Heat of Reaction
The objective of this lab is to investigate heat changes in chemical reactions and calculate the heat of reaction for two types of reactions: (1) dissolving a solid in water, and (2) an acid-base neutralization. Students will use various tools and equipment, interpret numerical and graphical data, demonstrate safe practices in the chemical laboratory, and apply their understanding of electrons in atoms, chemical bonding, chemical reactions, quantities in chemical reactions, molecules, compounds, and chemical composition. They will also demonstrate the proper use of exponential notation and significant figures, record their results, use scientific reasoning to evaluate physical and natural phenomena, and identify unifying themes in the scientific field of study.
The discussion provides an overview of energy changes during chemical reactions. It introduces the concept of heat of reaction, which represents the quantity of energy released or absorbed during a chemical reaction. Endothermic reactions absorb heat, while exothermic reactions release heat. The change in enthalpy (∆H) is used to quantify the heat change in a reaction, and it can be positive or negative depending on the direction of the heat flow.
In this experiment, two types of reactions are investigated: the dissolving of a solid in water and acid-base neutralizations. For the dissolving of a solid, the heat change is determined by the competition between lattice energy (energy required to break the ionic forces of the solid) and hydration energy (energy released when the ions become hydrated). The overall heat change is observed when two ionic solids are dissolved in water.
In acid-base neutralizations, heat is evolved during the reaction. The reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH) is used as an example. The net ionic equation shows that hydrogen ions (H+) and hydroxide ions (OH-) combine to form water, resulting in the evolution of heat. By measuring the temperature change when the acid and base are mixed, the heat of reaction can be calculated.
The lab procedure consists of three parts:
Part A involves the dissolving of ammonium nitrate (NH4NO3) in water. A Styrofoam cup is used as the calorimeter. Water is added to the cup, and its initial temperature is recorded. Ammonium nitrate crystals are weighed and added to the water. The temperature of the mixture is recorded at regular intervals for a total of 4 minutes. The temperature change is calculated by subtracting the lowest temperature recorded after mixing from the initial temperature. The heat change is determined by considering the heat lost by the cup and the heat lost by the water.
Part B focuses on the dissolving of anhydrous magnesium sulfate in water. The procedure is similar to Part A, with the only difference being the use of magnesium sulfate instead of ammonium nitrate. The temperature change is calculated by subtracting the initial temperature from the highest temperature recorded after mixing. The heat change is determined by considering the heat gained by the cup and the heat gained by the water.
Part C involves acid-base neutralizations using hydrochloric acid (HCl) and sodium hydroxide (NaOH). The concentrations of HCl and NaOH are known, and 50 mL of each solution is used. The solutions are mixed in the calorimeter, and the temperatures before mixing and the highest temperature after mixing are recorded. The heat change is calculated by considering the heat gained by the solution and the heat gained by the cup.
The calculations involve using specific heat equations and the heat capacities of the calorimeter cups. The heat change is determined by subtracting the heat lost or gained by the calorimeter from the heat lost or gained by the solution. The mass of the substances used is converted to moles, and the heat change is divided by the number of moles to obtain the heat of reaction per mole of the respective substance.
Throughout the lab, students are encouraged to record their measurements accurately, plot their data on graphs to visualize temperature changes over time, and make appropriate calculations using the provided equations and constants. The focus is on understanding heat changes in chemical reactions, interpreting data, and calculating the heat of reaction per mole of the substances involved.
Objective
· Use the tools and equipment necessary for basic scientific analysis and research
· Interpret the numerical and graphical presentation of scientific data
· Demonstrate safe practices in the Chemical Laboratory
· Demonstrate an understanding of Electrons in Atoms and the Periodic Table and Chemical Bonding
· Demonstrate an understanding of Chemical Reaction and Quantities in Chemical Reactions
· Demonstrate an understanding of Molecules, Compounds and Chemical Composition
· Demonstrate the proper use of Exponential Notation and Significant Figures
· Demonstrate an understanding of the composition of matter and energy
· Record the results of investigation through writing
· Use scientific reasoning to evaluate physical and natural phenomena
· Identify the unifying themes of the scientific field of study
Materials
· Styrofoam Cup
· Graduated Cylinder
· Ammonium Nitrate Crystals (NH4NO3)
· Thermometer
· Anhydrous Magnesium Sulfate
· 2M HCl
· 2M NaOH
Discussion
Every chemical change is accompanied by an energy transformation, typically in the form of heat. The amount of energy released or absorbed during a chemical reaction is known as the heat of reaction. Reactions that absorb heat are called endothermic, while those that release heat are called exothermic.
