Question

The heat of combustion of liquid hexane (C6H14) to carbon dioxide and liquid water at 298 K is -4215 kJ/mol .Find ΔE for this reaction.

Answer 1

Honestly I have my doubts, mostly with the expressions and data to use, but here's my attempt.

First, let's write and balance the reaction:

2C₆H_14(l) + 13O_2(g) -----------> 12CO_2(g) + 14H₂O_(l)

In a reaction with constant temperature (in this case, 298 K), the expression to use is the following:
ΔE = ΔH − RTΔn

Δn is the difference between the final moles and the innitial moles (in gaseous state only or aqueous) which are:

(12) - (13) = -1

ΔE = -4215 - (8.3144*298)(-1)
ΔE = -4215000 + 2477.6912
ΔE = -4212522.31 J/mol or -4212.52 kJ/mol

Hope this helps you.