# Calculate the enthalpy change for the reaction: C7H16 (l) + 11 O2 (g) → 7 CO2 (g) + 8 H2O (g) given the followingGasoline is not actually pure octane. There are some lighter combustible materials...

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

justaguide | College Teacher | (Level 2) Distinguished Educator

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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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