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Understanding Equilibrium Constants in Chemical Reactions, Study notes of Chemistry

Missouri State University (MSU)Chemistry

The concept of equilibrium constants in chemical reactions, focusing on forward and reverse reactions, rate constants, and le chatelier's principle. The lecture covers the relationship between concentration and rate constants, the effect of temperature, pressure, and concentration on chemical equilibrium, and how to write equilibrium expressions.

Typology: Study notes

Pre 2010

Uploaded on 07/23/2009

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CHM 105 & 106 MO1 UNIT FIVE, LECTURE FOUR 1

CHM 105/106 Program 44: Unit 5 Lecture 4

IN OR LAST LECTURE WE WERE LOOKING AT OR JUST BEGAN TALKING ABOUT REACTION RATES.

AGAIN RELATIVE TO THE FACT THAT IF A REACTION CAN GO OVER THE ACTIVATION ENERGY

BARRIER IN ONE DIRECTION IT MAY WELL BE ABLE TO GO OVER THE REACTION BARRIER OR

ACTIVATION ENERGY IN THE OTHER DIRECTION. AND AGAIN USING OUR BASIC REACTION A

PLUS B TO PRODUCE PRODUCT C WE SAID THAT INITIALLY HERE IF WE WERE TO TAKE OUR

CONTAINER AND WE WERE TO PUT SOME A PARTICLES IN HERE AND SOME B PARTICLES WE

COULD GET SOME AB COLLISIONS THAT WOULD OCCUR. AND IF THE AB COLLISION OCCURRED

WITH AT LEAST THE ACTIVATION ENERGY THEN IT COULD BECOME PRODUCT C. WELL AFTER

SOME LENGTH OF TIME THEN AS THE SYSTEM BEGINS TO OPERATE OF COURSE WE’RE GOING TO

FORM SOME C, SO WE’LL STICK SOME C’S IN THERE, AND IT WOULD NOT BE UNLIKELY THAT

OCCASIONALLY WE’RE GOING TO GET A C COLLIDING WITH A C, AND WHEN THAT HAPPENS OF

COURSE WE MIGHT GET WHAT WE CALL THE REVERSE REACTION. IN OTHER WORDS THE C’S NW

BREAKING APART ONCE AGAIN TO REFORM PRODUCT OR THE REACTANTS A AND B. BECAUSE

AGAIN IF WE LOOKED AT THE ACTIVATION DIAGRAM FOR THIS AND SO WE’RE STARTING WITH

OUR A AND B. WE HAVE SOME ACTIVATION ENERGY BARRIER AND WE GET C BUT THIS IS

REPRESENTATIVE OF THEN WHAT WE WOULD CALL AN EXOTHERMIC REACTION. ENERGY

BEING GIVEN OFF, AND SO IF WE HAVE ENERGY IN THE SYSTEM FOR A AND B TO GET OVER THE

HILL THEN IN THAT SAME SYSTEM WE MAY WELL HAVE ENOUGH ENERGY FOR WHEN TWO C’S

COLLIDE FOR THEM, LIKEWISE GET OVER THAT ENERGY BARRIER. SO WE TALKED ABOUT THEN

THAT WE REALLY HAVE TWO POSSIBLE REACTIONS OCCURRING HERE. WE HAVE A FORWARD

REACTION WHICH INVOLVES AND B COLLIDING WITH EACH OTHER AND WE DRAW THIS ARROW

SO WE’LL CALL THAT THE FORWARD AND OF COURSE THE FORWARD REACTION IN THIS CASE IS

DEPENDENT UPON OR PROPORTIONAL TO WE WOULD SAY THE CONCENTRATION OF A AND THE

CONCENTRATION OF THE B PARTICLES. WE SAW IN THE PREVIOUS CHAPTER THAT OF COURSE

ONE OF THE FACTORS CONTROLLING THE RATE OF A CHEMICAL REACTION IS IN FACT THE

CONCENTRATION WHICH AFFECTS THE NUMBER OF COLLISIONS AND SO WE HAVE THEN A

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CHM 105/106 Program 44: Unit 5 Lecture 4

