Conversion and reactor time [closed]








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Lets say it takes my batch reactor 3 hours for 35% conversion. 


If I wanted to find how long it would take for 15% then,


35-15 = 20,


then 3 hours/0.20 = 0.15 hours for 15% conversion.


I am not too sure if this is correct. Some guidance would be greatly appreciated




The result very depends on the reaction stoichiometry, kinetic order and the ratio of reagents, as "cheaper" reagents are often in excess.


Let suppose the conversion equation is $\ce{A + B -> C}$ and the reaction kinetics is of the second order.


Then 
$$\frac{\mathrm{d}[C]}{\mathrm{d}t}=k \cdot \ce{[A][B]}$$
$$\frac{\mathrm{d}[A]}{\mathrm{d}t}=-k \cdot \ce{[A][B]}$$
$$\frac{\mathrm{d}[B]}{\mathrm{d}t}=-k \cdot \ce{[A][B]}$$


If the stoichiometric ration is managed, then \ce{[A] = [B]}


$$\frac{\mathrm{d}[C]}{\mathrm{d}t}=k \cdot \ce{[A]}^2$$
$$\frac{\mathrm{d}[A]}{\mathrm{d}t}=-k \cdot \ce{[A]}^2$$


If we integrate the function over the time:
$$\int\_{a\_0}^{a}{\frac {\mathrm{d}[A]}{ -k \cdot \ce{[A]}^2 }}=\int\_0^{T}{ \mathrm{d}t}$$
$$\frac {1}{k} \cdot [\frac 1A - \frac 1{a\_0}]=T$$
$$\frac 1A =kT + \frac 1{a\_0}$$
$$A =\frac 1{kT + \frac 1{a\_0}}$$


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