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Equilibrium?

Equilibrium?. When we talk about equilibrium what we are talking about is the balance that a chemical reaction reaches. The balance is between how much product is produced and how much reactant is left behind and never reacts. The two are not always as balanced as the picture might show.

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Equilibrium?

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  1. Equilibrium? When we talk about equilibrium what we are talking about is the balance that a chemical reaction reaches. The balance is between how much product is produced and how much reactant is left behind and never reacts. The two are not always as balanced as the picture might show. Some product is produced and some reactants are left over unreacted.

  2. Chemical Amounts & Concentration So later in the unit we are going to calculate equilibrium. In order to calculate equilibrium we are going to need to be able calculate the concentration of a solution. And vice versa…..use the known concentration of a solution and calculate how much solid is dissolved in a certain volume of water. So lets review it.

  3. Molarity/amount concentration n = two moles V = one Litre of water C = 2 moles One Litre of water C= 2 mol/L

  4. MASS IN GRAMS MOLES Concentration OF SOLUTION mass molar mass (g/mol) n=m M Moles (n) Volume (L) Molarity (mol/L) C = n V

  5. Example: Find the molarity of a solution containing 75 g of MgCl2 in 250 mL of water. Problem 75 g MgCl2 moles MgCl2 0.7877mol MgCl2 = = Volume of Solvent 95.21 g/mol (MgCl2) 0.250 L water = C = 3.2mol/L of MgCl2

  6. Example: How many grams of NaCl are required to make a 1.54mol/L solution using 0.500 L of water? Need to solve for n n = CV n = (1.54 mol/L)(0.500L) = 0.77 mol

  7. = 0.77 mol Need to solve for mass m = nM = = 44.9988 g m = (n)(MNaCl) (0.77 mol)(58.44g/mol) ~ 45.0 g

  8. Complete The Table

  9. Acids/Bases & pH/pOH & [H3O+]/[OH-] So in order to calculate the equilibrium for an acid/base reaction you will need to know how to convert concentration (C) of solution into concentration of [H3O+]/[OH-]. And, if you have these concentrations how to use them to calculate pH and pOH.

  10. Mr. K’s pH/pOH Diamond of Awesomeness pH + pOH = 14 pH/ [H+] pOH/[OH-] pH = -log[H+] pOH = -log[OH-] [H+] = 1x10-pH [OH-] = 1x10-pOH Kw = [H+][OH-] = 1.0 x10-14

  11. Example A sample of tap water has a pOH of 6.3, what is the hydroxide ion concentration? [OH-] = 1 x 10-pOH [OH-] = 1 x 10-6.3 [OH-] = 5.011872336 x 10-7 mol/L Since the pOH has one decimal place we are allowed one significant digit. [OH-] = 5 x 10-7 mol/L

  12. Example: What is the amount concentration of HBr in a solution that has a pOH of 9.6? 1 1 1 HBr(aq)  H30+(aq) + Br -(aq) [HBr] = [H3O+] = [Br-] 1:1:1 Ratio Therefore: [HBr] = [H+] [H+] = 1x10-pH How can we find pH from pOH? pH + pOH = 14 pH = 14 - pOH [H+] = 1x10-4.40 pH = 14 – 9.6 [H+] = 3.981071706 x 10-5 mol/L [H+] = [HBr-] pH = 4.40 [HBr-] ~ 4 x 10-5 mol/L

  13. Complete The Table **** I want you to add another column at the end of the table, label it pOH***

  14. The easiest and simplest equilibrium systems are static (not moving). • Your book sitting on your desk is the best example • The downward pull of gravity is equal to and cancelled out by the normal force upward from the desk. • Most equilibrium systems we will look at will not be static. • These systems are said to be dynamic(always moving). • Rate of reactants turning into products and the rate products are breaking back down into reactants are equal.

  15. If you look at a solution that is at equilibrium it looks like nothing is happening. That’s because overall, nothing is changing. The rates of the solution crystalizing and settling to the bottom and solid on the bottom dissolving are equal. You can think of it like a tug of war between two equally matched teams. They both pull as hard as they can, but overall nothing moves…it’s a stalemate.

  16. Equal Rates The ions floating around in the solution are constantly crystalizing and sinking to the bottom. These molecules sitting on the bottom are constantly dissolving into solution. The rates of dissolving and crystalizing are equal.

  17. Problem… Eventually the water in the solution will evaporate and the solution will all be solid on the bottom. But what if the reaction occurred in a closed container like a pop can for example? The beaker is an open system and matter can escape and the equilibrium is always changing.

