- The Enthalpy Of Triethylamine Benzene Solutions At 298 15 K Are Given By Amix H X1 1 Xb A B 1 2xb J Mol Where 1 (64.68 KiB) Viewed 12 times
The enthalpy of triethylamine-benzene solutions at 298.15 K are given by: Amix H = x1(1 – XB)[A+ B(1 – 2xB)] J/mol where
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The enthalpy of triethylamine-benzene solutions at 298.15 K are given by: Amix H = x1(1 – XB)[A+ B(1 – 2xB)] J/mol where
The enthalpy of triethylamine-benzene solutions at 298.15 K are given by: Amix H = x1(1 – XB)[A+ B(1 – 2xB)] J/mol where A = 1418 J/mol and B = -482.4 J/mol where xb is the mole fraction of benzene. В — (a) Develop expressions for (HB - HB) and (HEA – HEA) (b) Compute values for (HB - HB) and (HEA – HEA) at IB 0.5 (c) One mole of a 25 mol % benzene mixture is to be mixed with one mole of a 75 mol % benzene mixture at 298.15 K. How much heat must be added or removed for the processed to be isothermal?