I need help with the answers.
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I need help with the answers.
Laboratory 9 CALORIMETRY: SPECIFIC HEAT OF A METAL AND HEAT OF REACTION INTRODUCTION Thermochemistry is the study of heat, which is defined as thermal energy flowing from an object at a higher temperature to one at a lower temperature. We can represent heat with the symbol, q. If an object absorbs heat, q is positive; if an object releases heat, q is negative. The sign convention for q is indicated below for a piece of hot metal dropped in a beaker of cold water: A piece of metal cools from 100 °C to 25 °C. niemen a The water absorbs the heat lost by the metal. metal water = + Heat flow is studied using a calorimeter, which is essentially a well-insulated container with an ac- curate thermometer. By assuming the calorimeter loses no heat to the surroundings and by knowing how much heat the calorimeter is expected to absorb, we can monitor temperature changes within the calorimeter and calculate the amount of heat absorbed or released within the calorimeter. Using a calorimeter, you can measure different properties of a substance, such as specific heat. Specific heat is defined as the amount of heat necessary to raise the temperature of 1 g of a substance by 1°C, so specific heat generally has units of J/g °C. Specific heat is relevant to all of us because tempera- tures on Earth are actually limited to a comfortably small range because of water's relatively high Adapted from Chemistry 1792 Lab Manual: General Chemistry I. Copyright © 2015 by Central New Mexico Community College. Additional duplication is prohibited without written permission of the authors. 95
Laboratory 9 specific heat. Water's specific heat is about seven times greater than that of rock, so a given amount of water can absorb seven times as much heat as the same quantity of rock before it experiences a tem- perature change of one degree Celsius. At night, the heat absorbed by the oceans is slowly released, so the water serves to regulate temperatures at night to some extent. Notice how temperatures fluctuate more in deserts compared to coastal regions. If the Earth were mostly covered with rock rather than water, daily temperature swings would be significantly greater than we are accustomed to. In this experiment, you will use a calorimeter made from two styrofoam coffee cups and a digital thermometer. Because all substances have a characteristic specific heat, you can use specific heat to identify an unknown substance. In the first part of the experiment, you will measure the amount of heat absorbed by a sample of water when it is heated by a piece of hot metal. You will also take into account any heat absorbed by the calorimeter. Since the metal and water are in contact with one another, the amount of heat absorbed by the water and the calorimeter is exactly equal to the amount of heat released by the metal. Knowing the amount of heat released by the metal, its mass, and the temperature change it experienced, you will calculate the specific heat of the metal and hence, determine its identity. All chemical reactions have heat associated with them. Because most reactions occur under atmo- spheric conditions, the heat of a reaction is the enthalpy of the reaction (AH). Reactions that release heat are exothermic and have negative values for AH Reactions that absorb heat are endothermic and have positive values for AH. The sign convention for AH is indicated below: ran S(s) + O₂(g) → SO₂(g) exothermic reaction: endothermic reaction: AH = -296.8 kJ AH = 114.2 kJ 2 NO₂(g) → 2 NO(g) + O₂(g) In practice, you experience an exothermic reaction in the lab if you observe that a beaker feels hot when a reaction occurs in it; the reaction is releasing heat to the surroundings, including your hand. In contrast, you experience an endothermic reaction in the lab if you observe that a beaker feels cold when a reaction occurs in it; the reaction is absorbing heat from the surroundings, including your hand. One example of an exothermic reaction is an acid-base neutralization reaction. Remember that in a neutralization reaction, an acid reacts with a base to produce water and a salt. Because the reaction occurs in water, the resulting solution will absorb much of the heat released by the reaction, just like the water absorbed the heat released by a piece of hot metal. If you carry out the reaction in a calo- rimeter, you can monitor the temperature change experienced by the solution to calculate the amount of heat absorbed by the solution. The heat of the reaction is essentially equal to the amount of heat absorbed by the solution. In the second part of this experiment, you will calculate the heat of neu- tralization for the reaction between hydrochloric acid, HCl(aq) and sodium hydroxide, NaOH(aq): AH HCl(aq) + NaOH(aq) → H₂O(l) + NaCl(aq) = ?
Laboratory 9 GRAPHING EXPERIMENTAL DATA In this experiment, you will take temperature readings as a piece of hot metal cools in a sample of water. You will plot the temperature data as a function of time, and from this plot, you will obtain the most accurate values for initial temperature, T, and final temperature, T. Consider the time- temperature data in Table 9-1: Table 9-1. Time-Temperature Data for the Specific Heat of an Unknown Metal Time -1:00 -0:45 -0:30 -0:15 0:00 0:15 0:30 Temperature (°C) 24.2 24.2 24.1 24.2 - 27.8 27.8 Time 0:45 1:00 1:15 1:30 1:45 2:00 2:15 Temperature (°C) 27.7 27.7 27.7 27.7 27.6 27.6 27.6 Time 2:30 2:45 3:00 3:15 3:30 3:45 4:00 Temperature (°C) 27.5 27.5 27.4 27.4 27.4 27.4 27.4 When plotting data, keep in mind the ranges you must plot for the x and y axes. Time is plotted on the x axis, and values range from -1:00 to 4:00 with 15-second intervals between each point. Temperature is plotted on the y axis, and values range from about 24 °C to about 28 °C. Since the temperature range is much shorter compared to the time range, time is plotted along the longest end of the graph paper and temperature along the shortest end (see Figure 9-1). This maximizes the graph paper used, so the plot extends to fill most of the page. The plot is prepared to determine the initial temperature, T, and final temperature, T., for the water in the calorimeter. These temperatures are used to calculate the amount of heat absorbed by the water and the calorimeter and ultimately, the specific heat and the identity of the piece of unknown metal. When plotting data, it is important to include a descriptive title for the plot and labels for both axes. You should also include important labels on the plot; for example, in Figure 9-1 the initial and final temperatures are obtained from the plot and indicated as T, and T. In this experiment, the purpose of plotting the data is to determine the initial and final temperatures of the water and the calorimeter, so you can calculate the amount of heat absorbed by both. Note that the points are not perfectly linear, so a best-fit line must be estimated. A best-fit line is a straight line that comes closest to all the points while balancing the number of points above and below the line. Note also that the final temperature as indicated on the plot, T, = 27.86 °C, is greater than the highest temperature recorded when data was collected during the experiment (27.8 °C from Table 9-1). This is because no data is taken right at Time=0:00, when the piece of metal is trans- ferred from the hot water bath to the calorimeter. This final temperature must be obtained from the plot by extrapolating from the data points back to Time=0:00. The final temperature obtained from the plot represents the highest temperature the system would have reached if the calorimeter were perfectly insulated and lost no heat to the surroundings.
