Determining Changes in Enthalpy

There are two ways in which chemists routinely determine ∆H values. One is experimentally using calorimetry; the other is theoretically using Hess’s Law.

Calorimetry

A calorimeter is a laboratory device used to measure the enthalpy change in a chemical or physical process. There are two types of calorimeters, one open, the other closed. In open calorimetry, the chemical system is open to the atmosphere and, as such, is subject to a constant pressure. In this case, the heat that flows into or out of a system is equal to the change in enthalpy. This is not the case in closed calorimetry. A simple open calorimeter consists of an insulating container (often just a Styrofoam cup) housing a known mass of water, along with a thermometer and stirring device. Since a calorie is a unit of energy (described in Chapter 1) and is defined as the amount of energy needed to raise the temperature of 1 gram of water by 1°C, a calorimeter, as its name implies, measures energy change. Based on this definition, the specific heat for water—that is, the particular amount of energy needed to raise a particular amount of water by a certain degree—is 1 cal/g°C. Since a joule, the SI unit for energy, is 4.186 times smaller than the calorie, the specific heat of water is also 4.186 J/g°C. Because of the known specific heat of water, the amount of energy change involved in a process can easily be calculated by monitoring the temperature of the water in the calorimeter present to the reaction to see how much it changes during the process. The calculation uses the formula

where q denotes the heat flow into or out of the water, m stands for the mass of the water, c represents the specific heat of the water, and ∆T its temperature change.

Example 1

Find the change in enthalpy, ∆H, in kJ/mol for the dissolution of NaOH in water; assume that when 0.10 moles (4.0 g) of the sodium hydroxide is dissolved in 96.0 g of water, the temperature of the water rises from 25.0°C to 34.9°C. Also, assume the specific heat of the overall solution is the same as water and that the overall solution (not just the water) will be what absorbs or releases energy from or to the system. This is typically done in fairly dilute solutions.

Using the equation to find the heat flow into the solution (this is the case since the temperature of the solution went up) we find that

This value represents the energy gained by the solution (mostly the water) and therefore the amount of energy lost by the chemical system. This value also represents the change in enthalpy for the reaction involving the 4.0 grams of NaOH used (which is only 0.10 moles of NaOH). Recall, however, that heat-of-reaction values are typically reported in units of kJ per mole. Therefore, it would be typical to change the sign on the heat flow, q, from positive to negative and the value from joules to kilojoules. Finally, dividing that amount of energy by 0.10 moles allows for representation of the reaction in a typical manner:

The making of this solution would be an exothermic process because the heat of reaction is negative. As 41 kJ of energy would be released if it was to occur, it is therefore shown as a product. The production of this solution would feel warm to the touch.

Example 2

0.050 L of a 1.0 M solution of NaOH is poured into 0.050 L of a 1.0 M solution of HCl already present in a calorimeter. The temperature of both solutions is 20.0°C initially, but once they are combined in the calorimeter, the temperature rises to 27.0°C. Find ∆H for this double replacement/acid-base/neutralization reaction. Assume the combination of both solutions produces a fairly dilute solution (mostly water) with a specific heat that would be the same as water. Also, assume that the density of all solutions is the same as water: 1 g/mL.

The mass of the total solution would be 100.0 grams, with the assumption above that each of the combined solutions is 50.0 g (as their volumes are 50.0 mL); therefore,

To associate this change in enthalpy with the chemical reaction, the sign would have to be changed to negative (as what the solution gained was lost by the chemical system), the units changed to kilojoules, and the value divided by the number of moles of NaOH or HCl used in the reaction (both of which are the limiting reactant). Recall from the previous chapter that according to the definition of molarity, the number of moles of a solute present in a solution is found by multiplying the molarity of the solution by its volume. Since each solution started out at 1.0 M and 0.050 L of it was present, the number of moles of either reactant used is 0.050 moles. Therefore,

Heating and Cooling Curves

Besides calorimetry, the equation q = mc∆T is used in determining the amount of energy needed to raise the temperature of any substance, not just water in a calorimeter. This is typically the case when evaluating the amount of energy needed to take a substance in the solid state, below its freezing point, to the gaseous state, above its boiling point. Figure. 8.2 provides a visual depiction of the process for water and is called a heating curve, as heat energy is constantly being supplied to the water as time goes by. If viewed in reverse, with energy being removed as time goes by and the gas caused to be turned into a solid, it would be called a cooling curve.

