kJ/mol or -35.1 kJ/mol?

Thanks,

David.

Customer: replied 6 years ago.
it was -35.1kJ/mol
the - sign was on the previous line
Expert:  David Scrafton replied 6 years ago.

Ok, Have to change the answer slightly. One second.

Expert:  David Scrafton replied 6 years ago.

Hi again,

Just need to change H1 to -35.1 kJ/mol.

H2 at the bottom is simply -3417.8 + 35.1 = 3382.7 kJ/mol

David.

Expert:  David Scrafton replied 6 years ago.

Hopefully it should all work out now. Let me know if you have any problems following the answer.

Thank you again for your question,

David.

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