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9/23/2008 Spontaneity of Chemical Reactions Chemistry 433 One might be tempted based on the results of thermochemistry to predict that all exothermic reactions would be spontaneous. The corollary this would be the statement that no endothermic reactions are spontaneous. However, this is not the case. There are numerous examples of p Of course,, endothermic reactions that are spontaneous. heat must be taken up from the surroundings in order for such processes to occur. Nonetheless, the enthalpy of the reaction does not determine whether or not the reaction will occur, only how much heat will be required or generated by the reaction. The observation that gases expand to fill a vacuum and that different substances spontaneously mix when introduced into the same vessel are further examples that require quantitative explanation. Lecture 9 Entropy and the Second Law NC State University Spontaneity of Chemical Reactions As you might guess by now, we are going to define a new state function that will explain all of these observations and define the direction of spontaneous processes. This state function is the entropy. py is related to heat and heat flow and yyet heat is not Entropy a state function. Recall that q is a path function. It turns out that the state function needed to describe spontaneous change is the heat divided by the temperature. Here we simply state this result. qrev T ΔS = We will prove that entropy is a state function in this lecture. A cyclic heat engine 2 1 I IV Phase I. II. III. IV. II 3 4 III Transition 1→2 2→3 3→4 4→1 Path Isothermal Adiabatic Isothermal Adiabatic Engines Historically, people were interested in understanding the efficiency with which heat is converted into work. This was a very important question at the dawn of the industrial revolution since it was easy to conceive of an engine powered by steam, but it turned out to be quite difficult to build one that was efficient enough to get anything done! In an engine engine, there is a cycle in which fuel is burned to heat gas inside the piston. The expansion of the piston leads to cooling and work. Compression readies the piston for the next cycle. A state function should have zero net change for the cycle. It is only the state that matters to such a function, not the path required to get there. Heat is a path function. As we all know in an internal combustion engine (or a steam engine), there is a net release of heat. Therefore, we all understand that δq ≠ 0 for the cycle. Work and Heat for the Cycle Condition w = -q ΔU = w w = -q ΔU = w The work is w = wI + wII + wIII + wIV q = qI + qIII = - wI - wIII For the adiabatic steps qII = qIV = 0 For the isothermal steps ΔU = 0 Neither the work nor the heat is a state function. Neither one is zero for the cycle as should be the case for a state function. The work is: w = wI + wII + wIII + wIV =-nRThotln(V2/V1)–Cv(Tcold–Thot)–nRTcoldln(V4/V3)–Cv(Thot – Tcold) w = -nRThotln(V2/V1) – nRTcoldln(V4/V3) [since wII = - wIV] w = -nRT nRThotln(V2/V1) – nRTcoldln(V1/V2) [since V4/V3 = V1/V2] w = -nRThotln(V2/V1) + nRTcoldln(V2/V1) [property of logarithms] The heat is: q = qI + qIII [since qII = qIV = 0 for adiabatic processes] = - wI - wIII [since dU = 0 for isothermal steps] q = nRThotln(V2/V1) + nRTcoldln(V4/V3) q = nRThotln(V2/V1) + nRTcoldln(V1/V2) [since V4/V3 = V1/V2] q = nRThotln(V2/V1) - nRTcoldln(V2/V1) [property of logarithms] 1 9/23/2008 A new state function: Entropy Thermodynamics of an Engine The heat is not a state function. The sum qI + qIII is not zero. From this point on we will make the following definitions: qI = qhot qIII = qcold The cycle just described could be the cycle for a piston in a steam engine or in an internal combustion engine. The hot gas that expands following combustion of a small quantity of fossil fuel drives the cycle. If you think about the fact that the piston is connected to the crankshaft you pressure on the p piston is will realize that the external p changing as a function of time and is helping to realize an expansion that as close to an ideal reversible expansion as the designers can get. If we ignore friction and assume that the expansion is perfectly reversible we can apply the above reasoning to your car. The formalism above for the entropy can be used to tell us the thermodynamic efficiency of the engine. V2 V – nRTcold ln 2 ≠ 0 V1 V1 V2 V2 = nRln – nRln =0 V1 V1 q = qhot + qcold = nRThotln qrev qhot qcold = + T Thot Tcold However, the heat divided by temperature is a state function. This reasoning leads to the idea of a state function called the entropy. We