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THE WORLD’S MOST FAMOUS LAST WORDS IN ANY LANGUAGE: “HEY, WATCH THIS.”

THE WORLD’S MOST FAMOUS LAST WORDS IN ANY LANGUAGE: “HEY, WATCH THIS.”.

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THE WORLD’S MOST FAMOUS LAST WORDS IN ANY LANGUAGE: “HEY, WATCH THIS.”

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  1. THE WORLD’S MOST FAMOUS LAST WORDS IN ANY LANGUAGE: “HEY, WATCH THIS.”

  2. LE CHATELIER’S PRINCIPLE: ‘IF A SYSTEM AT EQUILIBRIUM IS DISTURBED BY A CHANGE IN TEMPERATURE, PRESSURE, OR THE CONCENTRATION OF ONE OF THE COMPONENTS, THE SYSTEM WILL SHIFT ITS EQUILIBRIUM POSTION SO AS TO COUNTERACT THE EFFECT OF THE DISTURBANCE.” KEEP IN MIND THAT A SYSTEM IN EQUILIBRIUM IS IN A DYAMIC STATE – THE FORWARD AND REVERSE REACTIONS ARE EQUAL. THE SYSTEM IS IN BALANCE. IF WE ALTER THE CONDITIONS AS STATED ABOVE, WE UPSET THAT BALANCE. SO, THE SYSTEM IS GOING TO SHIFT TO A NEW STATE OF BALANCE – A NEW EQUILIBRIUM. CONSIDER THE FOLLOWING REACTION: N2 (g) + 3H2 (g) == 2NH3 (g)

  3. N2 (g) + 3H2 (g) == 2NH3 (g) Keq = [NH3]2 / [N2][H2]3 = [pNH3]2 / [pN2][pH2]3 IF WE ADD NITROGEN, THE SYSTEM IS GOING TO SHIFT TO REDUCE THE INCREASED CONCENTRATION OF NITROGEN, SO IT WILL SHIFT TO THE RIGHT. A SIMILAR SITUATION WOULD OCCUR IF WE INCREASED THE CONCENTRATION OF HYDROGEN – THE EQUILIBRIUM WOULD SHIFT TO THE RIGHT. YOU CAN THINK IN TERMS OF THE EQUILIBRIUM CONSTANT. IF WE INCREASE A CONCENTRATION TERM IN THE DENOMINATOR, FOR THE Keq TO REMAIN CONSTANT, THE CONCENTRATION TERM IN THE NUMERATOR WOULD HAVE TO INCREASE.

  4. N2 (g) + 3H2 (g) == 2NH3 (g) IN TURN, IF WE INCREASE THE CONCENTRATION OF AMMONIA, NH3, THE REACTION WILL SHIFT TO THE LEFT TO COME BACK INTO EQUILIBRIUM. AT A GIVEN TEMPERATURE, THE EQUILIBRIUM CONSTANT IS JUST THAT – CONSTANT. THE HABER PROCESS IS USED TO PRODUCE AMMONIA FOR USE IN FERTILIZER OR IN EXPLOSIVES. ONE WAY TO INCREASE THE PRODUCTION OF AMMONIA WOULD BE TO REMOVE IT FROM THE PROCESS AS IT IS PRODUCED. THIS CAN BE DONE BY LIQUEFYING IT. BP NH3 = -33o C BP N2 = -196o C BP H2 = -253o C

  5. EFFECT OF CHANGING THE PRESSURE AND VOLUME AGAIN, LET’S CONSIDER THE SAME REACTION N2 (g) + 3H2 (g) == 2NH3 (g) AT CONSTANT TEMPERATURE, WHAT HAPPENS IF WE INCREASE THE PRESSURE BY REDUCING THE VOLUME? IN THE ABOVE REACTION, WE HAVE 4 MOLES ON THE LEFT AND 2 MOLES ON THE RIGHT. IF WE INCREASE THE PRESSURE, THE REACTION WILL SHIFT IN A DIRECTION THAT REDUCES THE NUMBER OF MOLES OF GASES (REDUCES THE PRESSURE). SO, THE REACTION WOULD SHIFT TO THE RIGHT. INCREASING THE PRESSURE DOES NOT CHANGE Keq. IT CHANGES THE PARTIAL PRESSURES OF THE GASEOUS SUBSTANCES.