IN OR LAST LECTURE WE WERE LOOKING AT OR JUST BEGAN TALKING ABOUT REACTION RATES. AGAIN RELATIVE TO THE FACT THAT IF A REACTION CAN GO OVER THE ACTIVATION ENERGY BARRIER IN ONE DIRECTION IT MAY WELL BE ABLE TO GO OVER THE REACTION BARRIER OR ACTIVATION ENERGY IN THE OTHER DIRECTION. AND AGAIN USING OUR BASIC REACTION A PLUS B TO PRODUCE PRODUCT C WE SAID THAT INITIALLY HERE IF WE WERE TO TAKE OUR CONTAINER AND WE WERE TO PUT SOME A PARTICLES IN HERE AND SOME B PARTICLES WE COULD GET SOME AB COLLISIONS THAT WOULD OCCUR. AND IF THE AB COLLISION OCCURRED WITH AT LEAST THE ACTIVATION ENERGY THEN IT COULD BECOME PRODUCT C. WELL AFTER SOME LENGTH OF TIME THEN AS THE SYSTEM BEGINS TO OPERATE OF COURSE WE’RE GOING TO FORM SOME C, SO WE’LL STICK SOME C’S IN THERE, AND IT WOULD NOT BE UNLIKELY THAT OCCASIONALLY WE’RE GOING TO GET A C COLLIDING WITH A C, AND WHEN THAT HAPPENS OF COURSE WE MIGHT GET WHAT WE CALL THE REVERSE REACTION. IN OTHER WORDS THE C’S NW BREAKING APART ONCE AGAIN TO REFORM PRODUCT OR THE REACTANTS A AND B. BECAUSE AGAIN IF WE LOOKED AT THE ACTIVATION DIAGRAM FOR THIS AND SO WE’RE STARTING WITH OUR A AND B. WE HAVE SOME ACTIVATION ENERGY BARRIER AND WE GET C BUT THIS IS REPRESENTATIVE OF THEN WHAT WE WOULD CALL AN EXOTHERMIC REACTION. ENERGY BEING GIVEN OFF, AND SO IF WE HAVE ENERGY IN THE SYSTEM FOR A AND B TO GET OVER THE HILL THEN IN THAT SAME SYSTEM WE MAY WELL HAVE ENOUGH ENERGY FOR WHEN TWO C’S COLLIDE FOR THEM, LIKEWISE GET OVER THAT ENERGY BARRIER. SO WE TALKED ABOUT THEN THAT WE REALLY HAVE TWO POSSIBLE REACTIONS OCCURRING HERE. WE HAVE A FORWARD REACTION WHICH INVOLVES AND B COLLIDING WITH EACH OTHER AND WE DRAW THIS ARROW SO WE’LL CALL THAT THE FORWARD AND OF COURSE THE FORWARD REACTION IN THIS CASE IS DEPENDENT UPON OR PROPORTIONAL TO WE WOULD SAY THE CONCENTRATION OF A AND THE CONCENTRATION OF THE B PARTICLES. WE SAW IN THE PREVIOUS CHAPTER THAT OF COURSE ONE OF THE FACTORS CONTROLLING THE RATE OF A CHEMICAL REACTION IS IN FACT THE CONCENTRATION WHICH AFFECTS THE NUMBER OF COLLISIONS AND SO WE HAVE THEN A

DESCRIPTION OF DEFINING THE RATE OF THIS FORWARD REACTION, BUT THEN WE ALSO HAVE

THE POSSIBILITY AS WE SAID OF THIS REACTION HAPPENING WHICH IS THE REVERSE REACTION

NOW. SO WE’LL JUST PUT A LETTER R THERE FOR IT FOR REVERSE AND THE REVERSE REACTION

THEN IS GOING TO BE PROPORTIONAL TO THE CONCENTRATION OF THE C PARTICLES. BECAUSE

AS WE HAVE C PARTICLES BEING FORMED THERE IS MORE AND MORE PROBABILITY OF HAVING

A C-C PARTICLE COLLISION AND THEREFORE A GREATER CHANCE OF THE REACTION

PROCEEDING IN THE REVERSE DIRECTION. LET’S THEN STILL USING THE SAME EQUATION, LET’S

CHANGE OUR DESCRIPTION NOW FROM A PROPORTIONALITY TO AN EQUALITY AND WE DID

THIS WHEN WE LOOKED AT THE IDEAL GAS EQUATION. WETALKED ABOUT THE FACT THAT WE

COULD CHANGE FROM PRESSURE BEING PROPORTIONAL TO N AND R AND T AND B. ALRIGHT,

WE SAID THAT WE COULD MAKE THAT AN EQUALITY BY PUTTING IN A PROPORTIONALITY

CONSTANT K. IN THE CASE OF THE GASES OF COURSE WE REPLACED THAT K WITH THE LETTER

R BUT IT’S STILL A CONSTANT. SO WE CAN REWRITE THIS NOW, THE RATE OF THE FORWARD

REACTION IS EQUAL TO – SO NOW WE’RE TALKING ABOUT AN EQUALITY, NOT A

PROPORTIONALITY, MATHEMATICALLY. IT’S EQUAL TO A K, WHICH WE CALL A RATE

CONSTANT AND THAT IS GOING TO BE A NUMBER THAT IS SPECIFIC FOR THAT PARTICULAR

CHEMICAL REACTION. THIS IS NOT A UNIVERSAL K LIKE R WAS TO THE GASEOUS BEHAVIOR.

THIS K IS VERY SPECIFIC TO A PARTICULAR CHEMICAL PROCESS THAT’S OCCURRING. WELL WE

CAN DO THE SAME THING FOR THE REVERSE REACTION NOW AND WE CAN SAY THEN THE RATE

OF THE REVERSE REACTION IS EQUAL TO SOME RATE CONSTANT K TIMES THE CONCENTRATION

OF C. NOW, IF WE WERE TO TAKE A LOOK AT THESE TWO RATES OF THE REACTION HERE, SO

WE’RE GOING TO LOOK AT THE RATE NOW AND THE…TRYING TO GET THIS ON HERE RIGHT,

THERE WE GO…IF WE LOOK FIRST OF ALL AT THE FORWARD RATE AND THINK ABOUT IT FOR A

SECOND, KEEP IN MIND THAT WE HAVE A CLOSED CONTAINER. AS THE REACTION PROCEEDS TO

FORM C WHAT’S HAPPENING TO THE CONCENTRATION OF A AND B? ANYONE? IT DECREASES.

