Equations and constants for Exam 2, CHM 3400, Fall...
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Equations and constants for Exam 2, CHM 3400, Fall 2015 You are responsible for knowing the conditions under which the equations apply. 1 atm = 760 Torr = 1.01325 bar R = 8.31451 J K -1 mol -1 = 0.0831451 L bar K -1 mol -1 = 0.08206 L atm K -1 mol -1 dU = TdS – pdV ΔH = C p ΔT Δ r H(T’) = Δ r H(T) + ΔC p ΔT = ! = !"# !"## = − ∆ = ! ! ∆ = ! ∆ = ! ! ! ∆ = ∆ !"#$% ΔS = -nR(x A lnx A + x B lnx B ) dG = -SdT + Vdp ! = − ! = ∆ = ∆ − ∆ ΔG = VΔp ΔG = nRT(x A lnx A + x B lnx B ) = ∆ !"#$% ∆ !"#$% ! ! = ∆ !"# ! 1 ! − 1 ! f = c – p + 2 p A = x A p A * p B = x B K , p B = m B K’ = ! ! + ! ! ! = ! ∗ + ln ! ! = ! ∗ + ln ! a j = γ j x j Π = [B]RT ΔT f = K f m B ΔT b = K b m B ± = ! ! ! ! ! ! ! ! (ℎ = ! + ! ) !" ± = − ! ! !/! = 1 2 ! ! !
Transcript of Equations and constants for Exam 2, CHM 3400, Fall...
Equations and constants for Exam 2, CHM 3400, Fall 2015 You are responsible for knowing the conditions under which the equations apply. 1 atm = 760 Torr = 1.01325 bar
R = 8.31451 J K-1 mol-1 = 0.0831451 L bar K-1 mol-1 = 0.08206 L atm K-1 mol-1
dU = TdS – pdV
ΔH = CpΔT
ΔrH(T’) = ΔrH(T) + ΔCpΔT
𝑆 = 𝑘!𝑙𝑛𝑊
𝑑𝑆 =𝑑𝑞!"#𝑇
𝑑𝑆!"## =−𝑑𝑞𝑇
∆𝑆 = 𝑛𝑅𝑙𝑛𝑉!𝑉!
∆𝑆 =𝑛𝐶!𝑇 𝑑𝑇
∆𝑆 = 𝑛𝐶!𝑙𝑛𝑇!𝑇!
∆𝑆 =∆!"#$%𝐻
𝑇
ΔS = -nR(xAlnxA + xBlnxB)
dG = -SdT + Vdp 𝜕𝐺𝜕𝑇 !
= −𝑆
𝜕𝐺𝜕𝑃 !
= 𝑉
∆𝐺 = ∆𝐻 − 𝑇∆𝑆
ΔG = VΔp
ΔG = nRT(xAlnxA + xBlnxB)
𝑑𝑃𝑑𝑇 =
∆!"#$%𝐻𝑇∆!"#$%𝑉
𝑙𝑛𝑃!
𝑃! =∆!"#𝐻!
𝑅1𝑇! −
1𝑇!
f = c – p + 2
pA = xApA*
pB = xBK, pB = mBK’
𝐺 = 𝑛!𝜇! + 𝑛!𝜇!
𝜇! = 𝜇!∗ + ln𝑎!
𝜇! = 𝜇!∗ + ln𝑎!
aj = γjxj
Π = [B]RT
ΔTf = KfmB
ΔTb = KbmB
𝛾± = 𝛾!!!𝛾!!!
!!
(𝑤ℎ𝑒𝑟𝑒 𝜈 = 𝜈! + 𝜈!) 𝑙𝑜𝑔!"𝛾± = −𝐴 𝑧!𝑧! 𝐼!/!
𝐼 =12 𝑚!𝑧!!