Temperature (°C) 29.0 28.0 27.0 26.0 25.0 24.0 23.0 -2:00 Calorimetry: Specific Heat of a Metal and Heat of Reaction -1:00 Specific Heat of an Unknown Metal and em 2:00 Time (minutes: seconds) Figure 9-1. Plot of temperature vs. time for the specific heat of an unknown metal. 0:00 "Draw a "best-fit" line (approximately equal number of points above and below the line) T₁=27.86 °C =-urroundings T₁=24.18 °C 1:00 3:00 4:00 5:00 For the second part of this experiment, a similar Temperature versus Time plot is made for the reac- tion of hydrochloric acid, HCl(aq), with sodium hydroxide, NaOH(aq). Again, the plot is used to determine initial and final temperatures for the system, T, and T,, so you can calculate the amount of heat absorbed by the system and hence, determine the heat of reaction per mole of acid. CALORIMETRY W Let us define the system to be what we are studying and the surroundings to be everything else in the universe outside our system. The law of conservation of energy states that energy is neither created nor destroyed, but simply converted from one form to another. We can show this law as an equation in terms of a system and the surroundings: qystem + surroundings = 0 (1) This equation indicates that the heat for a system plus the heat of the surroundings equals zero. In other words, the overall change in heat for the universe is zero. Perhaps an easier way to represent this is using the following equation: (2)
Laboratory 9 Whatever heat is released by the system is absorbed by the surroundings, and whatever heat is ab- sorbed by the system is released by the surroundings. Specific Heat of a Metal In this first part of this experiment, you will heat a piece of unknown metal and determine its heat capacity. The sample of metal is transferred from a bath of boiling water to a coffee-cup calorimeter containing water at room temperature. Because the hot metal comes into contact with water at a much lower temperature, the metal cools, and the water heats up. In this example, the metal is the system and the calorimeter and the water are the surroundings. To calculate the specific heat of the metal, we need to determine the heat released by the metal, metal metal= (cal+qwater) (3) Equation (3) shows that the amount of heat released by the metal equals the amount of heat ab- sorbed by the calorimeter and the water in it. We can show the amount of heat absorbed by a calo- rimeter to be q 9% = C ΔT where C is the heat capacity of the calorimeter and AT is the change in temperature measured for the calorimeter. For this experiment, you will assume that most coffee-cup calorimeters have similar heat capacities, and you will use C = 21.0 J/°C. To calculate the heat absorbed by the water, qwater we use the following equation: water (water) (water)(AT) (5) where is the specific heat of water (4.184 J/g °C), mater is the mass of water, and AT is the change in temperature measured for the water. (4) Consider the example below: A student carries out a calorimetry experiment to determine the specific heat of a piece of un- known metal. She heats a 28.292 g sample of metal in a boiling water bath at 99.50 °C. She trans- fers the piece of metal to a calorimeter containing 50.0 mL of water. Her Time-Temperature data is shown in Table 9-1 and her plot of the data is shown in Figure 9-1. From her plot, she determined the initial temperature of the calorimeter and the water in it to be 24.18 °C and the final temperature of the calorimeter, water, and metal to be 27.86 °C. Using her data, she first calculates the heat absorbed by the calorimeter: q=CAT=(21.0J) (27.86 -24.18 °C) = (210) (3.68 °C) = 77.3 J (6) She then calculates the heat absorbed by the water, taking into account the density of water is 1.00 g/mL, so 50.0 mL of water has a mass of 50.0 g:
(ez) quater (C) (M) (AT) Cal Calorimetry: Specific Heat of a Metal and Heat of Reaction 4.184 J Cmetal= C(50.0 g) (27.86-24.18°C) 0000 4.184 J 8C(50.0 g) (3.68 °C) = 77.0 J She then uses equation (3) and the calculated values for q and quer to calculate the heat re- leased by the metal: qmetal (9+9)= -(77.3J+770.J) = -847J We can apply equation (5) to the metal and rearrange it to solve for the specific heat of the metal, metal qmetal (mm) (AT) Substituting for metal metal and AT metal you can solve for the specific heat of the metal: -847 J (28.292 g) (27.86-99.50 °C) = Molar Mass= (10) Note that qal is negative since the metal released heat, and AT is also negative since the final temperature was lower than the initial temperature for the metal. The two negative signs cancel, and the specific heat is positive. This makes sense, since heat must always be absorbed for the temperature of a substance to increase. Molar Mass -847 J (28.292 g) (-71.64 °C) = 0.418J/g. "C 25 J/mol C Cmetal THOSE Any sample of metal consists of large numbers of atoms packed very close to one another. When the metal is heated, the atoms vibrate more around their position in the lattice of metal at- oms. Since the structures of metal lattices are similar, the way they absorb heat is also similar. Therefore, like amounts of energy must be added per metal atom to increase the temperature by one degree Celsius. This means that the amount of heat energy per mole needed to cause a one degree increase in temperature is comparable for all metals. Note in Table 9-2 that as molar mass increases, the specific heat decreases. Note also that the molar heat capacities for these metals in J/mol C, fall within a narrow range. Hence, specific heat is related to molar mass using an approximation called the law of Dulong and Petit: bote ob shed griet) Less Using the specific heat for our metal, we get the following for molar mass: (7) 25 Cmetal (8) 25 J/mol. *C 0.418 J/g °C 60. g/mol (9) (11) (12) Looking at Table 9-2, which lists metals used in this experiment; the metal closest to both this molar mass and specific heat is nickel (Ni), so the metal must be nickel. 101
