Which diagram shows the effect of catalysis on chemical equilibrium?

My course book has the diagram in figure (1) to represent the effect of a catalyst on chemical equilibrium. I drew the second diagram in figure (2).


[![effect of catalyst on chemical equilibrium](https://i.stack.imgur.com/gKwCs.png)](https://i.stack.imgur.com/gKwCs.png)


When reviewing the diagram in figure (1), I observe the following mistakes:


1. The rate of catalyzed and uncatalyzed forward reaction at equilibrium will be the same, which seems to be wrong because the value of rate constant of catalyzed forward reaction ($k\_f$ catalyzed) more than the value of rate constant of uncatalyzed forward reaction ($k\_f$ uncatalyzed), so the rate of catalyzed forward reaction at equilibrium = ($k\_f$ catalyzed) $[A][B] > $ ($k\_f$ uncatalyzed) $[A][B]$, so it isn't the same as the diagram.
2. The rate of catalyzed forward reaction before equilibrium less than the rate of uncatalyzed forward reaction before equilibrium, which seems to be wrong too.


I need feed back about the diagram or both are wrong and about my reasoning.




If (as suggested in the comments) the y-axis is supposed to represent concentration, such that the upper curve is "concentration of $\mathrm{A+B}$" and the lower curve is "concentration of $\mathrm{C+D}$", then Figure 1 is correct, albeit horribly mislabeled.


Your Figure 2 is correct if we want the y-axis to represent a rate-of-conversion instead, i.e., if the upper curve is "number of parcels of $\mathrm{A+B}$ being converted to $\mathrm{C+D}$ per second", and the lower curve is vice-versa. As you point out, a catalyst *increases* the conversion rate, and does so in both directions. So, exactly as you have drawn in Figure 2, those catalyzed rate curves should both be *higher* than their uncatalyzed siblings.


("Rate of reaction" is an awfully ambiguous term here – since at equilibrium, the rate of reaction must be zero!)


 1

To know the exact diagram represent the effect of catalysis on chemical equilibrium, I searched on experimental data but I can't find suitable data.
 so I consider a hypothetical reversible first -order reaction in each direction where 
$$\ce{ A <=> B}$$
I choose those two equations for reversible first- order reaction to calculate the concentrations of product and reactant at a different time:
$$\ce{[A]}=\frac{k\_b +K\_fe^{-(k\_f+k\_b)t} }{k\_f+k\_b}{[A]\_0}$$ 


$$\ce{[B]}=\frac{k\_f -K\_fe^{-(k\_f+k\_b)t} }{k\_f+k\_b}{[A]\_0}$$


Starting with A at a concentration$[A]\_0={0.6}$ and the values of the rate constants of uncatalayzed reaction are $k\_f=3s^{-1}$ and $k\_b=1s^{-1}$.
and the values of the rate constants of catalayzed reaction are $k\_f=6s^{-1}$ and $k\_b=2s^{-1}$


I did the calculation using excel sheet and draw the graph of concentration Vs time :
[![enter image description here](https://i.stack.imgur.com/IGfrO.png)](https://i.stack.imgur.com/IGfrO.png)


Depending on the rate law expression$(r\_f=k\_f [A])\ and\ (\ r\_b =k\_b [B])$ I did the calculation using excel sheet and draw the graph of rate Vs time


[![enter image description here](https://i.stack.imgur.com/GPglg.png)](https://i.stack.imgur.com/GPglg.png)


 1

Figure 1 shows the rate as $k\_f [A][B]$ whereas Figure 2 shows only $k\_f$. This is why Figure 1 has the equilibrium at the same rate for the 4 reactions.


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