all right in this video we're looking at a review for chemical equilibrium this would be useful to do before a test or before the final exam it's aimed at chemistry 40s students but would also be a pretty good review for a piece chemistry students as well just one note if you are an AP chemistry student there won't be any references in this video to equilibrium constants calculated using pressures KP okay but that's not covered in the chemistry 40s course alright so the first question I'll give you a chance to read the question I'll pop and then I would suggest you pause the video try it yourself and then check your answers afterwards so in an aqueous solution of potassium chromate we have this equilibrium shown here existing and we're told that we want to use the shots I'll use principle to fill in the blanks below so sulfuric acid which has hydrogen ions in it is added to the system and the color changes well this is Lucia Talia's principle so why don't we look at the expression for Q q is gonna equal the concentration of dichromate divided by the concentration of chromate squared times the concentration of hydrogen ions squared now if we were to add sulfuric acid then we're gonna increase the concentration of those hydrogen ions which are in the denominator of Q so that's gonna make you get smaller if Q gets smaller than Q will be less than K see cuz KC is unaffected unless temperature changes if Q is less than K we know that causes a forward shift in the equilibrium so we're gonna see a forward shift as a result of adding those hydrogen ions so the color change would change from yellow it's started off as potassium chromate which is yellow and it'll change to dichromate which is orange so we're gonna see a change from yellow chromate to the orange dichromate so the concentration of the dichromate ions cr2 o7 - - is increasing while the concentration of chromate ions is decreasing in that forward shift things on the left side will decrease in concentration things on the right side will increase in concentration and we just said that a forward shift will occur questions 2 & 3 give them a try so question two a reaction that takes place in both a forward and reverse direction is said to be a reversible reaction products are formed when reactant particles collide but then those particles product particles will also collide and form the reactant particles again question three the symbol Q is for the reaction quotient so we talk we just reference that up above when Q is greater than kc a net reverse or backward reaction will occur until equilibrium is reached when Q was less than k we saw a forward shift when Q is equal to K means the system is at equilibrium alright now we have some true/false questions again I would suggest pausing the video and and seeing if you can do these i'm yourself so question for dynamic equilibrium rates of forward and reverse reactions are equal that's the definition of equilibrium so that is true number 5 you need a closed system for equilibrium to be established and a closed system just means reactants and products can't escape that is also true if reactant particles or product particles can escape then the forward or the reverse rate will be affected by that as a simple example to consider an open bottle of water and a closed bottle of water in the open bottle of water it just keeps evaporating evaporating until there's no more water left because as the water evaporates the water vapor escapes from the bottle and the reverse reaction the condensation reaction just doesn't happen but if you close the bottle if you put a lid on the bottle of water then now as the water evaporates the vapor can't escape so the the vapor will begin to condense again toward liquid water so the reverse reactions rate will build up and eventually the rate of evaporating than the rate of condensating condensing rather are equal and you've reached equilibrium so you need a closed system for equilibrium to be established number-6 equilibrium concentrations of reactants and products are equal that is false we said back to number four the rates of forward and reverse reactions are equal but it's rarely the case that the reactant and product concentrations would be equal they're almost never equal question seven when kc is very large products are favored in other words the reaction is said to be complete well if you think of a KC as being very simply a ratio of the concentration of the products divided by the concentration of the reactants now that is an oversimplification but it is a simplification when KC is large what does that mean it means that your numerator must be much larger than your denominator in order to get a large KC value that means you've got much greater concentrations of product than of reactants so yes the products are favored and because you have so much product in very little reactants left one of the reactants almost got used up or maybe all the reactants got almost used up therefore the reaction would be described as being almost complete question 8 if you hold the temperature constant increasing concentration and a reactant would cause KC to increase while the key there was