Question

C4H8 (g) + O2 (g) -> CO2 (g) + H2O (g)
Give ∆HR = -1255 KJ and
Hf of CO2 = -394 KJ/mol, H2O = -241 KJ/mol
a) Find Hf of C4H8
b) Find heat of reaction given by 51.9 g of C4H8 (g)
The thing I am not sure is about is the Delta Hr... Since it's given in Joules, should I also change it to KJ/mol or not?

Answer 1

are you okay with a)?

For b) -1255 kJ/mol is already in kJ/mol. you just need to convert the mass given to moles then multiply.

Answer 2

Well for a) I added the products up, and used Delta H = Products - reactants
So -635 - Hr = -1255
-Hr = -620
Hr = 620
The thing is, I am not sure if that's correct, considering -1255 is given in KJ and not in KJ/mol... Do I just have to divide it by the equation's moles, which totals to 4 or something?

for b)
isn't -1255 for the whole equation and not just C4H8? Or am I missing something?

Answer 3

i'm not sure what you're doing for a)

the equation is right but your set up is not.

it should be \(-1255 kJ= (H^{butane}_f)-[(4*(-394 ~kJ/mol))+(4*(-241~kJ/mol))]\)

then solve for \(H^{butane}_f\).

for b) yes it's for the whole equation, but it's in terms of the "normalized" moles, you would have to divide the moles of butane by it's stoichiometric coefficient before multiplying by -1255 kJ/mol, but since it's coefficient is 1, then you can skip that.

Answer 4

i just realized i put the reactants in place where the products should be and viceversa.

the equation should actually be:

\(−1255kJ=[(4∗(−394 kJ/mol))+(4∗(−241 kJ/mol))]-(H^{butane}_f)\)

Answer 5

Ah ha. But why doesn't it matter that the KJ is on one side and I am subtracting/adding the KJ/mol? Or... Did you multiply the 4 as in that's all the moles?
IF you did, aren't the Hf fo those specific substance? ARRGH... I am confused.
Whatever, so I can find the KJ of Butane by using that, which is
-1255 KJ = -2540KJ - Hf Butane
Hf butane = -1285KJ

and for b)
51.5g butane = 0.920 Moles of butane x -1255 KJ/mol = 1154 KJ produced by butane

So that's that, right?

Answer 6

it should be "kJ/mol" because \(H_{rxn}\) is in units per mole.

the 4's come from the stoichiometric coefficients in the balanced chemical equation.

The Hf's are specific to those substances, not sure what i wrote that would make you think otherwise.

b) is fine, except for the sign on your final answer (it should be negative).

Answer 7

Sorry, just kind of new to this concept... Although I learned what to do in class, I am not too sure what they mean individually. like Hf = heat formation, means that's the energy required for the equation to happen, right?
But then what's the KJ/mol suppose to represent? The amount of KJ needed to move them or something?

Why do I have to multiply the Hf by the stoichiometric coefficient? What's the reason for that?

Answer 8

the enthalpy of formation is the energy necessary to "make" the molecule essentially "from scratch", as in from elements, C, H etc.

These values are standards calculated, and as all standards are, they are in units of energy per mole.

You multiply by the coefficients because you need to account for the number of molecules/moles in the chemical equation since these values are per mole.

Answer 9

i should mention that the enthalpy values for the heat of formation are the bond energies that result form making said bonds.

Answer 10

Uh... I still can't intuitionally get why we need to multiply by the moles in the whole equation. I understand that the equation is describing what's happening with each mole of substance, and that the heat formation... Wouldn't the Heat Formation of the products just be for them? Why do I have to multiply by 4 if it's just for them?

Answer 11

C4H8 (g) + 4O2 (g) -> 4CO2 (g) + 4H2O (g)

because you're forming 4 moles and the value you have (Hf) is per one mole.

Answer 12

like if you had: 1 bag of flour + water -> 4 cupcakes + 4 cookies

if each cupcake took 50 kJ to make and you made 4, you'd need to multiply 4 by 50 kJ.

Answer 13

OH! So it wasn't balanced... Thank you very much! You are simply the best.