# Help. Enthalpy question

Q4.Use the information below to answer this question.
C(s) + O2(g) CO2(g) ΔH = −394 kJ mol−1
H2(g) + O2(g) H2O(l) ΔH = −286 kJ mol−1
4C(s) + 5H2(g) C4H10(g) ΔH = −126 kJ mol−1
The standard enthalpy of combustion of butane, in kJ mol−1
, is
A −2880
B −2590
C −806
D −554
Original post by TheJoke786
Q4.Use the information below to answer this question.
C(s) + O2(g) CO2(g) ΔH = −394 kJ mol−1
H2(g) + O2(g) H2O(l) ΔH = −286 kJ mol−1
4C(s) + 5H2(g) C4H10(g) ΔH = −126 kJ mol−1
The standard enthalpy of combustion of butane, in kJ mol−1
, is
A −2880
B −2590
C −806
D −554

Have attached my solution here!! The best way to do any question like this is using a Hess cycle!
Here is my working out - I prefer not to use a Hess Cycle.
(edited 3 years ago)
why did you work out the enthalpy of formation even though the question asked for the enthalpy of combustion?
it was two years ago bro xdd

Original post by y-habieb
why did you work out the enthalpy of formation even though the question asked for the enthalpy of combustion?
Original post by y-habieb
why did you work out the enthalpy of formation even though the question asked for the enthalpy of combustion?

The combustion of butane can be represented by the following equation:

C4H10(g) + 13O2(g) 4CO2(g) + 5H2O(l)

The standard enthalpy of combustion of butane can be calculated using the following equation:

ΔHcomb = ΣΔHf(products) - ΣΔHf(reactants)

where:

ΔHcomb is the standard enthalpy of combustion of butane
ΔHf(products) is the standard enthalpy of formation of the products
ΔHf(reactants) is the standard enthalpy of formation of the reactants
The standard enthalpies of formation of the products and reactants can be found in the table below:

Compound ΔHf(kJ mol−1)
CO2(g) −394
H2O(l) −286
C4H10(g) −126

Substituting the values into the equation, we get:

ΔHcomb = (4 × −394) + (5 × −286) - (1 × −126)

= −2880 kJ mol−1

Therefore, the standard enthalpy of combustion of butane is −2880 kJ mol−1.