The first order reversible liquid reaction takes place in a batch reactor at 338

Question

The first order reversible liquid reaction takes place in a batch reactor at 338 K with initial reactant and product concentrations C_A0 = 0.5 mol/litre and C_R0 = 0. After 8 minutes, conversion of A is 33.3%. It is found that equilibrium conversion rate X_Ae (338 K) = 66.7%. (a) Find the rate equation for this reaction. (b) If the reaction is carried out at 298K, it is found that equilibrium conversion rate X_Ae (298 K) = 90%, and after 8 minutes, the conversion of A is 20%. Calculate the activation energy values for forward and reverse reactions.

Explanation / Answer

For the reaction A<-> B, rate= -rA

-rA= K1CA-K2CB= K1*CAO*(1-XA)- K2(CBO+CAOXA), XA= conversion , CA= CAO*(1-XA)

=K1CAO*(1-XA)- K2CAO*(CBO/CAO+XA)

=K1CAO*(1-XA)- K2CAO*(M+XA), M= CBO/CAO= 0

=K1CAO*(1-XA)- K2CAO(M+XA) (1)

At equilibrium, the rate of forward reaction and back ward reaction are same.

Hence K1CAO*(1-XAe)= K2CAO*(M+XAe), XAe= equilibrium conversion

K1*(1-XAe)= K2*(M+XAe)

K2= K1*(1-XAe)/ (M+XAe)

Substituting the value of K2 in Eq.1,

K1CAO*(1-XA)- K1*CAO*(M+XA) *(1-XAe)/ (M+XAe)

K1CAO*{(1-XA)- (M+XA)*(1-XAe)/ (M+XAe}

=K1CAO*{(1-XA)*(M+XAe)- (M+XA)*(1-XAe)}/(M+XAe)

=K1CAO { M+XAe-MXA-XAXAe- ( M+ XA-MXAe-XAXAe)}/ (M+XAe)

=K1CAO*{ XAe-XA+M*(XAe-XA)}/ (M+XAe)

=K1CAO( M+1)*(XAe-XA)/ (M+XAe)

Hence –dCA/dt= K1CAO*(M+1)*(XAe-XA)/ (M+XAe)

-dCAO*(1-XA)/dt= K1CAO*(M+1)*(XAe-XA)/ (M+XAe)

dXA/dt= K1*(M+1)*(XAe-XA)/ (M+XAe)

dXA/(XAe-XA)= K1*(M+1)/(M+XAe)dt

when integrated noting that XA=0 at t= 0

-ln{ 1-XA/XAe)= K1*(M+1)/ (M+XAe)*t

Given M= CBO/CAO=0 and XA=0.333 at t= 8 min, XAe= 0.667 at 338K

Hence –ln(1-0.333/0.667)= (K1/0.667)*8

K1= 0.0576/min

But K1 and K2 are related as K2= K1*(1-XAe)/(M+XAe)= 0.0576*(1-0.667)/ 0.667= 0.0288

Hence the rate expression becomes –rA= 0.0576CA-0.0288 CB

At 298 K, XAe=0.9 and XA=0.2

-ln{ 1-XA/XAe)= K1*(M+1)/ (M+XAe)*t becomes

-ln(1-0.2/0.9)= K1*8/0.9= K1*8.9

K1= .25131/8.9= 0.028

K2=K1*(1-XAe)/(M+XAe) =0.028*(1-0.9)/0.9=0.00311

Rate at 298 K, -rA= 0.028CA-0.00311CB

Activation energy for the forward reaction can be calculated from Arhenius equation

Ln(K2/K1)= (Ea/R)*(1/T1-1/T2), Ea= activation energy for the given reaction R=8.314 J/mole.K

K2= rate constant at the given temperature and K1= rate cosntant at some other temperature.

For the forward reaction , the equation becomes

Ln(K338/K298)= (Ef/R)*(1/298-1/338)

Ln(0.0576/0.028)= (Ef/8.314)*0.000397

Ef= 14534 J/mole= 14.534 Kj/mole

For the reverse reaction, ln(0.0288/0.00311)= (Er/8.314)*0.000397

Er= 46597 J/mole= 46.597 Kj/mole

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