can write: q ΔS = rev T Question Thermodynamic Efficiency We define the efficiency as the work extracted divided by the total heat input. efficiency = work done heat used |w | |nR(Tcold – Thot)ln (V2 / V1)| Thot – Tcold = η = qtotal = nRThotln (V2 / V1) Thot hot Your car has an operating temperature of 400 K. If the ambient temperature is 300 K, what is the thermodynamic efficiency of the the engine? A. B. C. D. 75% 50% 25% 5% The efficiency defined here is the ideal best case. It assumes a reversible process with no losses due to friction. The temperature Thot is the temperature of the expansion in the engine. The temperature Tcold is the temperature of the exhaust. Tcold cannot be less than the temperature of the surroundings. Question Question Your car has an operating temperature of 400 K. If the ambient temperature is 300 K, what is the thermodynamic efficiency of the the engine? The thermodynamic cycle was derived for reversible expansions. What are the consequences if the cycle is not perfectly reversible? A. B. C. D. A. The work of expansion will decrease B. The work of compression will decrease C. There will be no adiabatic expansion D. There can be no cycle 75% 50% 25% 5% |w | T – Tcold T η = qtotal = hot = 1 – cold = 1 – 300K = 0.25 400K Thot Thot hot 2 9/23/2008 The Thermodynamic Temperature Scale Question The thermodynamic cycle was derived for reversible expansions. What are the consequences if the cycle is not perfectly reversible? A. The work of expansion will decrease B. The work of compression will decrease C. There will be no adiabatic expansion D. There can be no cycle The definition of entropy is qhot/Thot + qcold/Tcold = 0. We can write this as qhot/Thot = - qcold/Tcold Since qcold is negative we can combine the minus sign with qcold and write the expression as | hot|/Thot = |q |q | cold|/Tcold and finally |qhot|/|qcold| = Thot/Tcold The ratio of the heats is equal to the ratio of temperatures for two steps in a thermodynamic cycle. This defines a temperature scale and allows one to measure temperature as well (i.e. this scheme represents a thermometer). Both this expression and the thermodynamic efficiency further imply that there is an absolute zero of temperature. Heat Transfer To examine the function that we have just defined, let us imagine that we place to identical metal bricks in contact with one another. If one of the bricks is at equilibrium at 300 K and the other at 500 K, what will the new equilibrium temperature be? Intuitively, you would say 400 K and you would imagine g that heat flows spontaneously p y from the warmer brick to the colder brick. The entropy function makes these ideas quantitative. q q q1 q2 ΔS = 2 + 1 T2 T1 T1= 500 K T2= 300 K Using this definition of entropy change as the heat flow divided by the temperature. System and surroundings Up to this point we have considered the system, but we have not concerned ourselves with the relationship between the system and the surroundings. When we consider heat flow In the state function, entropy, we need to carefully account for how the heat flow takes place in order to determine the effect We will show that for reversible processes the effect. entropy change is zero. However, for irreversible processes the entropy change can be positive (spontaneous) or negative (reverse process is spontaneous). For isolated systems the entropy change is also zero. However, when there is heat flow between the system and surroundings the entropy can be non-zero. In the case that the heat flow is irreversible the entropy change will be non-zero. Heat Transfer Let’s assume that heat flows from the hot body to the cold body. Then q1 is negative (the flow from the hot body) and q2 is positive (the flow into the cold body). Moreover, q2 = -q1 = q This means that we can substitute in q to obtain: ΔS = q q – = q 1 – 1 = q 1 – 1 = 0.0013 q T2 T1 T2 T1 300 500 For this calculation it does not matter how big q is, but only that it is a postive number so that ΔS is positive. T2= 300 K T1= 500 K q Using this definition ΔS > 0, which says that the process is spontaneous. Calculating reversible and irreversible paths It is important to reiterate that the calculation of the entropy of the system always follows a reversible path. You might ask, well what happens if the process is not reversible? To consider this let us the example of expansion of gas in a cylinder. The process can occur along different paths. a. constant pressure expansion, w = -PextΔV = - Pext(Vf – Vi) b. reversible isothermal expansion the work w = -nRTln(Vf/Vi). For both a. (irreversible) and b. (reversible) we will calculate the entropy of the system along a reversible path. ΔSsys = q/T = -w/T = nRln(Vf/Vi) For the reversible path, we can use the fact that ΔSsurr = - ΔSsys to obtain ΔSsurr = - nRln(Vf/Vi). 