Lab 6. Calorimetry

Helen Tian CHEM 11200 2 Lab Partner: Jeff Baek Lab 6. Calorimetry Introduction Calorimetry is an experimental science that studies heat flow that may occur from either physical or chemical processes (i.e. melting, combustion, etc). Any processes that include heat generation and exchange can be a subject of calorimetric study. Such areas of study include the food industry, environmental science, the pharmaceutical industry, material science, and chemical engineering. For example, it can help determine the calorie content of food products, which is important for nutrition labeling and dietary planning, understand the thermal properties of materials, and study the heat effects of certain drug interactions. 1 In this experiment, we are measuring the temperature change from chemical reactions to find the enthalpy change of each reaction. To calculate enthalpy change, we use the following equation: . The heat transferred to the solution inside the calorimeter is given by ? = [? ?? ? ? + ? ?? ? ? ]∆? . The heat transferred to the physical calorimeter is . Thus, the total amount of heat [? ?? ? ? ∆?] [? ?? ? ? ∆?] freed from a reaction is the sum of these two parts. The calorimeter constant (c sc m c ) is derived from our calorimetry process with room temperature and ice-cold water; the equation is . We are given c sr and d r (which we will use in addition to the ? ?? ? ? = 4. 184[ 𝑉 ???? (? ? −? ???? ) (? ???ℎ −? ? ) ] − 𝑉 ???ℎ volume of each solution (~100 mL) to solve for m r ). Then, our main experiment will be analyzing the temperature change in each reaction. Using a stirring bar and plastic cover to get the most accurate data, we used a thermometer to record the temperature changes by every few seconds. We will be using the final temperature from this, in relation with the initial temperature of the original solution, to get to ∆? plug into our equation. Through all these variables and given that the pressure is constant, we find q which is - because heat is flowing out of the system into the surroundings (aka the bath system and ∆𝐻 ?𝑥? that’s why in the data section, you can see that the temperature is generally increasing). We will be calculating the enthalpy change and quantifying the relationship between heat flow and temperature change with the three following processes: 1. ???𝐻 (??) + 𝐻𝐶? (??) → ??𝐶? (??) + 𝐻 2 ? (?) 2. 𝐶𝐻 3 𝐶??𝐻 (??) + ???𝐻 (??) →𝐶𝐻3𝐶?? − (??) + 𝐻2? (?) 3. 2𝐹?𝐶? 3 (??) + ??𝐶? 2 (??) → 2𝐹?𝐶? 2 (??) + ??𝐶? 4 (??) Procedure Our procedure did not stray from the lab manual provided to us. We did, however, skip a couple of steps that were asking us to repeat testing the water. Additionally, the lab handbook provided intervals that didn't allow for us to see the detailed changes across the entire time period. Therefore, we recorded the temperature changes at intervals of 5 seconds for the first minute and then, every 30 seconds for the second minute; this was so we could easily observe the change in temperature.

Data Analysis (Tables + Graphs) Table 1. Temperature ( ℃ ) vs. Time (s) of H 2 O Initial Temperature of room-temperature water (T 0 = 22.0 ℃ ) and ΔT = 3.0 o C Time (s) Temp ( ℃ ) 0 22.0 5 19.0 10 19.0 15 19.0 20 19.0 25 19.0 30 19.0 35 19.0 40 19.0 45 19.0 50 19.0 55 19.0 60 19.0 90 19.0 120 19.0 Table 2. Temperature ( ℃ ) vs. Time (s) of HCl + NaOH Initial Temperature (T 0 = 21.5 o C) and ΔT = 3.5 o C Time (s) Temp ( ℃ ) 0 21.5 5 23.0 10 24.0 15 25.0 20 25.0 25 25.0 30 25.0 35 25.0 40 25.0 45 25.0 50 25.0 55 25.0 60 25.0 90 25.0 120 25.0

Calculations Part 1. Finding the Calorimeter Constant Quantity Experimental Value V bath (mL) 70 mL T bath (°C) 22.0°C V cold (mL) 30 mL T cold (°C) 3.8°C T final (°C) 19.0°C |ΔT| = |T bath - T final | 3.0°C c sc m C (J·deg -1 ) 343 J deg -1 Part 2. Calculating the Temperature Changes Reaction Initial T (°C) Final T (°C) ΔT (°C) HCl + NaOH 21.5 25.0 3.5 CH 3 COOH + NaOH 21.5 25.0 3.5 Fe 3+ + Sn 2+ 21.5 23 1.5 Part 3. Calculating ΔH rxn Discussion Questions 1. Every experiment has some specific errors associated with the measurements. How would you expect them to affect your results? Describe the possible sources of error entering into your determination of the heat of reaction. 2. Why is it important that the density of all solutions is approximately constant? 3. How would you suggest improving the experiment for the future?

Discussion This experiment contains three sections: calculating the calorimeter constant, the temperature differences, and finding the enthalpy changes. Based on our volumes and temperatures of the room-temperature (70 mL at 22 o C) and ice-cold (30 mL at 3.9 o C) water, we were able to calculate a calorimeter constant of 343 J deg -1 . This value, compared to the actual value of 19 J deg -1 , is at a 1710% error. Our experimental value was so far from the value provided, because of the simplicity of our calorimeter. The design and composition of calorimeters can vary, and the constant depends on factors like the materials used and the specific design of the calorimeter. Additionally, we simplified the procedure so that our calorimeter in this step had only one styrofoam cup and no stir bar. This means that the environment outside of the calorimeter has more of an effect on the inside and that heat can more easily escape/enter the system. The omission of the stir bar also causes problems with creating a balanced system; therefore, where the thermometer was touching could’ve been a different temperature than other parts of the solution. From here, we began to use a two styrofoam cup calorimeter with the stir bar inside. And for the HCl + NaOH solution, we recorded an initial temperature of 21.5 o C which increased to 25 o C after 3 minutes of adding HCl. This means there was a temperature change of +3.5 o C. Using ΔT=3.5 o C, the enthalpy change of this reaction is ΔH rxn =-59.4 kJ/mol. Similarly, the CH 3 COOH + NaOH solution also had a temperature change of +3.5 o C with its initial temperature at 21.5 o C and final temperature at 25 o C. Using ΔT=+3.5 o C, the enthalpy change of this reaction is ΔH rxn =-59.7 kJ/mol. Lastly, the redox reaction led to a temperature change of +1.5 o C. And unlike the other reactions (where the reactants had the same moles), the Sn 2+ solution was our limit reagent here, leading to ΔH rxn =-65.6 kJ/mol. Notice that all three enthalpy changes are negative, which combined with observing that the solutions all increased in temperature, means that the reaction is an exothermic process. For the exothermic processes, heat moves out of the system (i.e., q sys is negative) and warms up its surroundings (i.e., q surr is positive). Since the temperature of the system increases (T f > T i ), then some heat has been transferred to the system, and the value of q is positive. Additionally, through energy conservation, we can also write ΔH rxn =-q . All

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