section 6 - For a multi-step process we need to add the...

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1 For a multi-step process we need to add the steps. The implications of this calculation are that the change in entropy is always positive, and so the minimum possible value of S will occur at the lowest possible thermodynamic temperature. This statement is the bases of the third law of thermodynamics The entropy of all perfect crystalline substances is zero at T= 0 K. −Δ S must be positive when we raise the temperature Δ S must have its smallest possible value at 0 K. What is the absolute entropy of at 0 K? This is the… Recall that for each substance there is a reference state ost stable form of substance - most stable form of substance - Δ U = 0, Δ H = 0 The ‘reference state’ for entropy is not at STP or SATP, but at 0K. -atSTP Δ S o > 0 Nernst developed a heat theorem in 1906 that came to be recognized as the Third Law. “Entropy changes become zero at the absolute zero (of temperature), provided the states of the system are in thermodynamic equilibrium.” or or The entropy of all perfect crystalline substances is zero at T = 0 K. 0 As T 0, S 0 for every chemically homogeneous substance in the crystalline state. This has been found to work experimentally, but can also be theoretically derived using statistical mechanics. (CHM2330!)
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2 There’s a catch: It is impossible to decrease the temperature of any system down to 0K in a finite number of steps. (This is a corollary of the third law.) We have come close to zero though: Bose Einstein condensates are made at 10 -9 K. (Quantum gas - all the atoms have the same, very low velocity and are indistinguishable.) 2001 Nobel prize. Below 15 K we need to use Debye T 3 Law to calculate entropy: C P (T) T 3 as T 0K Cornell Ketterle Wieman S vs T Using Standard Entropies to Calculate Reaction Entropy Changes Knowing the standard entropy values for pure, separated products and reactants the standard entropy of the reaction can be calculated using: Products Reactants mm m Sv S v S °°° Δ= ∑∑ Analogous to the standard reaction enthalpy: Products Reactants oo o rm f m H vH Δ− Δ First we must write the balanced chemical equation r the formation of liquid water: Example of a standard entropy change calculation What is the standard change in entropy for the formation of liquid water? for the formation of liquid water: H 2 (g) + ½O 2 (g) H 2 O(l) ( ) -1 -1 2 ( ) 130.684 J K mol m SHg ° = ( ) -1 -1 2 ( ) 205.138 J K mol m SOg ° = ( ) -1 -1 2 () 69 .91 J K mo l SHO l ° = Look up S o m for reactants and products: m Products Reactants 1 22 2 2 -1 -1 1 2 -1 -1 (H O(l))- (H (g))- (O (g)) = 69.91 - 130.684 - 163.3 J K mol m mm m S v S SS S °° ° °° ° = × = −
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3 How can we determine whether a process will occur spontaneously/irreversibly? We can calculate total entropy: Δ S Total = Δ S Sys + Δ S Surr But what about energy changes in the system?
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This note was uploaded on 11/13/2010 for the course CHM CHM2132 taught by Professor Goto during the Fall '07 term at University of Ottawa.

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section 6 - For a multi-step process we need to add the...

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