Questions about equilibrium

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I just had a few questions about equilibrium, they're not really about how to answer the questions but more about intuition, anyway here are the questions.

1. Why don't catalysts influence equilibrium when temperature does. Even though people say catalysts lowers the activation energy equally doesn't a change in temperature also affect the proportion of successful collisions equally too since exothermic reaction still need an input of activation energy to undergo their reactions.

If it is because that a higher proportion of the reactants undergoing the endothermic reaction gain enough kinetic energy to undergo the reaction than the reactants on the exothermic side since it is likely that the exothermic have already reached the required energy due to lower activation energy then doesn't the same principle apply for catalysts? where decreasing the activation equally still proportionately results in more endothermic reactants gaining the required energy than the exothermic reactants.

2. Why does la chateliers principle hold? The Heinemann textbook and online resources just say what it is but its kind of like they want us to blindly follow this principle of 'opposing the change' without actually knowing why.

Thanks a lot appreciate any replies.

I'm actually going to start by answering 2, because it's easier that way. Let's work with the simple chemical equation:

So, we have some chemical, A, and it turns into another chemical, B. For simplicity let's assume this equation is at equilibrium, with concentrations of 0.5 M and 0.5 M - which means our equilibrium constant is K=1. For this explanation to work, you need to accept that all reactions have some K value they want to return to. The theoretical explanation for as to why requires you understand chemical kinetics and how reaction rates are calculated - if you think this is something you want to know for understanding purposes, I can do so, but it is not required at all. Instead, just know that you can prove experimentally that, over time, the backwards reaction rate will eventually equal the forwards reaction rate, and that point happens to be at equilibrium - when K equals some special number.

So, we've accepted that this equation is going to constantly change such that K=1. Well, I'm going to add enough of B to make the concentration of B now 1.0 M. Our new Q value (that is, the reaction quotient when the reaction is NOT at equilibrium) is:

Well, this is far bigger than 1! So, we want to decrease the value of Q - how do we decrease a number? We can either lower the numerator, or raise the denominator. So, I (the reaction) am going to lower the numerator by getting rid of B - but to get rid of B, I -HAVE- to make more A, because that's the only way to remove B from the equation as a chemical reaction. So, if I reduce the concentration of B to 0.9 M, then the concentration of A must also raise by 0.1 M, and so now Q is:

Well, that's a little better! But we need to remove more of B. Now, the reaction will never make so much A all at once such that we have MORE A than B, because it will reach equilibrium first. But, for the sake of learning, let's pretend that the reaction removes 0.2 M of B (again, by reacting B, so now we have another 0.2 M of A), this would now give us a Q value of:

Well, this is now LOWER than 1! Okay, so how do we increase a number? Raise the numerator, or reduce the denominator. So, let's reduce B by making more A... And we can do this forever, but eventually (and you may have guessed this) we're going to end up with 0.75 M of A and B, and we'll be back at equilibrium.

Okay, so what does this have to do with le Chatelier? Well, notice that by increasing the products, to get back to equilibrium, we had to "partially oppose" the reaction by making more reactants - and then, when we had more reactants, we had to "partially oppose" the reaction by making more products. All to get back to that magic K value. THAT is why le Chatelier's principle works - all it does is explain what you have to do to the chemicals to make the maths work. You can through similar explanations for everything else, not just adding more product/reactants, to explain changes in pressure, volume, etc. Let me know if this explanation didn't work for you (/what bits don't work), and I can try to explain in more detail.

So, why don't catalysts affect the constant? Well, look at the above - at no point did the rate of the reaction affect affect the maths above. The rate will certainly affect how fast the reaction reaches equilibrium, but a catalyst will increase the backwards reaction AS WELL AS the forwards reaction. So, why would the catalyst affect the equilibrium constant? It shouldn't at all!

Okay, so if catalysts don't affect equilibrium, why does temperature? If rate doesn't affect equilibrium, what does temperature do that ISN'T change the rate of reaction? Well, there's the basic explanation of, "well, if you imagine that heat is a product or reactant, then in an exothermic reaction, heat is a product, and in an endothermic reaction, heat is a reactant" - which, if that works for you, is great. No need to complicate things unnecessarily. But, if it doesn't work for you, you're right in being a little bit hesitant - heat isn't a chemical species, it's not in our equilibrium equation, so why does heat play a part?

Well, this is beyond VCE, but it all comes down to thermodynamics. Let's call the amount of energy in a system "G" (for "Gibbs' free energy". The reason is because some guy with the last name Gibbs came up with this stuff). The value of delta G depends on two different types of energy - the amount of energy from enthalpy, or heat, of the chemicals in question (the delta H value), as well as the amount of energy from the way in which the molecules arrange themselves (known as entropy, or the delta S value). Well, the Gibbs' free energy can be calculated like so:

Notice that delta G changes with temperature? Well, this isn't the ONLY way to calculate delta G. We can also use this equation:

You may not know what ln is - it's just a logarithm. The logarithms you're familiar with from calculating pH are logarithms to the base 10, this one is to the base e, which is approximately the base 2.7. Don't think too hard about it - it's just a fancy maths thing. All of these you already know - R is just the gas constant, T is temperature, K is the equilibrium constant.

So, from this equation, it isn't obvious that temperature also affects the equilibrium. But, what if you substitute the two above equations into each other? You get:

So from this equation, you can see that K depends on temperature! It's really obvious now - if you increase temperature, then K HAS to change - but whether it goes up or down isn't quite obvious... So, let's work at this step-by-step so focusing on the delta H bit. If T increases, then:

will get smaller. If T decreases, then A will get bigger. Okay, what about the following equation:

Well, now we need to know what delta H does. Firstly, if B is positive (and A is negative), then the equilibrium constant will get bigger - otherwise, it will get smaller. This is because:

So, let's say we have an endothermic reaction. Then, if you increase the temperature, A decreases, which means B will get LESS negative and MORE positive - so decreasing the temperature in an endothermic reaction causes the value of K to get bigger, so we need to make more reactants to make Q smaller. If you decrease the temperature, A increases, which means B will get MORE negative and LESS positive - so increasing the temperature in an endothermic reactions causes the value of K to get smaller, so we need to make more reactants to make Q bigger.

Exothermic reactions obey the same logic - however, since there's a negative sign outside the delta H, the two cancel out, and so increasing temperature will still cause B to get closer to 0 - but now it will become more negative instead of less negative.

And if I lost you in all that maths, all you need to know is this: temperature changes because it affects the amount of energy that single chemicals hold, it doesn't JUST change how fast they move around. (that is, it increases energy that isn't just the kinetic energy) A catalyst doesn't affect energy at all, and so the equilibrium constant doesn't change - in fact, a catalyst doesn't change the kinetic energy. In hindsight, maybe all I needed to explain to you was that a catalyst doesn't affect the kinetic energy of a system, just the pathway a system takes to get to the products, whereas temperature changes do change the energy, and the equilibrium constant is negatively proportional to energy... But oh well, you got the full thermodynamics explanation as well

So, the elephant in the room - why should you trust those equations? Well, that requires a full-on course in thermodynamics, so I can't exactly explain that to you in a short period of time... So, you kinda just have to trust those equations are true. If that explanation isn't good enough for you, then I have good news - you'd make a great scientist! Google is your friend to learn more.