WE HAVE TO USE A AND B UP TO FORM C RIGHT? WE DON’T JUST SPONTANEOUSLY GET C OUT

OF NOTHING. A AND B PARTICLES ARE BEING USED UP TO FORM THE C. WHICH TELLS US THEN

THAT THE RATE OF THE FORWARD REACTION IF A AND B ARE DECREASING WHAT’S GOING TO

CONDENSATION WE HAVE AN EQUILIBRIUM. AND WE TALKED ABOUT THE EQUILIBRIUM VAPOR

PRESSURE OF A LIQUID. AND HERE WE’RE TALKING NOW ABOUT AN EQUILIBRIUM AGAIN, TWO

OPPOSING PROCESSES HAVE BECOME EQUAL AND WE’VE REACHED EQUILIBRIUM, AND AS IN

THE PREVIOUS TWO EQUILIBRIUMS THAT WE TALKED ABOUT THIS IS A DYNAMIC EQUILIBRIUM.

BY THAT WE MEAN THAT AT ANY INSTANT IN TIME IF WE COULD STOP AND FREEZE THE

SYSTEM THE SAME PARTICLES ARE NOT NECESSARILY IN THE SAME FORM. IN OTHER WORDS

SOME OF THE PARTICLES THAT USED TO BE A AND B’S HAVE BECOME C AND SOME OF THE

PART ICLES THAT WERE C HAVE GONE BACK TO BEING A AND B. SO THERE’S A CONSTANT

PROCESS GOING ON, A DYNAMIC PROCESS GOING ON BUT AS FAR AS THE CONCENTRATIONS,

ONCE EQUILIBRIUM HAS BEEN REACHED WE WOULDN’T NOTICE ANY CHANGE IN

CONCENTRATION. IF ONE OF THESE SPECIES WERE COLORED FOR INSTANCE THE SOLUTION

WOULD REACH A MAXIMUM COLOR AND THAT WOULD BE IT, BUT IT DOESN’T MEAN THAT THE

CHEMICAL REACTION HAS STOPPED, IT ONLY MEANS THAT THE TWO PROCESSES HAVE BECOME

EQUAL AND SO IN CHEMICAL REACTIONS THAT WE CAN TALK ABOUT THIS EQUILIBRIUM. NOW

THIS EQUILIBRIUM IS IMPORTANT TO US FROM AN INDUSTRIAL STANDPOINT BECAUSE FOR

INSTANCE IF I WANT TO PRODUCE C HERE AS MY MONEY PRODUCT, IF I WANT TO PRODUCE C I

WOULD LIKE TO IF AT ALL POSSIBLE MAKE SURE THAT I GET A LOT OF C BEFORE EQUILIBRIUM

IS REACHED. AND THE QUESTION BECAME CAN ONE AFFECT THE RATE OF A CHEMICAL

REACTION? CAN, I SHOULD SAY IT DIFFERENTLY THAN THAT. CAN ONE AFFECT THE

CONDITIONS OF A CHEMICAL REACTION SUCH THAT YOU CAN FAVOR ONE SIDE OR THE OTHER

AT EQUILIBRIUM? IN OTHER WORDS CAN WE DO SOMETHING TO A REACTION TO GET A

GREATER AMOUNT OF C? CAN WE DO SOMETHING TO A REACTION TO PREVENT C FROM BEING

FORMED? THAT MIGHT BE JUST AS IMPORTANT TO US. WELL A FRENCH CHEMIST BY THE NAME

OF LECHATELIER STUDIED A LOT OF CHEMICAL SYSTEMS AT EQUILIBRIUM. IN OTHER WORDS,

HE ALLOWED THE SYSTEM TO REACH EQUILIBRIUM AND THEN HE SUBJECTED IT TO SOME

CHANGES IN CONDITIONS AND STUDIED WHAT EFFECT THOSE CHANGES HAD, AND

LECHATELIER’S PRINCIPLE THEN BASICALLY STATES THAT IF WE HAVE A SYSTEM WHICH HAS

REACHED EQUILIBRIUM…NOW WHEN WE TALK ABOUT SYSTEM HERE WE’RE TALKING ABOUT A

CHEMICAL REACTION…IF WE HAVE A SYSTEM AT EQUILIBRIUM AND SUBJECT IT TO A

CHANGE…NOW THAT CHANGE COULD BE A CHANGE IN TEMPERATURE, CHANGE IN

CONCENTRATION, CHANGE IN PRESSURE, CHANGE IN VOLUME, WE SUBJECT IT TO SOME

PARTICULAR CHANGE AND LECHATELIER SAID THE SYSTEM WILL ATTEMPT TO OFFSET THE

APPLIED CHANGE. NOW IN OTHER WORDS WHAT LECHATELIER FOUND WAS IF I HAVE A

CHEMICAL REACTION AND IT HAS REACHED EQUILIBRIUM AND I CHANGE THE PRESSURE ON

THAT SYSTEM THE SYSTEM WOULD TRY TO DO JUST THE OPPOSITE. IT WOULD TRY TO GET RID

OF THAT INCREASE IN PRESSURE. IT WOULD TRY TO DECREASE THE PRESSURE. IF I LOWERED

