How do catalysts provide alternative routes for reactions?

Question

According to collision theory, a reaction will take place between, say, two molecules, if the collision between the atoms has a sufficiently high kinetic energy (to meet the activation energy threshold) and have the correct orientation. If these two conditions are not met, the collision will not result in a reaction.

However, there is one way to make a reaction easier, and that's with the use of a catalyst. Catalysts do not lower the activation energy or increase the kinetic energy, as I've just learned trying to figure out this question, but they provide an alternative activation energy, lower than the initial one, that, if met, will allow the reaction to take place.

My question is this: what is the actual chemistry behind how catalysts provide this alternative route? What's happening with the molecules or atoms? How does it work?

For example, if I wrap zinc in oxidized copper wire and mix it with some hydrochloric acid, the oxidized copper will act as a catalyst for the reaction between zinc and hydrochloric acid. I'm not sure how this takes place, though.

Thanks.

Answer 1

So catalysts provide an alternative pathway by allowing a new intermediate to be formed. For example consider the reaction of $A$ and $B$ to form $C$. this means that $$\ce{A + B -> C}$$ Say this reaction has an activation energy of $x$ $kJmol^{-1}$.

This reaction could also be possible with a catalyst $X$. This means that a new possible route for the reaction could be (there a multiple routes the new reaction could take this is just a simple example): $$\ce{A + X->AX}$$ $$\ce{AX + B->C +X}$$ Now you have ended up in the same position using a catalyst and when not using a catalyst, seen as you add both reaction and cancel out common species then you get back to the equation of: $$\ce{A + B -> C}$$ These two reactions with the catalyst will have their own activation energies. Say these two reactions have activation energies $\pu{p}$ and $\pu{q}$ $\pu{kJmol^{-1}}$ and that $\pu{p>q}$ (doesn't matter if $\pu{p}$ applies to the first or second reaction) then if $x>p$ a particle needs less energy to proceed via the route with the catalyst compared to the route without the catalyst.

As a particle now needs less energy for the product $\pu{C}$ to be formed, more particles will now be able to take part in the reaction thus increasing the amount of successful collisions per second which is the same in an increase in rate. This can be seen nicely in a general Maxwell-Boltzmann distribution of molecular energies as the new route will mean there's a lower activation energy overall meaning a lot more particles can react.