Chem 106 Lab 17

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106.LB

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Chemistry

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Apr 3, 2024

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5/6/22 Experiment 17: Calorimetry and Thermochemistry – “Stealing Joules” Introduction In experiment 17, students will create and use a calorimeter for the addition of the following substances: strong base (NaOH) to a strong acid (HCl), a heated unknown metal to water, and ammonium chloride salt to water. By recording the temperature of the reaction and graphing the relationship between the time and temperature, the collected information can be used to understand the heat of the reaction, enthalpy change, and heat capacity. However, a calorimeter is needed to find these values because a calorimeter only allows heat to interact with a reaction’s components and prevents any heat from escaping. It also prevents any outside influence from entering the reaction. Within all parts of experiment 17, the First Law of Thermodynamics can be seen through the recorded temperature changes, where the heat lost from a system is absorbed by its surroundings. Materials - Lab manual – provides instructions for the experiment - Two Styrofoam cups with a lid – used to create the calorimeter - Beaker – holds the calorimeter in place and needed for the hot water bath in Part 2 - Thermometer – measures the temperature of the solution inside the calorimeter - Scale – used to measure the mass of the empty Styrofoam cups, the calorimeter with respective solutions, and the unknown metal - Stopwatch – needed to record the temperature of the calorimeter in 30-second intervals - 50 mL of HCl (2.0 M) and NaOH (2.0 M) – strong acid and base needed to test the heat of neutralization in Part 1 - Deionized water – needed in Part 2 and 3 for the solution in the calorimeter. It is also used for the hot water bath to heat the unknown metal in Part 2 - Hot plate – used to heat the water and unknown metal - Tongs – needed to remove the heated unknown metal - NH4Cl (s) – the salt used in Part 3 - Scoopula – used to take out 2.0 g of NH4Cl - Weighting paper – provides a protective surface for measuring the mass of NH4Cl - Excel – used to graph the relationship between the time after the addition of a substance and the temperature in the calorimeter. Y-intercept found from the line of best fit is used to find temperature differences. Observations and Experimental Mass of empty calorimeter: 7.27 g Part 1
HCl and NaOH have a molarity of 2.0 M Mass of calorimeter with 50.0 mL HCl: 56.96 g Mass of calorimeter with 50.0 mL HCl + 50 mL NaOH: 110.70 g Part 1 Data Table: Time After the Addition NaOH on the Temperature Seconds Temp (Cº) Seconds Temp (Cº) Seconds Temp (Cº) 0 24.4 300 37.2 600 36.4 30 37.9 330 37.1 630 36.4 60 37.9 360 37.1 660 36.3 90 37.8 390 37 690 36.2 120 37.7 420 36.9 720 36.1 150 37.7 450 36.9 750 36 180 37.6 480 36.7 780 36 210 37.5 510 36.6 810 35.9 240 37.3 540 36.5 840 35.9 270 37.3 570 36.4 870 35.9 After the addition of NaOH to the calorimeter with HCl, the temperature of the calorimeter decreased as time increased. Part 1: Calculations ∆T Calculation
Graph equation: y= -0.0026x+38.005 ∆T = y-intercept – initial temperature ∆T = 38.005ºC – 24.4ºC ∆T = 13.605ºC The Heat of Reaction Calculation qsoln= mC∆T qsoln= (110.70 g – 7.27 g)(3.89 J/gºC)(13.605 ºC) qsoln= 5473.87 J qsoln = -qrxn 5473.87 J = -qrxn -5473.87 J = qrxn The Heat of Neutralization Calculation nrxn= V•M nrxn= 0.05 L • 2M = 0.1 mol ∆H = ??𝑥? ??𝑥? ∆H = = -54738.72 J/mol = -54.74 kJ/mol −5473.87 𝐽 0.1 ??? ∆H = -54.74 kJ/mol Part 2 Mass of metal A: 70.80 g Mass of calorimeter + water: 53.46 g Part 2 Data Table: Time After the Addition of the Unknown Metal on the Temperature Seconds Temp (Cº) Seconds Temp (Cº) 0 22.3 300 31.8 30 30.3 330 31.8 60 32.3 360 31.7 90 32.4 390 31.6 120 32.3 420 31.6 150 32.2 450 31.5 180 32.1 480 31.4 210 32 510 31.3 240 32 540 31.3
270 31.9 570 31.3 300 31.8 With the addition of the unknown metal, the temperature inside the calorimeter increased for the first three points but steadily decreased for the following points. The slope of the graph is the steepest from the first to the second point, indicating that the temperature drastically increased at the beginning of the experiment. Part 2: Calculations The Heat of Reaction Calculation qwater= mC∆T qwater= (53.46 g –7.27 g)(4.184 J/gºC)(31.3ºC – 22.3ºC ) qwater= 1739.33 J qwater = -qmetal 1739.33 J = -qmetal -1739.33 J = qmetal Heat Capacity Calculation qmetal= mC∆T C= ??𝑒𝑡𝑎? ?∆𝑇 C= = 0.358 J/gºC −1739.33 𝐽 70.80 𝑔(31.3º𝐶–100º𝐶)
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