The standard heat of formation of a branched alkane is -35 kcal/mol; the standard heat of formation of the unbranched version of alkane (same molecular formula) is -28 kcal/mol. Finally, the standard enthalpy of combustion of the branched alkane is -632 kcal/mol. Given this information, what is the standard enthalpy of combustion of the unbranched alkane in terms of kcal/mol to the nearest ones?

The method to solve this question is based on Hess´s law of constant heat of summation which allows us to combine the enthalpies of individual reactions for which we know their enthalpy to obtain the enthalpy change for a desired reaction.

We are asked to calculate the standard enthalpy of formation of combustion of an unbranched alkane :

CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O ΔcHº = ?

where CnH2n+2 is the general formula for alkanes.

and we are given information for

n C+ (2n + n)/2 H₂ ⇒ CnHn+2 unbranched ΔfHº = -35 kcal/mol (1)

n C+ (2n + n)/2 H₂ ⇒ CnHn+2 branched ΔfHº = -28 kcal/mol (2)

CnHn+2 branched + O₂ ⇒ CO₂ + H₂O ΔcHº = -632 kcal/mol (3)

If we reverse (1) and add it to the sum (2) and (3) we get the desired equation for the combustion of the unbranched alkane:

CnHn+2 unbranched + O₂ ⇒ CO₂ + H₂O

Thus

ΔcHº unbranched = + 35 kcal/mol + (-28 kcal/mol) + (-632 kcal/mol)

= -625 kcal/mol