3 9/23/2008 System and surroundings The heat transfer example shows us that we always must consider the surroundings. Any time the system releases heat it goes into the surroundings and this contributes to the overall entropy change. Thus, the total entropy is: ΔStotal = ΔSsurr + ΔSsys For a reversible process the total entropy is zero. zero If a process is irreversible and spontaneous the entropy change is positive. This implies that we must treat the system and the surroundings differently when we calculate entropy. The rule is: always calculate the entropy of the system along a reversible path. If the process is truly reversible then: ΔSsurr = - ΔSsys ΔStotal = 0 and Understanding the irreversible path For the irreversible path we use the actual work of the constant pressure expansion, w = -PextΔV = - Pext(Vf – Vi) to calculate ΔSsurr = -q/T = w/T = -Pext(Vf – Vi)/T where T is the same temperature we used for the isothermal expansion. Note that the sign is opposite since the heat is flowing into the surroundings (and out of the system). We see that in this case the entropy change ΔSsurr is smaller in magnitude than ΔSsys. We know this from the first law where we saw that the irreversible work of expansion is always less than the reversible work of expansion. Thus, ΔStotal > 0 for the irreversible process. The dependence of the entropy on volume Statements of the second law ΔSrev = qrev/T = 0 for ΔSirr = qirr/T > 0 The entropy of an irreversible process is greater than zero If the process is spontaneous. Clausius: the entropy of the universe tends towards a maximum. For a constant temperature (isothermal)expansion we have: dS = δqrev/T = - δwrev/T. The logic behind this statement is that the internal energy change is zero for a constant temperature process and so δqrev = - δwrev. To calculate the reversible work we simply i l plug l iin δwrev = -PdV. PdV A According di to the h id ideall gas llaw P = nRT/V so dS = nRdV/V. S2 Famous words: “Die Entropie des Universums strebt ein Maximum zu.” The dependence of the entropy on temperature The entropy change as a function of the temperature is derived at constant volume using the fact that dU = δqV = nCVdT. The reversible heat in this case, qrev, is a constant volume heat and so it can be replaced by dS = δqrev/T = nC CvdT/T at constant volume l To obtain ΔS we need to integrate both sides S2 S1 dS = nCv T2 T1 dT T We obtain: ΔS = S2 – S1 = nCvln(T2/T1). Exactly the same reasoning applies at constant pressure, so that ΔS = nCpln(T2/T1). S1 dS = nR V2 V1 dV V The result of this equation is that ΔS = nRln(V2/V1) at constant temperature. History of Refrigeration In prehistoric times game was stored in a cave or packed in snow to preserve the game for times when food was not available. Later, ice was harvested in the winter to be used in the summer. Ice is still used for cooling and storing food. The intermediate stage in the history of cooling li foods f d was to t add dd chemicals h i l like lik sodium di nitrate it t or potassium nitrate to water causing the temperature to fall. Cooling wine via this method was recorded in 1550, as were the words "to refrigerate." The evolution to mechanical refrigeration, a compressor with refrigerant, was a long, slow process and was introduced in the last quarter of the 19th century. Source : U.S. Department of Agriculture 4 9/23/2008 Importance of Refrigeration Refrigeration slows bacterial growth. Bacteria exist everywhere in nature. They are in the soil, air, water, and the foods we eat. When they have nutrients (food), moisture, and favorable temperatures, they grow rapidly, increasing in numbers to the point where some types of bacteria can cause illness. Bacteria grow most rapidly in the range of temperatures between 40 and 140 °F, the "Danger Zone," some doubling in number in as little as 20 minutes. A refrigerator set at 40 °F or below will protect most foods. Coefficient of Performance The coefficient of performance, or COP (sometimes CP), of a heat pump is the ratio of the output heat to the supplied work or q COP = hot |wtotal| pp by y the condenser and w where q is the useful heat supplied is the work consumed by the compressor. (Note: COP has no units, therefore heat and work must be expressed in the same units.) According to the first law of thermodynamics, in a reversible system we can show that qhot = qcold + w and w = qhot − qcold, where qhot is the heat taken in by the cold heat reservoir and