holding the temperature constant the only experimental variable which would affect a KC is changing temperature if temperature is held constant then the kc value just doesn't change so in this case this would be a false statement okay a few more true/false questions pause the video and give those a try question nine says decreasing temperature for an endothermic reaction would cause the KC to increase well to answer that remember the graph that we saw with temperature and how it affects kc values there was a graph for endothermic and a graph for exothermic reactions for endothermic reactions the graph looks something like that now this said that decreasing temperature would cause KC to increase but as you can see in the graph if you were to lower the temperature the case he gets smaller so this statement is false question ten saturated salt solution would be an example of an equilibrium well earlier we used the example of a closed bottle of water with evaporating and condensing here we have a saturated solution where you have salt which is constantly dissolving or dissociating at the same time you have the ions which have already dissolved or dissociated they are precipitating back to the salt the solid salt so because you have those two opposite processes these salt ions are dissociating and the dissociated ions are precipitating they're happening at the same rate which is why it looks like the salt is no longer dissolving so yes a saturated solution is indeed at equilibrium number eleven a catalyst increases the percent yield in a reaction that's false the catalyst would make you reach an equilibrium system faster but the catalyst doesn't change the value of KC and therefore will not change the percent yield in the reaction number twelve when writing equilibrium expressions solids and liquids are left out because their concentrations are constant while we know that solids and liquids are left out but you remember why we leave them out this is exactly the reason this is true the concentration of a solid doesn't change if you use up half of a solid in a chemical reaction the concentration of the remaining solid is the same as the concentration of the starting solid and the same would be true for liquids so because their concentrations are constant we don't include them in the expression for KC all right we're not getting into some multiple choice questions number thirteen pause the video and give that a try so it says some so3 is put into an evacuated flask sulfur trioxide and you can see in the reaction that that decomposes to make sulfur dioxide and oxygen how do the rates of the forward and reverse reactions change as we approach aqua Librium well during the reaction the so3 that you put in is getting used up so it's concentration drops while it is decomposing it's producing the so2 in the o2 so their concentrations will increase now since concentration affects the rate of a reaction the drop in concentration of s o3 means the forward rate will decrease while the increasing concentrations of so2 and Oh to mean the reverse reaction would increase so we're looking for a decrease in the forward reaction and an increase in the reverse reaction number fourteen give that a try so the chemists put some ammonia into a two liter flask at equilibrium there was 0.06 zero moles of nitrogen in the flask question is what would be the concentration of hydrogen in the flask also at equilibrium well we look up at the balanced equation and we see that the nitrogen's coefficient is a 1 while the hydrogen's coefficient is a 3 that means there's a three to one ratio hydrogen to nitrogen so if we got point O six zero moles of nitrogen produced we would get three times that three times point zero six zero moles that's going to be 0.18 moles of hydrogen produced now the question wasn't asking for how many moles it asked for the concentration while concentration is moles divided by the volume so if we take 0.18 moles divided by the two point zero liters we get a concentration of zero point nine zero which is choice C question fifteen so for which of these reactions would KC simply equal the concentration of oxygen so in Part A we'll just go through each one of them in Part A the reactants and products are gaseous so that's a homogeneous gas phase equilibrium its KC is going to have ozone in the numerator and oxygen in the denominator it's not going to simply equal oxygens concentration so we can rule out a in Part B the products are solid the reactants are solid and gaseous so this is heterogeneous equilibrium we remember to omit the solids as well as liquids from a KC expression so for Part B the KC would equal products are all solid so we'll just put a 1 in the numerator as a placeholder in the denominator the only thing we can include is oxygen because it's a gas so KC would equal 1 over the oxygen constant tration but we wanted kc to equal oxygen not equal the reciprocal of oxygen so we can rule out choice B choice see the water reactant is a liquid so we would omit it from the kc expression so this case the expression has no denominator but on the right the hydrogen and oxygen are gases so we'll get hydrogen and