Ammonium Nitrate in Water Lab Report

d

How do we characterize thermochemical change?

In the study of thermochemistry, heat changes that accompany physical and/or chemical processes are measured with a calorimeter, a closed insulated container designed for this purpose and equipped with a thermometer.

# Constant-Pressure Calorimetry A constant-pressure calorimeter is used to measure the heat exchange between a system and the surroundings for a variety of processes such as acid-base neutralizations, heats of solution and heats of dilution. The constant-pressure calorimeter (Figure 1) consists of an insulated vessel that contains the reaction mixture or a liquid of known specific heat (usually water), a stirrer to ensure homogeneity in the liquid phase, and a thermometer to indicate the liquid phase temperature. While the vessel is closed to minimize heat loss, it is not sealed, consequently the pressure within the vessel remains constant and equal in magnitude to the ambient pressure.

Because the pressure is constant, the heat change for the process (𝑞) is equal to the enthalpy change (∆𝐻). In such experiments we consider the reactants and products to be the system, and the water in the calorimeter to be the surroundings. Typically, the small heat capacity

of the calorimeter is ignored. Figure 1. Constant-Pressure Calorimeter

In the case of an exothermic reaction, the heat released by the system is absorbed by the water (surroundings), thereby increasing its temperature. Knowing the mass (𝑚) of the water in the calorimeter, the specific heat (𝑠) of the water, and the change in temperature (∆𝑇), we can calculate 𝑞𝑝 (heat at constant pressure) of the system by

## 𝑞𝑠𝑦𝑠𝑡𝑒𝑚 = −𝑠𝑚∆𝑇

where ∆𝑇 is defined as 𝑇𝑓 − 𝑇𝑖. Note the minus sign makes 𝑞𝑠𝑦𝑠𝑡𝑒𝑚 a negative number if ∆𝑇 is positive (increasing temperature). This is in keeping with the sign convention of ∆𝐻. A negative ∆𝐻 or 𝑞 indicates an exothermic process, whereas a positive ∆𝐻 or 𝑞 indicates an endothermic process.

# Constant-Pressure Calorimetry Example: Determination of ΔHrxn

Assume we are interested in determining the ∆𝐻𝑟𝑥𝑛 for the acid-base neutralization between hydrochloric acid (HCl) and the base sodium hydroxide (NaOH). We start with 49.11 mL of 1.01 M HCl and 50.01 mL of 1.03 M NaOH. Both solutions are at room temperature, which we measure as 23.5 °C. The balanced chemical equation is:

## HCl + NaOH → NaCl + H2O

Both solutions are poured into a constant-pressure calorimeter and the calorimeter is closed. As the reaction proceeds, the temperature of the water increases as it absorbs energy released by the reaction. A maximum water temperature of 30.3 °C is recorded.

Assuming the density and specific heat of the HCl and NaOH solutions to be the same as those of water (1.00 g/mL,

4.184 J/g °C), we calculate 𝑞𝑠𝑜𝑙𝑛 (𝑞 for the solution) as follows:

### 𝑞𝑠𝑜𝑙𝑛 = 𝑠𝑚∆𝑇

𝑞𝑠𝑜𝑙𝑛 = (4.184 𝐽 ) (49.11 𝑔 + 50.01 𝑔)(30.3 ℃ − 23.5 ℃) = 2.82 × 103 𝐽 𝑔 ℃

Recall 𝑞𝑠𝑦𝑠𝑡𝑒𝑚 = −𝑞𝑠𝑢𝑟𝑟𝑜𝑛𝑑𝑖𝑛𝑔𝑠 and that we consider 𝑞𝑠𝑜𝑙𝑛 = 𝑞𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 with 𝑞𝑟𝑥𝑛 = 𝑞𝑠𝑦𝑠𝑡𝑒𝑚. Thus, assuming the heat capacity of the calorimeter is negligible, we know that 𝑞𝑟𝑥𝑛 = −𝑞𝑠𝑜𝑙𝑛, and can write

### 𝑞𝑟𝑥𝑛 = −2.82 × 103 𝐽

This is the heat of reaction when 0.0496 mol HCl reacts with 0.0515 mol NaOH. To determine ∆𝐻𝑟𝑥𝑛, we divide 𝑞𝑟𝑥𝑛 by the number of moles of the limiting reagent (in this case HCl),

𝑞𝑟𝑥𝑛

∆𝐻𝑟𝑥𝑛 =

𝑚𝑜𝑙𝐿𝑅

∆𝐻𝑟𝑥𝑛 = −2.82 × 103 𝐽 4 𝐽 𝑘𝐽

𝑚𝑜𝑙

=

56

.

