Answer:
-625 kcal/mol
Explanation:
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