  6. ANOTHER WAY OF CONSIDERING DIRECTIONS OF REACTION IS TO TAKE THE ACTUAL CONCENTRATIONS AND CALCULATE A REACTION QUOTIENT AND COMPARE THAT TO THE EQUILIBRIUM CONSTANT. WHEN WE SUBSTITUTE ACTUAL CONCENTRATIONS INTO THE EQUILIBRIUM CONSTANT EXPRESSION, THE RESULT IS THE REACTION QUOTIENT. AT EQUILIBRIUM, Q = Keq. IF Q>Keq, THE REACTION WILL SHIFT TO THE LEFT. IF Q<Keq, THE REACTION WILL SHIFT TO THE RIGHT.

  7. EFFECT OF TEMPERATURE ON EQUILIBRIUM LET’S CONSIDER HOW HEAT FLOWS IN THE TWO SITUATIONS WE’VE DISCUSSED. ENDOTHERMIC REACTION: REACTANTS + HEAT  PRODUCTS EXOTHERMIC REACTION: REACTANTS  PRODUCTS + HEAT IN AN ENDOTHERMIC REACTION, INCREASING THE TEMPERATURE WOULD SHIFT THE REACTION TO THE RIGHT. IN AN EXOTHERMIC REACTION, INCREASING THE TEMPERATURE WOULD SHIFT THE REACTION TO THE LEFT.

  8. TEMPERATURE CHANGES ACTUALLY CHANGE THE Keq. ENDOTHERMIC: INC. T RESULTS IN AN INCREASE IN Keq EXOTHERMIC: INC. T RESULTS IN A DECREASE IN Keq HOW DOES A CATALYST AFFECT EQUILIBRIUM??

  9. A CATALYST DOES NOT CHANGE THE POSITION OF EQUILIBRIUM. A CATALYST SPEEDS UP A REACTION. IN THE CASE OF A REVERSIBLE REACTION, IT WOULD SPEED UP THE FORWARD REACTION AND THE REVERSE REACTION, SO EQUILIBRIUM WOULD BE ACHIEVED FASTER. A CATALYST SPEEDS BOTH THE FORWARD AND REVERSE REACTIONS, BUT DOES NOT CHANGE THE POSITION OF EQUILIBRIUM.

  10. NO IS A POLLUTANT FROM COMBUSTION REACTIONS, ESPECIALLY AUTOMOBILES. IT IS INVOLVED IN THE FORMATION OF GROUND LEVEL OZONE AND ACID RAIN. ½ N2 (g) + ½ O2 (g)  NO (g) DH = 90.4 kJ AT THE CYLINDER TEMPERATURE, 2400o K, THE Keq = 5 X 10-2 AT 1200o K, THE Keq = 1 X 10-4 THE RATE OF REACTION IS ALSO DECREASED AT THE LOWER TEMPERATURE, AND AS THE GAS COOLS FURTHER, THE RATE OF THE REACTION IS EVEN SLOWER.

  11. REMEMBER THAT THE RATE OF REACTION DOUBLES WITH EACH 10o INCREASE IN TEMPERATURE. SO AT 300o K, THE CONCENTRATION OF NO WOULD BE “FROZEN.” CATALYSTS CAN BE USED TO CHANGE THIS, AND THAT IS WHAT IS DONE WITH CATALYTIC CONVERTERS ON AUTOMOBILES.

  12. GIVEN THE FOLLOWING REACTION: • 2SO2 (g) + O2 (g) == 2SO3 (g) DH < 0 • HOW WILL THE FOLLOWING AFFECT AN EQUILIBRIUM MIXTURE: • O2 IS ADDED TO THE SYSTEM • THE REACTION MIXTURE IS HEATED • THE VOLUME OF THE REACTION VESSEL IS DOUBLED • A CATALYST IS ADDED • SO3 IS REMOVED • THE TOTAL PRESSURE OF THE SYSTEM IS INCREASED BY ADDING AN INERT GAS

  13. GIVEN THE FOLLOWING REACTION: • 6C02 (g) + 6H2O (g) ==== C6H12O6 (s) + 6O2 (g) DH=2816 KJ • HOW IS THE EQUILIBRIUM AFFECTED BY: • INCREASING PCO2 • INCREASING TEMPERATURE • REMOVING CO2 • DECREASING THE TOTAL PRESSURE • REMOVING PART OF THE C6H12O6 • ADDING A CATALYST

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