pdf file

CEM 383 Equation Sheet, Final Exam - 2004ln [A] t = −kt+ ln[A] 0 K w = [H + ][OH - ]K w =1.0 x 10d[A]-14arate = − = k[A]pH = - log [H + ]dtpOH = -log[OH - ]pH = 14.0 - pOH−K w = K a × K b[ ] [ ]ktA A ett 1/2 =1[A]t1/2= 00.693k⎛ kln⎜⎝ k12⎞⎟⎠= −ER1= 2kt+t [A] k = Ae -E a / RT0=12k[A]0Z=a⎛⎜⎝1T1−1T2⎟ ⎞⎠2πNAσPvRT[[]t[ A] 0 k1( ) ( −kt k te e )1 −−k − kB 2t== [ A]2⎧⎨1+⎩11( ) ( −kt −ktk e − k e )12 1k − kC 2]012⎫⎬⎭K =kkfr