Question -4 following gas phase reaction takes place at

Question

Reaction engineering problems-

Question -4 following gas phase reaction takes place at 500^oc 4PH₃ (g) ----( P₄ (g) +6H₂ (g) With rPH = 85h^-1*CPH₃ What is the volume of the plug flow reactor operating at this temperature and 5.5 atm, giving 80% conversion for a feed of 2.5 kg-mole PH₃ /hr ?

Solution - Let xa = conversion of PH₃ For gas phase reaction - ?= moles products - moles reactants / moles reactants = 7-4/4 = 0.75 let Ca = concentration at any instant initial concentration, Cao = P/R.T = 5.5/0.08205 *(500+273) = 5.5/0.082*773 = =0.0867 kg-mole/m³ Ca/Cao = 1-xa/1+?.xa = 1-xa/1+0.75*xa Ca = Cao [1-xa/1+0.75xa] R =r = K.Ca = K.Cao [1-xa/1+0.75xa]

F= Feed rate = 2.5 kg-mole /hr V= volume of the reactor V/F =?dxa/r Volume of the reactor, V= F?( limit 0 to xa).dxa/r = F/K.Cao ?(limit 0 to xa).(1+0.75 xa/1-xa).dxa = F/K.Cao*[(1+0.75*ln1/(1-xa) -0.75.xa]( limit 0 To 0.8) = 2.5/85*0.0867 *[ (1+0.75)*ln 1/0.2 -- 0.75*0.8] = 0.75 m³ Reaction engineering problems-

Question -4 following gas phase reaction takes place at 500^oc 4PH₃

(g) ----( P₄ (g) +6H₂ (g) With rPH = 85h^-1*CPH₃ What is the volume of the plug flow reactor operating at this temperature and 5.5 atm, giving 80% conversion for a feed of 2.5 kg-mole PH₃ /hr ? Solution - Let xa = conversion of PH₃ For gas phase reaction - ?= moles products - moles reactants / moles reactants = 7-4/4 = 0.75

let Ca = concentration at any instant initial concentration, Cao = P/R.T = 5.5/0.08205 *(500+273) = 5.5/0.082*773 = =0.0867 kg-mole/m³ Ca/Cao = 1-xa/1+?.xa = 1-xa/1+0.75*xa Ca = Cao [1-xa/1+0.75xa] R =r = K.Ca = K.Cao [1-xa/1+0.75xa] F= Feed rate = 2.5 kg-mole /hr V= volume of the reactor V/F =?dxa/r Volume of the...