Question #75774, Chemistry / Physical Chemistry / Completed
The standard enthalpies of formation of CO2(g), H2O(l) and glucose (s) at 25°C are -400kJ, -300 kJ/mol and -1300kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is
(1) 2900kJ
(2) -2900kJ
(3) -16.11 kJ
(4) 16.11 kJ
Solution:
C6H12O6 + 6O2 = 6CO2 + 6H2O
6(-400) + 6(-300) - (-1300) = -2400 - 1800 + 1300 = -2900 kJ/mol
M = 180 g/mol
-2900/180 = -16.11 kJ/g
Answer: (3) -16.11 kJ/g.
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