Experiment 25: Calorimetry

Transcription

1 Aaron Bunch CHEM 111 Morning Lab 27 October 2014 Experiment 25: Calorimetry Conclusion: The unknown metal #14 has a specific heat of 0.36 J/g C; the heat of neutralization of HCl and NaOH is kj/mol H 2 O produced; and the heat of neutralization of HNO 3 and NaOH is kj/mol H 2 O produced.

2 Abstract The specific heat of unknown metal #14 was expected to be less than 0.50 J/g C, and the heats of neutralization of hydrochloric acid and nitric acid with sodium hydroxide were expected to lie between -57 and -58 kj/mol H 2 O produced. To test the first hypothesis, g of the metal was heated to thermal equilibrium with a boiling water bath and then combined with room temperature water in a calorimeter. The temperature change of the metal was C, and kj of energy were transferred from the metal to the water. The specific heat of the metal was then determined to be 0.36 J/g C, which is consistent with our expectations. To test the second hypothesis, each acid was combined with sodium hydroxide in a calorimeter in two separate trials. The temperature of the solution increased by C and by C for HCl and HNO 3, respectively. This corresponded to kj and kj of heat evolved by the reactions. The heats of neutralization were then determined to be kj/mol H 2 O produced and kj/mol H 2 O produced. These results fall just short of the expected values. Introduction In the first part of this experiment, the specific heat of an unknown metal is determined. In accordance with the experimental procedure outlined in Beran ( ), the metal is first brought to thermal equilibrium with a boiling water bath, and then added to water at room temperature in a double coffee cup calorimeter. The temperature of the water is recorded at regular intervals and plotted against time. A trend line fitted to the data estimates the maximum temperature of the water and metal in thermal equilibrium. The calorimeter is assumed to be an isolated system, since the heat lost to the surroundings is negligible. Therefore, the amount of heat absorbed by the water is equal to the amount of heat lost by the metal: q!"#$% = q!"#$% [1] The heat absorbed by the water is the product of the mass of the water, the specific heat of water, and the change in temperature of the water: q!"#$% = (m!"#$% )(s!"#$% )( T!"#$% ) [2] The specific heat of the metal is equal to the heat lost by the metal divided by the product of the mass of the metal and the change in temperature of the metal: s!"#$% =!!"#$% (!!"#$% )(!!"#$% ) [3] In the second part of this experiment, the heat of neutralization is determined for two strong acids and a sodium hydroxide solution. In accordance with Beran s procedure ( ), hydrochloric acid and nitric acid are combined in separate trials with sodium hydroxide in a double coffee cup calorimeter. The temperature of the mixture is recorded at regular intervals and plotted against time. A trend line fitted to the data estimates the maximum temperature of the mixture. The calorimeter is assumed to be an isolated system, since the heat lost to the surroundings is negligible. Therefore, the amount of heat evolved by the reaction is equal to the amount of heat absorbed by the mixture: q!"#$%&'( = q!"#$%&' [4] Since the mixture is mostly water, the heat absorbed by the mixture is given by equation [2] above. Dividing the heat evolved by the reaction by the moles of water produced gives the heat of neutralization in terms of Joules per mole water produced (J/mol H 2 O). 2

