WCLN - Equilibrium Calculations - 5
Equilibrium Calculations http://www.BCLearningNetwork.com. 0:00do 0:06in this example were given in equilibrium equation 0:09and only the initial concentrations and the value of k 0:13cue and her ass to calculate the concentrations of the reactants and 0:17products at equilibrium 0:18here's the example at 4:19 85 degrees Celsius 0:23the reaction h2 gas plus I to gas 0:26gives to HI gas has a key EQ value 49.8 0:311.5 malls have hiii is added to an empty 500 milliliter container at for an 85 0:38degrees 0:38and the system is allowed to reach equilibrium what are the concentrations 0:43are h2 0:44concentration divide to and concentration is hiii 0:48after the equilibrium has been established notice were given 0:52malls have hiii rather than concentration were also given the volume 0:57up the container 0:58remembering that molar concentration is equal to malls divided by leaders 1:02we can calculate the initial concentration have hiii 1:06it is 1.5 malls have hiii 1:09divided by the volume in liters the volume at the container is given as five 1:13hundred milliliters so we need to change it to leaders 1:16we take the five hundred milliliters and multiplied by the conversion factor 1:21one leader to 1,000 milliliters which gives us point 5000 leaders 1:27so the initial concentration is hiii 1:30has 1.50 moles divided by .5 000 leaders 1:35which comes out to 3.00 1:38moles per liter or 3.00 mauler 1:42so we'll make a noted that up here 1:45that the initial concentration abhi his three more 1:49because we're given initial information and were asked for equilibrium 1:53information 1:54we must use an a stable or an icebox 1:57has some teachers call in this particular type a problem 2:01where were given on the initial values in the cake you 2:04and no equilibrium values it's best to use three separate horizontal roles 2:10for equilibrium concentrations you see why 2:13as we work through this problem we can start by 2:17going to the cell which represents the initial concentration hiii 2:21and we write in the value for the initial concentration 2:24which is three mauler in this I stable 2:27will leave out the molarity unit to keep it simple if you read the question 2:33you'll see that it doesn't mention adding 2:35h2 RIT to the container it just says hiii is added to the container 2:41so we can assume that the initial concentrations a beach to an eye to 2:45are both 0 and we're 800 here 2:49our next step is to find the changes in concentration 2:53before we do that we need to know which way the reaction will move 2:57as it goes from its initial state to stay /div equilibrium 3:01an order to find out which way the reaction moods 3:05we insert the initial concentrations into decay 3:08cue expression in order to calculate a trial key kid 3:12so the trial KQ equals the concentration of H+ i squared divided by the product 3:19at the concentration at HD 3:20times concentration a bite to putting in 3d and for the concentration 3:26hiii n0n for the concentrations h2 an eye to 3:30we get three squared divided by 0 now anything divided by zero is undefined 3:36or infinitely large so the trout KQ 3:40which is infinitely large is much greater than 3:43the actual KQ which is given on the top of the screen 3:47as forty-nine point me because the trout K Q is much greater than the actual KQ 3:53initial ratio products to reactants is much too high 3:56so in order to achieve equilibrium the reaction will move to the left 4:01notice both h2 and I to you have initial concentrations have 0 4:06it is useful to know that a reaction will always move toward decide which has 4:12one or more 0 concentrations 4:14so in this case because %eh to an eye to both have initial concentrations have 0 4:19the reaction will move toward the left in cases where there is zero 4:24concentration on one side 4:26you don't need to calculate trout KQ 4:29you now know that the reaction will always move toward the side with the 4:32zero initial concentration 4:34because the reaction moves to the left the concentrations are both reactants 4:39h2 an eye to will increase so their changes in concentration 4:44will be positive and the concentration of the product 4:48hiii will decrease so which change will be negative 4:52were not given any concentrations in equilibrium so we don't know how much 4:57h2 will be increasing concentration so we'll call it X 5:03to find the changes in concentration at the other species