Laboratory studies of reactions are conveniently conducted under constant pressure conditions. In this context, the heat change in the system is referred to as the change in enthalpy, denoted as ΔH. In exothermic reactions, where heat is evolved, ΔH is negative. Conversely, in endothermic reactions, where heat is absorbed, ΔH is positive. Therefore, for the reaction:
(1) 2 H2 + O2 → 2 H20 + 68.3 kilocalories
ΔH = -68.3 kcal. The negative sign in exothermic reactions stems from the definition of ΔH as the difference between the enthalpy of the products and the enthalpy of the reactants.
(2) ∆H = (heat content of products) - (heat content of reactants)
Equation (2) can be rewritten as:
(3) ∆H = Hproducts - Hreactants
In this experiment, we will explore the heat of reaction for two types of reactions: (1) the dissolution of a solid in water, and (2) an acid-base neutralization. Rather than directly measuring the quantity of heat transferred, we will observe temperature changes as indicators of the heat effects and then calculate the heat change.
Dissolving of a Solid
When salt dissolves, two primary heat effects compete with each other. The first is the lattice energy, which represents the energy required to break the ionic forces holding the solid together. As the ions are released from the crystal lattice, they become hydrated, resulting in the liberation of energy known as hydration energy. If the lattice energy is greater than the hydration energy, the reaction will be endothermic. Conversely, if the hydration energy surpasses the lattice energy, the reaction will be exothermic. In this particular experiment, our focus will not be on measuring the specific values of lattice or hydration energies. Instead, we will examine the overall heat change that occurs when two ionic solids dissolve in water.
Neutralization Reactions
The addition of hydrochloric acid to a sodium hydroxide solution results in the release of heat. This heat is referred to as the heat of neutralization, as it corresponds to the reaction between an acid and a base. The equation representing this reaction is as follows:
(4) HCI + NaOH → NaCl + H2O + heat
Water is employed as a solvent to prepare NaOH and HCl solutions, resulting in the presence of ions for the acid, base, and salt in the solution.
(5) H+ + Cl- + Na+ + OH- → Na+ + Cl- + H2O + heat
Since sodium and chloride ions appear on both sides of Equation (5), they can be eliminated to derive a net ionic equation.
(6) H+ + OH- → H2O + heat
The net ionic equation demonstrates that a neutralization reaction between a strong acid and a strong base involves the combination of hydrogen ions (H+) and hydroxide ions (OH-) to produce water. This reaction is common to all strong acids and bases, and a fixed amount of heat is released for every mole of reacting hydrogen ion and hydroxide ion. Consequently, by utilizing different strong acids with a constant amount of base, an equivalent quantity of heat can be obtained for each neutralization reaction. However, this aspect falls outside the scope of the current experiment.
The heat generated during the neutralization reaction causes a temperature increase in the solution once the acid and base are mixed. By measuring the change in temperature using known volumes of NaOH and HCl with predetermined concentrations, it becomes possible to calculate the heat of the reaction.
Procedure
A. Dissolving of Ammonium Nitrate
1. Obtain a Styrofoam cup and ensure it is clean by rinsing it with water.
2. Using a graduated cylinder, measure 100 mL of water and carefully pour it into the cup.
3. Weigh approximately 5-7 grams of ammonium nitrate crystals, making sure to record the precise weight with an accuracy of +0.01 grams.
4. Record the temperature of the water, noting it to an accuracy of +0.1°C, every 30 seconds for a duration of 2 minutes.
5. Carefully add the weighed NH4NO3 crystals into the water and thoroughly mix until the solid is fully dissolved. Continue stirring the mixture.
6. Record the temperature of the mixture at 30-second intervals for a period of 4 minutes after mixing.
7. To determine the temperature change, calculate the difference between the initial temperature and the lowest temperature recorded after mixing.
8. For a visual representation of the temperature change, create a graph plotting temperature on the y-axis and time on the x-axis.
B. Dissolving of Magnesium Sulfate
Perform the same process for substance A once more, employing 100 mL of water. However, this time, incorporate 4-6 grams of anhydrous magnesium sulfate. Ensure to document temperature readings at 30-second intervals for a duration of 2 minutes before mixing, as well as 4 minutes after mixing. In this case, compute the temperature change as the highest temperature observed after mixing minus the initial temperature. As before, generate a graph displaying temperature variations over time and plot your recorded data.
C. Acid-Base Neutralizations
Using a graduated cylinder, measure 50 mL of a 2.0 M HCl solution and transfer it to the empty calorimeter cup. Similarly, take 50 mL of a 2.0 M NaOH solution and ensure that it has the same temperature as the HCl solution. Pour the NaOH solution quickly into the HCl solution while stirring rapidly with the thermometer. Take note of the initial temperatures of the solutions and record the highest temperature reached after mixing. By considering the temperature change, total volume of the mixed solution, and the specific heat of water, the amount of heat released can be determined. Express the heat of reaction in kilocalories per mole of water formed. Repeat this procedure exactly as described above.