02 Laboratory 9 Table 9-2. Specific Heats and Molar Heat Capacities of Some Metals Specific Heat 0.900 J/g. "C 0.523 J/g °C 58.69 g/mol 0.444 J/g °C 63.55 g/mol 0.387 J/g °C 65.39 g/mol 0.388 J/g °C 207.2 g/mol 0.129 J/g. "C "Note that the average molar heat capacity is about 25 J/mol °C. Element Aluminum Titanium Nickel Copper Zinc Lead Molar Mass 26.98 g/mol 47.88 g/mol Molar Heat Capacity 24.2 J/mol C 25.0 J/mol. C 26.1 J/mol. C 24.6J/mol. *C 25.4 J/mol C 26.7 J/mol. C Heat of Reaction In the second part of this experiment, you will use a calorimeter to determine the heat of reaction for the neutralization reaction between hydrochloric acid and sodium hydroxide. This time the calorim- eter and the resulting salt solution absorb the heat released by the reaction: action+q+qolution = 0 (13) Consider the example below: A student carries out an experiment in a coffee-cup calorimeter to determine the heat of reaction for the reaction between nitric acid and potassium hydroxide: HNO,(aq) + KOH(aq) → H₂O(1) + KNO,(aq) 9=CAT= He mixes 100.0 mL of 0.500 M HNO,(aq) with 100.0 mL of 0.500 M KOH(aq). During the experiment, he collects Time-Temperature data similar to that shown in Table 9-1, and he prepares a plot of the data similar to that shown in Figure 9-1. From his plot, the student obtains the initial and final temperature of the calorimeter and the resulting salt solution. He determines that the initial temperature of the acid is 24.25 °C and the final temperature of the solution is 27.53 °C. Assuming the resulting salt solution is sufficiently dilute and behaves like water, the student uses 4.184 J/g °C for the specific heat of the solution. Using his data, the student first calculates the heat absorbed by the calorimeter: (21.0J) (27.53-24.25 °C) = (210) (3.28 °C) = 68.9 J (14) Next, he calculates the heat absorbed by the solution, assuming the solution is sufficiently dilute, so its density is also 1.00 g/mL. Thus, the 200.0 mL of solution has a mass of 200.0 g:
Calorimetry: Specific Heat of a Metal and Heat of Reaction qolution (Csolacion) (min) (AT) (4.1841) 8 C(200.0 g) (27.53-24.25 °C) se u novos =(4:184) (200.0 g) (3.28 °C) = 2.74 × 10³ J He then uses equation (13) to get the heat of reaction: reaction (al+qolution) = -(68.9 +2.74 X 10' J) = -2.81 X 10'J (16) Finally, the enthalpy of reaction for a neutralization reaction is generally reported in kJ per mole of acid: AH= 1 kJ -5.62 x 10¹ J/mol X 1000 J reaction # of moles of acid -2.81 x 10¹ J (0.1000 L) 0.0500 mol HNO, = -5.62 x 10¹ J/mol -=-56.2 kJ/mol (15) mab (17) Thus, the enthalpy of reaction, AH for the neutralization reaction between nitric acid and potassium hydroxide is -56.2 kJ/mol. This value is very important. It applies to the net ionic equation for the neutralization of any strong acid by a strong base: H*(aq) + OH-(aq) → H₂O(1) AH = -56.2 kJ/mol Thus, you can compare your measured value to this accepted value to gauge the accuracy of your results. 103
104 Laboratory 9 PROCEDURE Always wear safety goggles when you are handling glassware or chemicals. Part I. Specific Heat of an Unknown Metal 1. Set up a ring stand with an iron ring and wire gauze. Add approximately 150 mL of deionized water to a 250 mL beaker, and place it on top of the wire gauze. Add a few boiling chips to the water. Heat the water to boiling using a Bunsen burner. 2. Obtain a piece of unknown metal, and weigh it. Record the mass (to 3 decimal places) on your Report Sheet. Place the metal carefully into the beaker of boiling water, using crucible tongs to lower the metal into the water. 3. Obtain a digital thermometer and two styrofoam cups for the calorimeter as shown in Figure 9-2. Use your graduated cylinder to measure out 50.0 mL of deionized water, and record the exact volume of water used on your Report Sheet. Add the water to the calorimeter, making sure that there is enough water to completely cover the metal. PE wire for stirring styrofoam cups reaction mixture 30,4 CHayden McNeil, LLC thermometer Figure 9-2. A coffee-cup calorimeter. 4. Allow the metal to remain in the boiling water for five minutes, to allow it to equilibrate. Get ready to take Time-Temperature measurements. One person should stir the solution the thermometer while the other person monitors the time and records the temperatures. 5. Use an alcohol thermometer to measure the temperature of the boiling water, and record the temperature on your Report Sheet as T (F G C C G G F U