Equilibrium Calculations http://www.BCLearningNetwork.com. 0:00do 0:06in this example were given in equilibrium equation 0:09and only the initial concentrations and the value of k 0:13cue and her ass to calculate the concentrations of the reactants and 0:17products at equilibrium 0:18here's the example at 4:19 85 degrees Celsius 0:23the reaction h2 gas plus I to gas 0:26gives to HI gas has a key EQ value 49.8 0:311.5 malls have hiii is added to an empty 500 milliliter container at for an 85 0:38degrees 0:38and the system is allowed to reach equilibrium what are the concentrations 0:43are h2 0:44concentration divide to and concentration is hiii 0:48after the equilibrium has been established notice were given 0:52malls have hiii rather than concentration were also given the volume 0:57up the container 0:58remembering that molar concentration is equal to malls divided by leaders 1:02we can calculate the initial concentration have hiii 1:06it is 1.5 malls have hiii 1:09divided by the volume in liters the volume at the container is given as five 1:13hundred milliliters so we need to change it to leaders 1:16we take the five hundred milliliters and multiplied by the conversion factor 1:21one leader to 1,000 milliliters which gives us point 5000 leaders 1:27so the initial concentration is hiii 1:30has 1.50 moles divided by .5 000 leaders 1:35which comes out to 3.00 1:38moles per liter or 3.00 mauler 1:42so we'll make a noted that up here 1:45that the initial concentration abhi his three more 1:49because we're given initial information and were asked for equilibrium 1:53information 1:54we must use an a stable or an icebox 1:57has some teachers call in this particular type a problem 2:01where were given on the initial values in the cake you 2:04and no equilibrium values it's best to use three separate horizontal roles 2:10for equilibrium concentrations you see why 2:13as we work through this problem we can start by 2:17going to the cell which represents the initial concentration hiii 2:21and we write in the value for the initial concentration 2:24which is three mauler in this I stable 2:27will leave out the molarity unit to keep it simple if you read the question 2:33you'll see that it doesn't mention adding 2:35h2 RIT to the container it just says hiii is added to the container 2:41so we can assume that the initial concentrations a beach to an eye to 2:45are both 0 and we're 800 here 2:49our next step is to find the changes in concentration 2:53before we do that we need to know which way the reaction will move 2:57as it goes from its initial state to stay /div equilibrium 3:01an order to find out which way the reaction moods 3:05we insert the initial concentrations into decay 3:08cue expression in order to calculate a trial key kid 3:12so the trial KQ equals the concentration of H+ i squared divided by the product 3:19at the concentration at HD 3:20times concentration a bite to putting in 3d and for the concentration 3:26hiii n0n for the concentrations h2 an eye to 3:30we get three squared divided by 0 now anything divided by zero is undefined 3:36or infinitely large so the trout KQ 3:40which is infinitely large is much greater than 3:43the actual KQ which is given on the top of the screen 3:47as forty-nine point me because the trout K Q is much greater than the actual KQ 3:53initial ratio products to reactants is much too high 3:56so in order to achieve equilibrium the reaction will move to the left 4:01notice both h2 and I to you have initial concentrations have 0 4:06it is useful to know that a reaction will always move toward decide which has 4:12one or more 0 concentrations 4:14so in this case because %eh to an eye to both have initial concentrations have 0 4:19the reaction will move toward the left in cases where there is zero 4:24concentration on one side 4:26you don't need to calculate trout KQ 4:29you now know that the reaction will always move toward the side with the 4:32zero initial concentration 4:34because the reaction moves to the left the concentrations are both reactants 4:39h2 an eye to will increase so their changes in concentration 4:44will be positive and the concentration of the product 4:48hiii will decrease so which change will be negative 4:52were not given any concentrations in equilibrium so we don't know how much 4:57h2 will be increasing concentration so we'll call it X 5:03to find the changes in concentration at the other species