THE TEMPERATURE OF A CHEMICAL REACTION IT WOULD TRY TO INCREASE ITS TEMPERATURE

BACK. IT WILL TRY TO DO EXACTLY OPPOSITE OF WHATEVER I DO TO THE SYSTEM. NOW IT

WILL ATTEMPT, HOWEVER THERE’S MANY CASES WHERE WE FIND THAT THE SYSTEM CAN DO

NOTHING ABOUT THE CHANGE THAT WE HAVE APPLIED, AND THEREFORE THE EQUILIBRIUM

CAN NOT BE AFFECTED. BUT LECHATELIER WAS SAYING THAT IF WE SUBJECT A SYSTEM TO A

CHANGE AND IF THE SYSTEM CAN IT WILL TRY TO OFFSET THAT CHANGE…IT WILL DO THE

OPPOSITE THING. THE BEST WAY TO GET A FEEL FOR LECHATELIER’S PRINCIPLE THEN IS TO

LOOK AT A PARTICULAR CHEMICAL PROCESS OR CHEMICAL REACTION AND MAKE SOME

PREDICTIONS AND SEE IF WE CAN UNDERSTAND WHAT IS OCCURRING. HERE NOW IS A

CHEMICAL REACTION BETWEEN GASEOUS WATER, SOLID CARBON, ENERGY IN THE FORM OF

HEAT, AND OUR PRODUCTS ARE CARBON DIOXIDE GAS AND HYDROGEN GAS. AND WE SHOW

EQUILIBRIUM IN A SYSTEM BY SHOWING A DOUBLE ARROW. SO WHEN WE WRITE A DOUBLE

ARROW IN A CHEMICAL REACTION IT INDICATES THAT THE CHEMICAL REACTION IS IN FACT A

REVERSIBLE REACTION MEANING THAT IT CAN REACH EQUILIBRIUM. NOW THIS SYSTEM HAS

REACHED EQUILIBRIUM. AND NOW WE’RE GOING TO APPLY LECHATELIER’S PRINCIPLE TO

MAKE SOME PREDICTIONS. THE FIRST IS LET’S SUPPOSE THAT WE DECREASE THE PRESSURE ON

THIS REACTION. WHAT WOULD THAT DO TO THE EQUILIBRIUM INVOLVED? IN OTHER WORDS

WILL IT NOW SHIFT TOWARDS MORE PRODUCTS OR SHIFT TOWARDS MORE REACTANTS –

WHATEVER THE CASE. NOW IF I DECREASE THE PRESSURE, IF I DECREASE THE PRESSURE THAT

SYSTEM WILL IF POSSIBLE TRY TO INCREASE THE PRESSURE. NOW WE HAVE TO GET A MENTAL

GASEOUS MOLECULES. IT MEANS IT WOULD SHIFT TOWARD THE REACTANT SIDE. ALRIGHT,

LET’S LOOK AT THIS ONE. TEMPERATURE IS DECREASED. NOW NOTICE THAT IN ORDER FOR

THIS REACTION TO BECOME PRODUCTS WE HAVE TO BE PUTTING ENERGY IN. IT’S AN

ENDOTHERMIC PROCESS. SO THEREFORE IF I WERE TO DECREASE THE AMOUNT OF ENERGY

AVAILABLE THE SYSTEM WILL TRY TO GIVE ME BACK THAT ENERGY. IF I START PULLING

ENERGY OUT IT WILL GIVE THE ENERGY BACK. THE REACTION IN THIS DIRECTION IS

EXOTHERMIC. IT’S ENDOTHERMIC AS IT’S WRITTEN FROM LEFT TO RIGHT BUT IT’S EXOTHERMIC

AS VIEWED FROM RIGHT TO LEFT – THE REVERSE REACTION. SO THE FORWARD REACTION IS

ENDOTHERMIC AND THE REVERSE REACTION IS EXOTHERMIC. SO IF I DECREASE THE

TEMPERATURE THE SYSTEM WILL TRY TO INCREASE THE TEMPERATURE. IT’LL TRY TO PUT

THAT HEAT ENERGY BACK AND TO PUT THAT HEAT ENERGY BACK IT WOULD TRY TO GO IN THE

EXOTHERMIC DIRECTION. SO IF I DECREASE THIS THE SYSTEM IS GOING TO GO THIS DIRECTION

TO TRY TO GIVE ME THAT ENERGY BACK. SO IT IS GOING TO MOVE TO THE LEFT OR TO THE

REACTANT SIDE. SO IN OTHER WORDS AGAIN IF I WANTED TO PRODUCE HYDROGEN GAS AS A

PRODUCT FROM AN EQUILIBRIUM STANDPOINT I WOULD WANT TO HAVE THE TEMPERATURE AS

HIGH AS POSSIBLE BECAUSE INCREASING THIS WOULD SHIFT EQUILIBRIUM THAT DIRECTION

GIVING ME A GREATER AMOUNT OF CARBON DIOXIDE AND WATER. ALRIGHT. NOW, LET'S JUMP

C A SECOND. LET’S GO DOWN TO D HERE. SOME HYDROGEN IS REMOVED. SOME HYDROGEN IS

REMOVED. IN OTHER WORDS BACK TO MY CONTAINER HERE. IF I SOMEHOW COULD PUT IN