Graph showing the change from ice to steam. On the vertical axis is temperature on the horizontal axis is time. As temperature increases ice transforms to water, and then water reaches its specific heat capacity to turn to steam. — Figure 8.2 Changing Ice to Steam

Ice changing to water and then to steam is not a process of continuous and constant change of temperature as time progresses; rather, it is accomplished in stages. Although there are really five stages to analyze when changing a solid below its freezing point to a gas above its boiling point, this graph supplies only numerical information to calculate the amount of energy needed to take ice at its freezing point to steam at its boiling point. Therefore, we will evaluate only these three energy changes for the system.

Starting at 0°C, water’s normal melting point, the ice begins to melt. Notice that the temperature is not increasing despite the constant supply of heat as time passes. The lack of temperature increase indicates that the kinetic energy of the system is not changing; this is so because temperature and kinetic energy are directly related to each other (see Chapter 7). If the energy is not being used to make the particles move more, what is it doing? The answer is that it is overcoming the attractive forces between the solid particles that hold them in fixed crystalline positions and thus allowing them to exist as non-rigid clusters of particles characteristic of the liquid phase. Consequently, the potential energy of the system must be on the rise, as the energy is being used to overcome the attractive forces between the particles and accomplish the phase change. The energy required to do this, the heat of fusion, was discussed in Chapter 7 as an important property of a solid. The heating curve shows various values, utilizing various units, for the heat of fusion of water. Different substances have values different from those shown for water. An upcoming calculation will use the value 6.01 kJ/mol as the heat of fusion of ice. In light of what has been discussed in this chapter, it is simply another particular ∆H value.

Once the entire sample of the solid has melted, the liquid water’s temperature begins to rise as more heat is pumped into the system. Since the temperature of the water is rising, the kinetic energy of the system is rising as well. The q = mc∆T equation can be used to calculate the energy required to do this. The temperature will continue to rise until it reaches 100.0°C and the liquid water begins to normally boil.

At the normal boiling point, liquid water turns into gaseous water (steam); once again we see a plateau associated with the temperature change. As before, KE (kinetic energy) is not increasing, and so PE (potential energy) must be going up. This time, the clusters associated with the liquid are being ripped apart into randomness. The energy required to do this, the heat of vaporization, was discussed as well in Chapter 7 as an important property of a liquid. From the heating curve, one value for the heat of vaporization of water is 40.79 kJ/mol.

Example

Find the total amount of energy needed to raise a 180. gram sample of ice at 0.0°C to steam at 100.0°C.

This problem involves the three stages of the heating curve described above. First, the ice must be melted. Second, the water must rise in temperature. Third, the water must vaporize to a gas.

Melting

Since the value used for the heat of fusion of ice is in kJ/mol, the amount of ice must be converted to moles. Via mental math, 180. grams converts to 10.0 moles because the molar mass of water is 18.0 grams. Using the heat of fusion, we find that

Raising the Temperature of the Liquid Water

Since the specific heat term has grams in it, the 180-gram value (not moles) should be used for the amount of water in this calculation. ∆T will be 100.0°C as the temperature changes from the normal melting point of water to the normal boiling point of water. Using the q formula, we find that

Boiling

Again, moles will be used for the amount of water because the heat of vaporization noted above incorporates that unit. Using heat of vaporization, we find that

To find the total energy required, simply add the energy associated with the various stages investigated. The total energy required to change ice at its melting point to steam at its boiling point is 60.1 kJ + 75.2 kJ + 401 kJ or 535 kJ.

Hess’s Law of Constant Heat Summation

As suggested previously, there is a theoretical way to determine the heat of reaction for a process. Hess’s Law takes advantage of the First Law of Thermodynamics, which states that like matter, energy is conserved in chemical and physical changes. Consequently, knowledge of ∆H values for certain reactions allows you to find the ∆H value for a reaction whose heat of reaction you don’t know. This is so because the change in enthalpy is constant—irrespective of whether a process takes place in one step or many—because energy is conserved. This idea complements previous ideas in this chapter, which stated that heat of reaction is pathway independent (and is called a state function) in that the total enthalpy change is dependent only on the initial and final enthalpies of the reactants and products.