3 Calorimetry is the measure of heat. Heat is the spontaneous transfer of energy from a system of higher energy to a system of lower energy through direct contact (Ebbing & Gammon, 231). The heat transfer stops when thermal equilibrium is reached, which means both systems have the same temperature. A calorimeter is an instrument used to measure heat transfer during a physical or chemical change (Ebbing & Gammon, 243). Two Styrofoam coffee cups, one inserted inside the other, and a loose-fitting Styrofoam lid with a hole for a thermometer, make an inexpensive constant-pressure calorimeter. The heat transfer occurs inside the coffee cups, and the temperature change is measured with the thermometer inserted through the lid. If the calorimeter is well insulated, and the heat transfer occurs relatively quickly, then heat loss to the surroundings can be neglected and the calorimeter can be treated as an isolated system. In an isolated system, the sum of the heat transfers within the system equals zero. In order to measure the heat of a chemical or physical change, one must determine the temperature when thermal equilibrium is first reached. This temperature is never directly measured. In the time it takes to reach thermal equilibrium, some small amount of heat is lost to the calorimeter (Beran 212). To account for this heat loss, the temperature at thermal equilibrium must be extrapolated from a trend line. In this procedure, the temperature of the mixture in the calorimeter is recorded as it cools well after the equilibrium point is first reached. A linear trend line is fit to the cooling part of the temperature data, and extended backwards to the time when the heat transfer began. This point gives a theoretical maximum temperature at thermal equilibrium, as if the heat were transferred instantly without any loss to the calorimeter. Examples of this procedure are given in Figures 1, 2, and 3 below. Acid-base neutralization reactions produce heat in addition to water. A strong acid and strong base neutralize according to the following net ionic equation (Beran 209): H + (aq) + OH - (aq) H 2 O (l) + heat [5] The heat of this reaction can be measured in a calorimeter. Since strong acid-base neutralization reactions happen quickly, and the calorimeter is well insulated, the amount of heat lost to the environment is negligible. Therefore, we may assume that the calorimeter is an isolated system. In an isolated system, the amount of heat evolved by the reaction is equal to the amount of heat absorbed by the mixture of solutions. Because the acid and base solutions are relatively dilute, we may assume that they have the same density (1 g/ml) and specific heat (4.18 J/g C) as water (Beran 209). Then the heat absorbed by the solutions can be calculated by equation [2] above. The heat of neutralization is usually given in terms of Joules per mole water produced by the reaction (Beran 209). According to equation [5], one mole of water is produced per mole reaction. Therefore, the heat per mole reaction is equal to the heat of neutralization per mole water produced. With only a few exceptions, the specific heats of metals lie below 0.50 J/g C ( Therefore, in the first part of this experiment, we expect the specific heat of the unknown metal to fall in this range. In the second part of this experiment, we expect the heats of neutralization to be between -57 and -58 kj/mol. Because strong acids and strong bases ionize completely in water, the net ionic equation for all strong acid-base neutralizations is the same (equation [5] above), and evolves the same quantity of heat. Other experiments have found this quantity to lie between -57 and -58 kj/mol. ( /Enthalpy/Enthalpy_Change_of_Neutralization). Results 1. Specific heat of an unknown metal 22.0 g of water at 22.8 C were put into a double coffee cup calorimeter. Meanwhile, g of the unknown metal were placed in a dry test tube and heated in a boiling water bath for approximately 15 minutes. 3

4 After 15 minutes, it was assumed that the metal and water bath had reached thermal equilibrium at 94.5 C. These initial conditions are summarized in the first four columns of Table 1. Mass and Temperature of Water and Metal Table 1: The mass and temperature of the water and metal before they were combined in the calorimeter, and the maximum temperature they reach when combined as extrapolated from their graph. mass of metal (g) initial temperature of metal ( C) mass of water (g) initial temperature of water ( C) maximum temperature of water and metal when combined ( C) The heated metal was quickly combined with the water in the calorimeter, and stirred with the bulb of a thermometer inserted through the lid of the calorimeter. The temperature of the water was recorded at irregular intervals for five minutes. These temperature measurements are summarized in Table 2. Time = 0 is the point at which the metal was added to the water in the calorimeter. Water Temperature over Time Table 2: Water temperature in a coffee cup calorimeter after adding an unknown metal heated to 94.5 C. The metal was added to the water at time = 0. time (s) water temperature ( C) trend line dataset ( C) In Figure 1 below, water temperature is plotted against the time since adding the metal to the water. The y-axis, t = 0, is the time at which the metal is added to the water. A linear trend line is fitted to the data for the time after the water has begun cooling. The y-intercept of the trend line gives the extrapolated maximum 4