SOME SORT OF A MATERIAL THAT WOULD ATTRACT HYDROGEN OVER HERE. SO SOME OF IT IS

REMOVED. OKAY. SO THAT’S WHAT WE’RE SAYING. WE’RE GOING TO REMOVE SOME OF THE

HYDROGEN. NOW AGAIN, ACCORDING TO LECHATELIER IF I REMOVE SOME HYDROGEN, IF I

DECREASE THE HYDROGEN THE SYSTEM WILL TRY TO INCREASE THE HYDROGEN. WHATEVER I

DO THE SYSTEM’S GOING TO TRY TO DO THE OPPOSITE. WELL THE ONLY WAY IT CAN INCREASE

THE HYDROGEN WOULD BE FOR MORE WATER TO REACT WITH CARBON TO FORM MORE

HYDROGEN. SO IF I PULL SOME OF THIS OUT THEN THE SYSTEM IS GOING TO SHIFT IN THAT

DIRECTION TO TRY TO PUT IT BACK. AND SO WE’RE GOING TO GET A SHIFT NOW TO THE RIGHT

FOR THAT PROCESS AND/OR TOWARD THE PRODUCT SIDE. SO IN OTHER WORDS IF I PULLED

SOME OUT THE SYSTEM WILL SHIFT. I WILL HAVE A DECREASE IN THE AMOUNT OF WATER NOW

BECAUSE SOME OF IT’S GOING TO BE USED UP. I’LL GET AN INCREASE IN THE CO 2 , AND I’LL GET

AN INCREASE BACK IN THE HYDROGEN. IT’S GOING TO TRY TO REPLACE THAT HYDROGEN

THAT I REMOVED. NOW, THE LAST QUESTION HERE DEALS WITH A CATALYST. A PALLADIUM

CATALYST IS ADDED. NOW WE KNOW THAT A CATALYST IS A CHEMICAL THAT SPEEDS UP A

CHEMICAL PROCESS AND IT’S ULTIMATELY REGENERATED IN THE FINAL STEPS OF THE

REACTION. CATALYST IS NOT USED UP IN THE PROCESS. IT CAN BE USED OVER AND OVER AND

OVER AND OVER AND OVER AND OVER. VERY SMALL AMOUNT CAN CATALYZE A LOT OF A

REACTION. ALRIGHT, THE QUESTION IS IF I ADD A CATALYST WILL THIS FAVOR ONE SIDE OR

THE OTHER? WELL IF WE GO BACK TO OUR ENERGY DIAGRAM, WE’LL JUST DRAW IT ON HERE.

IF WE LOOK AT AN ENERGY DIAGRAM ONCE AGAIN AND THIS WOULD BE OUR REACTANTS OVER

HERE. IN THIS CASE IT WOULD BE THE H 2 O AND THE CARBON BUT WE’LL JUST WRITE H 2 O AND

WE HAVE THIS BARRIER HERE AND WE HAVE OUR GASEOUS PRODUCTS AND I’LL JUST PUT THE

HYDROGEN HERE, BUT IF WE PUT IN A CATALYST WHAT DOES A CATALYST DO? IT LOWERS

THAT ACTIVATION ENERGY BARRIER RIGHT? IT CUTS THIS HILL DOWN IN SOME FASHION BY

PROVIDING A NEW PATHWAY. BUT NOTICE THAT IF I IN FACT CUT THIS HILL DOWN TO THIS

DIRECTION WHATEVER I’VE DONE TO DECREASE THE ACTIVATION ENERGY FOR THIS REACTION

I HAVE ALSO DECREASED THE ACTIVATION ENERGY FOR THE REVERSE REACTION BY THE SAME

AMOUNT. AND THEREFORE WHATEVER I HAVE DONE ENERGY-WISE TO ONE PROCESS I HAVE

DONE FOR THE OTHER PROCESS AND SO IN FACT ADDING A CATALYST TO A SYSTEM THEN HAS

NO EFFECT ON THE EQUILIBRIUM. ADDING A CATALYST DOES NOT AFFECT THE RATIO OF

PRODUCTS TO REACTANTS THAT WE GET. OKAY. SO CATALYST DOES NOT AFFECT

EQUILIBRIUM. NOW A CATALYST AFFECTS HOW FAST WE GET TO EQUILIBRIUM BUT A

CATALYST DOESN’T AFFECT AT WHAT POINT WE REACH EQUILIBRIUM. IN OTHER WORDS HOW

MUCH OF EACH OF THESE IS PRESENT IS INDEPENDENT OF WHETHER WE USE A CATALYST OR

NOT. IT WILL BE THE SAME WHETHER WE USE A CATALYST OR NOT BUT THE VALUE OF THE

CATALYST IS IT ALLOWS US TO GET TO THAT EQUILIBRIUM QUICKER AND FROM AN INDUSTRIAL

STANDPOINT THAT’S IMPORTANT BECAUSE IF I’M PUTTING ENERGY IN IN THE PROCESS TO

ON THE PISTON AND WHEN I DECREASE THE VOLUME NOW WE GO BACK TO OUR GAS LAWS.

