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Unformatted text preview: 1 K1 ) ln k = ln A + (E a /R)(1/T) ln( k 2 / k 1 ) = (E a /R)[(1/T 2 )(1/T 1 )] ln(C t /C ) = k t ln(C t ) = k t + ln(C ) t 1/2 = (ln 2)/ k (1/C t ) = (1/C ) + k t t 1/2 = 1/( k C ) C t = C k t t 1/2 = C /2 k pH = log[H 3 O + ] pOH = log[OH] pH + pOH = 14 [H 3 O + ][OH] = 1.0 x 1014 (@25°C) K a K b = 1.0 x 1014 pK a = log K a pK b = log K b pH = pK a + log{ [base] / [acid] } [base]/[acid] = 10 pH  pKa I = Q /t 1 A = 1 C/s F = 96,485 C/ mol eΔG° = nF Δ E ° 1 V = 1 J/C Δ E = Δ E ° – {( R T / nF ) ln Q } ln K = ( nF / R T) Δ E ° Standard Reduction Potentials ( E ° values are in V)...
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 Spring '09
 VUOCOLO
 Chemistry

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