Calculate the enthalpy change for the reaction: C7H16 (l) + 11 O2 (g) → 7 CO2 (g) + 8 H2O (g) given the following Gasoline is not actually pure octane. There are some lighter combustible materials like heptane that are mixed in with the octane. Using Hess's Law and the following four equations, find the change in enthalpy for the combustion of heptane: C7H16 (l) + 11 O2 (g) → 7 CO2 (g) + 8 H2O (g) . 1. CH4 (g) → C (s) + 2 H2 (g) H = +74.85 kJ/mol 2. CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (g) H = –802.3 kJ/mol 3. 7 C (s) + 8H2 (g) → C7H16 (l) H = –187.8 kJ/mol 4. 2 H2 (g) + O2 (g) → 2 H2O (g) H = –483.6 kJ/mol

Expert Answers

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The chemical equations for which the enthalpy change is given are:

CH4(g) `->` C(s) + 2H2 (g) `Delta H` = +74.85 kJ/mol ...(1)

CH4(g) + 2O2 (g) `->` CO2(g) + 2H2O (g) `Delta H` = -802.3 kJ/mol ...(2)

7C(s) + 8H2(g) `->` C7H16 (l) `Delta H` = -187.8 kJ/mol ...(3)

2H2 (g) + O2(g) `->` 2H2O (g) `Delta H` = -483.6 kJ/mol ...(4)

Using Hess' Law, `Delta H_(reaction) = Delta H_( p roducts) - Delta H_(reactants)`

The enthalpy to be determined is of the chemical reaction:

 C7H16(l) + 11O2 (g) `->` 7CO2(g) + 8H2O(g)

`(2) - (1)`

=> CH4(g) + 2O2(g) + C(s) + 2H2(g) `->` CO2(g) + 2H2O(g) + CH4 (g) `Delta H` = -877.15 kJ/mol

=> 2O2(g) + C(s) + 2H2(g) `->` CO2(g) + 2H2O(g)  `Delta H` = -877.15 kJ/mol ...(5)

`-(3) + 7*(5) - 3*(4)`

C7H16(l) + 14O2(g) + 7C(s) + 14H2(g) + 6H2O (g) `->` 7C(s) + 8H2 (g) + 7CO2 (g) + 14H2O(g)+ 6H2(g) + 3O2(g)

or C7H16(l) + 11O2(g) `->` + 7CO2(g) + 8H2O(g)

`Delta H` = 187.8 + 7*-877.15 - 3*-483.6 kJ/mol = -4876.25 kJ/mol

The enthalpy of change of the given reaction is -4875.25 kJ/mol

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