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
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!
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: