Δ G° = - n F E ° Q ln nF RT-E = E0 Q log n 0.0591-E = E0 Δ G° = - RT lnK 1 Faraday = 6.022 x 10 23 e-= 96,487 Coulombs k a 693 .0 t 1/2 =   k t a A 1 A 10 = − 0 1/2 A k a 1 t =  k t-a A A0 = − t 1/2 = [A0 ] 2ak k = Ae-E a /RT CH302 EXAM 4 EQUATION SHEET For all other constants, see the
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This note was uploaded on 06/13/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.