Calorimetry: Specific Heat of a Metal and Heat of Reaction 6. Stir the water in the calorimeter for a minute before taking any measurements. Begin recording Time-Temperature data for the water in the calorimeter one minute before you transfer the piece of hot metal. Record temperature starting with Time= -1:00. Record the temperature every 15 seconds. Right after recording the temperature for Time-0:15, the person monitor- ing the time should use crucible tongs to take the piece of hot metal from the boiling water, shake any excess water from it, and transfer it to the calorimeter right at Time=0:00. At this time, turn off your Bunsen burner. Do not take a temperature reading at Time=0:00. The next temperature reading should be taken at Time=0:15. 7. Continue stirring the water and recording measurements for the next 4 minutes. 8. Repeat the procedure with the same piece of metal to get a second set of data. 9. Discard the water in the sink, and return the metal sample to the appropriate unknown bottle. Part 11. Heat of Reaction Caution: Hydrochloric acid (HCI) can cause severe chemical burns and ruin clothes. Handle it with care. Any hydrochloric acid spilled on your skin should be rinsed immediately with water for 15 minutes. Any hydrochloric acid spilled on the lab benches should be neutralized with baking soda, the area rinsed with water and wiped clean. Inform your instructor of any acid spills. Caution: Sodium hydroxide (NaOH) can cause chemical burns and ruin clothes. Handle it with care. Any NaOH spilled on your skin should be rinsed immediately with water for 15 minutes. Any NaOH spilled on the lab benches should be neutralized with baking soda, the area rinsed with water and wiped clean. Inform your instructor of any NaOH spills. 1. Start with a clean, dry calorimeter. Use your graduated cylinder to measure out 50.0 mL of 1.0 M hydrochloric acid, HCl(aq), and transfer it to the calorimeter. Record the volume of acid on your report sheet. 2. Wash and dry your graduated cylinder, and use it to measure out 50.0 mL of 1.0 M sodium hydroxide, NaOH(aq). Record the volume of sodium hydroxide on your report sheet. 3. Stir your acid solution for a minute, then begin taking Time-Temperature data for the acid, beginning with Time = -1:00. At Time=0:00, pour the sodium hydroxide, NaOH(aq), into the calorimeter. Again, do not worry about taking a temperature measurement at Time=0:00. Continue taking measurements for the next 4 minutes. 4. Repeat the procedure to get a second set of Time-Temperature data for the Heat of Reaction of HCl(aq) and NaOH(aq). 5. Discard the reaction mixture in the waste container provided, and rinse out the calorimeter cups and return them for use by others. Note: At the end of every lab period, wash your lab bench with water and dry it to prevent chemical contamination and to eliminate potential hazards. 105
DC Laboratory 9 Part III. Preparing Your Plots for Time-Temperature Data 1. Take into account the range for Time to be plotted on the x-axis and the range for Temperature to be plotted on the y-axis. Choose appropriate increments for each axis to fill up as much of the graph paper as possible. 2. Choose an informative title for each plot, and label both axes, including units for each. 3. Plot your points with a dot, and circle each point. (See Figure 9-1.) Use a ruler to determine the best-fit line (approximately equal number of points above and below the line) and extrapolate back to time = 0:00 to get both the initial temperature, T, and the final temperature, T,. Indicate both the initial and final temperature on your plot using the labels T, AND T,. 4. You should prepare a total of four plots, two for determining the specific heat of an unknown metal and two for determining the heat of reaction for the neutralization reaction between HCl(aq) and NaOH(aq). Note that the two specific heat plots can be on the same piece of graph paper, if different notation is used for the two plots, e.g., dots for one set of data and squares for the other. 5. All of your plots should be neat and clear. They should be handed in with your Report Sheets in order for you to receive full credit for the experiment.
Name: Lab Partner(s): Time-Temperature Data Time (min: sec) -1:00 -0:45 -0:30 -0:15 0:00 0:15 0:30 0:45 1:00 1:15 1:30 1:45 2:00 2:15 2:30 2:45 3:00 3:15 3:30 3:45 4:00 Specific Heat Trial 1 Calorimetry: Specific Heat of a Metal and Heat of Reaction Temperature (°C) Specific Heat Trial 2 Temperature (°C) Date: Section: LABORATORY 9 REPORT SHEET Heat of Reaction Trial 1 Temperature (°C) Heat of Reaction Trial 2 Temperature (°C) 107
Laboratory 9-Report Sheet Part I. Specific Heat of an Unknown Metal Number of Unknown Metal mass of unknown metal, g volume of water, mL mass of water, g (d4,0 = 1.00 g/mL) T₁=T₁ (from your plot), "C L water Tical Tewater T, metal (from your plot), "C ΔΤ = ΔΤ °C J (Use C = 21.0J/°C) او qurta J Time (from boiling water), "C metal AT °C metal J/g °C Molar mass of metal, g/mol Average molar mass of metal, g/mol Identity of the metal* Trial 1 Show your calculations for Trials 1 and 2 below. Discuss reasons for differences. Trial 2 *Compare both your average specific heat and calculated molar mass to those listed in Table 9-2.
