We are asked to calculate the molar enthalpy of combustion to CO_{2} (g) and H_{2}O(l) for *n*-Butane.

$\overline{){\mathbf{\Delta H}}{{\mathbf{\xb0}}}_{{\mathbf{rxn}}}{\mathbf{=}}{\mathbf{\Delta H}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{prod}}{\mathbf{-}}{\mathbf{\Delta H}}{{\mathbf{\xb0}}}_{\mathbf{f}\mathbf{,}\mathbf{}\mathbf{react}}}$

Note that we need to *multiply each ΔH˚ _{f} by the stoichiometric coefficient* since ΔH˚

Also, note that ΔH˚_{f} for elements in their standard state is 0.

Balanced reaction:

C_{4}H_{10}_{(g) }+ 13/2 O_{2(g)}→ 4 CO_{2(g)} + 5 H_{2}O_{(g)}

Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation:

Hydrocarbon | Formula | H_{f}(kJ/mol) |

1,3-Butadiene | C_{4} H_{6} (g) | 111.9 |

1-Butene | C_{4} H_{8} (g) | 1.2 |

n-Butane | C_{4} H_{10} (g) | -124.7 |

For *n*-Butane calculate the molar enthalpy of combustion to CO_{2} (g) and H_{2} O(l).

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