qcold is the heat given off by the hot heat reservoir. Cyclic Refrigeration The refrigeration cycle uses a fluid, called a refrigerant, to move heat from one place to another. The key to understanding how it works is recognizing that at the same pressure, the refrigerant boils at a much lower temperature p , the refrigerant g commonlyy used in than water. For example, home refrigerators boils between 40 and 50°F as compared to water's boiling point of 212°F. Let's look at the process to see how boiling and condensing a refrigerant can move heat. The process is the same whether it is operating a refrigerator, an air conditioner or a heat pump. This example illustrates a closed-loop system. Cyclic Refrigeration This consists of a refrigeration cycle, where heat is removed from a low-temperature space or source and rejected to a high-temperature sink with the help of external work, and its inverse, the thermodynamic power cycle. l In I the th power cycle, l heat h t is i supplied li d from f a highhi h temperature source to the engine, part of the heat being used to produce work and the rest being rejected to a low-temperature sink. This satisfies the second law of thermodynamics. Coefficient of Performance Therefore, by substituting for w, q COPheating = q –hotq hot cold Therefore, COPheating = Thot Thot – Tcold temperature Thot is the temperature of the expansion in the engine. The temperature Tcold is the temperature of the exhaust. Tcold cannot be less than the temperature of the surroundings. For refrigeration, the COP is: q Tcold COPcooting = q –coldq = Thot – Tcold hot cold Cyclic Refrigeration We'll begin with the cool, liquid refrigerant entering the indoor coil, operating as the evaporator during cooling. As its name implies, refrigerant in the evaporator "evaporates." Upon entering the evaporator, the liquid refrigerant's temperature is between 40 and 50°F and without changing its temperature, it absorbs heat as it changes state from a liquid to a vapor. The heat comes from the warm, moist room air blown across the evaporator coil. As it passes over the cool coil, it gives up some of its heat and moisture may condense from it. The cooler, drier room air is re-circulated by a blower into the space to be cooled. 5 9/23/2008 Cyclic Refrigeration The vapor refrigerant now moves into the compressor, which is basically a pump that raises the pressure so it will move through the system. Once it passes through the compressor, the refrigerant is said to be on the "high" side of the system. Like anything that is put under pressure, the increased pressure from the compressor causes the temperature of the refrigerant to rise. As it leaves the compressor, the refrigerant is a hot vapor, roughly 120 to 140°F. It now flows into the refrigerant-to-water heat exchanger, operating as the condenser during cooling. Again, as the name suggests, the refrigerant condenses here. As it condenses, it gives up heat to the loop, which is circulated by a pump. Cyclic Refrigeration In summary, the indoor coil and refrigerant-to-water heat exchanger is where the refrigerant changes phase, absorbing or releasing heat through boiling and condensing. The compressor and expansion valve facilitate the pressure changes, increased by the compressor and reduced by the expansion valve. Cyclic Refrigeration The loop water is able to pick up heat from the coils because it is still cooler than the 120 degree coils. As the refrigerant leaves the condenser, it is cooler, but still under pressure provided by the compressor. It then reaches the expansion valve. The expansion valve allows the high pressure refrigerant to "flash" through becoming a lower pressure, cooled liquid. When pressure is reduced, as with spraying an aerosol can or a fire extinguisher, it cools. The cycle is complete as the cool, liquid refrigerant re-enters the evaporator to pick up room heat. In winter, the reversing valve switches the indoor coil to operate as the condenser and the heat exchanger as the evaporator. Vapor-compression cycle A simple stylized diagram of the refrigeration cycle: 1) condensing coil 2) expansion valve 3) evaporator coil 4) compressor. Cyclic Refrigeration A refrigeration cycle describes the changes that take place in the refrigerant as it alternately absorbs and rejects heat as it circulates through a refrigerator. Heat naturally flows from hot to cold. Work is applied to cool a space by pumping heat from a lower temperature heat source into a higher temperature heat sink. Insulation is used to reduce reduce the work and energy required to achieve and maintain a lower temperature in the cooled space. 6