oxygen in the kc expression so c is not the answer that leaves us with deemed e has to be the answer if you look at it the oxygen gas is on the right hand side as a product while oxygen liquid is as a reactant liquids are omitted so for this the kc would equal the product oxygen over 1 or just oxygens concentration which is what we were looking for questions 16 pause the video and give that a try so this question is going to require a little bit of calculations we have a 2.5 liter vessel nitrogen monoxide oxygen and nitrogen dioxide are present at equilibrium and we're told that we have a certain number of moles of each one we want to calculate KC so we're gonna use the formula C equals moles over volume and we can say at equilibria the concentration of the n-o gas will be two point eight three moles divided by two point five zero liters gives me a concentration of one point one three molarity doing the same thing for oxygen I won't show my work but oxygen is one point two zero molarity when you divide it's moles by the volume and the concentration of the no.2 the same way and over V is going to be seven point two zero molarity now that I have all three concentrations it's easy to calculate KC it's the concentration of the no.2 squared over the concentration of and squared times oxygen plugging those three concentrations into the expression and then evaluating we get KC is very close to 34 so choice D is the correct answer number 17 pause and give that a try so we're given the KC 0.0045 for one reaction that decomposition of n2o4 but we're looking for the KC of a different reaction and we compare that reaction to the original we see two things have changed we've reversed the original equation to get this one so the original equation has been reversed we've also multiplied the original equation by 1/2 where we used to have n2o4 now we have 1/2 and to go for so we've multiplied by 1/2 so we remember that the way you balance an equation changes the value of KC so KC here would equal reversing means we take the reciprocal of the old KC so we'd say 1 over 0.0045 because we reversed but if you multiply by a number then you raise KC ^ that number so in this case we multiplied by 1/2 so we have to also raise the KC to the power of 1/2 now raising something to the power of 1/2 is the same as saying square root so either one of these would be acceptable so grab your calculator and evaluate those and you'll find out the answer is 15 so choice B number 18 pause and give that a try ok that should be pretty quick we've discussed this already the value of KC changes if you change temperature any of these other changes have no effect on the value of KC which is why it's referred to as an equal constant but it does depend on temperature change the temperature you change the value of KC number 19 pause and give that a try so it's the same reaction we saw a moment ago n2o4 decomposing into no.2 gas and the kc is point zero zero four five the question says you put point two five moles of each of the two gases in a point five zero litre vessel as soon as I see moles and leaders the formula C equals n over V jumps into my head the question is which of the following would describe the change in concentrations as we approach equi Librium and we're looking for the n2o4 increasing or decreasing and the same for the no.2 so to answer that question we basically have to decide whether the equilibrium up above shifts in the forward direction or shifts in the reverse direction and to do that we need to calculate the reaction quotient Q G was calculated the same way as KC so concentration of no.2 squared over concentration of n2o4 but the difference is that for Q we can put any concentrations into the expression not just the equilibrium constant with concentrations so here we can take the 0.25 moles divided by zero point five zero liters and we get a concentration of 0.5 zero molarity for both the no.2 and the n2o4 which means the value for Q here is going to be 0.5 zero so right away we can see this was not at equilibrium because Q is not the same as the KC up above but more than that we can see that the Q is actually greater than the KC and as we said earlier that means a reverse shift will happen if the equilibrium shifts in the reverse direction then the no.2 concentration would decrease because in the reverse direction it's a reactant while the n2o4 s concentration would increase because in the reverse direction it would be a product so we're looking for the n2o4 to increase and the no.2 to decrease and that would be choice B question 20 pause and give it a try so this is ozone decomposing into oxygen gas and the kc is 36 what is the concentration of ozone if the equilibrium concentration of oxygen is 0.05 8 so the key there is that we're given the equilibrium concentration of the oxygen and we're simply looking for the concentration of ozone that would go with that so this is simply setting up the KC expression so KC is concentration of oxygen cubed over the concentration of ozone squared and we know two out of the three things in this expression the KC is 36 and the O is the oxygen is what we're trying to find sorry the