how to remove products from reaction space?(to apply LeChatelier's principle) I thought this question merited  a discussion of its own. In industrial applications, we talk of shifting equilibrium of reaction to the right(or desired products) by removing products quickly from reaction space? I want to know what this means? If you consider a simple fixed-bed/packed-bed reactor with gaseous reactant and product (say A-->B), does it mean I need to remove B quickly from the packed-bed reactor volume to push equilibrium to the right? My point is, isn't that already happening? By virtue of flow through the reactor isn't B already constantly being "removed" from the reactor(along with unreacted A), while the reactor keeps getting hit with fresh A? so what more effort is needed to push equilibrium to the right? are there any timescales involved? or does B have to be removed quicker than A? if so, how quicker? I need to understand this better,please help. This question can apply to other types of reactors as well like fluidized-bed or moving-bed and so on..
The scheme you ask about is effectively the same as $\ce{A &lt;=&gt;[k_1][k_{-1}] B -&gt;[k_2] C}$ where $k_2$ measures the effect of removing B making C the removed product. The actual reaction is much more complicated as presumably mixing/diffusion has to be taken into account. Leaving aside these complications  as $k_2$ increases then the amount of B in the reaction reduces and the amount of C increases if measured at a fixed time. As long as $k_2$ is not zero eventually all B is removed, you can see this in the figure which shows the amount of C for different rate constants $k_2$  assuming the initial amount of A = $1$ and $k_1= 1, \, k_{-1}=3$. When $k_2=0$ (top figure) no product is formed and A and B come to equilibrium. The lower figure show the effects on forming C by removing B.