Full transcript
0:01Such reactions with a double arrow are
0:03called reversible reactions or
0:05equilibrium reactions . In these , not
0:07only do the reactants react with each
0:09other to become products as in normal
0:11reactions , but the product can also
0:13become the reactant again . After some
0:15time , however , the ratio between the
0:16starting materials and final products
0:18remains constant . And to calculate the
0:20quantities of products and reactants
0:22you get in the end , you need the Law of
0:24Mass Action . In this video , I will show
0:27you what that is and how to apply it .
0:29To understand everything well , you
0:31should already know what a chemical
0:33equilibrium is . We will start with a
0:35simple example first . Hydrogen reacts
0:38with iodine to form hydrogen iodide .
0:41Once we have understood the basic
0:43principle , we will look at task B. Here
0:45, carbon dioxide reacts with ammonia
0:47and water to form ammonium carbonate .
0:51The Law of Mass Action describes the
0:54ratio between the multiplied
0:55concentrations of the final products
0:57and the concentrations of the reactants
0:59. The concentration of the two
1:02reactants , hydrogen and iodine ,
1:04therefore go into the denominator .
1:06Instead of C for concentration , you
1:08simply take the chemical formula of the
1:10substance and put it in square brackets
1:12. Then you enter the product of the
1:14final substances , i.e. , hydrogen iodide
1:16, in the numerator . However , we must
1:19note that it has the stoichiometric
1:20factor of 2 in the reaction equation .
1:23We include this as an exponent in the
1:25Law of Mass Action . Of course , we must
1:27also consider the stoichiometric
1:29factors of the other substances . But in
1:31this example , they are each one and
1:33therefore drop out . In the end , you
1:37determine the unit of the equilibrium
1:39constant K , because the special thing
1:41about the Law of Mass Action is that
1:42the unit of K changes depending on the
1:45reaction equation . For this , we insert
1:48the unit of concentration . That is
1:50moles per liter . In this simple example
1:52, all units cancel each other out . This
1:55always happens when the stoichiometric
1:57number of the reactants in the reaction
1:59equation is equal to that of the
2:00products . Now for task B. In the second
2:04equilibrium reaction , carbon dioxide ,
2:06ammonia , and water react to form
2:08ammonium carbonate . We write all
2:11reactants into the denominator again
2:13and pay special attention to the
2:14stoichiometric factor of the ammonia .
2:17At this point , we can simplify the Law
2:19of Mass Action for the reactants by
2:21crossing water out of the equation .
2:24This is possible in this example
2:25because the concentration of water
2:27practically does not change due to the
2:29reaction and is negligible . This is the
2:31case because everything takes place in
2:33an aqueous solution . There , the few
2:35water molecules that form during the
2:37course of the reaction do not make a
2:39big difference . As a product , one
2:41molecule of ammonium carbonate is now
2:43formed . We write its concentration
2:45again in square brackets in the
2:46numerator . In the last step , we
2:49calculate the unit of K again . This
2:51time , we have more concentrations in
2:53the denominator than in the numerator .
2:55So , as a result , we get liters squared
2:57per mole squared as the unit for k ,
2:59which is the inverse square of the
3:01concentration . Let's summarize the
3:03steps and solutions for both problems
3:05once again . The law of mass action is
3:08formed from the quotient of the
3:10concentrations of product and reactants
3:12. The stoichiometric factors are
3:15written as exponents of the
3:16concentrations . Concentrations of
3:19substances in excess can be omitted
3:20from the equation because their
3:22concentration does not change . The
3:25equilibrium constant K does not have a
3:27fixed unit . It can therefore also be
3:29dimensionless . Finally , here are the
3:31solutions to both problems once more .
3:33In the case of problem A , K is
3:35dimensionless , and in problem B , K has
3:38the inverse square of the concentration
3:40as its unit .
3:42Hi , I hope you liked the video and that
3:44we were able to help you .
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3:48material ,
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