oxygen we were given let me just erase that so the oxygen I didn't need to erase that the oxygen we actually know so the oxygen is point zero five eight molarity cubed divided by the ozone which is what we're trying to find sorry about that so now doing some algebra the Ozone's concentration squared will equal the 0.058 cubed divided by 36 and that gives me a value of five point four times 10 to the minus 6 now to get the Ozone's concentration we simply have to square root to get rid of that squared symbol so when you square root this number you get point zero zero two three which is choice a question twenty-one so we're looking for the mass of carbon dioxide that'll be present in equilibrium after you have a hundred grams of the solid calcium carbonate that was put in a 2-liter vessel and allowed to decompose at a high temperature and it reached equilibrium what mass of carbon dioxide would be present well we know the kc value while that kc value looking at the balanced equation is actually equal to the concentration of co2 the calcium oxide and the calcium carbonate are solids so they don't get included in the KC expression that means the concentration of carbon dioxide is point one zero zero molarity so since we know the volume is two liters we can use a couple of unit multipliers or you could use two formulas you could say C equals n over V and since we know the concentration of the carbon dioxide and the volume we could find the moles and then we can use moles as the mass divided by the molar mass since we know the moles and molar mass of carbon dioxide we can find the mass alternatively we could use a couple of unit multipliers we'll take the volume two point zero two two point zero zero liters and with one multiplier will convert liters into moles of carbon dioxide and then with the second multiplier will convert the moles of co2 to grams of co2 in the first case this is the concentration kc value is point one zero zero so point one zero zero moles per liter and then the second one is our molar mass of carbon dioxide one mol from a periodic table would waive forty four point zero one grams grabbing a calculator and evaluating that eight point eight zero grams of co2 would be present all right questions twenty-two and twenty-three before we start looking at Lachelle use principle pause the video read over the scenario and try question twenty-two so in an experiment we have four moles of so2 five moles of oxygen they're placed in a five liter vessel and allowed to reach equi Librium when equilibrium has reached ten percent of the so2 remains in the flask so we know initial conditions and we know something about the equilibrium conditions that should have been a clue to set up the case these write the ice table so I'm going to write the balanced equation again to so2 s react with oxygen to make to so3 s everything here was in the gas phase using our formula C equals and over V to get concentrations we can see the initial concentration of so2 was zero point eight zero zero molarity four moles divided by five liters the initial oxygen concentration was one point zero zero molarity five moles divided by five liters and there was no so3 in the container now we were told that at equilibrium so we go down to the equilibrium the third line of the ice table ten percent of the so2 remained in the flask while ten percent of 0.8 B means 10 over a hundred ten percent of 0.8 would be zero point zero eight zero zero molarity so at equilibrium there's no point zero eight zero zero molarity of the so2 now I can fill in the ice table because I know all of the initial numbers and I know one other number in this case equilibrium so the so2 if it began at point eight and ended at point O eight it lost zero point seven two zero molarity oxygen would also lose because it's also a reactant and it would lose half as much because the coefficient is a one while so2 has a coefficient of 2 so half of point seven two is zero point three six zero and the so3 is a product so it's going to gain it's on the other side of the equation and it would gain the same amount as the so2 lost because the ratio there is two to two in the balanced equation so it would gain 0.72 zero molarity so then we have zero point seven two zero molarity of so3 and one takeaway 0.360 we have zero point six four molarity of oxygen at equilibrium so from these choices point parts choice C is the correct answer for oxygen at equilibrium we want to calculate the Casey the Casey is going to equal the concentration of s o3 squared divided by the concentration of so2 squared times oxygen plugging in the equilibrium numbers from the ice table above into this expression will find that the KC value is very close to 130 so with two significant digits will write 130 as our answer all right so so far we've seen equilibrium theory definitions the reaction quotient and we've done some simple calculations with equilibrium another big topic in equilibrium is Lache ateliers principle and that's what we're gonna focus on the remaining questions so question number 24 give this a try 24 and 25 both deal with this equilibrium but triclinic Trustin 24 to begin with so