9 = −5.69 × 10

0.0496 𝑚𝑜𝑙 𝑚𝑜𝑙

# Constant-Volume Calorimetry

The heat of combustion is usually measured using constantvolume calorimetry. Typically, a known amount of sample is placed in a steel container called a constant-volume bomb, or simply a bomb, which is pressurized with oxygen (O2). The closed bomb is then immersed in a known amount of water within an insulated container, as illustrated in Figure 2. Together, the steel bomb and the water in which it is submerged constitute the calorimeter. The sample is ignited electrically, and the heat released by the combustion of the sample is absorbed by the bomb and the water and can be determined by measuring the increase in the water temperature. The special design of this type of calorimeter allow us to assume that no heat (or mass) is lost to the surroundings during the time it takes to carry out the reaction and measure the temperature change. Therefore, we call the bomb and the water in which it is submerged an isolated system. Because no heat enters or leaves the system during the process, the heat change of the system overall is zero (𝑞𝑠𝑦𝑠𝑡𝑒𝑚 = 0) and we can write: Figure 2. Constant-Volume Calorimeter

𝑞𝑐𝑎𝑙 = −𝑞𝑟𝑥𝑛

where 𝑞𝑐𝑎𝑙 and 𝑞𝑟𝑥𝑛 are the heat changes for the calorimeter and reaction, respectively. Thus,

𝑞𝑟𝑥𝑛 = −𝑞𝑐𝑎𝑙

To calculate 𝑞𝑐𝑎𝑙 we need to know the heat capacity of calorimeter (𝐶𝑐𝑎𝑙) and the change in temperature, that is,

𝑞𝑐𝑎𝑙 = 𝐶𝑐𝑎𝑙∆𝑇

And because 𝑞𝑐𝑎𝑙 = −𝑞𝑟𝑥𝑛 ,

𝑞𝑟𝑥𝑛 = −𝐶𝑐𝑎𝑙∆𝑇

The heat capacity of the calorimeter (𝐶𝑐𝑎𝑙) is determined by burning a substance (𝑋) with an accurately known heat of combustion (𝑞𝑐𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛) for a given mass and measuring the temperature increase ∆𝑇. Subsequently, the heat capacity of the calorimeter is given by:

𝑞𝑐𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛

𝐶𝑐𝑎𝑙 =

## ∆𝑇

Once 𝐶𝑐𝑎𝑙 has been determined, the calorimeter can be used to measure the heat of combustion of other substances. Because a reaction in a bomb calorimeter occurs under constant-volume rather than constant-pressure conditions, the measured heat change corresponds to the internal energy change (∆𝑈) rather than the enthalpy change (∆𝐻). It is possible to correct the measured heat changes so they correspond to ∆𝐻 values, but the corrections are usually quite small, so we will not concern ourselves with the details here.

# Constant-Volume Calorimetry Example: Determining the Energy Content per Mass and Calories per Serving

A sample of Post Grape-Nuts® cereal weighing 5.81 g is burned in a bomb calorimeter to determine its energy content. The heat capacity of the calorimeter is 43.7 kJ/°C. During the combustion, the temperature of the water in the calorimeter increases by 1.92 °C. Calculate the energy content (in kJ/g) and Calories per serving of Grape-Nuts®. First, we determine the heat released by the combustion using

𝑞𝑟𝑥𝑛 = −𝐶𝑐𝑎𝑙∆𝑇

𝑘𝐽

## 𝑞𝑟𝑥𝑛 = − (43.7 ) (1.92 ℃) = −83.9 𝑘𝐽

The negative sign in the result indicates heat is released by the combustion. To find the energy content per gram of Grape-Nuts®, we divide the heat released by the mass of the Grape-Nuts® sample.

