Calculate the enthalpy change for the combustion of 1 mole of ethanol.?
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Answer:
-------C2H6O(l) + 3O2(g) ---> 2CO2(g) + 3H2O(l) + heat mol = 1 => 3 ---> 2 => 3 + heat H = ( - 277) + (0) ---> ( - 393.52) + ( - 285.82) + heat mol*H = ( - 277) + (0) ---> ( - 393.52) + ( - 285.82) + heat heat = ( -393.52) + ( -285.82) - ( - 277) heat = - 1367.50 kJ/mol (exothermic-negative-heat released) ethanol mw = 46.05 g/mol heat = ( - 1367.50) / (46.05) = - 29.70 kJ/g
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Other answers
-------C2H6O(l) + 3O2(g) ---> 2CO2(g) + 3H2O(l) + heat mol = 1 => 3 ---> 2 => 3 + heat H = ( - 277) + (0) ---> ( - 393.52) + ( - 285.82) + heat mol*H = ( - 277) + (0) ---> ( - 393.52) + ( - 285.82) + heat heat = ( -393.52) + ( -285.82) - ( - 277) heat = - 1367.50 kJ/mol (exothermic-negative-heat released) ethanol mw = 46.05 g/mol heat = ( - 1367.50) / (46.05) = - 29.70 kJ/g
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