Entropy, Second Law: Chemical Reactions and Engine Thermodynamics - Prof. Stefan Franzen, Study notes of Physical Chemistry

Download Entropy, Second Law: Chemical Reactions and Engine Thermodynamics - Prof. Stefan Franzen and more Study notes Physical Chemistry in PDF only on Docsity! 1 Chemistry 331 Lecture 14 Entropy and the Second Law NC State University Spontaneity of Chemical Reactions One might be tempted based on the results of thermo- chemistry 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 endothermic reactions that are spontaneous. Of course, 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. 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. Entropy is related to heat and heat flow and yet heat is not 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. ∆S = qrevT 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, 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. A cyclic heat engine I II III IV 1 2 34 The work is w = wI + wII + wIII + wIV q = qI + qIII = - wI - wIII For the adiabatic steps qII = qIV = 0 For the isothermal steps ∆U = 0 Phase Transition Path Condition I. 1→2 Isothermal w = -q II. 2→3 Adiabatic ∆U = w III. 3→4 Isothermal w = -q IV. 4→1 Adiabatic ∆U = w Work and Heat for the Cycle 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 = -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 δwII = δwIV = 0] = - 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] 2 A new state function: Entropy 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 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 = qhot + qcold = nRThotln V2 V1 – nRTcoldln V2 V1 ≠ 0 qrev T = qhot Thot + qcoldTcold = nRln V2V1 – nRln V2V1 = 0 ∆S = qrevT Thermodynamics of an Engine 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 will realize that the external pressure on the piston is 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. Thermodynamic Efficiency We define the efficiency as the work extracted divided by the total heat input. 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. efficiency = work doneheat used η = |wtotal|qhot = |nR(Tcold – Thot)ln (V2 / V1)| nRThotln (V2 / V1) = Thot – TcoldThot 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? A. 75% B. 50% C. 25% D. 5% 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? A. 75% B. 50% C. 25% D. 5% η = |wtotal|qhot = Thot – Tcold Thot = 1 – TcoldThot = 1 – 300K400K = 0.25 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