Consider the unimolecular elementary reversible reaction: A B, with forward rate

Question

Consider the unimolecular elementary reversible reaction: A B, with forward rate constant K, and backward rate constant K - 1. The values of the rate constants are K_1 = 5.00 times 10^-3 s^-1 and K - 1 = 4.7 times 10^-4 s^-1 respectively. The concentration of A and B at times zero are 0.160M and OM, respectively. (a) Compute the equilibrium constant. (b) Compute the equilibrium Concentration of A and B. (C) Compute the relaxation time constant. (d) Compute the concentrations of A and B at = 150, and 900s.

Explanation / Answer

For a 1st order reverisble reactions

-ln(1-XA/XAe)= {(M+1)/(M+XAe)}*K1t

XA= conversion at anytime

M= CBO/CAO= ratio of intial concentrations of B and A= 0 ( for CBO=0)

K1= 5*10-3/sec

XAe= equilibrium conversion. At equilibrium K1CAO*(1-XAe)= K2(CBO+CAO*XAe)

Since CBO=0

K1*(1-XAe)= K2*XAe, XAe= K1/(K1+K2)= 5*10-3/(5*10-3+7*10-4) = 0.877

Hence –ln(1-XA/0.877)=(1/0.877)*5*10-3*t =0.005701*t

At t=150 sec, -ln(1-XA/0.877)= 0.005701*150=0.855

Ln(1-XA/0.877)= -0.855, 1-XA/0.877= 0.425, XA/0.877= 1-0.425=0.575

XA=0.575*0.877=0.5043

CA= CAO*(1-XA)=0.160*(1-0.5043)= 0.079, CA= CBO+CAOXA= 0+0.160*0.5043=0.081

At 900 dec, -ln(1-XA/0.877)= 0.005701*900 = 5.13, 1-XA/0.877= 0.006

XA= 0.877*(1-0.006)= 0.872, at 900 sec, CA= 0.16*(1-0.872) =0.020, CB= 0.16*0.872=0.14

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