Absorption of Heat. Internal energy is the appropriate energy variable to use at constant volume

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1 6 Absorption of Heat According to the First Law, E = q + w = q - P V, assuming P-V work is the only kind that can occur. Therefore, E = q V. The subscript means that the process occurs at constant volume. So, if we can control our system and maintain constant volume, we know that if we inject heat into a constant volume system then we know the change in internal energy. For an ideal gas, an increase in E results in a temperature increase. T q V So, what about constant pressure? E = q - P V; q = E + P V. Let s define H = E + PV, the enthalpy. P H = E + (PV) = E + P V + V P V So, H =q + w + P V + V P = q P at constant pressure. Thus, E = q V H = q P Internal energy is the appropriate energy variable to use at constant volume Enthalpy is the appropriate energy variable to use at constant pressure Note that at constant pressure, some of the heat goes into expanding (or compressing) the volume, so for a fixed amount of heat injection, the temperature increase should be smaller for the constant pressure case. T + work q P

2 7 This provides a nice lead in for the concept of heat capacity. Definitions of Heat Capacities Suppose we inject a given amount of heat q into a system consisting of one mole of a substance, and we observe a temperature rise. We say q = C T for one mole. C has units of Joules per mole per degree. This is the molar heat capacity. For n moles, we write q = nc T C is the molar heat capacity Sometimes we can express heat capacity per gram. We call these quantities specific heats. Specific always means per gram. We denote these quantities as c. For m grams, we write q = m c T. c has units of Joules per gram per degree. So, lets go back to heat absorption processes occurring at constant volume or pressure. At constant volume, Cv = q V / T = E / T At constant pressure, C P = q P / T = H / T C p vs C v Let s take two identical systems of gases, inject identical amounts of heat. Let one system be held at constant volume, the other at constant pressure. Measure T. What happens? At constant volume, all heat goes into increasing temperature. At constant pressure, some of the heat goes into increasing the volume of the gas, the rest into increasing the temperature. So, we expect T (constant V) > T (constant pressure). Since T = q/c, that means that C p > C v. At constant pressure, the gas must do work on the surroundings, or the surroundings do negative work on the system. For one mole of a gas, C P = C V + R For solids, C P C V Let s do some simple examples: Take one mole of helium gas, inject 100 J of energy, measure T

3 8 At constant volume, C V = 12.5 Joules per Kelvin per mole. T =Q/C = 100 J / [(12.5 J K -1 mol -1 ) ( 1 mol)] = 8.0 degrees. Note that the degree size is the same for Celsius and Kelvin, so if the gas is initially at 0 C, T goes to 8.0 C. The gas is also at K, so its new temperature is K. At constant pressure, C P = C V + R = J K -1 mol -1 = 20.8 J K -1 mol -1 Then T =Q/C = 100 J / [(20.8 J K -1 mol -1 ) ( 1 mol)] = 4.8 degrees. The difference arises from the fact that the container expands, so some of the energy is converted into work. This amount of work steals 3.2 degrees of the temperature rise. What does heat capacity mean?: Heat capacity tells us about the atomic and molecular motions of the system. In a gas, atoms can absorb heat in their kinetic energy. Remember from kinetic theory that for a monatomic gas, KE = 3 kt This is the result for three dimensions. 2 Molecules can also rotate: and vibrate: Rotations generally add ½RT of energy per mole per rotation. Linear molecules contribute two rotations, nonlinear molecules 3 rotations. Vibrations in molecules in the gaseous state generally do not make a full contribution of RT to the internal energy. Vibrations in solids are the only modes of internal energy available to absorb heat. A solid generally has a heat capacity of 3R J K -1 mol J K - 1 mol -1 Calorimetry

4 9 We can use the ideas of the First Law to understand how two objects at different temperatures reach equilibrium. A styrofoam coffee cup makes a pretty good calorimeter. In this case, q = 0, since no heat is exchanged with the surroundings. Example: Place 100 g of H 2 O( l ) at 25 C in an insulated cup. Drop 30.0 g of Fe at 150 C in the cup. Calculate the final state of the system. q(fe) + q(h 2 O) = 0 q(fe) = moles Fe heat capacity of Fe (T final T initial ) q(h 2 O) = moles H 2 O heat capacity of H 2 O (T final T initial ) Work with molar heat capacities T final = 28.9 C C(Fe) = 25 J K -1 mol -1 C(H 2 O) = 75.3 J K -1 mol -1 Moles Fe = 30.0 g / 55.8 g mol -1 = 0.54 mol Moles H 2 O = 100 g / 18 g mol -1 = 5.55 moles T initial (H 2 O) = 25 C T initial (Fe) = 150 C (0.54)(25)(T final 150) + (5.55)(75.3)(T final 25) = T final T final 2025 = T final = In a laboratory calorimeter, combustion is often carried out. The T is measured, and the heat evolved when M grams of substance undergoes combustion is calculated from the heat capacity of the calorimeter and its contents. Note that the units are J per degree. No moles involved in this calculation. Example: Suppose 1.00 grams of a substance with a formula weight of 100 g /mole undergoes combustion in a calorimeter. The

5 10 heat capacity of the calorimeter and its contents is 19,000 J per degree. A T of 2.20 C is observed. What is the E for combustion? q V = 19,000 J/deg 2.20 C = 41,800 J, so this is a change in internal energy. This heat is per 1g/100 g/mol = 0.01 moles of compound So E = -q V /moles = kj/mole. Why is the sign for the reaction negative? Positive q is absorbed by the surroundings. Therefore negative q is absorbed by the system. Properties of Enthalpy The enthalpy is an especially useful quantity for calorimetry, since most chemical reactions occur at constant pressure, 1 atm. The units of H are J mol -1. So, for the reaction CO(g) + ½O 2 (g) CO 2 (g) H = kj kj of energy are liberated when one mole of CO 2 is formed. So, = kj is liberated when 2 moles of CO 2 are formed. H is an extensive thermodynamic variable. H = E + PV. E, P, V are thermodynamic functions of state (they are not interactions between the system and the surroundings) so H is also a state function. H is independent of path. This is a very important concept. C(s) + ½O 2 (g) H 1 CO(g) ½O 2 (g) ½O 2 (g) H 2 CO 2 (g) H 3 We can make CO 2 directly via the lower pathway, or indirectly, via the upper, stepwise pathway. H 1 = kj mol -1 H 2 = kj mol -1 H 3 = kj mol -1