CHE134 Summer Lecture 2

CHE134 Summer Lecture 2 - T H+(aq) + OH- * H2O q H...

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Click to edit Master subtitle style 1/10/11 Calorimetric Determination of Reaction Enthalpies T H q H+(aq) + OH- * H2O CH3COOH + OH- CH3COO- + H2O HCl NaOH CH3COOH
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1/10/11 Concepts: Strong/Weak Acids and Bases Dissociation Neutralization Limiting Reagent Heat Heat Capacity Enthalpy Calorimetric Constant Endothermic Exothermic Hess’ Law 1st & 2nd Laws of Thermodynamics Purpose: Determine @ Hdis, the enthalpy of dissociation of Acetic Acid. I.e., ÿ H for the reaction: Techniques: Calorimetry Temperature Measurement vs Time Apparatus: Calorimeter Precision Thermometer CH3COOH * CH3COO- + H+
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1/10/11 Review: Measurement of Heat and Temperature Cannot measure heat , q , directly, but can measure its effect on a substance – namely temperature change * , T , when heat flows into , or from , the substance. For reasonable changes in temperature , for a given mass of a substance, m, q is proportional to the mass, m and the temperature change, T . * Or other changes, such as melting T = Tfinal - Tinitial * or boiling q = C m T
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1/10/11 Proportionality constant, C , is called the Heat Capacity per gram of the substance (specific heat). q = C m T By convention, absorbed heat is positive . ( liberated heat is negative ) Calorimeter – a device that measures temperature changes accompanying heat changes in its contents T > 0 ; q > 0 T < 0 ; q < 0 For water(l ), C = 4.18 J / g K For Fe, C = 0.45 J / g K
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1/10/11 Calorimeters - constructed to minimize heat exchange with environment so effects of heat are entirely on the calorimeter Must be able to separate effects of heat on from effects on its contents . ( Use Covered, Nested Styrofoam Cups ) I.e., determine the Calorimetric Constant for the calorimeter, Ccal Environment Calorimeter Contents and its contents. INSULATE ! CALIBRATE ! Determine heat absorbed by calorimeter when known amount of heat is added to its contents.
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1/10/11 High precision thermometer 0- 50oC $$
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1/10/11 CALIBRATION How do we calculate Tmix when two samples of water of known mass and temperature are mixed? E.g., suppose we add: 60 g of hot water at 50oC 30 cold 25oC Thot Tcold 30 g *4.18 J/g * ( Tmix 25 ) = - 60 Tmix - 50 ) 30 * ( Tmix 25 = - 60 * ( Tmix 50 ) Tmix = ( 60 * 50 + 30 25 ) / ( 60 30 = 42oC Ideally heat absorbed by cold water = lost hot water qcold = mc Cc Tc - qhot - mh Ch Th = q = m C T to - - Tmix 90 g of water at Tmix - > 0 <0 qabs > 0 Cc = Ch = 4.18
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1/10/11 Thot Tcold Thot Tcold Tmix The Ideal Case qhot - qhot = qcold 60 g 50oC 30 g 25oC 90 g 42oC
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1/10/11 Now, let’s perform the same operation in a real calorimeter. We can insure the cold water and calorimeter have the same initial temperature. Can we calculate the final temperature of the water mixture (and the calorimeter)? by letting the cold water and calorimeter come to thermal equilibrium
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1/10/11 Actual Tmix will differ from ideal because of heat exchange with container – i.e., calorimeter Assume calorimeter is initially at the temperature of the cold water and it experiences same temperature change as its cold water contents. Some heat lost by the hot water
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This note was uploaded on 01/09/2011 for the course CHE 134 taught by Professor Staff during the Summer '08 term at SUNY Stony Brook.

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CHE134 Summer Lecture 2 - T H+(aq) + OH- * H2O q H...

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