there's two ways to approach lechatelier's principle some people like to approach you with kinetics considering how the rates of forward and reverse reactions are affected I prefer usually to use the reaction quotient to think of it so looking at this question we can say the reaction quotient up above is the concentration of carbon dioxide cubed divided by the concentration of carbon monoxide cubed we leave out the IRA knock side because it's solid and we leave out iron because it's liquid they don't appear in the kidney expression for Q or KC so the question is which of the following would cause the equilibrium to shift to the right while equilibrium shifts to the right shifts forwards if Q is less than K C so anything that would make you get smaller or cute KC get bigger will result in Q being less than KC so we'd like you to decrease or we'd like KC to increase in order to achieve Q being less than KC so let's see the first choice says add more iron oxide while looking at the expression for Q ayran oxide is not there so that would actually have no effect on the equilibrium choice B says add more ayran by the same logic I drain is not in the expression for Q solids and liquids don't affect the equilibrium so adding adding iron has no as no effect but removing carbon monoxide choice C that would definitely change Q carbon monoxide is in the denominator if Q if you were to remove some of it then the denominator of cube would get smaller if you have a smaller denominator you're dividing by a smaller number Q but actually get bigger but we want Q to decrease so we know the answer is not a B or C then what about choice D if you removed carbon dioxide well that's in the numerator of Q that would make the numerator smaller if you have a smaller numerator your fraction is smaller than cube will decrease which is what we were looking for so choice D is the correct answer number 25 says when the system was cooled the mass of iron oxide increased and then we want to choose which of the statements below is true and looking at the choices we have to decide whether KC would increase or decrease and whether the reaction is endothermic or exothermic just to remind you of the balanced equation up above so we're cooling the system and it resulted in an increase in the mass of iron oxide does that cause KC to increase or decrease is the reaction exothermic or endothermic pause and see if you can figure that out well the first thing I would note is if the iron oxide were increasing then there was a reverse shift in the equilibrium we've got more iron oxide by the reaction shifting reverse now the reason a reaction would shift reverse would be if Q were greater than KC now we know that we're changing temperature so that has no effect on Q but it does have an effect on KC so if Q is greater than KC KC must have decreased at equilibria in the two of them are equal so if Q is now greater than K k must have decreased now looking at the graphs we have one for exothermic situations and we have another one for endothermic situations let's just take a look at one of them we'll look at the endothermic graph KC versus temperature the graph does this while exothermic oh I guess I'll draw them both exothermic KC versus T the graph looks like this so we know that when we cooled the system down KC decreased which of these two graphs shows that lowering temperature causes KC to get smaller but I hope you focused on the first one when you lower the temperature the KC's there will get smaller so that means we have an endothermic reaction and KC decreased which means of those choices we're looking at choice a questions 26 and 27 these guys refer to methanol being produced methanol ch3oh by reacting carbon monoxide and hydrogen with the help of a catalyst so there's the balanced equation pause the video and see if you can answer question 26 so this is again letelier's principle which of the following would describe the effect of reducing of the volume of the vessel which creates an increase in pressure we want to see what would happen to the yield the amount of product being produced and the rate of the reaction well this is letelier's principle so let me write the expression for Q Q for this reaction would be concentration of methanol divided by the concentration of carbon monoxide and hydrogen squared now we reduced the volume of the vessel we know that concentration is moles divided by volume so if you reduce the volume a smaller volume you get bigger concentrations so the concentrations in both the numerator of Q and the denominator of Q would increase because the volume has been lowered but what happens to Q the numerator increases the the denominator increases but do they increase equally we look at the exponents in the expression and in the numerator we see the only exponent is a 1 so we would say the numerator is first-order while in the denominator the exponents are a 1 and a 2 which means the denominators order is third order so since you increased all of the concentrations all three of them equally the numerator and denominator increase but the denominator which is third order it's cubed it would increase even more than the