Example: The Metabolism of Glycine
The biochemical reactions necessary to sustain life in a person produce about 6000 kJ/day of heat at constant pressure. This is the basal metabolic rate. Depending on level of activity 8000 to 12,000 kJ/day may be required in the form of food to replenish a person’s energy requirement (a nutritionist's Cal is a chemist’s kcal = 4.184 kJ). Each gram of protein or carbohydrate provides about 15 kJ/g and fat provides about 35 kJ/g.
Let us consider the energy that can be extracted from the metabolism of the amino acid glycine (NH2CH2COOH). By metabolism we mean its breakdown into the physiologically relevant products, carbon dioxide and urea.
The oxidation of solid glycine at 25o C to form CO2, NH3, and H2O is given below:
(1) 3 O2(g) + 2 NH2CH2COOH(s)
® CO2(g) + 2 H2O(l) + 2 NH3(g) DH1 = -1163.5 kJ/molThe standard enthalpy for the hydrolysis of urea is:
(2) H2O(l) + H2NCONH2(s)
® CO2(g) + 2 NH3(g) DH2 = 133.3 kJ/molIf we subtract these two reactions, treating them as thermochemical reactions we obtain
(3) 3 O2 + 2 NH2CH2COOH(s)
® H2NCONH2(s) + 3 CO2(g) + 3 H2O(l) DH3 = -1296.8 kJ/molThe above equation is of biochemical interest, because urea rather than ammonia is the main oxidative product of amino acid metabolism. However, the biological reaction does not involve solid glycine and solid urea but rather aqueous solutions. Therefore, we note that standard enthalpy of solution of urea and glycine.
(4) NH2CH2COOH (s)
® NH2CH2COOH (aq) DH4 = +15.7 kJ/mol(5) H2NCONH2 (s)
® H2NCONH2 (aq) DH5 = +13.9 kJ/molNote that we have used the limiting enthalpy of solution which implies that the solid substances are dissolved in an infinite amount of liquid water. It is assumed that these enthalpies do not depend strongly on concentration.
We now subtract two times equation (4) and add equation (5) to equation (3). Note that there are two moles of glycine on the left-hand side of the equilibrium in (3) so that to apply Hess’s Law we must multiply the molar reaction in Eqn. 4 by the factor 2.
Solvated Equation for Biological Oxidation of Glycine = Eqn. (3) – 2 x Eqn. (4) + Eqn. (5)
(6) 3 O2 + 2 NH2CH2COOH (aq)
® H2NCONH2 (aq) + 3 CO2 (g) + 3 H2O (l)D
H6 = DH3 –2 DH4 + DH5 =(-1296.8 kJ/mol) – 2(15.7 kJ/mol) + (13.9 kJ/mol)D
H6 = -1314.3 kJ/molDo not be confused by the stoichiometric coefficients in the reactions. The enthalpies given refer to the heat released or absorbed per mole for the reaction as written. For example, Eqn. 3 is written in terms of moles of urea produced and therefore two moles of glycine are consumed in this reaction. We could have written it instead in terms of the number of moles of glycine consumed, however, we would need to modify the enthalpy by the appropriate factor in order to obtain the correct amount of heat released.
3/2 O2 + NH2CH2COOH(s)
® ½ H2NCONH2(s) + 3/2 CO2(g) + 3/2 H2O(l)½
DH6 = -648.4 kJ/mol