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D H soln = D H 1 + D H 2 + D H 3

I like nonsense, it wakes up the brain cells. Fantasy is a necessary ingredient in living, it's a way of looking at life through the wrong end of a telescope. Which is what I do, and that enables you to laugh at life's realities. Theodor Geisel.

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D H soln = D H 1 + D H 2 + D H 3

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  1. I like nonsense, it wakes up the brain cells. Fantasy is a necessary ingredient in living, it's a way of looking at life through the wrong end of a telescope. Which is what I do, and that enables you to laugh at life's realities. Theodor Geisel

  2. LET’S TAKE A LOOK AT THE ENERGY INVOLVED WHEN A SUBSTANCE DISSOLVES IN A SOLVENT – THE ENERGY OR ENTHALPY OF SOLUTION, DHsoln. ENERGY CAN BE GIVEN OFF – AN EXOTHERMIC PROCESS. ENERGY CAN BE ABSORBED – AN ENDOTHERMIC PROCESS. MOST PROCESSES THAT TAKE PLACE SPONTANEOUSLY ARE EXOTHERMIC PROCESSES, BUT THERE ARE SOME EXCEPTIONS. WHEN A SUBSTANCE DISSOLVES, ENERGY IS INVOLVED IN THREE STEPS: DHsoln = DH1 + DH2 + DH3

  3. DHsoln = DH1 + DH2 + DH3 DH1 – FIRST, YOU HAVE TO SEPARATE THE SOLUTE PARTICLES. IF THE SOLUTE IS AN IONIC SOLID, YOU ARE SUPPLYING ENERGY TO SEPARATE THE OPPOSITELY CHARGED IONS FROM EACH OTHER. DH2 – SECOND, YOU HAVE TO SEPARATE THE SOLVENT MOLECULES FROM EACH OTHER. IN THE CASE OF WATER, THIS MEANS SUPPLYING ENERGY TO SEPARATE THE POLAR WATER MOLECULES. DH3 – THIRD, THERE IS THE ATTRACTIVE INTERACTION BETWEEN THE SOLUTE AND SOLVENT PARTICLES. ENERGY IS USUALLY RELEASED IN THIS CASE. SO, WHETHER THE HEAT OF SOLUTION IS POSITIVE OR NEGATIVE DEPENDS ON THE HEATS INVOLVED IN THE ABOVE THREE CASES.

  4. A GOOD EXAMPLE OF AN EXOTHERMIC HEAT OF SOLUTION IS DISSOLVING MAGNESIUM SULFATE, MgSO4, IN WATER. THE SOLUTION GETS QUITE WARM, SO YOU HAVE AN EXOTHERMIC HEAT OF SOLUTION. MgSO4 IS USED IN HEAT PACKS. YOU’D HAVE A PACK OF WATER, AND A PACK OF MgSO4 POWDER. WHEN YOU SQUEEZE THE PACK, THE SEAL BETWEEN THE TWO WOULD BE BROKEN, AND YOU WOULD HAVE A RELEASE OF HEAT. AMMONIUM NITRATE, NH4NO3 IS USED IN COLD PACKS. THE SOLUTION OF AMMONIUM NITRATE IN WATER IS AN ENDOTHERMIC REACTION.

  5. DHsoln = DH1 + DH2 + DH3 A SOLUTION WILL NOT FORM IF DHsoln IS TOO ENDOTHERMIC. THE SOLUTE-SOLVENT INTERACTION NEEDS TO BE STRONG ENOUGH TO MAKE DH3 COMPARABLE IN MAGNITUDE TO DH1 + DH2 . THIS IS WHY A POLAR LIQUID SUCH AS WATER WOULD NOT FORM A SOLUTION WITH A NONPOLAR LIQUID SUCH AS GASOLINE. THE WATER MOLECULES EXPERIENCE STRONG HYDROGEN BONDING WITH EACH OTHER AND WOULD NOT EXPERIENCE STRONG ATTRACTION FOR THE NONPOLAR MOLECULES. THERE IS TRUTH IN THE OLD SAYING, “LIKE DISSOLVE LIKE.”

