Factors that affect the state of equilibrium: Changes in temperature

1. Factors that affect the state of equilibrium: Changes in temperature

2. We decided to throw in a quick review, because Temperature is where it gets a bit complicated. s.O.S

3. Reactions are reversible when they can go forwards or backwards without any extra energy being used. A reversible reaction is labeled with the two arrows in the above equation Review starts here

4. When the forward reaction (reactants forming products) and reverse reaction(products changing back into the reactants) have perfectly equal reaction rates, the reaction has reached equilibrium • Question: Is this reaction at Equilibrium? • Forward Rate: 1.79Reverse Rate: 1.77 What is equilibrium? Answer: No! The rates are not perfectly Equal!

5. Example: The reaction with Nitrogen and Hydrogen forming Ammonia, NEVER STOPS. It's like when you're trying to stand on the balance board in the gym: the board goes left, you have to go right, the board goes right, you have to go left, until you reach an equilibrium. However unlike us on a balance board, the reaction keeps on going, forever. It's just that we no longer notice changes occurring in the reaction. Help! I Still don’t understand equilibrium!

6. Here’s a visual example As the concentration of Hydrogen and Nitrogen drop, making them less likely to collide and keep reaction, therefore the rate of the forward reaction slows down!At the same time the concentration of Ammonia rises, more Ammonia molecules are available to break up into the reactant gases (hydrogen and Nitrogen), and the rate of the reverse reaction speeds up! Eventually they both happen at the same rate, where Hydrogen and Nitrogen form Ammonia, and Ammonia breaks up to form Hydrogen and Nitrogen at the same time.

7. A reaction will shift to the right if reactants are being added (meaning that products are being formed), and to the left if products are being added (meaning that reactants are being formed)Before we get into the example, think of a See-saw How does shifting work

8. So if we have N2(g) + 3H2(g) <=> 2NH3(g) • If we add N2(g) or H2(g) the reaction will shift to the right, because we are adding reactants, which results in the formation of products • If we add NH3(g) the reaction will shift to the left, because we are adding products, which means more reactants are being formed as well. Example of adding products/reactants Kind of like a See-Saw right? The more of something you add to one side, the more the other side produces, therefore it will shift to the other side!

9. Now, what happens if we decide to remove instead of add? Well it's simple • If we remove N2(g) or H2(g), the reverse reaction will have to try and produce more of these reactants, so in order to balance out it'll shift to the left (towards the reactants). • If we remove NH3(g), the product, the direct reaction will have to try and produce more product in order to bring the reaction back to equilibrium, therefore shifting it to the right (Towards the products). WAIT! What if we remove instead of add?

10. This is all summarized with Le Chatelier's principle which states: If stress is placed on a system at Equilibrium, the system will proceed in a direction that minimizes the stress. So in ‘non-scientific-ky’ language, if you try and alter a reaction at equilibrium, the reaction will keep trying to balance itself out, eventually resulting in equilibrium again. Le Chatelier

11. Back to the previous reaction we had going for us, the Haber process, (Don't worry Haber was just a guy who managed to use high pressure and a catalyst to directly react nitrogen gas and hydrogen gas to form ammonia, don't worry this isn't important for us now, but random fact: it was used in the making of bombs to kill millions of people in WWI and WWII). Haber process

12. Temperature is one of the three ways to affect the position of Equilibrium (The other two being concentration and pressure). • Endothermic reactions, are reactions which consume/take in heat, are favored if heat is being added, therefore higher temperatures tend to feed into Endothermic reactions. • Exothermic reactions, are reactions which release heat, commonly function better at low temperatures. And the heat the Exothermic releases tends to favor the reverse Endothermic reaction. • Therefore if heat is added to the reaction, it forces the reaction back to the left and if that heat is removed, the reaction is forced back to the right. Temperature

13. We have: N2(g) + 3H2(g) <=> 2NH3(g) This is an exothermic process since heat is being produced over the course of the reaction, the delta H is -92.4KJ/mol (Remembering that delta H = delta H of products - delta H of reactants) let's bring back Le Chatelier's principal and mix it in a bit with the Haber process.

14. Let's forget about what's on each side for a second and look at it like • this: Left-side <=> Right-side • Now we just said that the delta H = -92.4KJ/mol, meaning that it is exothermic, allowing us to add it into the equation on the right hand side, so now we have this: • Left-side <=> Right-side + heat • Now imagine heat as a product of chemical nature, so as we previously stated, adding a product will shift the reaction to the left, because more reactants are being formed in order to reach equilibrium. Temperature shifts

15. Why this works is because the reaction wants to consume heat as we add the heat to it, shifting it to the left, and if we were to remove heat from the right-side, the reaction would compensate for that loss, by shifting it back to the right side. More temperature shifts

16. We know our reaction rate equation, so I'll just skip right into it, I'll write this on the board (Anyone stop me if you don't understand how I got here): • We have: N2(g) + 3H2(g) <=> 2NH3(g) • Therefore the following is true: • K = (NH3)^2/(H2)^3(N2) • By adding heat we determined that we are shifting the equation to the left, meaning that the NH3 goes down (decreases), and the N2 and H2 goes up (increases). • Therefore if the numerator decreases, and the denominator increases, the K (constant) will decrease. • So if we add heat in an exothermic reaction, the K value decreases. Mathematical type-stuff

17. Now if we were to remove that heat, the reaction would shift back to the right, meaning an increase in NH3, decrease in N2 and H2, therefore the K value would increase • Now I'll set up the same reaction, pretending that it was endothermic and we were adding and removing heat (even though this isn't true), and now what will happen to the K value? More Mathematical type-stuff

18. Increasing the Volume of the container for a reaction results a decrease in pressure. • When this occurs the equilibrium will shift to the side with the most number of moles, so back to our Haber process for a second: • N2(g) + 3H2(g) <=> 2NH3(g) • Left hand side has 4 moles, Right hand side has 2, therefore it will shift to the left, if the volume is increased. • A decrease in pressure will allow it to shift to the side with the least number of moles, therefore shift to the right. Intro to pressure