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Chemistry 18 Online
OpenStudy (anonymous):

Predict the change in enthalpy due to combustion of 10.0g of propane used in a camp stove

OpenStudy (anonymous):

@aaronq @zepdrix

OpenStudy (aaronq):

Use Hess's Law

OpenStudy (anonymous):

cannt i use caloirmetry ?

OpenStudy (anonymous):

we didnt get to hesses law yet so my teacher will be pissed

OpenStudy (aaronq):

you can use calorimetry if you know other data; like the change in temperature of the calorimeter system

OpenStudy (anonymous):

this is an excample on my book and it says the molar entlapy is -2043.9kj i dont get where they got the number from

OpenStudy (aaronq):

Okay, so you can use that value? \(\Delta H_{rxn}=H^0_{combustion}*n\) so convert the 10 g to moles and multiply by the molar enthalpy

OpenStudy (anonymous):

but how do i know the molar entlahy

OpenStudy (anonymous):

its not given in the question

OpenStudy (aaronq):

you often have to look these up; these are standard values in any textbook, on a test or exam they would be given to you.

OpenStudy (aaronq):

unless they give you other information to work it out from

OpenStudy (anonymous):

like in my data table its is given for propnae -103.8

OpenStudy (aaronq):

hm what are the units on both values?

OpenStudy (anonymous):

but that way off of what they put in for the molar entaloy

OpenStudy (anonymous):

u mean answer ?

OpenStudy (aaronq):

no for the standard molar enthalpies you're giving; what are the units, it's possible one is in kJ/mol and the other in kJ/g

OpenStudy (anonymous):

kj/mol

OpenStudy (anonymous):

for propane it says -103 kj/mol

OpenStudy (aaronq):

That's the enthalpy of formation; the other is the enthalpy of combustion. To use the enthalpy of formation you need to use Hess's Law

OpenStudy (anonymous):

ok , can u help me with another one

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