## CHEMISTRY THE CENTRAL SCIENCE

**5 THERMOCHEMISTRY**

**5.6 HESS'S LAW**

It is often possible to calculate the Δ*H* for a reaction from the tabulated Δ*H* values of other reactions. Thus, it is not necessary to make calorimetric measurements for all reactions.

Because enthalpy is a state function, the enthalpy change, Δ*H*, associated with any chemical process depends only on the amount of matter that undergoes change and on the nature of the initial state of the reactants and the final state of the products. This means that whether a particular reaction is carried out in one step or in a series of steps, the sum of the enthalpy changes associated with the individual steps must be the same as the enthalpy change associated with the one-step process. As an example, combustion of methane gas, CH_{4}(*g*), to form CO_{2}(*g*) and H_{2}O(*l*) can be thought of as occurring in one step, as represented on the left in **FIGURE 5.21**, or in two steps, as represented on the right in Figure 5.21: (1) combustion of CH_{4}(*g*) to form CO_{2}(*g*) and H_{2}O(*g*) and (2) condensation of H_{2}O(*g*) to form H_{2}O(*l*). The enthalpy change for the overall process is the sum of the enthalpy changes for these two steps:

**GO FIGURE**

**What process corresponds to the** –**88 kJ enthalpy change?**

**FIGURE 5.21 Enthalpy diagram for combustion of 1 mol of methane.** The enthalpy change of the one-step reaction equals the sum of the enthalpy changes of the reaction run in two steps: –890 kJ = –802 kJ + (–88 kJ).

The net equation is

**Hess's law** states that *if a reaction is carried out in a series of steps*, **Δ***H for the overall reaction equals the sum of the enthalpy changes for the individual steps*. The overall enthalpy change for the process is independent of the number of steps and independent of the path by which the reaction is carried out. This law is a consequence of the fact that enthalpy is a state function. We can therefore calculate Δ*H* for any process as long as we find a route for which Δ*H* is known for each step. This means that a relatively small number of experimental measurements can be used to calculate Δ*H* for a vast number of reactions.

Hess's law provides a useful means of calculating energy changes that are difficult to measure directly. For instance, it is impossible to measure directly the enthalpy for the combustion of carbon to form carbon monoxide. Combustion of 1 mol of carbon with 0.5 mol of O_{2} produces both CO and CO_{2}, leaving some carbon unreacted. However, solid carbon and carbon monoxide can both be completely burned in O_{2} to produce CO_{2}. We can therefore use the enthalpy changes of these reactions to calculate the heat of combustion of carbon.

**GIVE IT SOME THOUGHT**

What effect do these changes have on Δ*H* for a reaction:

**a.** reversing the reaction,

**b.** multiplying the coefficients of the equation for the reaction by 2?

**SAMPLE EXERCISE 5.8 Using Hess's Law to Calculate** Δ *H*

The enthalpy of reaction for the combustion of C to CO_{2} is −393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO_{2} is −283.0 kJ/mol CO:

Using these data, calculate the enthalpy for the combustion of C to CO:

**SOLUTION**

**Analyze** We are given two thermochemical equations, and our goal is to combine them in such a way as to obtain the third equation and its enthalpy change.

**Plan** We will use Hess's law. In doing so, we first note the numbers of moles of substances among the reactants and products in the target equation, (3). We then manipulate equations (1) and (2) to give the same number of moles of these substances, so that when the resulting equations are added, we obtain the target equation. At the same time, we keep track of the enthalpy changes, which we add.

**Solve** To use equations (1) and (2), we arrange them so that C(*s*) is on the reactant side and CO(*g*) is on the product side of the arrow, as in the target reaction, equation (3). Because equation (1) has C(*s*) as a reactant, we can use that equation just as it is. We need to turn equation (2) around, however, so that CO(*g*) is a product. Remember that when reactions are turned around, the sign of Δ*H* is reversed. We arrange the two equations so that they can be added to give the desired equation:

When we add the two equations, CO_{2}(*g*) appears on both sides of the arrow and therefore cancels out. Likewise, is eliminated from each side.

**PRACTICE EXERCISE**

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is –393.5 kJ/mol, and that of diamond is −395.4 kJ/mol:

Calculate **Δ***H* for the conversion of graphite to diamond:

** Answer:** +1.9 kJ

**SAMPLE EXERCISE 5.9 Using Three Equations with Hess's Law to Calculate Δ H**

Calculate Δ*H* for the reaction

given the following chemical equations and their respective enthalpy changes:

**SOLUTION**

**Analyze** We are given a chemical equation and asked to calculate its Δ*H* using three chemical equations and their associated enthalpy changes.

**Plan** We will use Hess's law, summing the three equations or their reverses and multiplying each by an appropriate coefficient so that they add to give the net equation for the reaction of interest. At the same time, we keep track of the Δ*H* values, reversing their signs if the reactions are reversed and multiplying them by whatever coefficient is employed in the equation.

**Solve** Because the target equation has C_{2}H_{2} as a product, we turn the first equation around; the sign of Δ*H* is therefore changed. The desired equation has 2 C(*s*) as a reactant, so we multiply the second equation and its Δ*H* by 2. Because the target equation has H_{2} as a reactant, we keep the third equation as it is. We then add the three equations and their enthalpy changes in accordance with Hess's law:

When the equations are added, there are 2 CO_{2}, O_{2}, and H_{2}O on both sides of the arrow. These are canceled in writing the net equation.

**Check** The procedure must be correct because we obtained the correct net equation. In cases like this you should go back over the numerical manipulations of the Δ*H* values to ensure that you did not make an inadvertent error with signs.

**PRACTICE EXERCISE**

Calculate Δ*H* for the reaction

given the following information:

**Answer:** –304.1 kj

The key point of these examples is that *H* is a state function, so *for a particular set ofreactants and products*, Δ*H is the same whether the reaction takes place in one step or in a series of steps*. We reinforce this point by giving one more example of an enthalpy diagram and Hess's law. Again we use combustion of methane to form CO_{2} and H_{2}O, our reaction from Figure 5.21. This time we envision a different two-step path, with the initial formation of CO, which is then combusted to CO_{2} (**FIGURE 5.22**). Even though the two-step path is different from that in Figure 5.21, the overall reaction again has Δ*H*_{1} = **–** 890 kJ. Because *H* is a state function, both paths *must* produce the same value of Δ*H*. In Figure 5.22, that means Δ*H*_{1} = Δ*H*_{2} + Δ*H*_{3}. We will soon see that breaking up reactions in this way allows us to derive the enthalpy changes for reactions that are hard to carry out in the laboratory.

**GO FIGURE**

**Suppose the overall reaction were modified to produce 2 H**_{2}**O( g) rather than 2 H**

_{2}**O(**Δ

*l*). Would any of the values of

*H*in the diagram stay the same?**FIGURE 5.22 Enthalpy diagram illustrating Hess's law.** The net reaction is the same as in Figure 5.21, but here we imagine different reactions in our two-step version. As long as we can write a series of equations that add up to the equation we need, and as long as we know a value for Δ*H* for all intermediate reactions, we can calculate the overall **Δ***H*.