CHAPTER 6 THERMOCHEMISTRY SYSTEM: THE PART OF THE WORLD IN WHICH WE HAVE INTEREST. E.G. REACTION VESSEL, CALORIMETER, AND SO ON. SURROUNDINGS: COMPRISE THE REGION OUTSIDE THE SYSTEM. HERE WE MAKE OUR MEASUREMENTS.

OPEN SYSTEM: MATTER CAN BE TRANSFERRED THROUGH THE BOUNDARY BETWEEN THE SYSTEM AND SURROUNDING. CLOSED SYSTEM: MATTER CAN NOT PASS THROUGH THE BOUNDARY. ISOLATED SYSTEM: A CLOSED SYSTEM THAT HAS NEITHER MECHANICAL OR THERMAL CONTACT WITH ITS SURROUNDINGS.

WORK: FUNDAMENTAL PHYSICAL PROPERTY IN THERMODYNAMICS WORK: FUNDAMENTAL PHYSICAL PROPERTY IN THERMODYNAMICS. WORK IS MOTION AGAINST AN OPPOSING FORCE. E.G. EXPANSION OF A GAS THAT PUSHES OUT A PISTON AND RAISES A WEIGHT. ENERGY: IT IS THE CAPACITY OF A SYSTEM TO DO WORK. WHEN A SYSTEM WORKS THE ENERGY OF THE SYSTEM IS REDUCED. E.G. UNWINDING OF A SPRING REDUCES ITS ENERGY. HEAT: WHEN ENERGY OF A SYSTEM CHANGES AS A RESULT OF TEMPERATURE DIFFERENCE BETWEEN SYSTEM AND SURROUNDING , WE SAY ENERGY TRANSFERRED AS HEAT.

EXOTHERMIC PROCESS: RELEASES ENERGY AS HEAT INTO ITS SURROUNDINGS EXOTHERMIC PROCESS: RELEASES ENERGY AS HEAT INTO ITS SURROUNDINGS. ALL COMBUSTION REACTIONS ARE EXOTHERMIC. E.G. BURNING OF PETROL OR COAL. ENDOTHERMIC PROCESS: ENERGY IS GAINED FROM THE SURROUNDINGS AS HEAT. E.G.VAPORIZATION OF WATER.

ADIABATIC PROCESS: A SYSTEM CHANGES FROM ONE STATE TO ANOTHER WITHOUT CHANGE OF HEAT. I.E. ∆ H = 0, TEMPERATURE MAY CHANGE. ISOTHERMAL PROCESS: A SYSTEM CHANGES FROM ONE STATE TO ANOTHER WITHOUT CHANGE OT TEMPERATURE. I.E. ∆T=0, HEAT MAY BE CHANGED. INTERNAL ENERGY: IN THERMODYNAMICS, THE TOTAL ENERGY OF SYSTEM IS CALLED INTERNAL ENERGY, U. IT IS THE TOTAL KINETIC AND POTENTIAL ENERGY OF THE MOLECULES IN THE SYSTEM. ∆U = U – U; IT IS A STATE FUNCTION. ANY CHANGE OF THE STATE VARIABLES, LIKE PRESSURE, RESULTS IN CHANGE OF INTERNAL ENERGY. IT IS AN EXTENSIVE PROPERTY. U, H AND W ARE ALL MEASURED IN THE SAME UNITS, JOULE(J). 1J = 1 Kg m2 s-2.

FIRST LAW OF THERMODYNAMICS: “THE INTERNAL ENERGY OF AN ISOLATED SYSTEM IS CONSTANT”. MATHEMATICAL STATEMENT OF FIRST LAW OF THERMODYNAMICS: ∆U = q +w, w IS THE WORK DONE ON A SYSTEM AND q IS THE ENERGY TRANSFERRED AS HEAT TO THE SYSTEM. THE STUDY OF ENERGY AND ITS INTERCONVERSIONS IS CALLED THERMODYNAMICS.

THE LAW OF CONSERVATION OF ENERGY IS OFTEN CALLED THE FIRST LAW OF THERMODYNAMICS AND STATED AS FOLLOWS: “THE ENERGY OF THE UNIVERSE IS CONSTANT” THERMODYNAMIC QUANTITIES ALWAYS CONSISTS OF TWO PARTS: A NUMBER, GIVING THE MAGNITUDE OF THE CHANGE, AND A SIGN, INDICATING THE DIRECTION OF FLOW. FOR EXAMPLE, IF A QUANTITY OF ENERGY FLOWS INTO THE SYSTEM VIA HEAT (ENDOTHERMIC PROCESS), q IS POSITIVE, i.e. THE SYSTEM’S ENERGY IS INCREASING. ON THE OTHER HAND, WHEN ENERGY FLOWS OUT OF THE SYSTEM VIA HEAT (EXOTHERMIC PROCESS), q IS NEGATIVE, i.e. THE SYSTEM’S ENERGY IS DECREASING.

