# Hess's Law

Question states, "Construct a suitable cycle and apply Hess’s law to calculate the enthalpy of formation of butane (C4H10) using the following data:

Enthalpy of combustion of butane = - 2877.1 kJmol-1
Enthalpy of combustion of graphite = - 393.5 kJmol-1
Enthalpy of combustion of hydrogen = - 285.9 kJmol-1 "

Ok, could someone solve this, step-by-step for me please. Rep to the most helpful person, thanks!
Jalapeno
Question states, "Construct a suitable cycle and apply Hess’s law to calculate the enthalpy of formation of butane (C4H10) using the following data:

Enthalpy of combustion of butane = - 2877.1 kJmol-1
Enthalpy of combustion of graphite = - 393.5 kJmol-1
Enthalpy of combustion of hydrogen = - 285.9 kJmol-1 "

Ok, could someone solve this, step-by-step for me please. Rep to the most helpful person, thanks!

i did chemistry but i cant remember, but my main point is your rep isnt worth anything. you need at least 100 points to give 1 point of rep so anyone that helps you will just be doing it out of helpfulness.
olliemccowan
i did chemistry but i cant remember, but my main point is your rep isnt worth anything. you need at least 100 points to give 1 point of rep so anyone that helps you will just be doing it out of helpfulness.
Psh, points!? A rep from me, regardless of the point effect, is something to be valued!
Any help would be greatly appreiciated!!
hess' law states that overall enthalpy change in a reaction is the sum of the reaction enthalpies of each step of the reaction.

Target equation: 4C(s) + 5H2 (g) = C4H10

Data equations for enthalpy of combustion:
1) C(s) + O2(g) = CO2 (g) -393.5 x 4
2) H2(g) + ½O2(g) = H2O(l) -285.9 x5
3) 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3

Answer: 4(-393.5) + 5(-285.9) + 2877.1 = ............ (work it out urself)

all values in kj/mol

makes sense?
_jennifer_
hess' law states that overall enthalpy change in a reaction is the sum of the reaction enthalpies of each step of the reaction.

Target equation: 4C(s) + 5H2 (g) = C4H10

Data equations for enthalpy of combustion:
1) C(s) + O2(g) = CO2 (g) -393.5 x 4
2) H2(g) + ½O2(g) = H2O(l) -285.9 x5
3) 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse

To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3

Answer: 4(-393.5) + 5(-285.9) + 2877.1 = ............ (work it out urself)

all values in kj/mol

makes sense?
Crystal. Thankyou so much! You very helpful person! Rep for you.
One thing, isn't the H2O going to be vapour? So gas?