Part II. Heat of Reaction Calorimetry: Specific Heat of a Metal and Heat of Reaction Report Sheet volume of HCl, mL volume of NaOH, mL volume of solution, mL. mass of solution, g (dion = 1.00 g/mL) solution = T₁, (from your plot), "C T=Tu (from your plot), "C = AT solution C J (Use C 21.0J/°C) T 'Leal AT Lolution solution F reaction J reaction J moles of HCl used, mol AH kJ/mol Average AH kJ/mol Trial 1 Trial 2 Show your calculations for Trial 1 below. Compare your result to that on pages 100-101, and discuss reasons for differences. 109
Calorimetry: Specific Heat of a Metal and Heat of Reaction Report Sheet 111
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Laboratory 9 specific heat. Water's specific heat is about seven times greater than that of rock, so a given amount of water can absorb seven times as much heat as the same quantity of rock before it experiences a tem- perature change of one degree Celsius. At night, the heat absorbed by the oceans is slowly released, so the water serves to regulate temperatures at night to some extent. Notice how temperatures fluctuate more in deserts compared to coastal regions. If the Earth were mostly covered with rock rather than water, daily temperature swings would be significantly greater than we are accustomed to. In this experiment, you will use a calorimeter made from two styrofoam coffee cups and a digital thermometer. Because all substances have a characteristic specific heat, you can use specific heat to identify an unknown substance. In the first part of the experiment, you will measure the amount of heat absorbed by a sample of water when it is heated by a piece of hot metal. You will also take into account any heat absorbed by the calorimeter. Since the metal and water are in contact with one another, the amount of heat absorbed by the water and the calorimeter is exactly equal to the amount of heat released by the metal. Knowing the amount of heat released by the metal, its mass, and the temperature change it experienced, you will calculate the specific heat of the metal and hence, determine its identity. All chemical reactions have heat associated with them. Because most reactions occur under atmo- spheric conditions, the heat of a reaction is the enthalpy of the reaction (AH). Reactions that release heat are exothermic and have negative values for AH Reactions that absorb heat are endothermic and have positive values for AH. The sign convention for AH is indicated below: ran S(s) + O₂(g) → SO₂(g) exothermic reaction: endothermic reaction: AH = -296.8 kJ AH = 114.2 kJ 2 NO₂(g) → 2 NO(g) + O₂(g) In practice, you experience an exothermic reaction in the lab if you observe that a beaker feels hot when a reaction occurs in it; the reaction is releasing heat to the surroundings, including your hand. In contrast, you experience an endothermic reaction in the lab if you observe that a beaker feels cold when a reaction occurs in it; the reaction is absorbing heat from the surroundings, including your hand. One example of an exothermic reaction is an acid-base neutralization reaction. Remember that in a neutralization reaction, an acid reacts with a base to produce water and a salt. Because the reaction occurs in water, the resulting solution will absorb much of the heat released by the reaction, just like the water absorbed the heat released by a piece of hot metal. If you carry out the reaction in a calo- rimeter, you can monitor the temperature change experienced by the solution to calculate the amount of heat absorbed by the solution. The heat of the reaction is essentially equal to the amount of heat absorbed by the solution. In the second part of this experiment, you will calculate the heat of neu- tralization for the reaction between hydrochloric acid, HCl(aq) and sodium hydroxide, NaOH(aq): AH HCl(aq) + NaOH(aq) → H₂O(l) + NaCl(aq) = ?
Laboratory 9 GRAPHING EXPERIMENTAL DATA In this experiment, you will take temperature readings as a piece of hot metal cools in a sample of water. You will plot the temperature data as a function of time, and from this plot, you will obtain the most accurate values for initial temperature, T, and final temperature, T. Consider the time- temperature data in Table 9-1: Table 9-1. Time-Temperature Data for the Specific Heat of an Unknown Metal Time -1:00 -0:45 -0:30 -0:15 0:00 0:15 0:30 Temperature (°C) 24.2 24.2 24.1 24.2 - 27.8 27.8 Time 0:45 1:00 1:15 1:30 1:45 2:00 2:15 Temperature (°C) 27.7 27.7 27.7 27.7 27.6 27.6 27.6 Time 2:30 2:45 3:00 3:15 3:30 3:45 4:00 Temperature (°C) 27.5 27.5 27.4 27.4 27.4 27.4 27.4 When plotting data, keep in mind the ranges you must plot for the x and y axes. Time is plotted on the x axis, and values range from -1:00 to 4:00 with 15-second intervals between each point. Temperature is plotted on the y axis, and values range from about 24 °C to about 28 °C. Since the temperature range is much shorter compared to the time range, time is plotted along the longest end of the graph paper and temperature along the shortest end (see Figure 9-1). This maximizes the graph paper used, so the plot extends to fill most of the page. The plot is prepared to determine the initial temperature, T, and final temperature, T., for the water in the calorimeter. These temperatures are used to calculate the amount of heat absorbed by the water and the calorimeter and ultimately, the specific heat and the identity of the piece of unknown metal. When plotting data, it is important to include a descriptive title for the plot and labels for both axes. You should also include important labels on the plot; for example, in Figure 9-1 the initial and final temperatures are obtained from the plot and indicated as T, and T. In this experiment, the purpose of plotting the data is to determine the initial and final temperatures of the water and the calorimeter, so you can calculate the amount of heat absorbed by both. Note that the points are not perfectly linear, so a best-fit line must be estimated. A best-fit line is a straight line that comes closest to all the points while balancing the number of points above and below the line. Note also that the final temperature as indicated on the plot, T, = 27.86 °C, is greater than the highest temperature recorded when data was collected during the experiment (27.8 °C from Table 9-1). This is because no data is taken right at Time=0:00, when the piece of metal is trans- ferred from the hot water bath to the calorimeter. This final temperature must be obtained from the plot by extrapolating from the data points back to Time=0:00. The final temperature obtained from the plot represents the highest temperature the system would have reached if the calorimeter were perfectly insulated and lost no heat to the surroundings.
