chemistry lab

CHEM 1411 Online Calorimetry Lab

Objectives:

1. To introduce constant pressure calorimetry

2. To experimentally determine the enthalpy change of a chemical reaction

Materials:

· Styrofoam lid · 2 styrofoam cups · 3 M HCl · NaOH solid · Thermometer · Balance · Weigh boat

· Graduated cylinder · 250 mL beaker · 250 mL Erlenmeyer flask · Stirring rod · Timer (student provides) · Distilled water (student provides)

Introduction:

Heat can be absorbed or given off during a chemical reaction. Reactions where heat is absorbed are called endothermic. Those reactions that release heat are called exothermic. The first law of thermodynamics states that energy cannot be created or destroyed. Therefore, any heat released by a chemical reaction (the system) is transferred to the surroundings. In this lab, the heat released by a neutralization reaction between aqueous reactants HCl and NaOH will be absorbed by the water and calorimeter. We will monitor the heat absorbed by measuring the temperature increase of the water.

Calorimetry is the process of measuring heat exchange. In order to measure the heat of a reaction, the reaction must be isolated so that no heat is lost to the environment. This is achieved by use of a calorimeter, a device used to measure the amount of heat released or absorbed by a chemical reaction. The calorimeter used in this lab is a simple thermally insulated device. It consists of two nested styrofoam cups with a lid or cardboard top and thermometer (Figure 1.)

Figure 1: Calorimeter

In this experiment, we will determine the heat or enthalpy change that occurs when HCl and NaOH undergo neutralization. The molar enthalpy, H, is the amount of heat released or absorbed per mole of reactant consumed or product produced in a chemical reaction. In this experiment, we will determine the heat released per mole of water formed when solutions of HCl and NaOH are added together.

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

In our experiment, known amounts of HCl and NaOH will be mixed together in a calorimeter. Since this reaction is exothermic, heat will be released. The heat released by the reaction will be absorbed by the aqueous solution in the calorimeter as well as the calorimeter. The solution’s temperature will increase and this increase will be measured by a thermometer. Since the mass of the solution and the specific heat of the solution are known, you will be able to calculate the heat absorbed by the solution. The following equation gives the total energy change of the solution in the calorimeter, qsolution.

qsolution = m x SH x ΔT

· m = mass of solution

· SH = specific heat of solution

· ΔT = Tf - Ti

Since the calorimeter itself will also absorb heat, this must be accounted for if we are to accurately measure the total heat released by the neutralization reaction. We can account for this if we know the heat capacity (C) of the calorimeter.

The heat capacity of a calorimeter is the amount of heat required to raise the temperature of the calorimeter by one degree Celsius. The heat capacity is used to calculate the amount of heat absorbed by the calorimeter, qcalorimeter. This is given by the following equation:

qcalorimeter = C x T

· C = the heat capacity of the calorimeter

· ΔT = Tf - Ti

In this experiment, we will assume that the heat capacity of the calorimeter is 10.0 J/oC. In other words, for every 1 oC rise in temperature, the Calorimeter is absorbing 10.0 J.

The total heat of the reaction, qrxn, will be the sum of the heat absorbed by the solution and the heat absorbed by the calorimeter:

qrxn = − (qsolution + qcalorimeter)

The negative sign indicates this is an exothermic reaction. A positive sign indicates the reaction is endothermic.

You must remember to include the negative sign in your answer.

Procedure:

Part I: Build a Calorimeter

1. Build the calorimeter by nesting two styrofoam cups together (one inside the other) and putting the lid cover on top. Put your clean thermometer through the hole.

See Figure 1.

Lab Technique Tip: Be sure to remove the plastic sheath covering the metal thermometer. Also be careful not to poke a hole in the Styrofoam cups with the pointed metal end of the thermometer.

2. Place the calorimeter in the 250 mL beaker for stability or you can place the calorimeter in a sturdy mug.

Part II: Prepare NaOH and HCl Solutions

1. Using the graduated cylinder place 50.0 mL of distilled water into the Erlenmeyer flask. Be exact with the volume.

2. Using the weigh boat, weigh out as exactly as you can 2.00 g of the solid NaOH. If your weight of the NaOH is very different from 2.00 g, then calculate the molarity of your NaOH solution. Do this by dividing your grams by the molar mass of NaOH (39.997 g/mol) to obtain moles. Then divide the moles by the 0.0500 L.

