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?

-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