Temperature (°C) 29.0 28.0 27.0 26.0 25.0 24.0 23.0 -2:00 Calorimetry: Specific Heat of a Metal and Heat of Reaction -1:00 Specific Heat of an Unknown Metal and em 2:00 Time (minutes: seconds) Figure 9-1. Plot of temperature vs. time for the specific heat of an unknown metal. 0:00 "Draw a "best-fit" line (approximately equal number of points above and below the line) T₁=27.86 °C =-urroundings T₁=24.18 °C 1:00 3:00 4:00 5:00 For the second part of this experiment, a similar Temperature versus Time plot is made for the reac- tion of hydrochloric acid, HCl(aq), with sodium hydroxide, NaOH(aq). Again, the plot is used to determine initial and final temperatures for the system, T, and T,, so you can calculate the amount of heat absorbed by the system and hence, determine the heat of reaction per mole of acid. CALORIMETRY W Let us define the system to be what we are studying and the surroundings to be everything else in the universe outside our system. The law of conservation of energy states that energy is neither created nor destroyed, but simply converted from one form to another. We can show this law as an equation in terms of a system and the surroundings: qystem + surroundings = 0 (1) This equation indicates that the heat for a system plus the heat of the surroundings equals zero. In other words, the overall change in heat for the universe is zero. Perhaps an easier way to represent this is using the following equation: (2)
Laboratory 9 Whatever heat is released by the system is absorbed by the surroundings, and whatever heat is ab- sorbed by the system is released by the surroundings. Specific Heat of a Metal In this first part of this experiment, you will heat a piece of unknown metal and determine its heat capacity. The sample of metal is transferred from a bath of boiling water to a coffee-cup calorimeter containing water at room temperature. Because the hot metal comes into contact with water at a much lower temperature, the metal cools, and the water heats up. In this example, the metal is the system and the calorimeter and the water are the surroundings. To calculate the specific heat of the metal, we need to determine the heat released by the metal, metal metal= (cal+qwater) (3) Equation (3) shows that the amount of heat released by the metal equals the amount of heat ab- sorbed by the calorimeter and the water in it. We can show the amount of heat absorbed by a calo- rimeter to be q 9% = C ΔT where C is the heat capacity of the calorimeter and AT is the change in temperature measured for the calorimeter. For this experiment, you will assume that most coffee-cup calorimeters have similar heat capacities, and you will use C = 21.0 J/°C. To calculate the heat absorbed by the water, qwater we use the following equation: water (water) (water)(AT) (5) where is the specific heat of water (4.184 J/g °C), mater is the mass of water, and AT is the change in temperature measured for the water. (4) Consider the example below: A student carries out a calorimetry experiment to determine the specific heat of a piece of un- known metal. She heats a 28.292 g sample of metal in a boiling water bath at 99.50 °C. She trans- fers the piece of metal to a calorimeter containing 50.0 mL of water. Her Time-Temperature data is shown in Table 9-1 and her plot of the data is shown in Figure 9-1. From her plot, she determined the initial temperature of the calorimeter and the water in it to be 24.18 °C and the final temperature of the calorimeter, water, and metal to be 27.86 °C. Using her data, she first calculates the heat absorbed by the calorimeter: q=CAT=(21.0J) (27.86 -24.18 °C) = (210) (3.68 °C) = 77.3 J (6) She then calculates the heat absorbed by the water, taking into account the density of water is 1.00 g/mL, so 50.0 mL of water has a mass of 50.0 g:
(ez) quater (C) (M) (AT) Cal Calorimetry: Specific Heat of a Metal and Heat of Reaction 4.184 J Cmetal= C(50.0 g) (27.86-24.18°C) 0000 4.184 J 8C(50.0 g) (3.68 °C) = 77.0 J She then uses equation (3) and the calculated values for q and quer to calculate the heat re- leased by the metal: qmetal (9+9)= -(77.3J+770.J) = -847J We can apply equation (5) to the metal and rearrange it to solve for the specific heat of the metal, metal qmetal (mm) (AT) Substituting for metal metal and AT metal you can solve for the specific heat of the metal: -847 J (28.292 g) (27.86-99.50 °C) = Molar Mass= (10) Note that qal is negative since the metal released heat, and AT is also negative since the final temperature was lower than the initial temperature for the metal. The two negative signs cancel, and the specific heat is positive. This makes sense, since heat must always be absorbed for the temperature of a substance to increase. Molar Mass -847 J (28.292 g) (-71.64 °C) = 0.418J/g. "C 25 J/mol C Cmetal THOSE Any sample of metal consists of large numbers of atoms packed very close to one another. When the metal is heated, the atoms vibrate more around their position in the lattice of metal at- oms. Since the structures of metal lattices are similar, the way they absorb heat is also similar. Therefore, like amounts of energy must be added per metal atom to increase the temperature by one degree Celsius. This means that the amount of heat energy per mole needed to cause a one degree increase in temperature is comparable for all metals. Note in Table 9-2 that as molar mass increases, the specific heat decreases. Note also that the molar heat capacities for these metals in J/mol C, fall within a narrow range. Hence, specific heat is related to molar mass using an approximation called the law of Dulong and Petit: bote ob shed griet) Less Using the specific heat for our metal, we get the following for molar mass: (7) 25 Cmetal (8) 25 J/mol. *C 0.418 J/g °C 60. g/mol (9) (11) (12) Looking at Table 9-2, which lists metals used in this experiment; the metal closest to both this molar mass and specific heat is nickel (Ni), so the metal must be nickel. 101