numerator so the denominator increases more than the new now you have a bigger denominator than numerator that makes Q gets smaller so Q will decrease if Q decreases q is going to now be less than KC which creates a forward shift in equilibrium that means we're going to make more methanol if you have a forward shift so the yield will increase so we're looking for choice C or D the equilibrium yield have has increased but what about the reactions rate in our kinetics unit we remember increasing concentrations would cause the rate to increase also so because we decreased the volume of the vessel concentrations went up that means the rate will also go up so choice C where the yield increased because of the chatelier's and the rate increased because of rates dependence on concentration question 27 which of the following combinations describe the effect of removing the catalyst remember that that reaction that we just discussed was helped with the with by adding a catalyst well we know that catalysts increase reaction rates so if you were to remove the catalyst the reaction rate would decrease so choice a or B looks good and we also know that the catalyst has no effect on KC and no effect on Q so the catalyst has no effect on the yield the amount of produce that we the amount of product will get so we're looking at choice B there'd be no change in the product the amount of product but the rate at which you're producing it would decrease questions 28 and 29 deal with hydrogen gas reacting with iodine vapor and we're told that the iodine vapor is a purple colored gas while the other two gases are colorless so which of the following graphs would show what happens if H I was removed and then a new equilibrium established pause the video and see if you can figure that out so all four graphs show that the H eyes concentration on the y-axis is decreasing we were told H I was removed so there'd be a sudden drop in H I looking up at the expression for Q for this we get Q is H I squared over hydrogen times iodine if you were to remove some H I the numerator would get smaller and that means Q would be less than KC we'd see a forward shift in equilibrium in other words removing some H I the system is going to shift forwards and more H I would be produced but here's the key idea although more H I is being produced it will never achieve the original concentration before we removed some so we removed some H I that caused a stress the system shifts forwards to make more H I and relieve the stress but it can't completely relieve the stress so in the end there will still be less H I than there was at the beginning the graph that shows that is choice B where the H I increased but didn't reach the concentration that it began with before when the stress was applied question 29 what would happen to the color in the vessel as a result of removing H I well we know that the reaction was shifting in the forward direction as it shifted forwards the amount of iodine vapor in the flask would decrease because iodine would be a reactant in that forward shift that means we have less purple gas and so the color of the gas would get lighter right it would fade a bit so we're looking for something that says the color fades which would be a choice let's see choice B or choice D and it fades because of a forward shift that we just discussed so that was choice B question 30 31 for more questions with Leisha tell you and then we're done a chemist is studying the equilibrium nitrogen and oxygen make nitrogen monoxide at constant temperature and there's a graph here that shows what happens over time based on the graph which of the following would have occurred at time T I'll let you just look at the graph what occurred at time T during the equilibrium pause the video and see if you can figure that out so your choices nitrogen was added is that true a and B both say nitrogen was added so looking up at the graph the nitrogen concentration suddenly spiked while the other two concentrations didn't so that means twice a and B would say nitrogen was added those are both reasonable choices choice C and D which said the system was moved to a smaller flask doesn't make as much sense because if you had moved it to a smaller flask all three of the concentrations would have suddenly spiked not just the nitrogen so we're looking now at choices a or B because C and D didn't make sense so will there be a forward or reverse shift well we could answer that the same we were doing earlier by setting up the reaction quotient and looking at that but we have this graph right in front of us so looking up the graph we can see that after we added the nitrogen the nitrogen concentration fell and the oxygens concentration fell while nitrogen and oxygen are both reactants so if those are decreasing in concentration this must be a forward shift and that means the arrows concentration should have gone up and sure enough it did there's also some stoichiometry involved in this if you were to look closely you should find that the NO increased by twice as much as both the n2 and o2 co2 fell the balanced evasion ratios are one to two so looking at this we're looking for nitrogen