CHM1045 Lecture 19Outline of Last LectureI. HeatII. Constant-Volume CalorimetryIII. Constant-Pressure CalorimetryOutline of Current LectureI. Example 1II. Standard EnthalpyIII. Example 2IV. Example 3Current LectureExample 1: A quantity of 1.00 × 102 mL of 0.500 M HCl was mixed with 1.00 × 102 mL of 0.500 M NaOH in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the HCl and NaOH solutions was the same, 22.50°C, and the final temperature of the mixed solution was 25.86°C. Calculate the heat change for the neutralization reaction on a molar basis:Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively).Qsystem=qsolution + qreaction=0qsolution= ms∧Tqsolution= (200g)(4.184)(25.8-22.5)qsolution= 2812J or 2.812 kJqreaction r*n = -2.812kJ.500mol/1L * .100L= .0500 mol HCl2.812kJ = -56.12 kJ/mol.0500 mol HClStandard EnthalpyBecause there is no way to measure the absolute value of the enthalpy of substance, must I measure the enthalpy change for every reaction of interest?- standard enthalpy of formation (∧H0f) is as a reference point for all enthalpy expressions.- Standard enthalpy of formation (∧H0f ) = heat change that results when one moleof a compound is formed from its elements at a pressure of 1 atm.- The standard enthalpy of formation of any element in its most stable form is zero.(∧H0f )(O2) = 0 ∧H0 (O3) = 142 kJ/mol∧H0 (C, graphite) = 0 ∧H0 (C, diamond) = 1.90 kJ/molstandard enthalpy of reaction (∧H0f ) is the enthalpy of a reaction carried out at 1 atm.Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.Example 2 :Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements:The equations for each step and the corresponding enthalpy changes are 2* (Cgraphite+O2CO2∧H= -393.5)H2+1/2 O2 H2O ∧H= -285.8 flip and multiply by 1/2 2C2H2 + 5O2 4CO2 + 2H2O ∧H= -2598.8 2Cgraphite+O22CO2∧H= -787 H2+1/2 O2 H2O ∧H= -285.82CO2 + H2O 2C2H2 + 5/2O2∧H= 1299.4-787+(-285.8)+ 1299.4= 226.6 kJ/molExample 3:The thermite reaction involves aluminum and iron(III) oxideThis reaction is highly exothermic and the liquid iron formed is used to weld metals. Calculate the heat released in kilojoules per gram of Al reacted with Fe2O3. The for Fe(l) is 12.40 kJ/mol.∧H= sum of products – sum of reactants (-1669.8+ (2*12.40)- (2*0+ (-822.2) ∧Hr*n= -822.8 kJ/mol-822.8 kJ/mol * 1 mol Al = - 822.8 = -15.25kg/g 2mol Al 26.98g Al
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