02 Laboratory 9 Table 9-2. Specific Heats and Molar Heat Capacities of Some Metals Specific Heat 0.900 J/g. "C 0.523 J/g °C 58.69 g/mol 0.444 J/g °C 63.55 g/mol 0.387 J/g °C 65.39 g/mol 0.388 J/g °C 207.2 g/mol 0.129 J/g. "C "Note that the average molar heat capacity is about 25 J/mol °C. Element Aluminum Titanium Nickel Copper Zinc Lead Molar Mass 26.98 g/mol 47.88 g/mol Molar Heat Capacity 24.2 J/mol C 25.0 J/mol. C 26.1 J/mol. C 24.6J/mol. *C 25.4 J/mol C 26.7 J/mol. C Heat of Reaction In the second part of this experiment, you will use a calorimeter to determine the heat of reaction for the neutralization reaction between hydrochloric acid and sodium hydroxide. This time the calorim- eter and the resulting salt solution absorb the heat released by the reaction: action+q+qolution = 0 (13) Consider the example below: A student carries out an experiment in a coffee-cup calorimeter to determine the heat of reaction for the reaction between nitric acid and potassium hydroxide: HNO,(aq) + KOH(aq) → H₂O(1) + KNO,(aq) 9=CAT= He mixes 100.0 mL of 0.500 M HNO,(aq) with 100.0 mL of 0.500 M KOH(aq). During the experiment, he collects Time-Temperature data similar to that shown in Table 9-1, and he prepares a plot of the data similar to that shown in Figure 9-1. From his plot, the student obtains the initial and final temperature of the calorimeter and the resulting salt solution. He determines that the initial temperature of the acid is 24.25 °C and the final temperature of the solution is 27.53 °C. Assuming the resulting salt solution is sufficiently dilute and behaves like water, the student uses 4.184 J/g °C for the specific heat of the solution. Using his data, the student first calculates the heat absorbed by the calorimeter: (21.0J) (27.53-24.25 °C) = (210) (3.28 °C) = 68.9 J (14) Next, he calculates the heat absorbed by the solution, assuming the solution is sufficiently dilute, so its density is also 1.00 g/mL. Thus, the 200.0 mL of solution has a mass of 200.0 g:
Calorimetry: Specific Heat of a Metal and Heat of Reaction qolution (Csolacion) (min) (AT) (4.1841) 8 C(200.0 g) (27.53-24.25 °C) se u novos =(4:184) (200.0 g) (3.28 °C) = 2.74 × 10³ J He then uses equation (13) to get the heat of reaction: reaction (al+qolution) = -(68.9 +2.74 X 10' J) = -2.81 X 10'J (16) Finally, the enthalpy of reaction for a neutralization reaction is generally reported in kJ per mole of acid: AH= 1 kJ -5.62 x 10¹ J/mol X 1000 J reaction # of moles of acid -2.81 x 10¹ J (0.1000 L) 0.0500 mol HNO, = -5.62 x 10¹ J/mol -=-56.2 kJ/mol (15) mab (17) Thus, the enthalpy of reaction, AH for the neutralization reaction between nitric acid and potassium hydroxide is -56.2 kJ/mol. This value is very important. It applies to the net ionic equation for the neutralization of any strong acid by a strong base: H*(aq) + OH-(aq) → H₂O(1) AH = -56.2 kJ/mol Thus, you can compare your measured value to this accepted value to gauge the accuracy of your results. 103
104 Laboratory 9 PROCEDURE Always wear safety goggles when you are handling glassware or chemicals. Part I. Specific Heat of an Unknown Metal 1. Set up a ring stand with an iron ring and wire gauze. Add approximately 150 mL of deionized water to a 250 mL beaker, and place it on top of the wire gauze. Add a few boiling chips to the water. Heat the water to boiling using a Bunsen burner. 2. Obtain a piece of unknown metal, and weigh it. Record the mass (to 3 decimal places) on your Report Sheet. Place the metal carefully into the beaker of boiling water, using crucible tongs to lower the metal into the water. 3. Obtain a digital thermometer and two styrofoam cups for the calorimeter as shown in Figure 9-2. Use your graduated cylinder to measure out 50.0 mL of deionized water, and record the exact volume of water used on your Report Sheet. Add the water to the calorimeter, making sure that there is enough water to completely cover the metal. PE wire for stirring styrofoam cups reaction mixture 30,4 CHayden McNeil, LLC thermometer Figure 9-2. A coffee-cup calorimeter. 4. Allow the metal to remain in the boiling water for five minutes, to allow it to equilibrate. Get ready to take Time-Temperature measurements. One person should stir the solution the thermometer while the other person monitors the time and records the temperatures. 5. Use an alcohol thermometer to measure the temperature of the boiling water, and record the temperature on your Report Sheet as T (F G C C G G F U