5 temperature of the water and metal in thermal equilibrium. This value is given in the final column of Table 1 above, and can be read from the equation of the trend line in the figure below (36.11 C). Figure 1: Heated Unknown Metal Combined with Water. The y-axis, time = 0, is the time at which the metal is added to the water. The trend line is fitted to the data for the time after the water has begun cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the water and metal in thermal equilibrium Heated Unknown Metal Combined with Water water temperature ( C) y = x R² = time (s) The specific heat of the metal is calculated from the temperature change of the metal, the heat lost by the metal, and the mass of the metal (from Table 1 above). Since we assume the calorimeter to be an isolated system, the amount of heat lost by the metal is equal to the amount of heat gained by the water. The heat gained by the water is calculated from the temperature change of the water, the mass of the water (from Table 2 above), and the specific heat of water (4.184 J/g C). These results are summarized in Table 1 above and Table 3 below. 5

6 Specific Heat of the Metal Table 3: The specific heat of the metal is calculated from the temperature change of the metal, the heat lost by the metal, and the mass of the metal (from Table 1). The amount of heat lost by the metal is equal to the amount of heat gained by the water. The heat gained by the water is calculated from the temperature change of the water, the mass of the water (from Table 1), and the specific heat of water (4.184 J/g C). temperature change of water ( C) heat gained by water (kj) heat lost by metal (kj) temperature change of metal ( C) specific heat of metal (J/g C) Heat of neutralization of two strong acids with NaOH In trial #1, 50.0 g of 1.1 M HCl solution at 22.0 C was added to 50.0 g of standardized M NaOH solution at 22.0 C in a double coffee cup calorimeter. The density of both solutions is assumed to be that of water (1 g/ml). The mixture was stirred with the bulb of a thermometer inserted through the lid of the calorimeter, and the temperature of the mixture was recorded at irregular intervals for five minutes. These measurements are given in Table 4. Temperature of Solution over Time Table 4: Temperature of solution in a coffee cup calorimeter after combining equal volumes of HCl and NaOH solutions. The solutions were mixed at time = 0. time (s) temperature ( C) trend line dataset ( C)

7 In Figure 2, the temperature of the mixture is plotted against the time since combining the solutions in the calorimeter. The y-axis, time = 0, is the time when the solutions are mixed. A linear trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the mixture. Figure 2: HCl Combined with NaOH. The y-axis, time = 0, is the time when the solutions are mixed. The trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the mixture HCl Combined with NaOH temperature of solution ( C) y = x R² = time (s) In trial #2, 50.0 g of 1.1 M HNO 3 at 21.5 C was added to 50.0 g of standardized M NaOH solution at 22.0 C in a double coffee cup calorimeter. The density of both solutions is assumed to be that of water (1 g/ml). The mixture was stirred with the bulb of a thermometer inserted through the lid of the calorimeter, and the temperature of the mixture was recorded at irregular intervals for five minutes. These measurements are given in Table 5 on the next page. 7

8 Temperature of Solution over Time Table 5: Temperature of solution in a coffee cup calorimeter after combining equal volumes of HNO 3 and NaOH solutions. The solutions were mixed at time = 0. time (s) temperature ( C) trend line dataset ( C) In Figure 3 on the next page, the temperature of the mixture is plotted against the time since combining the solutions in the calorimeter. The y-axis, time = 0, is the time when the solutions were mixed. A linear trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the mixture. 8

