Từ phản ứng 2 : DeltaG20 = -2,303RTLgKp,2
Từ phản ứng 3 : DeltaG30 = -2,303RTLgKp,3
Từ phản ứng 1 : DeltaG10 = -2,303RTLgKp,1
Mà DeltaG10 = DeltaG20 + DeltaG30
=> -2,303RTLgKp,1 = -2,303RTLgKp,2 + -2,303RTLgKp,3
=> LgKp,1 = LgKp,2 + LgKp,3 = -4984/T + 12,04
=> (dlnKp,1)/T = d/dT(2,303(-4984/T +12,04)) = 2,303.4984/T2
=> Hiệu ứng nhiệt của phản ứng
Delta H = 2,303.8,314.4984 = 95429 J/mol
ta thấy pư(2) + pư(3) = pư (1)
=>\(\bigtriangleup G\) o1 = \(\bigtriangleup G\) o2 + \(\bigtriangleup G\) o3
<=>RTlnKP1 =RTlnKp2 +RTlnKp3
=> \(\bigtriangleup G\)o1 = 8.314 (-3149 + 5.43T) + 8.314 (-1835+6.61T) = -41436.975 + 100.1T
=> \(\bigtriangleup\)H = -41436.975
\(2H_2+CH_3COOH=2CH_3OH\)(2)
\(CH_3OH+CO=CH_3COOH\)(3)
Lấy (2) + (3), ta có:
\(2H_2+CO=CH_3OH\)(1).
\(\Rightarrow\Delta G_1=\Delta G_2+\Delta G_3\)
\(\Leftrightarrow\)ln Kp1 = lnkp2 + lnkp3
\(\Leftrightarrow\)logkp1 = logkp2+ logkp3 = \(\frac{-4984}{T}+12,04\).
\(\Rightarrow\Delta G_1=\) -R.T.lnkp1=-R.ln10.(-4984+12,04.T)
Ta lại có:
\(\frac{\partial}{\partial T}\left(\frac{\Delta G}{T}\right)_p=-\frac{\Delta H}{T^2}\)\(\)
\(\Leftrightarrow\) \(\Delta H=T^2.\left(\frac{R.ln10.4984}{2.T^2}\right)\) = 47706,1.
Ta có: \(\Delta\)G= - RT ln\(K_p\)
=> \(\Delta\)\(G_2\) = - RT (\(\frac{-3149}{T}\) + 5.43) ln10
\(\Delta G_3\) = - RT (\(\frac{-1835}{T}\) + 6.61) ln10
\(\Delta G_1\) = \(\Delta G_2\) + \(\Delta G_3\) = -RT.ln10 (\(\frac{-4984}{T}\) + 12.04 )
mà: d(\(\frac{\Delta G}{T}\)) = -\(\frac{\Delta H}{T^2}\)dT
=> -R.ln10 (-4984.lnT) .dT = -\(\frac{\Delta H}{T^2}\)dT
=> \(\Delta H\) = R.ln10. (-4984). \(T^2\).lnT = -95,4.\(10^3\).T2.lnT