Calorimetry: Specific Heat of a Metal and Heat of Reaction 6. Stir the water in the calorimeter for a minute before taking any measurements. Begin recording Time-Temperature data for the water in the calorimeter one minute before you transfer the piece of hot metal. Record temperature starting with Time= -1:00. Record the temperature every 15 seconds. Right after recording the temperature for Time-0:15, the person monitor- ing the time should use crucible tongs to take the piece of hot metal from the boiling water, shake any excess water from it, and transfer it to the calorimeter right at Time=0:00. At this time, turn off your Bunsen burner. Do not take a temperature reading at Time=0:00. The next temperature reading should be taken at Time=0:15. 7. Continue stirring the water and recording measurements for the next 4 minutes. 8. Repeat the procedure with the same piece of metal to get a second set of data. 9. Discard the water in the sink, and return the metal sample to the appropriate unknown bottle. Part 11. Heat of Reaction Caution: Hydrochloric acid (HCI) can cause severe chemical burns and ruin clothes. Handle it with care. Any hydrochloric acid spilled on your skin should be rinsed immediately with water for 15 minutes. Any hydrochloric acid spilled on the lab benches should be neutralized with baking soda, the area rinsed with water and wiped clean. Inform your instructor of any acid spills. Caution: Sodium hydroxide (NaOH) can cause chemical burns and ruin clothes. Handle it with care. Any NaOH spilled on your skin should be rinsed immediately with water for 15 minutes. Any NaOH spilled on the lab benches should be neutralized with baking soda, the area rinsed with water and wiped clean. Inform your instructor of any NaOH spills. 1. Start with a clean, dry calorimeter. Use your graduated cylinder to measure out 50.0 mL of 1.0 M hydrochloric acid, HCl(aq), and transfer it to the calorimeter. Record the volume of acid on your report sheet. 2. Wash and dry your graduated cylinder, and use it to measure out 50.0 mL of 1.0 M sodium hydroxide, NaOH(aq). Record the volume of sodium hydroxide on your report sheet. 3. Stir your acid solution for a minute, then begin taking Time-Temperature data for the acid, beginning with Time = -1:00. At Time=0:00, pour the sodium hydroxide, NaOH(aq), into the calorimeter. Again, do not worry about taking a temperature measurement at Time=0:00. Continue taking measurements for the next 4 minutes. 4. Repeat the procedure to get a second set of Time-Temperature data for the Heat of Reaction of HCl(aq) and NaOH(aq). 5. Discard the reaction mixture in the waste container provided, and rinse out the calorimeter cups and return them for use by others. Note: At the end of every lab period, wash your lab bench with water and dry it to prevent chemical contamination and to eliminate potential hazards. 105
DC Laboratory 9 Part III. Preparing Your Plots for Time-Temperature Data 1. Take into account the range for Time to be plotted on the x-axis and the range for Temperature to be plotted on the y-axis. Choose appropriate increments for each axis to fill up as much of the graph paper as possible. 2. Choose an informative title for each plot, and label both axes, including units for each. 3. Plot your points with a dot, and circle each point. (See Figure 9-1.) Use a ruler to determine the best-fit line (approximately equal number of points above and below the line) and extrapolate back to time = 0:00 to get both the initial temperature, T, and the final temperature, T,. Indicate both the initial and final temperature on your plot using the labels T, AND T,. 4. You should prepare a total of four plots, two for determining the specific heat of an unknown metal and two for determining the heat of reaction for the neutralization reaction between HCl(aq) and NaOH(aq). Note that the two specific heat plots can be on the same piece of graph paper, if different notation is used for the two plots, e.g., dots for one set of data and squares for the other. 5. All of your plots should be neat and clear. They should be handed in with your Report Sheets in order for you to receive full credit for the experiment.
Name: Lab Partner(s): Time-Temperature Data Time (min: sec) -1:00 -0:45 -0:30 -0:15 0:00 0:15 0:30 0:45 1:00 1:15 1:30 1:45 2:00 2:15 2:30 2:45 3:00 3:15 3:30 3:45 4:00 Specific Heat Trial 1 Calorimetry: Specific Heat of a Metal and Heat of Reaction Temperature (°C) Specific Heat Trial 2 Temperature (°C) Date: Section: LABORATORY 9 REPORT SHEET Heat of Reaction Trial 1 Temperature (°C) Heat of Reaction Trial 2 Temperature (°C) 107
Laboratory 9-Report Sheet Part I. Specific Heat of an Unknown Metal Number of Unknown Metal mass of unknown metal, g volume of water, mL mass of water, g (d4,0 = 1.00 g/mL) T₁=T₁ (from your plot), "C L water Tical Tewater T, metal (from your plot), "C ΔΤ = ΔΤ °C J (Use C = 21.0J/°C) او qurta J Time (from boiling water), "C metal AT °C metal J/g °C Molar mass of metal, g/mol Average molar mass of metal, g/mol Identity of the metal* Trial 1 Show your calculations for Trials 1 and 2 below. Discuss reasons for differences. Trial 2 *Compare both your average specific heat and calculated molar mass to those listed in Table 9-2.
Part II. Heat of Reaction Calorimetry: Specific Heat of a Metal and Heat of Reaction Report Sheet volume of HCl, mL volume of NaOH, mL volume of solution, mL. mass of solution, g (dion = 1.00 g/mL) solution = T₁, (from your plot), "C T=Tu (from your plot), "C = AT solution C J (Use C 21.0J/°C) T 'Leal AT Lolution solution F reaction J reaction J moles of HCl used, mol AH kJ/mol Average AH kJ/mol Trial 1 Trial 2 Show your calculations for Trial 1 below. Compare your result to that on pages 100-101, and discuss reasons for differences. 109
Calorimetry: Specific Heat of a Metal and Heat of Reaction Report Sheet 111