5
$\begingroup$

I have looked for quite a long time on the site however all the answers involve mathematical explanations involving Arrhenius's equation. I would really like an intuitive reason (or one using collision theory?) as to why adding a catalyst should increase both reactions equally.

It seems that if you add a catalyst it should decrease the activation energy by a constant for both forward and reverse reactions. Thus it should decrease the forward activation energy proportionately more than the reverse reaction then the forward reaction should go forward, and shift equilibrium. But it doesn't. Why not?

enter image description here

$\endgroup$
  • 2
    $\begingroup$ Well, if the catalyst shifted equilibrium, you would be able to make a perpetual motion machine out of that. As to me, that's a pretty solid reason to review one's assumptions. $\endgroup$ – Ivan Neretin Mar 20 at 13:32
  • $\begingroup$ What do you mean by " increase both reactions equally": the rates? or the energy barriers? $\endgroup$ – Buck Thorn Mar 20 at 19:12
  • $\begingroup$ The equilibrium constant is fundamentally set by the energies of the reactant and product (and not rate constants) so the reaction pathway cannot matter to the equilibrium, just how fast it is reached. A catalyst causes the reaction to follow a different and faster pathway. As shown in the answer by @Andrew the change in activation cause by the catalyst energy cancels out. $\endgroup$ – porphyrin Mar 21 at 8:35
4
$\begingroup$

I don't know if this qualifies as an intuition answer or not, but it's very easy to show mathematically. The key is that the rate constant is not linearly proportional to the activation energy. By the Arrhenius equation, $k = Ae^{-E_\mathrm{A}/(RT)}$, so the rate constant is dependent on an exponential of the activation energy. The equilibrium constant $K$ is a function of the ratio of the forward and reverse rate constants. For a simple reaction $\ce{A <=>[k_1][k_{-1}] B}$, the equilibrium constant $K=\frac{k_1}{k_{-1}}$.

If we assign the activation energy $E_\mathrm{A,fwd}$ to the forward reaction and $E_\mathrm{A,rev}$ to the reverse, we have

$$k_1 = Ae^{-E_\mathrm{A,fwd}/(RT)} \quad \text{and} \quad k_{-1} = Ae^{-E_\mathrm{A,rev}/(RT)},$$

so

$$K=\frac{Ae^{-E_\mathrm{A,fwd}/(RT)}}{Ae^{-E_\mathrm{A,rev}/(RT)}}.$$

Using simple rules of exponents, this reduces to

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}}.$$

Now if we subtract a quantity $E_\mathrm{cat}$ from both activation energies, the same derivation gives

$$K = e^{E_\mathrm{A,rev} - E_\mathrm{cat} - (E_\mathrm{A,fwd} - E_\mathrm{cat})} = e^{E_\mathrm{A,rev} - E_\mathrm{A,fwd}},$$

so $K$ is unchanged.

Intuitively, it may help to think about the fact that the reverse reaction is necessarily composed of exactly the same microscopic states as the forward reaction, just happening in reverse order (by the principle of microscopic reversibility). The catalyst can stabilize one of more of the states on that reaction pathway, but cannot influence what direction that stabilized state reacts in. So making a microscopic state more easily achievable in the forward direction necessarily makes it more easily achievable in the reverse direction.

$\endgroup$
1
$\begingroup$

You are asking a quantitative question, but you are hoping for an answer besides the "mathematical explanations" already on the site. That might be impossible, but here is an attempt to talk about it without explicitly using the Arrhenius relationship.

It seems that if you add a catalyst it should decrease the activation energy by a constant for both forward and reverse reactions.

If you clarify this a bit and say the difference in activation energy with or without the catalyst is the same for the forward and the reverse reaction, then this is correct.

Thus it should decrease the forward activation energy proportionately more than the reverse reaction then the forward reaction should go forward, and shift equilibrium.

This contains a correct statement and and incorrect conclusion. It is true that when I subtract some value from a larger value, and then subtract the same value from a smaller value, the smaller value changes by a larger factor. So the first part of that statement is correct.

However, that does not mean that the rate constant changes by a larger factor (which is what the second part of your statement implies). The change in rate constant depends on the difference between activation energies (catalyzed or not), not the quotient of activation energies. Stating the Arrhenius relation in words, the rate constant changes by a factor that is proportional to the difference in activation energies. So forward and reverse rate constants change by the same factor, meaning that if the forward and reverse rates were the same without the catalyst, they are the same with the catalyst as well.

I would really like an intuitive reason (or one using collision theory?) as to why adding a catalyst should increase both reactions equally.

Collision theory is good to explain the concentration dependence of rates. It is less useful for explaining what fraction of the collisions leads to a reaction. It is easy to explain why a catalyst would be helpful for reactions in both directions. It is hard to make quantitative predictions based on a non-quantitative model. Even in collision theory, you have to know the distribution of the energies of collisions to determine how the fraction of successful collisions changes when you lower the activation energy by a certain amount. If you know the distribution and do the analysis, you arrive at the Arrhenius relationship.

Why not?

In his comments, Ivan Neretin had an excellent argument about how our world would change if catalysts could change the equilibrium constant. His argument works like a mathematician's proof of contradiction: It has to be this way because if it were any different, it would contradict something else we know (well, if we are talking about math, contradict an axiom we would like to keep). So here is my version of the contradiction: We could stick in catalyst into the reaction mix, the reaction would go in one direction until it reaches equilibrium, we would pull it out again, the reaction would go in the other direction to reach equilibrium. Imagine these are reactions in a battery. Batteries go toward equilibrium when we use them as a source of energy, and to charge them going away from equilibrium, it takes work. In our hypothetical scenario where catalysts change equilibrium constants, we would never have to charge the battery again. That is not our collective experience.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.