Consider a reaction in which 2 molecules of A react with B to form a trimeric co

Question

Consider a reaction in which 2 molecules of A react with B to form a trimeric complex A_2B. Suppose the rate of this reaction is first order with respect to A and B. Assume the volume is constant and the reagents are provided initially in stoichiometric amounts. No A_2B is present initially. Develop an equation for the rate of change in the extent of reaction, zeta. Put the differential equation in a form amenable to integration, i.e., zeta should be the only dependent variable in the differential equation. You need not solve the differential equation. Develop an equation for the rate of change in the concentration of A. Again, put this equation in a form that can be integrated. Show your work and not just the final answer. Derive the dependence of concentration of A on time. That is, determine A(t). Suppose the initial concentration of A_2B is not zero, but equal to the amount of A that was initially present. How does this affect your answer to 4c? Explain

Explanation / Answer

The reaction is 2A+B--àA2B

The rate of reaction is –rA= K[A] [B]

Let XA= conversion, XA= 1-CA/CO

CAO*2XA= CBO*XB, where XB= Conversion with respect to B

-rA= K1CAO*(1-XA) *(CBO-CBOXB)

Hence –dCA/dt = K1*CAO*(1-XA)*(CBO-CAO2XA)

=K1CAO2*(1-XA)*(M-2XA), where M= CBO/CAO

But -dCA/dt= K1CAO2*(1-XA)*(M-2XA)

But CA= CAO*(1-XA), -dCA/dt= -CAO*XA

Hence –d(CAO*(1-XA)/dt= K1CAO2*(1-XA)*(M-2XA)

dXA/dt= K1CAO*(1-XA)*(M-2XA)

dXA/ {(1-XA)*(M-2XA)}= K1CAO*dt

when this equation is integrated

ln [(M-2XA)/{M*(1-XA)]=CAO*(M-2)*Kt,

when there is no A -dCA/2dt = dA2B/dt

when A2B is initially present

[A2B]= [A2B]0+ CAO*2XA =CAO*(1+2XA)

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