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Question
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is ______.
Options
\[\ce{2C4H10 (g) + 1302 (g) -> 8CO2 (g) + 10H2O (l) ∆_cH = - 2658.0 kJ mol^{-1}}\]
\[\ce{ C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (g) ∆_CH = -1329.0 kJ mol^{-1}}\]
\[\ce{C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (l) ∆_cH = - 2658.0 kJ mol^{-1}}\]
\[\ce{C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (l) ∆_cH = + 2658.0 kJ mol^{-1}}\]
Solution
During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is \[\ce{C4H10 (g) + 13/2 O2 (g) -> 4CO2 (g) + 5H2O (l) ∆_cH = - 2658.0 kj mol^{-1}}\].
Explanation:
Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.
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