Question

The changes in internal energy for the oxidation of glucose and stearic acid at 310K are -2.9E3 kJ/mol and -11.36E3 kJ/mol, respectively a) Calculate the change in enthalpy for each reaction. b) Which substance is more likely to be useful for physiological energy storage?

Answer 1

As we know, Change ij enthalpy (H) = change in internal energy (U)+ pV

or, H=U+RT

for glucose, U=-2.9E3 kJ/mol and for stearic acid U=-11.36E3 kJ/mol

given T=310K and R=8.314 J/K mol for both glucose and stearic acid

now, H for glucose=-2.9E3 kJ/mol + 8.314 J/ K mol X 310 K

=-2.9E3 kJ/mol + 2577 J/mol

=-2900 kJ/mol + 2.577 kJ/mol

=-2897.423 kJ/mol

=-2.897E3 kJ/mol

similarly for stearic acid, H =-11.36E3 kJ/mol + 8.314 J/ K mol X 310 K

=-11.36E3 kJ/mol + 2577 J/mol

=-11360 kJ/mol + 2.577 kJ/mol

=-11357 kJ/mol

=-11.357E3 kJ/mol

Since stearic acid yields much more energy than glucose so it is better suited for physiological energy storage.