Thermochemistry is the subject that deals with the heats involved in chemical reactions. A typical chemical reaction might have a form similar to the following hypothetical chemical reaction:
a A + b B → c C + d D. (1)(In Equation (1) the upper case letters stand for elements or compounds and the lower case letters stand for small whole numbers which balance the reaction. You would read this as saying, "a moles of A reacts with b moles of B to give c moles of C and d moles of D.")
A chemical reaction is a process just like any other thermodynamic process. It has an initial state (the reactants) and a final state (the products). We can calculate the changes in internal energy, enthalpy, and so on for the reaction. For example,
ΔU = U products− U reactantsand
ΔH = H products− H reactants.One thing is sometimes not made very clear. "Reactants" and "products" in these equations means that the reactants and products are separated, isolated, and pure. Furthermore, the reactants and products are all at the same temperature and pressure. So, for example, the ΔH above is the enthalpy of c moles of C (isolated and pure in its own container at temperature, T, and pressure, p) plus the enthalpy of d moles of D (isolated and pure in its own container at temperature, T, and pressure, p) minus the enthalpy of a moles of A (isolated and pure in its own container at temperature, T, and pressure, p) minus the enthalpy of b moles of B (isolated and pure in its own container at temperature, T, and pressure, p).
From time to time we will add a superscript o to H or U to indicate that reactants and products are in their "standard states." That is, they are in their most stable state at T and p. For example, the standard state of water at 25oC and 1 atm pressure is liquid water.
If our reaction takes place at constant V, as in a bomb calorimeter, dV =0 and
ΔUV = qV.If the reaction takes place at constant p, as in open to atmospheric pressure, dp = 0 and
ΔHp = qp.If ΔHp < 0 we say that the reaction is exothermic. That is, the system gave heat to the surroundings. On the other hand, if ΔHp > 0 we say that the reaction is endothermic. The system absorbed heat from the surroundings.
From the definition of enthalpy we find that
ΔH = ΔU + Δ (pV), (2)where
Δ (pV) = (pV) products − (pV) reactants.For liquids and solids Δ(pV) is quite small. Δ(pV) is not necessarily small for gases, but we can get a reasonable estimate for this quantity by approximating the gases as ideal. Then
Δ (pV)gas ≈ (pV)gas products − (pV)gas reactants ≈ n gas products RT − n gas reactants RT = RTΔ n gas,where Δn gas is the difference in the number of moles of gaseous products and reactants. Using this approximation we get
ΔH = ΔU + RTΔ n gas. (3)However, we have to be careful how we understand this equation because the conditions of the reaction must be the same on both sides of the equation. Since ΔH is the heat we measure if the reaction is run and constant pressure (ΔHp = qp) and ΔU is the heat we measure if the reaction is run at constant volume (ΔUV = qV), it is tempting (and common) to write Equation (3) as
qp = qV + RTΔ n gas. (4)However, Equation (4) cannot be rigorously true since the q's refer to different conditions, one at constant p and one at constant V. We can get an indication whether or not Equation (4) is a good approximation with the following example:
Consider the reaction.
2 C(s) + O2(g) → 2 CO(g), (5)run at dV = 0.
Δ (pV) ≈ Δ (pV)gas ≈ RTΔ n gas = RT.The heat measured is qV. Now let us find a way to measure qp Consider the following two steps
2 C(s) + O2(g) → 2 CO(g) → 2 CO(g) (6)
p1, V1, T, qV p2, V1, T, ΔH2 p1, V2, T.
The first step is the constant volume reaction we had before with ΔUV = qV. Notice that the pressure increases. The second step takes the product of the constant volume reaction and reduces the pressure back to the original pressure. We call the heat for this step ΔH2. So it is rigorously true that
ΔHp = ΔHV + ΔH2 = qV + RT + ΔH2. (7)However, the ΔH2 term is the enthalpy for the expansion of a gas at constant temperature. If the gas is ideal this term is zero. For real gases this term would be very small, so we make a negligible error is neglecting it. So to pretty good approximation we can use the equation,
qp = qV + RTΔ n gas. (4)
Hess' law states that if you add or subtract chemical reaction equations you can (must) add or subtract their corresponding ΔH's or ΔU's to get ΔH or ΔU for the overall reaction. For example, if we add the reaction
a A + b B → c C + d D. ΔH1, ΔU1to the reaction
e E + f F → g G + h H ΔH2, ΔU2to get
a A + b B + e E + f F → c C + d D + g G + h H,Then, for the overall reaction
ΔH = ΔH1 + ΔH2 and ΔU = ΔU1 + ΔU2.The great utility of Hess' law is that we don't have to tabulate ΔH for every possible reaction. We can get ΔH for a particular reaction by adding and subtracting ΔH's for a much smaller set of reactions, called formation reactions. We define ΔfH o for a compound to be the enthalpy of the reaction:
pure, isolated elements, in their standard
(most stable) states
→ one mole of compound in its standard state.
For example, the heat of formation of liquid water is defined as ΔH o for the reaction,
H2(g) + 1/2 O2(g) → H2O(l).By definition, then, the heat of formation for an element in its most stable state is zero.
