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5.5: Calorimetry AP Chemistry

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5.5: Calorimetry AP Chemistry How can we go about measuring enthalpy changes? For a constant pressure reaction, the enthalpy change is equal to the heat evolved or absorbed by the system (∆H = q). The heat evolved/absorbed will cause a measurable change in temperature of the system/surroundings. The measurement of heat flow is called calorimetry and is performed using a device called a calorimeter. As substances absorb heat, their temperatures increase. However, substances change temperature at different rates. If you have ever been to the pool or beach on a sunny summer day, you probably noticed the sand or concrete was much hotter than the water. This is because water maintains its temperature much better than sand. Heat capacity, C, is the amount of heat required to raise the temperature of a substance by 1 K (or 1 °C). The greater the heat capacity, the more heat will be required to change the temperature. Usually, we are more interested on the amount of heat required for a given amount of a substance so we use molar heat capacity, or more commonly, specific heat capacity. Molar heat capacity, Cmol, is the amount of heat required to raise the temperature of 1 mole of a substance by 1 K (or 1 °C), so its units are J/mol-K (or J/mol-°C). Specific heat capacity, Cs, is the amount of heat required to raise the temperature of 1 gram of a substance by 1 K (or 1 °C), so its units are J/g-K (or J/g-°C). Specific heat capacity is often shortened to simply specific heat. Knowing the specific heat of a substance allows us to solve for the heat evolved or absorbed in a chemical reaction. To do so, we use the equation: 푞 = 푚 ∙ 퐶푠 ∙ ∆푇 Where q = heat absorbed/evolved m = mass of substance Cs = specific heat of substance ∆T = temperature change of substance (Tf – Ti) Suppose you wanted to measure the heat given off by the neutralization reaction between aqueous solutions of hydrochloric acid and sodium hydroxide. You could place the solutions in a simple calorimeter made of Styrofoam cups, as shown in the diagram to the right. Here, the system would be the reactants and products, and the surroundings would be the water they are reacting in, as well as the foam cups. However, since the foam cups are good insulators, you could assume all heat given off by the reaction would be absorbed by the water. So by measuring the temperature change of the water, you can determine the amount of heat lost by the system. Some calorimeters are designed to withstand large pressure changes. These are called bomb calorimeters (see lower diagram). Reactions that take place inside bomb calorimeters occur at constant volume, so the heat transfer is actually a measure of the internal energy change, not the enthalpy change (∆E, not ∆H). However, remember that for most reactions the two values are very similar. Solved Examples 1) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 °C. To calculate the heat absorbed we can use the equation: 푞 = 푚 ∙ 퐶푠 ∙ ∆푇. We must be careful that our units match the units in the specific heat. In this case, specific heat is given with units of J/g-K, so our units of mass must be converted from kg to g. While the temperature in the specific heat is K, a change of 1 K is the same as 1 °C, so we don’t need to convert the temperature in this case. q = m•Cs•∆T = (50,000g)(0.82J/g-K)(12.0K) = 492,000 J or 492 kJ 2) What temperature change would these rocks (question 1) undergo if they emitted 450 kJ of heat? Here we will use the same equation and same specific heat, but this time we have the heat (q) and are solving for the temperature change. We can rearrange the equation to solve for ∆T. 