Caution: Always wear gloves when working with acids or bases.

1. Carefully add the NaOH to the distilled water in the flask and stir well. This should be a 1.0 M NaOH solution.

2. Record the molarity of the NaOH solution on the lab report.

3. Using the graduated cylinder, transfer 50.0 mL of NaOH solution to the calorimeter and record the exact volume to the correct number of sig figs as this will affect the moles of base calculation.

4. Let the NaOH solution sit for 10 minutes in the calorimeter to be sure it is at room temperature while you prepare the HCl solution.

Lab Technique Tip: Remember to always use clean equipment. Clean and rinse your 100 mL graduated cylinder before using it to measure the HCl solution.

5. Rinse the out the Erlenmeyer flask with plenty of water.

6. Measure out 34.0 mL of the distilled water and place it in the flask.

7. Measure out 17.0 mL of the 3 M HCl solution and pour it into the flask with the water.

8. Mix the solution well with the stirring rod. This will give you a 1.0 M solution of the HCl.

9. Record the molarity of the HCl on the lab report.

Lab Technique Tip: Remember to always follow safe procedures. Acid should be added to water and never the other way around.

10. Measure 50.0 mL HCl solution in the graduated cylinder and record the exact volume to the correct number of sig figs as this will affect the moles of acid calculation.

11. Let the HCl sit for a few minutes to be sure it is at room temperature.

12. Take a picture of the calorimeter with the graduated cylinder next to it and insert this in your lab report. The lab report must be saved as a doc or docx file with the clear, but small picture inserted. Be sure the picture is not too large of a file. The image should not be uploaded separately.

Part III: Mix Solutions and Record Temperatures

1. Take the temperature of the NaOH solution inside the calorimeter to the nearest 0.1°C every minute for exactly 4 minutes. The temperature should remain fairly constant and should be at room temperature.

2. Exactly, on the 5th minute, quickly pour the HCl solution into the calorimeter and replace the cover. Be sure the thermometer is inserted in the lid.

3. Swirl the calorimeter to mix and start recording the temperature again, beginning with the 6th minute

4. Keep recording the temperature every 30 seconds ending with the 10th minute.

5. Graph the data as shown in the example graph. Follow the instructions below on how to graph the data and determine the temperature change.

6. Label your graph showing the change in temperature.

7. When submitting the lab report, insert a clear picture of the graph. The scanned image or the picture must be inserted into the lab report and the lab report saved as a word doc or docx file. The images should not be uploaded separately. Be sure your picture does not have too high of a resolution. A high resolution will make the picture too large.

WASTE DISPOSAL: Dispose of all solutions down the drain with lots of water. Wash your equipment with soap and water as well as clean your work area.

Calculations:

Heat absorbed by solution, qsolution, is calculated from the solution’s mass, specific heat and temperature increase.

qsolution = m x SH x ΔT

· Assume that the specific heat of the solution is 4.184 J/g oC.

· Assume that the density of the solution is 1.00 g/mL. The masssolution can be calculated from the volume and density of the solution according to D = m/V.

· Determine ΔTsolution from graph. See instructions under the Example Graph (Figure 2).

Heat gained by calorimeter is calculated from the Heat Capacity (C) of the calorimeter and the temperature increase. Assume C = 10.0 J/oC.

qcalorimeter = C x T

Heat evolved in the reaction = − (heat absorbed by solution + heat absorbed by calorimeter)

qrxn = − (qsolution + qcalorimeter)

Molar enthalpy of reaction is the amount of heat released or absorbed per mole of reactant consumed or product produced in a chemical reaction. We will determine the heat released per mole of water formed.

H =

Moles of water is determined from the limiting reactant and the mole-to-mole ratio. The limiting reactant is the one that is used up first based on the moles of each (see lecture notes on limiting reactants and stoichiometry). If both of the reactants have the exact same moles, and the mole-to-mole ratio is one to one, then either one can be used as the limiting reactant. Using stoichiometry, we can then determine the moles of water from the moles of the limiting reactant and the mole-to-mole ratio (from the balanced equation).

