Jun 14, 2009
45
0
0
Status
Pre-Medical
Ok, so I'm working on a problem that is asking me to find the heat of formation of Acetylene. I'm given the heat of formation of both CO2 and O2. From that I'm assuming I am suppose to use the heat of combustion of acetylene (given) to determine the heat of formation of acetylene. I got it up to there.

My problem is when you balance the equation, you get:

C2H2 + 2.5O2 --> 2CO2 + H2O

That too makes sense, but why do you not balance it out so that all the coefficients are whole numbers? For example why wouldn't it be:

2C2H2 + 5O2 --> 4CO2 +2H2O

Solving for the heat of formation gives different results depending on which equation you use. The book I'm using uses the first equation with the 2.5O2 coefficient.
 

ksmi117

GEAUX TIGERS!!!
Moderator Emeritus
Lifetime Donor
10+ Year Member
7+ Year Member
Mar 16, 2008
21,960
179
481
Status
Resident [Any Field]
I'm not sure I'm exactly following your process here. They should give the same result, but realize that if you use the second reaction, then you'd have to multiply the given heat of combustion of C2H2 by 2 for the problem to work out right. Did you do that?
 

BerkReviewTeach

Company Rep & Bad Singer
Vendor
10+ Year Member
May 25, 2007
3,884
655
281
Ok, so I'm working on a problem that is asking me to find the heat of formation of Acetylene. I'm given the heat of formation of both CO2 and O2. From that I'm assuming I am suppose to use the heat of combustion of acetylene (given) to determine the heat of formation of acetylene. I got it up to there.

My problem is when you balance the equation, you get:

C2H2 + 2.5O2 --> 2CO2 + H2O

That too makes sense, but why do you not balance it out so that all the coefficients are whole numbers? For example why wouldn't it be:

2C2H2 + 5O2 --> 4CO2 +2H2O

Solving for the heat of formation gives different results depending on which equation you use. The book I'm using uses the first equation with the 2.5O2 coefficient.
First off, are they asking for heat of combustion or heat of formation? The reaction you've shown is a combustion reaction. It's a little thing that is often overlooked for some reason, so they can trick you here if you're not careful.

The reason it is the first version, with the non-whole number of 2.5, is because by definition, the numbers are listed in terms of kJ/mole (or kcal/mole) of the reactant of interest. You always put a 1 in front of what every compound they are asking about.

I'm not sure I'm exactly following your process here. They should give the same result, but realize that if you use the second reaction, then you'd have to multiply the given heat of combustion of C2H2 by 2 for the problem to work out right. Did you do that?
You are correct that in the solution process, you'd need to use new coefficients (such as 2 X heat of formation of acetylene), but the two different questions (using 1 mole of C2H2 versus using 2 moles C2H2) won't give the same amount of total energy. If you use 2 moles of reactant, then you will generate twice as much heat in the reaction as using 1 mole of reactant (Hess' law). This can easily be remedied if you pay attention to the units at the very end of the question, where you've ultimately determined the energy per 2 moles of acetylene. Take the number you get with the two moles of C2H2 reactant and divide by 2, and you're back to the correct answer.
 
Jun 14, 2009
45
0
0
Status
Pre-Medical
First off, are they asking for heat of combustion or heat of formation? The reaction you've shown is a combustion reaction. It's a little thing that is often overlooked for some reason, so they can trick you here if you're not careful.

The reason it is the first version, with the non-whole number of 2.5, is because by definition, the numbers are listed in terms of kJ/mole (or kcal/mole) of the reactant of interest. You always put a 1 in front of what every compound they are asking about.
They are asking for the heat of formation.

The question is related to a passage and they give the heat of combustion for acetylene as -1255.5 kj/mol. They give the heat of formation of both CO2 and H20. Which are -393.5 kj/mol and -241.8 kj/mol respectively.

The heat of formation of acetylene should thus be -226.6 kj/mol.

But, if you choose to use whole numbers in the balanced equation you get -401.05 kj/mol. Using this way has a 2 coefficient in front of the acetylene instead of a 1.

But, I see what you are saying. Always use the coefficient with a 1 in front of what you are trying to find.
 

htbruin

7+ Year Member
Aug 27, 2009
49
3
141
Status
Medical Student
They are asking for the heat of formation.

The question is related to a passage and they give the heat of combustion for acetylene as -1255.5 kj/mol. They give the heat of formation of both CO2 and H20. Which are -393.5 kj/mol and -241.8 kj/mol respectively.

The heat of formation of acetylene should thus be -226.6 kj/mol.

But, if you choose to use whole numbers in the balanced equation you get -401.05 kj/mol. Using this way has a 2 coefficient in front of the acetylene instead of a 1.

But, I see what you are saying. Always use the coefficient with a 1 in front of what you are trying to find.
Heat of formation is the heat needed to form one mole of that compound.

I'm curious about the answer being -226.6 kj/mol as I got positive 226.6kj/mol.

heat of combustion = heat of formation (products) - heat of formation (reactants)

-1255.5kj/mol = (-241.8kj/mol + (2*-393.5)) - X
X = 226.6kj/mol.

Can you show me how you calculated yours? thanks
 

askamsky51

7+ Year Member
Feb 16, 2010
339
40
161
Status
Resident [Any Field]
Can't you conceptually determine that the heat of combustion is going to be a negative number, due to that fact that it's an exothermic reaction?