To obtain ΔH o for the hypothetical reaction, Equation (1), we add and subtract the appropriate heats of formations,
ΔrH o = cΔfHCo + d ΔfHDo− a ΔfHAo− b ΔfHBo. (8)For example, ΔH o for the reaction,
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l), (9)is given by
(We don't always write out explicitly that the coefficients which balance the reaction have units, but we have done so here to make it clear that these numbers have units. This will become an issue later.)
ΔH at Other Temperatures
Tables of heats of formation usually give data for reactions at 25oC. We frequently need to know the heat of a reaction at a temperature other than 25oC. If we know the heat capacities at constant pressure we can calculate the heat of reaction at a temperature other than 25oC. We use the following chain of reasoning. We know that,
(10a, b, c)where ΔCpo is defined for our hypothetical chemical reaction, Equation (1) as,
(11)Prepare Equation 10c for integration as
(13)For very accurate work we will have to use the temperature dependent heat capacities in Equations (11) and (13), but more often than not we can regard the heat capacities as approximately constant over the temperature range, so that Equation (13) becomes,
(14)As an example, let's calculate the heat of reaction for the reaction in Equation (9) at 95oC.
ΔCpo for the reaction in Equation (9) is given by,
(15)Then, Equation (14) gives,
(16)(Note: There is another way to do this problem. Calculate ΔH o for cooling the reactants down to 25oC, calculate ΔH o for the reaction at 25oC, calculate ΔH o for heating the products back up to 95oC, and then add them up. Since H is a state function ΔH is independent of path. This method will also work if one of the components of the reaction has a phase change somewhere in the temperature range.
For example, if we let the new temperature be over 100oC we would have to account for the vaporization of the liquid water product. This is not hard to do, but requires some extra steps, including also the use of the heat capacity of water vapor. If the upper temperature in this problem is above 100oC we carry out the reaction in several steps:
Step 1 - Cool the reactants from the upper temperature to 25oC,The heat of reaction at the upper temperature is the sum of the ΔH's for all six steps. Again we have taken advantage of the fact that ΔH is independent of path.)
Step 2 - Run the reaction at 25oC,
Step 3 - Heat the product CO2 from 25oC to the upper temperature,
Step 4 - Heat the liquid water from 25oC to 100oC,
Step 5 - Vaporize the water at 100oC,
Step 6 - Heat the water vapor from 100oC to the upper temperature.
Δ H as Making and Breaking Chemical Bonds
Breaking a chemical bond is an endothermic process. That is, you must put energy into the system to break the bond.
Forming a chemical bond is an exothermic process. The energy released in forming the bond goes into the surroundings.
We can make an estimate of the ΔH for a chemical reaction by adding the bond energies of all the bonds broken and subtracting the bond energies of all the bonds formed.
ΔH ≈ BE bonds broken − BE bonds formed.Let's try it on the gas phase reaction,
N2 + 3 H2 → 2 NH3.(There are tables of bond energies in physical chemistry texts and in data handbooks. From the tables we find the following bond energies:
N−N 945 kJIn the reaction we break 1 N−N bond and 3 H−H bonds. We form 6 N− H bonds. The approximate ΔH is then,
H−H 436 kJ
N−H 388 kJ.
ΔH ≈ 1 × 945 + 3 × 436 − 6 × 388 = − 75 kJ.This can be compared to the actual value of − 92 kJ. The method is not super accurate, but it gives a ball-park answer and might be useful in cases where other data are not available. Further, however it demonstrates graphically that the heat of a reaction is related to the making and breaking of chemical bonds.
Heats of Formation of Ions in Water Solution
When you look in a table of heats of formation you find values listed for ions in water solutions. That is, you will find an entry for species such as Na+(aq). It is fair to ask where these numbers come from. We know that in equilibrium chemistry it is impossible to prepare Na+(aq) ions in solution all by themselves. It is possible to prepare a solution that has both Na+(aq) and Cl− (aq) ions, but not a solution that has only ions of one charge.
The heats of formation of solutions of soluble ionic compounds can be measured. That is, we can measure the heat of formation of HCl(aq). The heat of formation of HCl(aq) is defined as ΔH o for the reaction,
1/2 H2(g) + 1/2 Cl2(g) → HCl(aq). (17)Since ionic compounds in solution are completely dissociated, it must be true that
(18)We cannot know the heat of formation of either of the ions in solution, but we do know their sum.
By convention we arbitrarily set the heat of formation of the H+(aq) ion equal to zero. That is,
(19)Then the heats of formation of all other aqueous ions can be determined relative to the heat of formation of the H+(aq) ion. As a start, we see that
(20a, b, c)This convention allows us to build up a table of heats of formation of aqueous ions.
For example, from the measured value of the heat of formation of NaCl(aq) and the knowledge that
(21)we find that
(22a, b)Continuing these procedures we can define the heats of formations of other aqueous ions,
(23a, b, c)We now have enough data in our table to calculate the heat of formation of aqueous NaBr without measuring it.,
(24a, b)If the heat of formation of H+(aq) were set to some value other than zero, it would have cancelled out of Equation (24b). In this manner an entire table of heats of formation of aqueous ions can be built with all values relative to the heat of formation of the H+(aq) ion.
Copyright 2004, W. R. Salzman
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