Question

A 1.738 g sample of a component of the light petroleum distillate called naphtha is found to yield 5.327 g CO2(g) and 2.544 g H2O(l) on complete combustion.

This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula.

The compound also has the following properties:

melting point of −154 ∘C ,

boiling point of 60.3 ∘C ,

density of 0.6532 g/mL at 20 ∘C ,

specific heat of 2.25 J/(g⋅∘C) , and

ΔH∘f=−204.6kJ/mol .

Part C

Calculate the enthalpy of combustion, ΔH∘comb, for C6H14. You'll first need to determine the balanced chemical equation for the combustion of C6H14.

Express your answer to four significant figures and include the appropriate units.

Answer 1

(a)complete combustion of C6H14 is given as

2C₆H_{14 (g)}+19O₂ -12CO₂ +14H₂O+energy

from the balanced chemical equation we can calculate standard heat of combustion for C₆H₁₄ is :

heat of formation of C₆H₁₄ =12(heat of formation of CO₂)+14(heat of formtion of water)+heat of combustion

standard heat of formation of CO₂=-94.0 kcal

standard heat of formation of water=-68 kcal

heat of formation of C6H14

C(g)-c(s)=-172kcal/mol

6C(g) - 6C(s) -1032kcal/mol

H(g) -1/2H(g) -52.09kcal/mol

14H(g) -7H2 - 729.26kcal/mol

adding above equation we get

6C+14H -C₆H₁₄    -1761.26

now heat of combustion is given as

-H =12(-94.0)+14(-68)+1761.26

=-2080+1761.26

=-318.74kcal/mol =-1333.6kj/m0l