If given $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{20} (g)$ $2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)$ and $DeltaH$ = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?

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If given $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{20} (g)$ $2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)$ and $DeltaH$ = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?

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Answer 1 · 2016-09-09T07:40:06

$DeltaH""^@""_"combustion"$ $=$ $-2875*kJ*mol^-1$

Explanation:

The $"molar enthalpy of combustion"$ is the energy associated with the complete combustion of 1 mole of hydrocarbon to give carbon dioxide and water. The given equation quotes enthalpy output,

$"PER MOLE OF REACTION AS WRITTEN"$

And thus it represents TWICE the enthalpy of combustion of butane because TWO moles of butane were combusted.

$DeltaH""^@""_"combustion"$ $=$ $(-5750*kJ*mol^-1)/2= -2875*kJ*mol^-1$

Of course, certain reference standards are specified (the most annoying of which is $100$ $kPa$ , but that's another story!).

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If given $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{20} (g)$ $2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)$ and $DeltaH$ = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?

1 Answer

Explanation:

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If given 2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)2C4H10(g)+13O2(g)→8CO2(g)+10H20(g)2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g) and DeltaHDeltaH = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?

1 Answer

Explanation:

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If given $2 C_{4} H_{10} (g) + 13 O_{2} (g) \to 8 C O_{2} (g) + 10 H_{20} (g)$ $2C_4H_10 (g) + 13O_2 (g) -> 8CO_2 (g) + 10H_20 (g)$ and $DeltaH$ = -5750 kJ. How do you determine the molar enthalpy of combustion of butane?