Equations and Constants for FINAL EXAM 09-106 S 11

# Equations and Constants for FINAL EXAM 09-106 S 11 - -1 K-1...

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Equations and Other Information Provided on Final Exam 09-106 Spring 2011 PV= n RT ( R = 0.082 L•atm•mol -1 •K -1 ) K = °C + 273 1 L•atm = 101.325 J ΔE = q + w w = -P ext ΔV w = - n RT ln(V f /V i ) q = mc S ΔT = n cΔT q = n ΔH fus q = n ΔH vap q = n ΔH reaction ΔE = n c V ΔT ΔH = n c P ΔT c V = 1.5 R c P = 2.5 R c P = c V + R ( R = 8.31 J•mol -1 •K -1 ) ΔH = ΔE + Δ(PV) = ΔE + RTΔ n gas P i V i γ = P f V f γ (γ = c P / c V ) ΔS = q /T ΔS = n c P ln(T f /T i ) = n c V ln(T f /T i ) ΔS = n R ln(V f /V i ) = n R ln(P i /P f ) ΔS° vap = ΔH° vap /T b ΔS° fus = ΔH° fus /T f ΔS tot = ΔS sys + ΔS surr ΔS surr = -ΔH sys /T ΔS surr = - q sys /T ΔG = ΔH – Δ(TS) ΔG = ΔH – TΔS ΔGº = ΔHº – TΔSº ln K = [- H°/R](1/T) + [ S°/R] ln( K 2 / K 1 ) = ( rx /R)[(1/T 1 )-(1/T 2 )] ln(P 2 /P 1 ) = ( vap /R)[(1/T 1 )-(1/T 2 )] ΔG = ΔGº + RT ln Q ΔGº = -RT ln K Average Bond Enthalpies (in kJ/mole)

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-E a /(RT) k = A e ( R = 8.31 J•mol
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Unformatted text preview: -1 K-1 ) 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 10-14 (@25°C) K a K b = 1.0 x 10-14 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 e-ΔG° = -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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## This note was uploaded on 08/25/2011 for the course CHEMISTRY 09-106 taught by Professor Vuocolo during the Spring '09 term at Carnegie Mellon.

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Equations and Constants for FINAL EXAM 09-106 S 11 - -1 K-1...

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