  18. Types of Equilibrium Solubility Equilibrium Chemical Reaction Equilibrium

  19. Equilibrium Equations The two arrows show that the reaction can occur in both directions. The forward reaction is reading the equation from left to right. The reverse reaction is when the products turn back into the original reactants, the equation is read right to left.

  20. Hydrogen-Iodine Reaction

  21. Describing an Equilibrium Chemists describe an equilibrium in two ways: In terms of Percent Reaction/Percent Yield. Measure as a Equilibrium Constant. ***Only describe the equilibrium for a specific example…one single experiment. ***Can be applied to allequilibrium experiments with the same chemicals done at the same temperature.

  22. Percent Reaction/Percent Yield Looking at the percent yield of a reaction gives scientists a good idea about how far the reaction occurred to completion. By seeing how much products was produced compared to how much could have been made using stoichiometry we can tell how far the reaction occurred before it stopped at equilibrium.

  23. % Yield and Equilibrium Classification No Reaction <1%

  24. Equilibrium Shifts (changes) When chemical reactions occur, the amount of original reactants and products change because some of the reactants combine to form products. So if the amount of products increases then the amount of reactants must decrease, but by how much? How can we know how much of the reactants it takes to make a certain amount of product? Well when did stoichiometry we learned that the amount of products and reactants are related and can be calculated using balancing coefficients (R/G) (N/G)…….(2/1) or (1/2) for example. 1

  25. ICE Tables An ICE table uses the idea of the amount of reactants and products being related by balancing ratios (1:2, 1:3, 2:5, whatever). So by measuring the change of either reactants or products, we can calculate the change of all the other changes by multiplying by the ratio. ICE tables stand for: Initial Change Equilibrium By knowing the amount of something before and after a reaction we can calculate the change (the difference). Or vice versa.

  26. Lets Try One I2(g) (mmol/L) 2HI(g) (mmol/L) H2(g) (mmol/L) 0 1.00 1.00 +1.6 -0.78 -0.78 0.22 1.6 0.22 H2(g) (mmol/L) I2(g) (mmol/L) 2HI(g) (mmol/L) [H2(g)]=[I2(g)] x (1/1) [H2(g)]=(-0.78mmol/L)(1/1)= -0.78mmol/L 0 1.00 1.00 [[HI(g)]=[I2(g)] x (2/1) [[HI(g)]=(-0.78)(2/1)= +1.6mmol/L -0.78 0.22

  27. Equilibrium Constant (Kc) Equilibrium constants are just like the name implies CONSTANT….this value can be used for any equilibrium experiment. After scientists did hundred of equilibrium experiments with all kinds of different concentrations of reactants and products they calculated the constant. They studied the results of the reaction and used a lot of math to come up with a relation…..well use it.

  28. Equilibrium Constant (Kc) Concentration of products multiplied together divided by the concentration of the reactants multiplied together. ***Remember! Kc has NO UNITS…its just a number. Kcis NOT affected by concentration changes of products or reactants, it’s a CONSTANT. The only thing that can change the Kcvalue of an experiement is a change in temperature. The [C] and [D]represent the concentrations of the products in the equilibrium. The [ [A] and [B] represent the concentration of the reactants in the equilibrium. The balancing coefficients a, b, c, d from the balanced chemical reaction become the exponents in the equilibrium constant. This is the general formula for a chemical reaction. A and B represent the reactants….(chemical formulas) C and D represent the products…(chemical formulas) a, b, c, drepresent the balancing coefficients you would use to balance it. Equilibrium Constant Expression/Formula

  29. Solving Questions Using Kc When solving a problems using the Equilibrium constant expression the first thing you need to do is write a balanced chemical reaction equation. 2ndthing you need to do is to manipulate (re-arrange) the expression so that whatever your trying to find out is on one side of the = and everything else is on the other side. (*** If your unsure how to do this…practice, ask a friend, ask me, whatever. Just find out how.***) After that, just put in the values from the question (with units so you can cancel out!) into the expression and abbrakadabra…. Kabamm? Capowee? AllaKazamm! Old School Batman! Anybody??....No?...Just me.

  30. Practice With Kc(Easy) Calculate K c when 0.200mol of hydrogen and iodine react to produce 0.400mol of hydrogen iodide. [0.400mol/L]2 = [0.200mol/L]1[0.200mol/L]1 0.16mol/L2 = 0.04mol/L2 = 4 ~ 4.00

  31. Practice With Kc(Hard) The question doesn’t give us enough information to manipulate and solve the equilibrium expression. 1 ??? 1 1 1 1 CO2(g) + H2(g) CO(g) + H2O(g)  2 ??? But! Since its an equilibrium question we can always try using an ICE table. This problem is a little different, it doesn’t give us the concentration of anything at equilibrium.