IF THE SYSTEM DOES WORK ON THE SURROUNDING(ENERGY FLOWS OUT OF THE SYSTEM), w IS NEGATIVE. IF THE SURROUNDINGS DO WORK ON THE SYSTEM(ENERGY FLOWS INTO THE SYSTEM), w IS POSITIVE. PROBLEM: CALCULATE ∆U FOR A SYSTEM UNDERGOING ENDOTHERMIC PROCESS IN WHICH 15.6 kj OF HEATFLOWS AND WHERE 1.4 Kj WORK DONE ON THE SYSTEM. WE HAVE ∆U = q + w= 15.6 + 1.4 =17.0 kJ

PV WORK: WORK = P∆V THIS GIVES US THE MAGNITUDE (SIZE) OF THE WORK REQUIRED TO EXPAND A GAS ∆V AGAINST A PRESSURE P. SIGN OF THE WORK: THE GAS (SYSTEM) IS EXPANDING AGAINST THE PRESSURE. THUS THE SYSTEM IS DOING WORK ON THE SURROUNDINGS, SO FROM THE SYSTEM’S POINT OF VIEW THE SIGN OF THE WORK SHOULD BE NEGATIVE. FOR AN EXPANDING GAS, ∆V IS A POSITIVE QUANTITY BECAUSE THE VOLUME IS INCREASING. THUS ∆V AND w MUST HAVE OPPOSITE SIGNS. i.e. w = - P∆V

NOTE THAT FOR A GAS EXPANDING AGAINST PRESSURE P, w IS A NEGATIVE QUANTITY AS REQUIRED, SINCE WORK FLOWS OUT OF THE SYSTEM. WHEN A GAS IS COMPRESSED, ∆V IS A NEGATIVE QUANTITY, WHICH MAKES w A POSITIVE QUANTITY (WORK FLOWS INTO THE SYSTEM). PROBLEM: CALCULATE THE WORK ASSOCIATED WITH THE EXPANSION OF A GAS FROM 46 L TO 64 L AT A CONSTANT EXTERNAL PRESSURE OF 15 atm. Solution: W = -P∆V = -15X(64-46) =-270 L.atm NOTE THAT SINCE THE GAS EXPANDS, IT DOES WORKON ITS SURROUNDING.

ENTHALPY AND CALORIMETRY: H = U + PV ( H = qp) THUS ENERGY FLOWS OUT OF THE GAS, SO w IS A NEGATIVE QUANTITY. FOR AN IDEAL GAS, WORK CAN OCCUR ONLY WHEN ITS VOLUME CHANGES. THUS , IF A GAS IS HEATED AT CONSTANT VOLUME, THE PRESSURE INCREASES BUT NO WORK OCCURS. ENTHALPY AND CALORIMETRY: H = U + PV ( H = qp) SINCE INTERNAL ENERGY, PRESSURE AND VOLUME ARE STATE FUNCTIONS, ENTHALPY IS ALSO A STATE FUNCTION. ∆H = ∆U + P∆V (P IS CONSTANT) THUS ∆H = ∆Hpr - ∆Hreact, ∆Hpr > ∆Hreact, ∆H= +VE (ENDOTHERMIC); ∆HPr< ∆Hreact, ∆H = -VE (EXOTHERMIC)

CALORIMETRY CALORIMETRE-DEVICE USED EXPERIMENTALLY TO DETERMINE THE HEAT ASSOCIATED WITH A CHEMICAL REACTION. HEAT CAPACITY=(HEAT ABSORBED)/ (INCREASE IN TEMPERATURE) SPECIFIC HEAT CAPACITY – HEAT CAPACITY PER g OF SUBSTANCE (UNIT : J/K.g) MOLAR HEAT CAPACITY – HEAT CAPACITY PER MOLE OF A SUBSTANCE (UNIT- J/K.mol) ∆H(NEUTRALISATION)-HEAT EVOLVED WHEN ONE MOLE AN ACID IS COMPLETELY NEUTRALISED WITH A BASE OR VICE VERSA. ∆H = s X m X ∆T ∆H(SOLUTION)- HEAT EVOLVED OR ABSORBED WHEN ONE MOLE OF A SOLUTE IS DISSOLVED COMPLETELY IN A SOLVENT.

