T - Calorimetry (Lesson)

Calorimetry

Thermal energy changes are an integral part of all chemical reactions. For example, it is very important to know enthalpy changes associated with reactions with industrial applications so reaction systems that can handle large heat loads for highly exothermic reactions can be designed. Because knowing the magnitude of these changes is so important, it is therefore important to be able to experimentally determine them. A technique referred to as calorimetry is utilized to determine these changes in thermal energy. Using a calorimeter, scientists can measure temperature changes associated with both physical and chemical changes and then use that data to calculate changes in enthalpy. The image on the right depicts a diagram of a simple calorimeter comprised of insulating cups wherein a chemical reaction occurs. A thermometer has also been inserted to measure the temperature change during the reaction.

Specific Heat Capacity

But how exactly is a change in temperature related to an exchange of heat between the system and surroundings? How can this information be used to calculate a change in enthalpy? Scientists recognize that different substances store heat energy to different degrees. In other words, to increase the heat content of different substances requires different amounts of temperature change. This property of substances is referred to as heat capacity and is defined as the quantity of heat necessary to increase the temperature of a substance by 1 K. A more related, and more important, value is specific heat capacity which is the amount of heat required to increase the temperature of 1 g of a substance by 1 K and has units of J/g $LaTeX: \cdot$ $\cdot$ K. Some heat capacities of various substances are shown in the table below. The higher the heat capacity, the more energy is required to increase the temperature of that substance. For example, it takes roughly ten times more energy to increase the temperature of 1 gram of ethanol by 1 K compared to silver metal.

Table
Substance	Specific Heat Capacity (J/g $LaTeX: \cdot$ $\cdot$ K)
Silver	0.235
Copper	0.385
Iron	0.449
Aluminum	0.903
Ethanol	2.42
Water	4.184

Furthermore, the relationship between heat (q) and temperature change ( $LaTeX: \Delta$ $\Delta$ T) is as follows:

$LaTeX: q=mc_s\Delta T$ $q=mc_s\Delta T$

In this relationship,

q is the heat energy expressed in Joules (J)
c_s is the specific heat capacity of the substance being studied expressed in J/g $LaTeX: \cdot$ $\cdot$ K
$LaTeX: \Delta$ $\Delta$ T is the change in temperature

For example, suppose a 90.5 g sample of iron is heated from 53.2 to 65.4 $LaTeX: ^{^{^{\circ}}}$ $^{^{^{\circ}}}$ C. What quantity of heat would be required to accomplish this? Using the equation and specific heat capacities above it can be determined as follows:

**Note that no unit conversion from Celsius to Kelvin is required because a change by 1 $LaTeX: ^{^{\circ}}$ $^{^{\circ}}$ C is the same as a change by 1 K.

$LaTeX: q=mc_s\Delta T$ $q=mc_s\Delta T$

$LaTeX: \Delta T=T_{final}-T_{initial}$ $\Delta T=T_{final}-T_{initial}$

$LaTeX: c_s=0.449\left(\frac{J}{g\cdot K}\right)$ $c_s=0.449\left(\frac{J}{g\cdot K}\right)$

$LaTeX: q=\left(90.5\right)\left(0.449\right)\left\lbrack\left(65.4-53.2\right)\right\rbrack=496J$ $q=\left(90.5\right)\left(0.449\right)\left\lbrack\left(65.4-53.2\right)\right\rbrack=496J$

The sample problem above is an example of a physical change. There is no chemical change involved and the mass of iron is simply being heated. What about a chemical reaction? How are energy changes for chemical reactions determined? Assume a 0.896 g sample of magnesium reacts with 300 g of dilute hydrochloric acid according to the equation shown below. Because the acid solution is so dilute the specific heat capacity of water can be utilized from the table shown above. As the reaction proceeds, temperature of the acid solution increases from 23.8 $LaTeX: ^{\circ}$ $^{\circ}$ C to 31.3 $LaTeX: ^{\circ}$ $^{\circ}$ C.

Mg (s) + 2 HCl (aq) $LaTeX: \longrightarrow$ $\longrightarrow$ MgCl₂ (aq) + H₂ (g)

The calculations are similar to the previous example with two notable differences described below:

$LaTeX: q=mc_s\Delta T$ $q=mc_s\Delta T$

$LaTeX: \Delta T=T_{final}-T_{initial}$ $\Delta T=T_{final}-T_{initial}$

$LaTeX: c_s=4.184\left(\frac{J}{g\cdot K}\right)$ $c_s=4.184\left(\frac{J}{g\cdot K}\right)$

$LaTeX: q=\left(300.896\right)\left(4.184\right)\left\lbrack\left(31.3-23.8\right)\right\rbrack=9442J$ $q=\left(300.896\right)\left(4.184\right)\left\lbrack\left(31.3-23.8\right)\right\rbrack=9442J$

As mentioned above, there are a few things to note. First of all, it is necessary to recognize that the mass used in the equation is the total mass. The 300.896 grams of mass used in the equation represents the mass of both the Mg metal and the hydrochloric acid. The second item is associated with the direction of heat flow. According to the calculations, a positive q value was determined, indicating an endothermic process. However, it is necessary to understand what temperatures are actually being measured. The chemical reaction itself should be considered the system, and the surroundings would be the solution that the reaction is taking place in. For this process, it is the temperature of the surroundings that is actually being measured and not the system; therefore, there is a decrease in the heat content of the system as the energy is transferred from the system to the surroundings causing the increase in temperature that is measured. Because of this, the sign for this process is negative because it is an exothermic process. In other words:

$LaTeX: q_{system}=-q_{solution}$ $q_{system}=-q_{solution}$

$LaTeX: q_{system}=-(9442J)$ $q_{system}=-(9442J)$

$LaTeX: q_{system}=-9442J$ $q_{system}=-9442J$

You Try It!

In the following self-assessment activity, determine how much heat energy is absorbed. Click on the plus sign to check your answer!

Determination of Enthalpy

In the previous lesson, the concept of enthalpy was discussed including the incorporation of enthalpy into chemical equations transforming them into thermochemical equations. This lesson has focused primarily on how to calculate the energy change, q, associated with chemical and physical processes. The question remains as to how to take a calculated q value obtained from calorimetry experiments and convert that to a $LaTeX: \Delta$ $\Delta$ H value for that process.

A $LaTeX: \Delta$ $\Delta$ H_rxn can be calculated from calorimetry data quite simply. By dividing the q value calculated using mass, specific heat capacity, and temperature change by the number of moles of the limiting reactant, a q value (in J or kJ) can be converted into a $LaTeX: \Delta$ $\Delta$ H value (in kJ/mol) for that particular reaction. For example, returning to the example above involving the reaction between magnesium metal and hydrochloric acid, the q value obtained there can be converted to $LaTeX: \Delta$ $\Delta$ H_rxn by assuming that magnesium is the limiting reactant (Note: this cannot always just be assumed and would have to be determined via calculation). First, the mass of magnesium must be converted into moles:

$LaTeX: 0.896gMg(\frac{1 mol Mg}{24.31 g Mg})=0.0369molMg$ $0.896gMg(\frac{1 mol Mg}{24.31 g Mg})=0.0369molMg$

Dividing the q value of -9442 J by the number of moles of this limiting reactant provides the $LaTeX: \Delta$ $\Delta$ H_rxn:

$LaTeX: \frac{-9442J}{0.0369mol Mg}=-256,000Jor-256kJ$ $\frac{-9442J}{0.0369mol Mg}=-256,000Jor-256kJ$

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