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).



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


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] &gt; $ ($k_f$ uncatalyzed) $[A][B]$, so  it isn't the same as the diagram.
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!)