푞 −450,000 ∆푇 = = = −ퟏퟏ. ퟎ푲 푚 ∙ 퐶푠 (50,000) ∙ (0.82) Here the negative sign in the numerator is because the rocks are emitting the heat, and as a result, the answer is negative. This means the rocks will decrease in temperature by 11.0 K or 11.0 °C. 3) When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 °C to 23.11 °C. The temperature increase is caused by the following reaction: AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq) Calculate ∆H for this reaction in kJ/mol AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g °C. Because this is a constant-pressure reaction, we can assume the enthalpy change is equal to the heat transfer (∆H = q). We can use the specific heat equation to solve for q. 4.18퐽 푞 = 푚 ∙ 퐶 ∙ ∆푇 = (100.0푔) ( ) (23.11 − 22.30) = 340 퐽 푠 푔℃ We must be careful at this point for two reasons. First, we have just calculated the heat gained by the surroundings, because the thermometer would be measuring the solution temperature. We are interested in the system though. So the heat gained by the solution was lost by the solute. Therefore, q = -340J. Second, we were not just asked for q or ∆H. We were asked to solve for the enthalpy change per mole of AgNO3. Thus, we must solve for the number of moles of silver nitrate, and then divide the enthalpy change by the moles. 0.0500 L AgNO3 0.100 mol AgNO3 = 0.00500 mol AgNO3 1 L AgNO3 −340 퐽 = −ퟔퟖ, ퟎퟎퟎ푱/풎풐풍 풐풓 − ퟔퟖ풌푱/풎풐풍 0.00500 푚표푙 4) A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/°C. The temperature increases from 23.10 °C to 24.95 °C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole. a) Here we can use the calorimeter’s heat capacity and the temperature change to determine the heat absorbed by the calorimeter (surroundings). This will be equal in magnitude, but opposite in sign to the heat given off by the system. Then we can divide the heat by the grams of the sample to determine the heat of combustion per gram. 푘퐽 푞 = 퐶 ∙ ∆푇 = (4.812 ) (24.95 − 23.10℃) = 8.90 푘퐽 ℃ −8.90푘퐽 = −ퟏퟓ. ퟐ 풌푱/품 0.5865 푔 b) We can use the same heat transfer as in part (a). This time we will convert the mass of the sample to moles and divide the heat by the number of moles to get the heat of combustion per mole. 0.5865 g HC3H5O3 1 mol HC3H5O3 = 0.006510 mol HC3H5O3 90.09 g HC3H5O3 −8.90 푘퐽 = −ퟏ, ퟑퟕퟎ 풌푱/풎풐풍 0.006510 푚표푙 Name: __________________________________ Period: ________ Date: _______________ Calorimetry AP Chemistry – 5.5 Directions: Answer the following questions concerning specific heat and calorimetry. Show your work for all calculations. 1) Which will release more heat as it cools from 50 °C to 25 °C, 1 kg of water or 1 kg of aluminum? How do you know? 2) What are the units of… a. Molar heat capacity? b. Specific heat capacity? c. If you know the specific heat of copper, what additional information do you need to calculate the heat capacity of a particular piece of copper pipe? 3) Two solid objects, A and B, are placed in boiling water and allowed to come to temperature there. Each is then lifted out and placed in separate beakers containing 1000 g water at 10.0 °C. Object A increases the water temperature by 3.50 °C; B increases the water temperature by 2.60 °C. a. Which object has the larger heat capacity? b. What can you say about the specific heats of A and B? 4) What is the specific heat of liquid water? 5) What is the molar heat capacity of liquid water? 6) What is the heat capacity of 185 g of liquid water? 7) How many kJ of heat are needed to raise the temperature of 10.00 kg of liquid water from 24.6 °C to 46.2 °C? 8) Refer to the substances in Table 5.2 (on the top of the previous page). a. Which substance requires the smallest amount of energy to increase the temperature of 50.0 g of that substance by 10 K? b. Calculate the energy needed for this temperature change. 9) The specific heat of ethylene glycol is 2.42 J/g-K. How many joules of heat are needed to raise the temperature of 62.0 g of ethylene glycol from 13.1 °C to 40.5 °C? 10) When a 9.55-g sample of solid sodium hydroxide dissolves in 100.0 g of water in a coffee-cup calorimeter, the temperature rises from 23.6 °C to 47.4 °C. Calculate ∆H (in kJ/mol NaOH) for the solution process NaOH(s) Na+(aq) + OH-(aq) Assume that the specific heat of the solution is the same as that of pure water. 11) A 2.200-g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/°C.
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