  32. Practice With Kc(Hard) If we represent the change of one reactant with –X we can use stoichiometry and balancing coeficicient (R/G) to find the change in the other compounds in the eaction. Because the balancing cratio was 1:1:1:1, using R/G will always be 1/1….so the change for all other compounds will also be X……-X for reactants.. But + X for the products cause their going to increasse. Kc = 4.20 The only other information the question gives is the Kc But! We know that if only reactants are present their concentration is going to decrease…so we can say the change is –X. Nothing else is initially in the reaction vessel.

  33. Practice With Kc(Hard) Kc = 4.20 3 (x)(x) (4.00-x)(4.00-x)

  34. Practice With Kc(Hard)

  35. Practice With Kc(Hard)

  36. So when reactants are mixed they produce products, but not completely, it only happens until equilibrium is reached. Well in industry this is just annoying because the product is worth a lot of money, the reactants are cheap…..so we want to create as much product as possible. N2(g) + H2(g)  NH3(g) Ammonia used in fertilizers and cleaning products Worth: $575/tonne. Price of BOTH…about $240/tonne.

  37. Predicted that once a chemical reaction reaches equilibrium it will stay at equilibrium unless you disturb it. You can disturb the equilibrium of a system by changing 3 things: 1. Reactant/Product Concentration. 2.Temperature of the system. 3. Volume (Pressure) of the system. French chemist and engineer

  38. Chemical Reactions andCollision Reaction Theory So equilibrium is when the rates of the forward and reverse reactions is equal. So what determines the rates? What causes two reactants to combine to form a compound (forward reaction)? And vice versa, what causes a compound to break back up into it elements(reverse reaction)? Chemical reactions are caused by collisions.

  39. Chemical Reactions andCollision Reaction Theory Vs Low Concentration High Concentration So the Collision Reaction Theory states that reaction are caused by molecules colliding…. So which solution do you think will have more collisions? The one with only a few molecules (Low Concentration) or the one with lots of molecules (High Concentration)?

  40. Chemical Equilibrium Shifts Totter Analogy” “The Teeter Le Chatelier came up with a set of rules that could be used to predict how an equilibrium would change if you disturbed it. By change it really means would more product or more reactant be produced as a result of the change. In order to predict how a change would affect the equilibrium you must first write a balanced chemical equation for the reaction.

  41. 1. Le Chatelier’s Principle states the increasing the concentration of a species in a chemical reaction will cause the reaction to shift the opposite direction. Example: What about concentration decreases? Shift Right (Products) Shift Left (Reactants) Decrease SO2(g) Increase SO2(g) Increase SO3(g)

  42. 1. Rules: Concentration increases cause the reaction to shift to the opposite side of the equation. Concentration decreases cause the reaction to shift to the same side of the equation.

  43. 2. Le Chatelier’s principle also states that an increase in temperature (Thermal Energy) will cause the reaction to shift in the opposite direction. Increase Temperature? So where’s the energy go? Shift Left (Reactants) Problem: ***You can’t predict temperature shifts with knowing which side the energy is on (Endotermic/Exothermic)*** Where’s energy? Increase Thermal Energy ***YOU HAVE TO KNOW WHETHER ITS EXO OR ENDO TO PREDICT SHIFT***

  44. 2. Rules? The rules are the same as for concentration changes. REMEMBER to put the energy in the equation. Exothermic…energy goes on the _________side of the equation. Endothermic…energy goes on the ________side of the equation.

  45. 3. Le Chatelier’s principle states that a change in the volume of the container (vessel) that a reaction happens in will also the equilibrium to shift. To predict how pressure/volume changes will affect a reaction you need to know how many mol are on each side….you need to have a BALANCED chemical equation. ***Remember from Chem 20 that an increase in volume causes a decrease in pressure….and vice versa. (Pressure) Increase in pressure. Shift Right (Products) (2) (3) An increase in pressure/decrease in volume will shift the equilibrium to the side with less moles.

  46. 3. (Pressure) Decreasein pressure. Shift Left (Reactants) (2) (3) A decrease in pressure/increase in volume will shift the equilibrium to the side with moremoles.

  47. 3. (Pressure) Rules:

  48. Catalysts are compounds added to a reaction that speed up the reaction by decreasing the activation energy. Big money in discovering and using catalysts…. CATALYSTS DO NOT AFFECT EQUILIBRIUM! DO NOT CAUSE ANY SHIFT, JUST MAKE REACTION HAPPEN FASTER.

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