Calculations
A. Dissolving of ammonium nitrate
This reaction is endothermic, so the temperature will drop after mixing, and heat is absorbed by the solution.
(7) Heat of Reaction + Heat lost by cup + Heat lost by water = 0
(8) Heat Lost (Total) = Heat of Reaction
(9) Heat Lost = heat lost by solution + heat lost by calorimeter
(10) Heat Lost = ( Msoln) (specific heat of water) (∆T) + (Heat capacity of cup) (∆T)
The specific heat of water has a value of 1.00 cal/g ∙ °C, and in the experiment, Heat of Fusion of Ice, you determined the heat capacity of the cup to be 14.0 cal/ºC. Substituting these values into Equation (10) gives:
(11) Heat Lost = (Msoln) (1.00 cal/g ∙ °C) (∆T) + (14.0 cal/°C) (∆T)
Convert the mass of ammonium nitrate that was used in the experiment to moles, and then divide Equation (11) by this number to obtain the heat of reaction per mole of ammonium nitrate.
B. Dissolving of magnesium sulfate
This reaction is exothermic, so the temperature will rise after mixing, and heat is liberated by the reaction.
(12) Heat of Reaction = Heat Gained
(13) Heat Gained = heat gained by solution + heat gained by calorimeter
(14) Heat Gained = (Msoln) (specific heat of water) (∆T) + (heat capacity of calorimeter) (∆T)
Once again, the specific heat of water is 1.00 cal/g ∙ °C and the heat capacity of the cup is 14.0 cal/°C, so substituting these values into Equation (14) gives:
(15) Heat Gained = (Msoln) (1.00 cal/g ∙ °C) (∆T) + (14.0 cal/°C) (∆T)
Convert the mass of magnesium sulfate dissolved in the experiment into moles of magnesium sulfate, and divide the value obtained in Equation (15) by this number of moles in order to obtain the heat of reaction per mole of magnesium sulfate.
C. Heat of Neutralization
(16) HEAT GAINED = HEAT of REACTION
(17) Heat Gained = heat gained by solution + heat gained by the calorimeter
(18) Heat Gained = (Msoln) (specific heat of water) (∆T) + (heat capacity of calorimeter) (∆T)
(19) Heat Gained = (Msoln) (1.00 cal/g ∙ °C) (AT) + (14.0 cal/°C) (AT)
Calculate the number of moles of water formed in Equation (6), and divide the heat gained in Equation (19) by this number to calculate the heat of reaction per mole of water formed.
Report Sheet for HEAT of REACTION
Name_______________________________________ Date___________________________
Lab / Section_________________________________
Questions for HEAT of REACTION
1. In this experiment, HEAT of REACTION, list the two types of reactions and give their definition
a)
b)
2. List what type of reactions below when using the following chemicals
a) MgSO4______________________________
b) NH4NO3_____________________________
3. During the heat of neutralization reaction, please list the two types of chemical solutions being added to yield NaCl and H20.
Define what are the properties (acid or base) of each solution to perform this neutralization
a)
b)
4. The stoichiometry of the reactants for the neutralization experiment are
a) 1:1
b) 2:1
c) 1:2
Review Questions
1. What is the purpose of this lab?
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2. How are endothermic and exothermic reactions different?
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3. Define the term "heat of reaction."
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4. Explain the concept of lattice energy.
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5. What is hydration energy?
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6. How does the heat of reaction relate to the change in enthalpy (∆H)?
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7. Describe the process of dissolving a solid in water.
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8. What factors determine whether a reaction is endothermic or exothermic when a solid is dissolved in water?
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9. What is the net ionic equation for an acid-base neutralization reaction?
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10. What does the heat of neutralization represent?
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11. How is the heat change in a reaction related to the temperature change of the solution?
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12. What is the specific heat of water and its significance in the calculations?
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13. Explain the concept of heat capacity and its role in calorimetry.
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14. How can you determine the heat lost or gained by a calorimeter?
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15. How do you calculate the heat change for the dissolving of ammonium nitrate?
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16. What is the significance of the temperature change in the dissolving of magnesium sulfate?
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17. How is the heat change calculated in an acid-base neutralization reaction?
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18. What is the importance of converting mass to moles in the calculations?
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19. How does plotting temperature versus time data help analyze the heat changes in the reactions?
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20. What are the key skills and knowledge demonstrated in this lab related to scientific analysis, data interpretation, and the understanding of chemical reactions?
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