Stated in a useful manner, Hess’s Law allows for reactions about which you know ∆H values to be added to find ∆H values for reactions about which you don’t know. This is seen in the examples below.

Example

Find the heat of reaction, ∆H, for the process

when the heats of reaction for the following processes are provided:

Addition of the chemical processes above results in the reaction for which the heat of reaction is desired. The CO seen in both the products of the first reaction and the reactants of the second cancel out upon the addition of the reactions. Furthermore, the ½ O₂(g) reactant, seen in both reactions, adds to make present 1 O₂(g), as is found in the target reaction (i.e., the reaction for which the heat of reaction is not given). Carbon, C, is found in the reactants upon addition, and CO₂ is likewise found in the products. Hence, the addition of the two “steps” produces the target reaction. As stated before, energy change is constant; it is not contingent upon the number of steps it takes to accomplish a process. Since the two reactions add up to the target reaction, their ∆H values should add up to that of the overall reaction. The ∆H for the target reaction is then –393.5 kJ.

Sometimes reactions with known ∆H values are given as steps for an overall process but cannot be added together directly to achieve the target reaction. These may require some preliminary mathematical manipulation prior to the addition of the steps. The types of mathematical manipulation include

FLIPPING A REACTION (i.e., looking at the reaction in reverse). This will cause the substances previously in the reactants to now be in the products and vice versa. As a result, the enthalpy change will become the opposite of what it previously was from a thermodynamic perspective. This is easily shown by changing the sign on ∆H.
MULTIPLYING OR DIVIDING A REACTION. Since enthalpy is an extensive property, in which the amount of substance involved in a reaction is important to consider, proper numbers of moles of reactants and products need to be seen in the steps to achieve the target reaction upon addition. Whatever value the reaction is multiplied or divided by should also be applied to the ∆H value.

Example

Find the heat of reaction, ∆H, for the process

with the heats of reaction for the following processes provided:

Addition of the reactions above, as written, would not yield the target reaction, and so appropriate mathematical manipulation must be done. Analysis of the reactions shows that the pertinent substance in both the target reaction and the first step is tungsten, W. Since the target reaction indicates only 1 mole of tungsten reacting but the first step indicates 2 moles of tungsten reacting, the first step must be divided by 2 and rewritten as

Further analysis of the reactions shows that the pertinent substance in both the target reaction and the second step is carbon, C. Since the target reaction indicates 1 mole of carbon is reacting and the second step indicates the same, no manipulation of step 2 is needed.

Final analysis of the reactions shows that the pertinent substance in both the target reaction and the third step is tungsten carbide, WC. Since the target reaction indicates 1 mole of tungsten carbide is produced and the third step shows 2 moles of tungsten carbide reacting, the third step must be divided by 2 and flipped.

Addition of the three manipulated equations produces the target reaction (after the CO₂, O₂, and WO₃ all cancel) and so the addition of the ∆H values associated with those reactions will give the ∆H for the target reaction.

Standard heats of formation (introduced at the beginning of this chapter) are values typically found in the appendix of most chemistry textbooks. They serve as a resource to predict the spontaneity of reactions like synthesis, decomposition, and single replacement. They can also be used as a resource to help determine ∆H° values for reactions as a less cumbersome alternative to the method shown above. Rooted in Hess’s Law, the technique is based on the concept that a standard heat of reaction, ∆H°, is equal to the difference between the total enthalpy of the reactants and the total enthalpy of the products. This can be expressed as follows:

where

∑ = the summation of

n = the number of moles of the substances involved in the reaction

∆ = the standard heat of formation for the substance found in Table A in the Appendix.

Example

Having found the heat of reaction for the decomposition of sodium chlorate,

find the total amount of energy released when a 106.5 gram sample of sodium chlorate is decomposed.

Since the reaction equation describes the amount of energy change associated with 2 moles of NaClO₃ decomposing, you must first find the number of moles of NaClO₃ represented by 106.5 grams of the substance. A quick check of the Periodic Table reveals that the molar mass of sodium chlorate is 106.5 grams. Therefore, the amount of NaClO₃ decomposing is simply 1.000 mol. Employing dimensional analysis, complete the calculation using the ∆H value associated with the reaction equation:

Since the energy change is shown as a negative value, the information should be interpreted to mean that 52.3 kJ of energy will be released upon the decomposition of 106.5 g of sodium chlorate.