Determining Change in Temperature from the Graph:

In order to find the final temperature of the solution after the reaction occurs we must graph the data of temperature vs. time. We have to do it this way since the change in temperature after you add the HCl is not instantaneous. There is a lag time between adding the HCl, the reaction occurring, the water heating, and the thermometer monitoring the change. If we could instantly record this temperature, we would not have to graph.

The data we take between 6 and 10 minutes will give us a line that we can extrapolate, or extend back to 5 minutes. Remember that the 5th minutes is the time when we added the HCl and we cannot instantly record this temperature. After the reaction occurs and heats up the solution, it will eventually cool down. We continue to take the temperature during the time interval of 6 to 10 minutes to measure the cooling of the solution.

Even though we are unable to take a temperature reading at exactly 5 minutes, we can still determine this temperature by extrapolating the data we take from 6 to 10 minutes. See the example graph in Figure 2. The blue part of the line is extrapolated data. The red line is experimental data, that is, the data we collected during the experiment. Follow the steps to determine the change in temperature.

1. Plot temperature vs time on the graph provided.

2. Draw the “best straight line” through the data points from the interval of 6 to 10 minutes. (This is the red line on the graph.) The “best straight line” goes through the most number of points. Any points not falling on the line will lie above or below the line. Arrange the line so that about half of the points not on the line will be above the line and half will be below the line.

3. Extend this “best straight line” from the 6 to 10 minute interval back to 5 minutes or before. (This is the blue line.)

4. Determine the temperature at 5 minutes, the time when the HCl was added.

· Draw a vertical line from 5 minutes on the x-axis up to the extrapolated line.

· From there, draw a horizontal line to the y-axis.

· This point on the y-axis gives the final T. This example has a Tf = 30.5 C.

Figure 2: Graph of Temperature vs. Time

Sample Problem:

In an neutralization experiment similar to the one you are performing today, a student mixes 22.3 mL of a 1.30 M solution of KOH with 24.0 mL of a 1.10 M HNO3 solution. After plotting the temperature vs time data, she found that ∆T was 6.9 oC. Assume that the density of the solution is 1.00 g/mL, the specific heat of the solution is 4.184 J/g oC, and the heat capacity of the calorimeter is 10.0 J/oC. Calculate the following:

a. qsolution, the heat absorbed by the solution

b. qcal, the heat absorbed by the calorimeter

c. qrxn, the heat evolved in the reaction in Joules

d. the molar heat of neutralization using the moles of water produced

Part a: Heat absorbed by solution

qsolution = m x SH x ΔT

msolution = mL of total solution x 1.00 g/mL= (22.3 +24.0) mL x 1.00 g/mL = 46.3 g

qsolution = 46.3 g x 4.184 J/g oC x 6.9 oC

= 1336.7 J

(Limited to two sig figs since 6.9 oC is two sig figs, but do not round yet)

Part b: Heat gained by calorimeter:

qcalorimeter = C x T

= 10.0 J/ oC x 6.9oC

= 69 J

(Limited to two sig figs since 6.9 oC is two sig figs, but do not round yet)

Part c: Heat of the reaction:

qrxn = − (qsolution + qcalorimeter)

= − (1336.7 J + 69 J)*

= − 1405.7 J

= − 1.4057 kJ

*underlined numbers are the significant digits

Part d: Molar heat of neutralization:

H =

The reaction is: HNO3 (aq) + KOH (aq) → KNO3 (aq) + H2O (l)

Since this reaction has a 1 to 1 stoichiometry of all the species, the reactant with the smallest number of moles will be limiting. You can calculate the number of moles of each reactant and compare. Remember to change the mL to L.

# moles of HNO3 = MHNO3 x V HNO3 = 1.10 M x 0.0240 L = 0.02640 mol HNO3

# moles of KOH = MKOH x VKOH = 1.30 M x 0.0223 L= 0.02899 mol KOH

We use the limiting reactant to determine the moles of water produced using our mole-to-mole ratio and stoichiometry.

Therefore, H =

=

= − 53.246 kJ/mol

= − 53 kJ/mol

North Lake College CHEM 1411 Online Calorimetry Lab p. 7 Sum 2018