# 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:

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.

• OK, I'm bailing on the full answer. I'm 99% sure there is a mistake in the figure you were given and that the y axis should really be concentration, not rate (it was actually Mithoron who figured this out when I brought it up in chat after I found myself stumped). If this is the case, then figure 1 makes sense. To answer your questions in the comment above: the rate of catalyzed forward rxn is always greater than the rate of the uncatalyzed forward rxn. Of course the same goes for the rates of the reverse rxns (catalyzed is faster). Sorry I can't answer your full Q more directly. Oct 30, 2017 at 1:35
• @airhuff thank you , 1- I gave figure1 from course book and from internet, which i consider it wrong because it has the equilibrium at the same rate for the 4 reactions. Oct 30, 2017 at 5:15
• @airhuff 2- what about the answer of the second Q in the comment . Does the rate of catalyzed forward reaction before equilibrium will be less or more than the rate of uncatalyzed forward ? Oct 30, 2017 at 5:18
• The catalyzed forward reaction rate before equilibrium will be greater than that of the uncatalyzed forward reaction. This is how the catalyzed reaction reaches equilibrium faster than the uncatalyzed reaction. Oct 30, 2017 at 6:09
• Can you please include the source of the original figure? Oct 31, 2017 at 5:57

## 3 Answers

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!)

• That can't be right. A catalyst does not change the location of the equilibrium. By definition, the rate of the forward reaction has to equal the rate of the reverse reaction. Oct 31, 2017 at 5:04
• Yes, but the catalyst here (in my reading of Figure 2) isn't changing the position of the equilibrium, right? The catalyst increases the rate-of-conversion-of-A-to-C, and also increases the rate-of-conversion-of-C-to-A. Our uncatalyzed rates at equilibrium might be one-parcel-per-second, then with the catalyst they might be two-parcels-per-second.
– gone
Oct 31, 2017 at 5:45
• Yes, you're right. I was mixing rates with concentrations. This is confusing. Oct 31, 2017 at 5:49
• I think your second assumption is the one we need.
– Jan
Oct 31, 2017 at 6:15
• @Jan you mean Figure 2 is correct Oct 31, 2017 at 15:21

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 :

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

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.

• Figure 1 has the equilibrium at the same rate for the 4 reactions, and Figure 2 show the rate at equilibrium more when catalyzed.so :Which Figure shows the effect of catalysis on chemical equilibrium? Oct 28, 2017 at 18:57
• Can you please explain how" Figure 2 shows only $k_f$" Nov 2, 2017 at 23:28
• @AdnanAL-Amleh $k_f$ is how fast the reaction occurs. Therefore, the rate of reaction with a catalyst will always be faster than without a catalyst (curves 4 and 5 must be greater than curves 1 and 2 at all times). When you multiply $k_f$ with $[A][B]$ and call that the rate, then the figure looks like Figure 1.
– LDC3
Nov 4, 2017 at 1:41
• As you said curve 4 must be greater than curve 1 at all times,can I conclude Fig1 is not correct , because it shows curve 4 less than curve 1. Nov 4, 2017 at 5:57
• @AdnanAL-Amleh Figure 1 is correct if you take the product $k_f [A][B]$, not just $k_f$ alone.
– LDC3
Nov 5, 2017 at 14:51