WHEN I DECREASE THE VOLUME WHAT HAPPENS TO THE PRESSURE? IT INCREASES. OKAY.

VOLUME AND PRESSURE INVERSELY PROPORTIONAL. SO WE SAID VOLUME IS DECREASED

THEREFORE PRESSURE IS INCREASED. NOW IF THE PRESSURE IS INCREASED THE SYSTEM

SHOULD TRY TO DO WHAT? WE’VE INCREASED THE PRESSURE, THE SYSTEM WILL TRY TO?

DECREASE IT. ALRIGHT, NOW THE ONLY WAY THE SYSTEM CAN DECREASE THE PRESSURE IS TO

DECREASE THE NUMBER OF GASEOUS PARTICLES. WELL, ALRIGHT, SO WE WANT TO DECREASE

THE ONLY GASEOUS PARTICLE WE HAD IN THE WHOLE THING, IN THE WHOLE EQUATION WAS

SO 2. AND WE WANT TO DECREASE IT BECAUSE THE PRESSURE IS INCREASED. SO LET’S GO BACK

AND LOOK AT THE CHEMICAL EQUATION. IN ORDER TO DECREASE THE SO 2 WHICH WAY DOES

THE REACTION HAVE TO GO? TOWARD THE REACTANT RIGHT? IT NEEDS TO GO THIS WAY TO

USE THE SO 2 UP. IF THE SO 2 AND WATER IN THIS REACT TO FORM THIS IT WILL DECREASE THE

SO 2. SO THEREFORE IF VOLUME IS DECREASED PRESSURE IS INCREASED. THE SYSTEM’LL TRY

TO DECREASE THE PRESSURE. IT’LL SHIFT TO THE SIDE WITH THE SIDE WITH THE FEWEST

NUMBER OF GASEOUS PARTICLES. SO IT’LL SHIFT IN THIS CASE TO THE LEFT TOWARDS

FORMING OR KEEPING MORE REACTANTS. OKAY. SOME CALCIUM CARBONATE IS REMOVED.

WHOOPS, I THINK I MISWROTE THAT. I THINK THAT MAYBE THIS SHOULD BE CALCIUM SULFITE

BECAUSE THAT’S THE ONLY SOLID. AND SOME CALCIUM SULFITE SOLID IS REMOVED. WHAT

WILL THIS DO ON THE EQUILIBRIUM? AGAIN, IF I THINK ABOUT OUR PICTURE HERE. IF WE HAD

A CHUNK OF CALCIUM SULFITE DOWN HERE, THAT’S THE SOLID, CaSO 3 SOLID. LET’S SUPPOSE THAT I REMOVED HALF OF IT. WENT IN AND I JUST PICKED OUT HALF OF THE SOLID THAT WAS IN THERE. IS THAT GOING TO AFFECT THE EQUILIBRIUM? THE QUESTION IS BY TAKING SOME OF THE SOLID OUT DID I CHANGE THE CONCENTRATION OF THE SOLID? NO. BECAUSE CONCENTRATION OF A SOLID IS ONLY DEPENDENT UPON ITS DENSITY AND WHETHER I HAVE A LITTLE TINY CHUNK OR A GREAT BIG BLOCK IN THERE IT DOESN’T MAKE ONE BIT OF DIFFERENCE. IT’S THE SAME CONCENTRATION OF CALCIUM SULFITE IN THERE. SO THEREFORE ADDING OR REMOVING A SOLID DOES NOT CHANGE CONCENTRATION AND THEREFORE IF WE’RE NOT CHANGING CONCENTRATION OF ANYTHING WE’RE OBVIOUSLY NOT CHANGING THE RATE

OF A NYTHING. AND SO ADDING OR REMOVING A SOLID FROM A SYSTEM AT EQUILIBRIUM DOES

NOT AFFECT THE EQUILIBRIUM. SO THIS ONE THEN WOULD HAVE NO EFFECT. NO EFFECT.

