LECTURE13 - Last time - Equilibrium Tables aA bB Initial...

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1 Last time - Equilibrium Tables aA bB A B Initial conc [A] [B] I nitial moles [A] V [B] V C hange in moles. –ax +bx E quilibrium moles. [A] V – ax [B] V+ bx Equilibrium conc [A] – ax/V [B]+ bx/V find x K c = [B] b [A] a = ([B]+bx/V) b ([A]-ax/V) a Last time: Equilibrium Tables using Pressures aA(g) bB(g) Initial P P Ai P Bi Initial moles. P Ai V i /RT P Bi V i /RT Change in moles. –ax +bx Equilibrium moles. P Ai V i /RT – ax P Bi V i /RT + bx Equilibrium P P Ai V i /V e –ax×RT/V e P Bi V i /V e + bx×RT/V e V Constant Easier: Consider x as a change in P: aA(g) bB(g) Initial P P A i P B i Change in P –ax +bx Equilibrium P P A i – ax P B i + bx K p = P B b P A a
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2 From last time: Example14F: Equilibrium Table Equal amounts of hydrogen and iodine gas are placed in a container at 1atm pressure and 445°C. What partial pressure of hydrogen iodide gas exists at equilibrium? K p = 50.2 at 445°C Initial amounts 0.5atm 0.5atm 0atm Changes -x -x +2x Equil. Amounts 0.5-x 0.5-x 2x  22 2 2 2 (2 ) 50.2 0.5 HI p HI Px K PP x  P HI = 2HI(g) (g) I (g) H 2 2 2 7.085 2x = 7.085 (0.5 ) 0.5 9.085x = 7.085 0.5 x = 7.085 0.5/9.085 x x x   0.39atm
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LECTURE13 - Last time - Equilibrium Tables aA bB Initial...

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