CHEM 1332 1st Edition Lecture 12 Outline of Last Lecture - Activation energy- Constant equilibrium- Constant pressure equilibrium Outline of Current Lecture - Reaction quotient - Le Chatelier’s Principle- Finding Kc value- Reaction tableCurrent Lecture- Q is the reaction quotient is product of righties to the coefficient over the product of lefties to the coefficient.- If Q is less than k the reaction goes to the right - If Q is greater than k the reaction goes to left - If Q=k the reaction is at equilibrium- Le Chareliers Principle you consider a system that is already at equilibrium - If you add more reactant Q is less than k is uses up the reactant and makes product - If you add product Q is greater than k and it uses up product to make more reactant- If you remove reactant Q is greater than k it uses up product and makes reactant - If you remove product Q is less than k it uses up the reactant and makes product- Le Chatelier’s Principle if you change something that takes the reaction out of equilibrium it will change and adapt to reach equilibrium once more - If you increase temperature the reaction is not at equilibrium anymore the reaction wants to remove heat so it uses up heat through an endothermic direction- If you decrease the temperature the Q does not equal k the reaction goes an exothermicdirection - If you increase pressure it decreases the volume of the reaction and the number of moles of gas molecules - If you decrease the pressure the reaction goes in a different direction- It is possible to find kc if you are given the initial concentrations and the amount at equilibrium These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.- An example problem is you are given the balance equation 3 H2+ N2⇄2 NH3 and theproblem is I mix in 1L container 3 mol H2 and 2 mol N2 at equilibrium I have 0.5 mol of NH3. Calculate kc.- We can setup the kc equation as NH32eqN2H23- So we know that if we have 0.5 mol of NH3 we need 0.25 mol of N2and 0.75 mol of H2- So at equilibrium H2 is 3mol-0.75 mol so 2.25 mol and N2 is 2 mol- 0.25mol so 1.75 mol - If we put this in the previous equation we get 0.521.75 2.253 from this we solve and get the kc value - There is alternative method we can make a reaction table of the GeneralN2H2NH3Start 2 3 0Change -x -3x +2xEquilibrium 2-x 3-3x 2x- This chart gives the k equal to a function-NH3= 0.5mol=2x so x=0.25- We plug in x to get the rest of the values-N2 is 2-x=2 mol- 0.25mol so 1.75 mol-H2 is 3-3x is 3mol-3*0.25 mol is 2.25 mol- We end up with the same answer from before.Key TermsLe Chatelier’s Principle- when a chemical system at equilibrium is disturbed, it attains equilibrium by undergoing a net reaction that reduces the effect of the disturbanceReaction table- this table uses the initial quantities and change to solve for the