ALRIGHT, LET’S LOOK AT THIS ONE. SOME SO 2 IS REMOVED. NOW IF I PULL SOME OF THIS OUT I

COULD PUT SOMETHING IN THAT MIGHT ABSORB THE SO 2. IF I TAKE SOME OF THIS OUT, IF I

DECREASE IT, ACCORDING TO LECHATELIER’S PRINCIPLE IF A SYSTEM CAN IT WOULD INCREASE

IT. IN ORDER TO INCREASE THE SO 2 , THE ONLY WAY WE CAN GET MORE SO 2 IN THE SYSTEM IS

FOR MORE REACTION TO OCCUR. MORE OF THIS HAS TO BE CONVERTED TO THAT SIDE TO GIVE

US BACK THE SO 2. SO THEREFORE WE WOULD PREDICT THAT IT’S GOING TO SHIFT TOWARDS

THE RIGHT OR WE WOULD SAY TOWARD THE PRODUCT SIDE. WE’RE GOING TO GET THEN MORE

CALCIUM SULFATE FORMED AND MORE WATER FORMED AND MORE ENERGY GIVEN OUT IF WE

DECREASED THE SO 2 BECAUSE THE SYSTEM TO PUT IT BACK HAS TO FORM MORE OF THOSE

OTHER THINGS AS WELL. THE LAST ONE HERE, PRESSURE IS DECREASED. WELL THIS IS JUST

THE OPPOSITE OF WHAT WE JUST SAID A MOM ENT AGO WHEN WE TALKED ABOUT VOLUME

BECAUSE IF PRESSURE IS DECREASED THEN IT MEANS THAT IN ORDER TO BRING THE PRESSURE

BACK IT WOULD HAVE TO DO WHAT? INCREASE THE NUMBER OF GASEOUS PARTICLES. WHICH

SIDE HAS THE GREATEST NUMBER OF GASEOUS PARTICLES? THE PRODUCT SIDE. MATTER OF

FACT, IT’S THE ONLY SIDE THAT HAS A GASEOUS PARTICLE AND SO IF I DECREASE THE

PRESSURE IN ORDER FOR THE SYSTEM TO INCREASE THE PRESSURE IT WOULD HAVE TO

INCREASE THE NUMBER OF GASEOUS PARTICLES AND SO IT IS GOING TO SHIFT TOWARDS THE

RIGHT OR TOWARDS THE PRODUCT SIDE. SO, LECHATELIER’S PRINCIPLE BASICALLY AGAIN

SAYS THAT IF WE HAVE A SYSTEM AT EQUILIBRIUM, INDICATED USUALLY BY THE DOUBLE

ARROW, AND IF WE SUBJECT THAT SYSTEM TO A CHANGE, WHATEVER CHANGE –

TEMPERATURE, PRESSURE, VOLUME, CONCENTRATION, THE SYSTEM WILL TRY TO OFFSET THAT

CHANGE. CAN’T ALWAYS DO SO. IN THE CASE OF A SOLID IT CAN’T CHANGE IT. ALRIGHT,

SOMETIMES IN THE CASE OF A PRESSURE OR VOLUME IT COULDN’T CHANGE IT. IF WE HAD THE

SAME NUMBER OF GASEOUS MOLECULES ON BOTH SIDES. IF WE HAD TWO MOLECULES ON THE

REACTANT AND TWO ONE THE PRODUCT SIDE THEN CHANGING PRESSURE OR VOLUME CAN’T

CHANGE EQUILIBRIUM…BECAUSE THERE WOULDN’T BE ANY REASON FOR IT TO SHIFT ONE

THIS IS DEFINED AS THE EQUILIBRIUM CONSTANT THEN FOR THE SYSTEM. THIS IS DEFINED AS

AN EQUILIBRIUM EXPRESSION FOR THE SYSTEM. NOW IN GENERAL FOR EQUILIBRIUM

REACTIONS WE CAN WRITE IN A GENERAL FASHION THAT THE EQUILIBRIUM CONSTANT K IS

EQUAL TO, AND THIS IS AN EQUALITY SO THIS IS A NUMERIC VALUE THAT WE’RE TALKING

ABOUT, THE EQUILIBRIUM CONSTANT EXPRESSION IS EQUAL TO PRODUCTS’ CONCENTRATION –

THAT’S WHAT THOSE LITTLE BRACKETS MEAN AND WE TALKED ABOUT THAT A LITTLE EARLIER

IN A CHAPTER WHEN WE TALKED ABOUT Ph FOR INSTANCE. BRACKETS MEANS MOLES PER LITER, CONCENTRATION OF THE PRODUCTS MOLES PER LITER RAISED TO THE POWER X MULTIPLIED BY THE CONCENTRATION OF THE REACTANTS RAISED TO THE POWER Y – WHERE X AND Y ARE THE COEFFICIENTS FROM THE BALANCED CHEMICAL EQUATION. THE COEFFICIENTS FROM THE BALANCED CHEMICAL EQUATION. ALRIGHT, SO NOW WE CAN GO AHEAD AND BEGIN WRITING THEN EQUILIBRIUM CONSTANT EXPRESSIONS FOR CHEMICAL REACTIONS. AND LET’S JUST TAKE A LOOK AT DOING A COUPLE OF THESE. PRODUCTS OVER REACTANTS RAISED TO THE POWER THAT THEY APPEAR IN THE BALANCED CHEMICAL EQUATION. THE FIRST ONE, WE HAVE THE CHEMICAL REACTION NO 2 + SO 2 YIELDS SO 3 + NO, AND WE WANNA WRITE THE EQUILIBRIUM EXPRESSION. SO KEQ IS GOING TO BE EQUAL TO THE CONCENTRATION OF SO 3 TIMES THE CONCENTRATION OF NO. NOW I DON’T PUT THE POWER 1 DOWN. IF NOTHING IS SHOWN IT AUTOMATICALLY IMPLIES THAT IT’S TO THE FIRST POWER. DIVIDED BY CONCENTRATION OF NO 2 TIME THE CONCENTRATION OF SO 2. PLEASE NOTE THAT THESE ARE PRODUCTS. THESE ARE MULTIPLIED NOT ADDED TOGETHER. I’M NOT PUTTING A MULTIPLICATION SIGN BETWEEN THEM, JUST PUTTING THE BRACKETS THAT WAY IMPLIES THAT THEY ARE MULTIPLIED TIMES EACH OTHER. PRODUCTS OVER REACTANTS, EQUILIBRIUM EXPRESSION. LET’S LOOK AT ANOTHER ONE HERE. HERE WE HAVE A REACTION OF PHOSPHORUS PENTACHLORIDE WITH HYDROGEN WE USE TO PRODUCE PHOSPHORUS TRICHLORIDE AND HYDROGEN CHLORIDE. SO WE’RE GOING TO WRITE KEQ, THE EQUILIBRIUM EXPRESSION WILL BE EQUAL TO THE CONCENTRATION OF THE PCl 3 MULTIPLIED BY THE CONCENTRATION OF THE HCL QUANTITY SQUARED. ALL DIVIDED BY CONCENTRATION OF PCl 5 AND THE CONCENTRATION OF H 2. SO AGAIN PRODUCTS OVER REACTANTS, EACH RAISED TO