being added and a forwarded shift occurring number 31 pause and give that a try so if you recall the question began by telling us that this was happening at constant temperature if temperature is being held constant then KC does not change right KC is only changed if the temperature changes if temperature were held constant KC should not change so choice C is the best answer there number 32 pause would give that a try so what is the effect of increasing temperature on this equilibrium looking at the balanced equation up above we see we're also told the enthalpy change the heat of reaction Delta H and I see that it's negative a negative Delta H means that this was an exothermic reaction so right away I'm going to recall that graph we saw earlier exothermic KC changes with temperature and it changes like this it drops as temperature is increased so here we had an increase in temperature which means KC should have dropped so KC Falls when the temperature increases if the case he fell q is unaffected that means Q is now bigger than KC KC got smaller and that caused a reverse shift to occur when Q is bigger than K so the equilibrium will shift reverse or left and KC gets smaller so choice B is the correct answer and that brings us to the last question which is not a multiple-choice question we have the harbor process were named after fritz haber a german chemist who figured out how to synthesize ammonia nh3 from nitrogen and hydrogen gases listen we have an equilibrium system with all three of those gases inside a sealed syringe so you can see here the syringe with all three gases present there's a plunger which can move in and out of the syringe but the rubber cap prevents the gas from escaping so we also notice that the reaction for the synthesis of ammonia was exothermic the Delta H is less than zero which means it's negative so this is an exothermic situation just like we saw earlier so let's just sketch the graph again KC in temperature for exothermic situations does something like that so we want to put check marks beside any of any modification that would increase the yield a greater yield of ammonia so we want this equi Librium to shift forwards to make more ammonia so let's write the Q expression Q is going to equal the concentration of ammonia squared over concentration of hydrogen cubed and nitrogen so but the Q expression and the graph for KC and temperature see if you can figure out which of these changes would cause a greater yield of ammonia in other words a forward shift in equilibrium so the first two choices we have to decide whether we should add or remove nitrogen well if we want the reaction to shift forwards then we want Q to become less than KC which causes a forward shift and more ammonia B to be produced so if we want to get Q to become a smaller so Q is less than K should we add nitrogen or should we remove nitrogen well nitrogen ism is in the denominator of Q so if you were to add more nitrogen the denominator would get bigger and Q would get smaller so the first choice there to add more nitrogen looks like a good choice hydrogen should be added or removed well this is very similar if you were to add more hydrogen hydrogen is also in the denominator of Q so adding hydrogen would make the denominator bigger that would make you get smaller she would be less than K and a forward shift occurs so adding hydrogen also looks good should be warm or cool the system while we want Q to become less than KC so far we've been doing that by making Q get smaller but if we change the temperature here we're gonna warm or cool it temperature affects the value of KC looking up at the graph above we want KC to be bigger than Q so we want KC to get bigger you want KC to increase well in this graph if you were to lower the temperature the KC gets bigger so we want to cool the system not heat it should we reduce the volume or should we increase the volume push the plunger in or pull the plunger out well this one's a little bit more complicated why don't we just try one let's reduce the volume if we reduce the volume as we saw earlier all of the concentrations would increase so both the numerator and the denominator would increase but just like before the denominator here has a bigger overall order its overall order would be 3 plus 1 looking at those two exponents so it's fourth order in the denominator while it's only second order the ammonia is squared in the numerator so while both would increase the denominator would increase even more and Q would get smaller which is what we wanted so yes reduced the volume should work and the last choice adding a catalyst would that increase the yield well we discussed that earlier adding a catalyst has no effect on Q and no effect on KC so the catalyst has no effect on the yield but it would make the system recheck the Librium faster but it won't change the yield of the ammonia alright so there's that's it for a review of equi Librium this should be a good review for both tests and for the final exam if you're an AP chemistry student be sure you also check out my youtube video which has equilibrium review questions specifically for Advanced Placement chem