9 Figure 3: HNO 3 Combined with NaOH. The y-axis, time = 0, is the time when the solutions are mixed. The trend line is fitted to the data for the time after the mixture has begun cooling. The y-intercept of the trend line gives the extrapolated maximum temperature of the mixture. HNO 3 Combined with NaOH temperature ( C) time (s) y = x R² = The results of both trials are summarized in Table 6. The heat of neutralization (kj/mol H 2 O) of the strong acid with NaOH in each trial is the heat evolved by the mixture (kj) per mole of H 2 O formed. By the stoichiometry of the reaction, we know that one mole of H 2 O is produced for every mole of OH - reacted. The heat evolved by the mixture is calculated from the total mass of the mixture, the change in temperature of the mixture, and the specific heat of water (4.184 J/g C). Heat of Neutralization of a Strong Acid and a Strong Base Table 6: The heat of neutralization (kj/mol H 2 O) of a strong acid with NaOH is the heat evolved by the mixture (kj) per mole of H 2 O formed. One mole of H 2 O is produced for every mole of OH - reacted. The heat evolved by the mixture is calculated from the total mass of the mixture, the change in temperature of the mixture, and the specific heat of water (4.184 J/g C). acid total mass of mixture (g) initial temperature of mixture ( C) maximum temperature of mixture ( C) change in temperature ( C) heat evolved by reaction (kj) moles of OH - reacted (mol) moles of H 2 O formed (mol) heat of neutralization (kj/mol H 2 O) HCl HNO

10 Discussion 1. Specific heat of an unknown metal When the heated metal was combined with the room temperature water in the calorimeter, heat spontaneously transferred from the hotter metal to the cooler water until the metal and the water reached thermal equilibrium. The recorded value of this temperature is 35.8 C, but the extrapolated value is C. The extrapolated value corrects for the heat lost to the calorimeter before the equilibrium temperature reached its maximum. Since the calorimeter is assumed to be an isolated system, the amount of heat lost by the metal is equal to the amount of heat gained by the water. The heat gained by the water was determined to be kj, so the heat lost by the metal was kj. The specific heat of the metal was then determined to be J/g C using equation [3]. As expected, this value (below 0.5 J/g C) is typical of the specific heat of most metals. 2. Heat of neutralization of two strong acids with NaOH. When the acids were combined with the NaOH solution in the calorimeter, the H + ions and the OH - ions combined to form liquid water. The reaction also released an amount of heat. This heat was absorbed by the mixtures (mostly water) in the calorimeter, and raised their temperature. The peak temperature was measured to be 28.2 C for HCl and 27.8 C for HNO 3, but the extrapolated values are C for HCl and C for HNO 3. The extrapolated values correct for the heat lost to the calorimeter before the maximum temperature was reached. Since the calorimeter is assumed to be an isolated system, the amounts of heat evolved by the reactions are equal to the amounts of heat absorbed by the mixtures. The heats evolved by the reactions were determined to be kj for HCl and kj for HNO 3. Dividing by moles of H 2 O produced gives kj/mol H 2 O and kj/mol H 2 O, respectively. These values for the heats of neutralization are kj/mol lower than the expected value. This error can be accounted for by some combination of 1) underestimating the temperature change of the mixture and 2) overestimating the moles of H 2 O produced by the reaction. The temperature change of the mixture would be underestimated if its rate of cooling were underestimated. In that case, the trend line would be flatter, and it would extrapolate a lower maximum temperature. It is possible that this occurred in the present experiment. The HNO 3 and NaOH mixture cooled only 0.3 C over three minutes, which seemed suspiciously slow. It is also possible that the amount of H 2 O produced was overestimated. This could happen if some of the solution were inadvertently spilled. Conclusion The unknown metal #14 has a specific heat of 0.36 J/g C; the heat of neutralization of HCl and NaOH is kj/mol H 2 O produced; and the heat of neutralization of HNO 3 and NaOH is kj/mol H 2 O produced. References Beran, J. A. Laboratory Manual for Principles of General Chemistry, 9E. John Wiley & Sons, Inc: USA, Reprinted in Citrus Lab Manual: CHEM 111/112. John Wiley & Sons, Inc: USA, Ebbing, Darrell D. and Steven D. Gammon. General Chemistry, 10E. Brooks/Cole, Cengage Learning: USA,

11 The Engineering ToolBox. Web Resource. < Accessed on October 26, UC Davis ChemWiki. Web Resource. < > Accessed on October 26,