Physical Biochemistry Kwan Hee Lee, Ph.D. Handong Global University
Week 3
CHAPTER 2 The Second Law: Entropy of the Universe increases
What is entropy Definition: measure of disorder The greater the disorder, the greater is the entropy. Disorder occurs spontaneously and thus entropy tends to increase. In an isolated system, every change within the system increases its entropy. Fluctuations may decrease the entropy of an isolated system slightly for a short time.
Concepts If the system is not isolated from the surroundings, then the entropy of the system can decrease. This decrease is accompanied by a larger increase in entropy of the surroundings. So the entropy of the universe always increases. Gibbs free energy always decreases or does not change, in a process that occurs spontaneously at constant temperature and pressure.
Applications The purpose of science is the prediction of the future. However, the first laws does not tell us whether a particular reaction can spontaneously occur. Key questions: which reactions are impossible under given conditions and how the conditions can be changed to make impossible reactions probable. Thermodynamics does not tell us how fast a reaction occur or how the rate depends on the reaction conditions.
Carnot cycle Carnot cycle: an ideal engine that undergoes a particular cycle of four steps to return to its original state. All steps are here are reversible. Steps I and III are isothermal. Steps II and IV are adiabatic
Carnot cycle In step I, a hot ideal gas at T hot expands isothermally and reversibly in a cylinder. In step II, the gas expands adiabatically and reversibly. The engine does work (w 2 is negative) without heat input, its energy drops and the temperature of the gas reduced to T cold. In step III, the gas is compressed isothermally and reversibly. In step IV, the gas is compressed adiabatically and reversibly (w 4 is positive, q 4 =0) to return to its original condition.
Carnot cycle Total work (w) = w1 + w2 + w3 + w4 Total heat absorbed (q) = q1 + q2 + q3 + q4 Since the initial and final states are the same and the energy is a state function, ΔE = o = q + w So it is important to know the work and heat in each step.
Carnot cycle heat engine PV diagram
Work and heat in Carnot cycle For step I( isothermal) w 1 = - PdV = -nrt hot ln (V 2 /V 1 ) E 2 -E 1 = 0, q 1 + w 1 = E 2 -E 1 = 0 Therefore q 1 = nrt hot ln (V 2 /V 1 ) For step II (adiabatic) q 2 = 0 (adiabatic) w 1 = E 2 -E 1 = C V (T hot -T cold )
Work and heat in carnot cycle dv will result in dt The energy change ΔE is C V dt In adiabatic, energy change is equal to the work. C V dt = - PdV = - nrt/v dv C V dt/t = -nr dv/v (from V 2 -V 3 ) C V ln T cold /T hot = -nr lnv 3 /V 2 = nr lnv 3 /V 2
Work and heat in carnot cycle q 3 = -w 3 = nrt cold lnv 4 /V 3 q 4 = 0, w 4 = C V (T hot -T cold ) -CV ln T cold /T hot = nr lnv 4 /V 1 Total heat absorbed (q) = q 1 +q 2 +q 3 +q 4 = nrt hot lnv 2 /V 1 + 0 + nrt cold lnv 4 /V 3 + 0 Total work (w): -w = -(w 1 +w 2 +w 3 +w 4 ) =nrt hot lnv 2 /V 1 + nrt cold lnv 4 /V 3 => w= -q
New state function, Entropy New finding: sum of q for the reversible is not zero, but the sum of q rev /T is zero In step I, dq rev /T = 1/T hot dq rev = q 1 /T hot In step III, dq rev /T = 1/T cold dq rev = q 3 /T cold In step II, IV, dq rev /T =0 The sum of dq rev /T = q 1 /T hot + q 2 /T cold = nrlnv 2 /V 1 + nrlnv 4 /V 3 = nrlnv 2 V 4 /V 1 V 3 = 0 (3.1a, 3.1b) q 1 /T hot + q 3 /T cold = 0
Entropy ΔS= ds = dq rev /T = q rev /T Efficiency = -w/q hot = -w/q 1 q 1 /T hot = -q 3 /T cold -w = q 1 + q 3 Efficiency = -w/q hot = (q 1 +q 3 )/q1 = 1+ q 3 /q 1 q 3 /q 1 = -T cold /T hot Efficiency = 1 + q 3 /q 1 = 1- T cold /T hot
Second Law of Thermodynamics: Entropy is not conserved Entropy is an extensive variable Entropy is a state function ΔS (system) + ΔS (surrounding) 0 For an isolated system, ΔS 0
Molecular interpretation of Entropy The more disorder, the higher is the entropy. The universe is becoming more disordered. Biological processes involve decreases of entropy for the organism itself but they are always coupled to other processes that increase the entropy, so the sum is always positive. Entropy depends on the phase of the materials.
Molecular interpretation of Entropy Liquid molecules have larger entropy than solid ones. For monoatomic elements. entropy can be qualitatively related to the hardness of an element (diamond is more ordered than graphite) All entropies increase as the temperature go up.
Molecular interpretation of Entropy In the gas phase, if the number of product molecules is less than the number of reactant molecules, entropy decreases. For reactions in solution, it is more difficult to predict the entropy change because of the large effect of the solvent.
Molecular interpretation of Entropy For the first four reactions, neutralization of charges occurs. Because a charged species tends to orient the water molecules around it, charge neutralization results in disorientation of some of the solvent molecules and an increase in entropy
The next two reactions involve transfer of charge but no neutralization. The entropy changes are small as a consequence. The last reaction involves the transfer of a nonpolar methane molecule from a polar solvent to a nonpolar solvent. The large positive ΔS is primarily the result of the ordering of water molecules around a nonpolar molecules.
Fluctations In figure a, pressures tend to be uniform. In figure b, compositions tend to be uniform. These processes can take place without changing the surroundings. The reverse of them can only be done if the surroundings are also changed.
Fluctuations In a system of uniform pressure, there will be very slight increases of pressure on one side and decreases on the other side, owing to the random motion of the molecules. A system of uniform composition will have fluctuations in composition in any volume of the system. The second law of thermodynamics does not work in fluctuations.
Measurement of Entropy ΔS = S 2 -S 1 = ds = dq rev /T Entropy is a state function which depends on the initial and final states. For irreversible path, the entropy change is not equal to the heat absorbed divided by T. It is greater ΔS = S 2 -S 1 > dq rev /T
Chemical reactions n A A + n B B n C C + n D D ΔS = n C S C + n D S D n A S A n B S B The entropy of each reactant or product depends on T and P, so ΔS of a given reaction is dependent on T and P. For convenience and comparison, standard state (reference) is defined as 1 atm and 25. Standard molar entropy
Third law of thermodynamics The entropy of any pure, perfect crystal is zero at 0K. S A (0K) = 0 Near 0K, the disorder of a substance can approach zero, the number of microstate approaches 1. For this to happen, the substance must be pure.