Question

The elementary liquid-phase series reaction
A ----k1------> B ----k2-----> C
is carried out in a 500-dm^3 batch reactor. The initial concentration of A is 1.6 mol/dm^3. The desired product is B, and separation of the undesired product C is very difficult and costly. Because the reaction is carried out at a relatively high temperature, the reaction is easily quenched.
(a) Plot and analyze the concentrations of A, B, and C as a function of time. Assume that each reaction is irreversible, with k1 = 0.4 h^-1 and k2 = 0.01 h^-1.

Answer 1

Answer:

Concentration of A: $\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Concentration of B: $\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t } -e^{-k_{2}t } )$

Concentration of C: $\frac{C_{C} }{C_{Ao} } =1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2}t } -\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1} t}$

the image shows the graphs of the three concentrations

Explanation:

We have the reaction:

A ------->k1--------->B------------->k2--------->C

Each reaction:

$r_{A} =-k_{1} C_{A} \\r_{B} =k_{1} C_{A} -k_{2} C_{B} \\r_{C} =k_{2} C_{C}$

Where Cn is the concentration of each specie (A,B,C)

The mass balance for A:

$-\frac{dC_{A} }{dt} =-r_{A} \\-\frac{dC_{A} }{dt}=k_{1} C_{A} \\-\int\limits^y_x {\frac{dC_{A} }{dt} } \,=k_{1} t\\\frac{C_{A} }{C_{Ao} } =e^{-k_{1}t }$

Where x=CAo and y=CA

The mass balance for B:

$-\frac{dC_{B} }{dt} =-r_{B} \\-\frac{dC_{B} }{dt}=k_{2} C_{B} -k_{1} C_{A} \\\frac{dC_{B} }{dt}+k_{2} C_{B}=k_{1} C_{A}\\\frac{C_{B} }{C_{Ao} } =\frac{k_{1} }{k_{2}-k_{1} } (e^{-k_{1}t }-ex^{-k_{2}t } )$

The mass balance for C:

$\frac{C_{C} }{C_{Ao} } =1-\frac{C_{A} }{C_{Ao} } -\frac{C_{B} }{C_{Ao} } \\\frac{C_{C} }{C_{Ao} }=1+\frac{k_{1} }{k_{2}-k_{1} } e^{-k_{2} t}-\frac{k_{2} }{k_{2}-k_{1} } e^{-k_{1}t }$

The maximum concentration of C is:

$C_{Cmax} =C_{Ao} (\frac{k_{2} }{k_{1} } )^{\frac{k_{2} }{k_{2}-k_{1} }} =1.6(\frac{0.01}{0.4} )^{\frac{0.01}{0.01-0.4} } =1.76mol/dm^{3}$

and the maximum time is:

$t_{max} =\frac{ln\frac{k_{2} }{k_{1} } }{k_{2}-k_{1} } =\frac{ln\frac{0.01}{0.4} }{0.01-0.4} =9.4 h$