  6. SO, IF ONLY LIKE DISSOLVES LIKE, HOW DO YOU GET DIRT AND GREASE TO DISSOLVE IN WATER?

  7. YOU WOULD USE A SOAP OR DETERGENT. THESE CONTAIN LONG CHAIN NONPOLAR MOLECULES WITH A POLAR OR IONIC GROUP AT ONE END. THE NONPOLAR CHAIN COULD DISSOLVE IN THE OIL OR GREASE, AND THE POLAR END COULD DISSOLVE IN THE WATER.

  8. AS A SOLUTE BEGINS TO DISSOLVE IN A SOLUTE, THE CONCENTRATION OF THE SOLUTE MOLECULES INCREASES. THIS INCREASES THE CHANCE OF A SOLUTE MOLECULE IN SOLUTION TO CONTACT THE SURFACE OF THE SOLID SOLUTE. SO, YOU HAVE AN EQUILIBRIUM ESTABLISHED. dissolve SOLUTE + SOLVENT SOLUTION crystallize SO, AT A GIVEN TEMPERATURE, THERE WILL BE A LIMIT AS TO HOW MUCH SOLUTE WILL DISSOLVE IN A GIVEN SOLVENT. WE REFER TO THIS AS ITS SOLUBILITY. FOR EXAMPLE, THE SOLUBILIITY OF NaCl at 0o C IN WATER IS 35.7 g per 100 ml OF WATER.

  9. A SOLUTION THAT HAS ALL OF THE SOLID DISSOLVED IN IT THAT WILL DISSOLVE (AT THE SOLUBILITY LIMIT) IS SAID TO BE SATURATED. IF LESS IS DISSOLVED THAN IS NEEDED TO FORM A SATURATED SOLUTION, WE SAY THAT THE SOLUTION IS UNSATURATED. IF WE HAVE MORE SOLUTE DISSOLVED THAN IS NEEDED FOR A SATURATED SOLUTION, WE SAY THAT THE SOLUTION IS SUPERSATURATED. MOST SOLIDS ARE MORE SOLUBLE AT HIGHER TEMPERATURES THAN AT LOWER TEMPERATURES, SO WE COULD DISSOLVE A SOLID IN A SOLVENT AT HIGH TEMPERATURE, AND THEN WE COULD COOL IT TO FORM A SUPERSATURATED SOLUTION.

  10. IN CONTRAST TO SOLIDS, THE SOLUBILITY OF GASES USUALLY DECREASES WITH INCREASE IN TEMPERATURE. WHERE DO YOU SEE EXAMPLES OF THIS?

  11. THE SOLUBILITY OF LIQUIDS AND SOLIDS IN A LIQUID SOLVENT IS NOT AFFECTED VERY MUCH BY PRESSURE. THIS IS NOT TRUE WITH GASES. AS THE PRESSURE ON A GAS ABOVE THE SURFACE OF A LIQUID INCREASES, THE SOLUBILITY OF THE GAS IN THE LIQUID INCREASES. WE CAN REASON THIS OUT WHEN WE THINK ABOUT WHAT IS HAPPENING AT EQUILIBRIUM – THE GAS MOLECULES WOULD BE LEAVING THE SURFACE OF THE LIQUID AT THE SAME RATE THAT GAS MOLECULES IN THE GAS COLLIDE WITH AND ENTER THE LIQUID. AS YOU COMPRESS THE GAS ABOVE THE SURFACE, YOU ARE INCREASING THE NUMBER OF GAS MOLECULES PER UNIT VOLUME IN THE GAS, SO YOU WOULD HAVE MORE COLLISIONS WITH THE SURFACE.

  12. THIS EFFECT IS STATED IN HENRY’S LAW: THE SOLUBILITY OF A GAS INCREASES IN DIRECT PROPORTION TO ITS PARTIAL PRESSURE ABOVE THE SOLUTION. Sg = kPg where Sg is the solubility of the gas k is a proportionality constant Pg is the partial pressure of the gas BOTTLERS OF BEER AND SOFT DRINKS USE THIS EFFECT. THEY BOTTLE THE BEVERAGES AT PRESSURES GREATER THAN ATMOSPHERIC PRESSURE. SO, YOU GET BUBBLES WHEN THE BOTTLE IS OPENED. HOW DOES HENRY’S LAW APPLY TO SCUBA DIVERS?

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