E.g. REACTION BETWEEN NITROGEN AND OXYGEN TO GIVE NITROGEN DIOXIDE. HESS’S LAW “THE ENTHALPY CHANGE OF A CHEMICAL REACTION IS SAME WHETHER IT TAKES PLACE IN ONE STEP OR MORE THAN ONE STEPS.” E.g. REACTION BETWEEN NITROGEN AND OXYGEN TO GIVE NITROGEN DIOXIDE. e.g.N2(g)+O2(g)=2NO2(g), ∆H1=68kJ THE REACTTION CAN ALSO BE CARRIED OUT TWO DISTINCT STEPS, WITH ENTHALPY CHANGES, ∆H2 AND ∆H3.

N2(g) + O2(g) = 2NO2(g) ∆H2 = 180 Kj 2NO(g)+ O2(g) = 2NO2(g), ∆H2= -112 Kj NET REACTION : N2(g)+O2(g) =2NO2(g) ∆ H2 + ∆ H3 =68 kJ THUS ∆H1= ∆H2 + ∆H3 ENTHALPY DIAGRAM FOR THE REACTION

ENTHALPY DIAGRAM FOR REACTION BETWEEN METHANE AND OXYGEN

STANDARD ENTHALPY OF FORMATION(∆Hf): THE CHANGE IN ENTHALPY THAT ACCOMPANIES THE FORMATION OF ONE MOLE OF A COMPOUND FROM ITS ELEMENTS WITH ALL SUBSTANCES IN THEIR STANDARD STATES. STANDARD STATES: THE STANDARD STATE OF A GASEOUS SUBSTANCE IS A PRESSURE OF EXACTLY 1 atm. FOR A PURE SUBSTANCE IN A CONDENSED STATE (LIQUID OR SOLID), THE STANDARD STATE IS THE PURE SOLID OR LIQUID. FOR A SUBSTANCE PRESENT IN A SOLUTION, THE STANDARD STATE IS A CONCENTRATION OF EXACTLY 1M. FOR AN ELEMENT, THE STANDARD STATE OF AN ELEMENT IS THE FORM IN WHICH THE ELEMENT EXISTS UNDER CONDITIONS OF 1 atm AND 25 Oc. E.g. THE STANDARD STATE OF OXYGEN IS O2(g) AND THAT OF SODIUM IS Na(s) AND MERCURY IS Hg(l).

Table : Standard Enthalpies of Formation for Several Compounds at 25oC Formula Name ΔfH o (kJ/mol) H2 Hydrogen 00.0 CH4 Methane -74.9 C2H6 Ethane -83.7 C2H2 Acetylene 226.8 C3H8 n-Propane -104.6 C4H10 n-Butane -125.5 C5H12 n-Pentane -146.9 C6H14 n-Hexane -167.4 C7H16 n-Heptane -187.9 C8H18 n-Octane -208.4 C9H20 n-Nonane -229.3 C10H22 n-Decane -249.4 Table : Standard Enthalpies of Formation for Several Compounds at 25oC Compound Hf (kJ/mol) NH3(g) -46 NO2(g) -34 H2O(l) -286 Al2O3(s) -1676 Fe2O3(s) -826 CO2(g) -394 CH3OH(l) -239 C8H18(l) -269

ENTHALPY OF REACTION FROM STANDARD ENTHALPY OF FORMATION THE ENTHALPY OF A CHEMICAL REACTION CAN BE CALCULATED BY SUBTRACTING THE ENTHALPY OF FORMATION OF THE REACTANTS FROM THE ENTHALPY OF FORMATION OF THE PRODUCTS. REMEMBER TO MULTIPLY THE ENTHALPY OF FORMATION BY THE INTEGERS AS REQUIRED BY THE BALANCED EQUATION. ∆H(REACTION) = ∆Hfo(PRODUCT) - ∆Hfo(REACTANT)

∆H(REACTION) = ∆Hfo(PRODUCT) - ∆Hfo(REACTANT) PROBLEM: USING THE STANDARD ENTHALPY OF FORMATION, CALCULATE THE ENTHALPY CHANGE FOR THE OVERALL REACTION THAT OCCURS WHEN AMMONIA IS BURNT IN AIR TO FORM NITROGEN DIOXIDE AND WATER. SOLUTION: ∆H(REACTION) = ∆Hfo(PRODUCT) - ∆Hfo(REACTANT) =[4X(34) + 6X(-286)] – [4X(-46) + 7X0] =[136 – 1716] –[-184 + 0] = -1580 + 184 = -1396 kJ