# Weight of Food and Energy Production

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**Thermochemical equations** are used to relate energy changes to the chemical reactions that produce them. For example, we've already seen in Metabolism of Dietary Sugar that sugar is metabolized according to the equation^{[1]}:

_{6}H

_{12}O

_{6}(s) + 6 O

_{2}(g) → 6 CO

_{2}(g) + 6 H

_{2}O(l) (25

^{o}, 1 Atm) Δ

*H*= –2808 kJ (1) Here the sign of Δ

_{m}*H*(delta

_{m}*H*subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and the value enables us to find the actual quantity of energy involved. By convention, if Δ

*H*is

_{m}*positive*, heat is

*absorbed*by the reaction; i.e., it is endothermic. More commonly, Δ

*H*is

_{m}*negative*as in Eq. (1), indicating that heat energy is

*released*rather than absorbed by the reaction, and that the reaction is

**exothermic**. This convention as to whether Δ

*H*is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (1), the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of Δ

_{m}*H*.

_{m}It is important to notice that Δ*H _{m}* is the energy change for the equation as written. This is necessary because the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (1) tells us that 2805 kJ of heat energy is given off

*for every mole*

*of C*for every 6 mole

_{6}H_{12}O_{6}which is consumed. Alternatively, it tells us that 2808 kJ is released*of*

*H*H

_{2}O produced, i.e., 468 kJ is produced for every mol H_{2}O. Δ_{m}for Equation (1) also tells us that 2808 kJ of heat is released when 6 mol of carbon dioxide is produced, or 6 mol of oxygen is consumed. Seen in this way, ΔH

_{m}is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q

*is the quantity of heat absorbed and*n

*is the amount of substance involved, then*

**EXAMPLE 1** How much heat energy is obtained if we assume that the eagle's diet of 250-550 g includes 350 g of glucose, C_{6}H_{12}O_{6}, which is burned in oxygen according to the equation:

C_{6}H_{12}O_{6}(s) + 6 O_{2}(g) → 6 CO_{2}(g) + 6 H_{2}O(l)

*H*= –2808 kJ (3)

_{m}

**Solution** The mass of C_{6}H_{12}O_{6} is easily converted to the amount of C_{6}H_{12}O_{6} from which the heat energy *q* is easily calculated by means of Eq. (2). The value of Δ*H _{m}* is –2805 kJ per mole of C

_{6}H

_{12}O

_{6},

so that

Note: By convention a negative value of*q*corresponds to a release of heat energy by the matter involved in the reaction.

The quantity Δ

*H*is the

_{m}**enthalpy change for the reaction equation as written**. In this context the symbol Δ (delta) signifies change in” while

*H*is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Enthalpy For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy.

It is important to realize that the value of Δ*H _{m}* given in thermochemical equations like (1) or (3) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. (1), the value of Δ

*H*would be different from -2808 kJ. It is also necessary to specify both the temperature and pressure since the value of Δ

_{m}*H*depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure.

_{m}Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that *writing an equation in the reverse direction changes the sign of the enthalpy change. *For example,

_{2}O(

*l*) → H

_{2}O(

*g*) Δ

*H*= 44 kJ (4

_{m}*a*) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H

_{2}O(

*g*) → H

_{2}O(

*l*) Δ

*H*= –44 kJ (4

_{m}*b*) To see why this must be true, suppose that Δ

*H*[Eq. (4a)] = 44 kJ while Δ

_{m}*H*[Eq. (4b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for Δ

_{m}*H*of the reverse reaction to be equal in magnitude but opposite in sign from Δ

_{m}*H*of the forward reaction. That is, Δ

_{m}*H*forward = –Δ

_{m}*H*reverse

_{m}## References

- ↑ Atkins, P. Physical Chemistry, 6th Ed., W.H. Freeman &Co. New York, 1998, p. 69

### Contributors

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.