CHM1 12 Thermochemical Equations Collection

Thermochemical Equations

Energy changes which accompany chemical reactions are almost always expressed by thermochemical equations, such as

`CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)`     (25°C, 1 atm pressure)

`DeltaH_m = –890 kJ`      (1)

Here the `DeltaH_m` (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs as written, and also enables us to find the actual quantity of energy involved. By convention, if `DeltaH_m` is positive, heat isabsorbed by the reaction; i.e., it is endothermic. More commonly, `DeltaH_m` is 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 `DeltaH_m` 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 `DeltaH_m`.

It is important to notice that `DeltaH_m` is the energy for the reaction as written. In the case of Equation (1), that represents the formation of 1 mol of carbon dioxide and 2 mol of water. 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 890.4 kJ of heat energy is given off for every mole of CH₄ which is consumed. Alternatively, it tells us that 890.4 kJ is released for every 2 moles of H₂O produced. Seen in this way, `DeltaH_m` is a conversion factor enabling us to calculate the heat absorbed or released when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed or released and n is the amount of substance involved, then

      (2)

EXAMPLE 6 How much heat energy is obtained when 1 kg of ethane gas, C₂H₆, is burned in oxygen according to the equation:

`2C_2H_6(g) + 7O_2(g) -> 4CO_2(g) + 6H_2O(l)`

`DeltaH_m = –3120 kJ`      (3) 

 

The quantity `DeltaH_m` is referred to as an enthalpy change for the reaction. In this context the symbol `Delta` (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 Chap. 15. 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 `DeltaH_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 `DeltaH_m` would be different from -890.4 kJ. It is also necessary to specify both the temperature and pressure since the value of `DeltaH_m` 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.

Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is thatwriting an equation in the reverse direction changes the sign of the enthalpy change. For example,

`H_2O(l) -> H_2O(g)`

`DeltaH_m = 44 kJ`      (4a)

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_2O(g) -> H_2O(l)`

`DeltaH_m = –44 kJ`      (4b)

To see why this must be true, suppose that `DeltaH_m` [Eq. (4a)] = 44 kJ while `DeltaH_m` [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 `DeltaH_m` of the reverse reaction to be equal in magnitude but opposite in sign from `DeltaH_m` of the forward reaction. That is,

`DeltaH_m "forward" = –DeltaH_m` reverse 

Subpages (1): Example 6