## |𝑞𝑟𝑥𝑛| 83.9 𝑘𝐽

𝐸𝑛𝑒𝑟𝑔𝑦 𝐶𝑜𝑛𝑡𝑒𝑛𝑡 𝑝𝑒𝑟 𝑔𝑟𝑎𝑚 = = = 14.4 𝑘𝐽/𝑔 𝑚𝑠𝑎𝑚𝑝𝑙𝑒 5.81 𝑔

(Note, because energy content must be a positive quantity, we write 83.9 kJ as a positive value. In the above, we use the absolute value notation, |𝒒𝒓𝒙𝒏|, to remind us of this.)

According to the nutrition facts label on the Grape-Nuts® package, a serving size is 58 g. The dietary Calorie (Cal) is 4.184 kJ. Hence, converting the energy per gram to Calories per serving gives,

(14.4 𝑘𝐽) ( 1 𝐶𝑎𝑙 ) ( 58 𝑔 2 𝐶𝑎𝑙/𝑠𝑒𝑟𝑣𝑖𝑛𝑔

) = 2.0 𝑥10 𝑔 4.184 𝑘𝐽 𝑠𝑒𝑟𝑣𝑖𝑛𝑔

Which, ignoring proper sig figs, is the value given on the Grape-Nuts® nutrition facts label.

Lab Assignment 2

All work must be very neat and organized. To receive credit, you must show all work in a legible, highly organized manner. Responses that are illegible and/or lack organized supporting work will be scored a “zero.” Significant figures must be reasonable, and correct units (where applicable) must be present.

Problem 1. Assume you are interested in exploring the use of ammonium nitrate (NH4NO3) as the active ingredient in an inexpensive home-made cold pack. You decide that a practical cold pack should be able to depress the temperature of 1 kg of human muscle tissue by 5 oC. Now you need to know the amount of NH4NO3 required to achieve this. Ammonium nitrate is a highly water soluble solid. You believe the enthalpy of dissolution for NH4NO3 must be determined. To this end, you add 2.339 g of NH4NO3 to 99.7 mL of water at 23.7 °C in a constant-pressure calorimeter and close the calorimeter. The temperature of the water decreases to a minimum of 21.9 °C.

1a (5p). Assume the density and specific heat of water is 1.00 g/mL and 4.184 J/g °C, respectively. Using the data above, determine the for ammonium nitrate in water. Assume the heat capacity of the calorimeter is negligible. To receive credit, you must show all work in a legible, highly organized manner.

1b (2p). Is the dissolution of ammonium nitrate in water an endothermic or exothermic process? Briefly state your reasoning.

1c (5p). The specific heat of human muscle tissue (smuscle) is 3.47 kJ/kg oC. Given this value, how many grams of ammonium nitrate (NH4NO3) is required to depress the temperature of 1.00 kg of human muscle by 5.00 oC? Please remember, you must show all work in a legible, highly organized manner.

Problem 2. You wish to determine the energy content per gram and Calories per serving of a chocolate chip cookie and a banana bread slice from the Student Union to decide which is less “fattening.” The intact cookie and banana bread slice you are analyzing weigh 72.501 g and 69.003 g, respectively. Before you can analyze the samples, the calorimeter heat capacity () must be determined. It is known that combustion of 1.000 g of benzoic acid releases 26.44 kJ of heat (= 26.44 kJ). Thus, you combust 1.000 g of benzoic acid in a constant-volume calorimeter and record a temperature increase of 2.52 °C.

2a (2p). From the data given above, determine the heat capacity of the constant-volume calorimeter (). To receive credit, you must show all work in a legible, highly organized manner.

2b (4p). Now you take a 5.688 g sample of the chocolate chip cookie and burn it in the same calorimeter. During the combustion, the temperature of the water in the calorimeter increases by 11.81 °C. For the chocolate chip cookie, calculate the energy content in kJ/g and the Calories per serving. Please remember, you must show all work in a legible, highly organized manner.

2c (4p). Finally, you take a 4.904 g sample of the banana bread and run it in the same calorimeter. During the combustion, the water temperature in the calorimeter increases by 12.29 °C. For the banana bread slice, calculate the energy content in kJ/g and the Calories per serving. Don’t forget, you must show all work in a legible, highly organized manner.

2d (3p). Based on the above Problem 2 data, is the chocolate chip cookie or banana bread slice less “fattening” in terms of the energy content in kJ/g? What about in terms of the Calories per serving? In complete sentences, clearly justify your reasoning citing the evidence.