THE POWER THAT THEY APPEAR IN THE BALANCED STOICHIOMETRIC EQUATION. ALRIGHT,

LET’S LOOK AT A THIRD EXAMPLE HERE. WRITING AN EQUILIBRIUM EXPRESSION. SO WE

WOULD WRITE KEQ IS EQUAL TO CONCENTRATION OF THE S 2 - , AQUEOUS, CONCENTRATION OF

H 2 O, DIVIDED BY CONCENTRATION OF HS-, AND THE OH-^ - ALL OF THESE ARE AQUEOUS, I JUST

DIDN’T WRITE IT…EXCEPT FOR WATER WHICH IS A LIQUID. NOW, SEEING THAT THIS IS A WATER

SYSTEM, WATER SOLUTION, AQUEOUS IMPLIES THAT THESE ARE WATER SOLUTIONS, THE

CONCENTRATION OF WATER ITSELF CAN’T BE CHANGED. AND SO THIS IS IN FACT A CONSTANT

IN ITSELF. THE CONCENTRATION OF H 2 O – DUMPING IN MORE H 2 O DOESN’T CHANGE ITS

CONCENTRATION. IT’S CONCENTRATION IS DEPENDENT UPON ITS DENSITY. AND SO IT DOESN’T

MAKE ANY DIFFERENCE HOW MUCH WATER WE HAVE AS A LIQUID. IT’S CONCENTRATION IS

ALWAYS THE SAME. SO THEREFORE IF THIS IN FACT IS A CONSTANT THEN WHY INCLUDE IT IN

THE EXPRESSION? IT COULD BE JUST INCORPORATED OVER HERE, AND SO OUR FIRST RULE ON

WRITING EQUILIBRIUM EXPRESSIONS IS THAT WE DO NOT INCLUDE, WE DO NOT INCLUDE A

LIQUID IN AN EQUILIBRIUM EXPRESSION…THAT’S BECAUSE THE CONCENTRATION OF A LIQUID

IS IN FACT A CONSTANT IN ITSELF. SO WE WOULD JUST MERELY WRITE THE S 2 -^ DIVIDED BY HS-

AND OH-. ALRIGHT, AND VERY QUICKLY HERE LOOK AT TWO MORE EXAMPLES. THE NEXT ONE

WE HAVE KEQ IS EQUAL TO CONCENTRATION OF THE GOLD 3+^ ION TIMES THE CONCENTRATION

OF THE PHOSPHATE 3-^ ION, ALL DIVIDED BY THE CONCENTRATION OF AuPO 4 SOLID PLUS H 2 LIQUID, WHOOPS, NOT PLUS – MULTIPLIED BY. OKAY, NOW WE JUST GAVE THE RULE. WE DO NOT INCLUDE A LIQUID IN AN EQUILIBRIUM EXPRESSION SO IT SHOULDN’T BE THERE. BUT ALSO, WE CAN’T CHANGE THE CONCENTRATION OF A SOLID. THE CONCENTRATION OF A SOLID IS A CONSTANT IN ITSELF. AND SO AGAIN IF WE CAN’T CHANGE IT THERE’S NO REASON TO SHOW IT IN THERE BECAUSE THERE’S NOTHING WE CAN DO TO IT TO AFFECT THIS NUMERIC VALUE. AND SO THEREFORE WE DO NOT INCLUDE A SOLID IN AN EQUILIBRIUM EXPRESSION. NOW IN ACTUALITY WE’RE INCLUDING THEM AND I SHOULDN’T SAY THAT WE’RE NOT INCLUDING THEM. THE ACTUAL WAY THAT WE STATE IT IS THAT THE CONCENTRATION OF A SOLID IS CONSIDERED A UNITY – AS ONE. THE CONCENTRATION OF A LIQUID IS CONSIDERED OR DEFINED AS UNITY – AS ONE. WELL OF COURSE OBVIOUSLY IF WE HAVE THIS AND THIS OVER