Question

Part A

The change in enthalpy (ΔrHo) for a reaction is -32 kJmol−1 . The equilibrium constant for the reaction is 4.6×103 at 298 K. What is the equilibrium constant for this reaction at 661 K?

Part B

Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g)

Calculate ΔrG for this reaction at 125 ∘C under the following conditions: PCH3OH= 0.890 bar PCO= 0.145 bar PH2= 0.195 bar

Answer 1

A) We have relation   ln ( K2/K1) = ( dH / R) ( 1/T1 -1/T2)

T1 = 298 K , T2 = 661 K ,   dHo = -32 KJ/mol= -32000 J/mol ,

ln ( K2/4.6x10^3) = ( -32000 /8.314) ( 1/298 1-/661)

K2 = 3.82

B) for the given reaction Kp = (pCO x (pH2)^2 ) / ( pCH3OH)

Kp = ( 0.145 x 0.195^2) / ( 0.89)

           = 0.006195

dG reaction = dGo reaction + RT ln K

dGo rxn = dH CO (g) + 2dGo H2 - dGo CH3OH

               = -137.16 + 2(0) - (-166.4)

                  = 29.24 KJ/mol = 29240 J/mol

dG rxn = 29240 + (8.314 x 398 x ln ( 0.006195)                           ( where T = 125C = 125+273 = 398 K)

dG rxn = 12417 J/mol = 12.4 KJ/mol