Question

Consider an irreversible 2nd-order reaction (A + B → C) in a batch reactor with the initial molar ratio of ΘB=CB0/CA0. Find the relation between time (t) and conversion (XA). (Hint: You need to consider cases of both CA=CB and CA≠CB.)

Answer 1

Irreversible second order reaction in batch reactor.

A + B --> C

Initial concentration of A =  Cao

Initial concentration of B = Cbo

Let us assume elementary reaction,

Rate of reaction of A (-rA) = kCaCb

-rA = kCaCb

-dCa/dt = kCaCb

Ca = Cao(1- Xa)

Cb = Cbo - CaoXa = Cao(Cbo/Cao - Xa) = Cao(M - Xa)

Case1 : Cao =/= Cbo, Taken M = Cbo/Cao

-dCao(1-Xa)/dt = kCao(1-Xa)Cao(M - Xa)

CaodXa/dt = kCao^2(1-Xa)(M - Xa)

dXa/dt = kCao(1-Xa)(M - Xa)

Taking partial fraction,

1/(1-Xa)(M-Xa) = A/(1-Xa) + B/(M - Xa)

1 = A(M - Xa) + B(1- Xa)

Put M = Xa,

B = 1/(1-M)

Put Xa = 1,

A = 1/(M-1)

1/(1-Xa)(M-Xa) = 1/(M - 1) * [ 1/(1-Xa) - 1/(M - Xa) ]

kCao*(M-1) *t =[ - ln(1-Xa) + ln(M - Xa) ]0Xa

ktCao(M-1) = ln[(M-Xa)/(1-Xa)] - lnM

ktCao(M-1) = ln[(M-Xa) /M(1-Xa) ]

this is the relation between Xa vs t.  

case 2: Cao = Cbo , M = 1

dXa/dt = kCao(1-Xa)(1- Xa)

ktCao = [ 1/(1-Xa)]0Xa

ktCao = [ 1/(1- Xa) - 1]

ktCao = Xa/(1-Xa)

when Cao = Cbo then,

ktCao = Xa/(1-Xa)

this is the relation between t vs Xa.