CHEM 150 1nd Edition Lecture 15Outline of Last Lecture • Calorimeter• Heat Capacity• Calorimeter constants Outline of Current Lecture • Cont. Calorimeter • Measuring ΔH (or ΔE)Current LectureQ: If 4.45kJ of heat are applied to 85.00g of copper at 23.0C, what is the final temperature of the copper? Cs(cu,solild) = 0.387J/gCA: q= Cs.m.ΔTΔT= q/Cs.mΔT= 4450J/0.387J/C =135.2789Tf= Ti +ΔT= 32.0+135.2789C =158.2789C• Calorimetry: Measuring ΔH (or ΔE)• Bomb calorimeter: Constant volumeif ΔV = 0, then w = PΔV= 0 and ΔE = qQ: A bomb calorimeter has a heat capacity (C) of 4.1356 kJ/C. The combustion of 1.500 g of C6H5COOH raises the temperature of the calorimeter from 22.50 C to 31.69 C. What is ΔE for the combustion of benzoic acid in kJ/mol?A: 1.500g x 1mol/122.118g =0.0122832molQcal =4.1356KJ/C (31.69C - 22.50C) =38.006164KJQrxn= -Qcal = -38.006164 KJΔE =Qxrn/n = -38.00616KJ/0.0122832mol =-3094.1582KJ/molΔE= -3.09x10^2 KJ/molQ: The heat of combustion of camphor (C10H16O) is -5903.6 kJ/mol. If the combustion of 4.682 g of camphor raises the temperature of a bomb calorimeter from 23.073 EC to 38.554 EC, what is the heat capacity of the calorimeter?A: 4.682g x 1 mol/162.228g = 0.030756495molQrxn =0.0307565 mol x -5903.gKJ/1mol =-181.574KJQcal =-Qrxn = 181.574KJCp = Qcal/ΔT =181.574KJ /(38.554C-23.073) =11.72883KJ• Coffee cup calorimetry:[Constant pressure calorimetry ΔH=q/n]Q: 200.0mLof 0.150M H2SO4 and 200.0mL of 0.250M NaOH, both at an initial temperature of 22.80EC, are mixed. The temperature of the resulting solution is 30.50C. What is ΔH of the neutralization reaction? [We will assume that dilute aqueous solutions have the same density and specific heat as water.]A: H2SO4(aq) + 2NaOH(aq) -----> 2H2O(l) + Na2SO4(aq)desnity = 1.000g/mol Cs = 4.184J/gCqsol = qH2o = 4.184J/gC x 400.0g (30.50C -22.80C) = 12888.72JQrxn = -qH2O = -12..88872KJ0.2000L x 0.150M H2SO4 = 0.0300mol H2SO40.2000L x 0.250M NaOH = 0.0500mol NaOH0.0300 mol H2SO4 x 2 mol H2O/ 1 mol H2SO4 =0.0600 mol H2O0.0500 mol NaOH x 2mol H2O/ 2mol NaOH = 0.0500mol H2OΔH = -12.88872KJ/ 0.0500mol H2O x 2 mol H2O/rxn = -515 KJ• The heat of solution of sodium hydroxide is -44.505 kJNaOH (s) -----> Na+(aq) + OH-(aq) ΔH =-44.505KJQ: If 30.00g of sodium hydroxide are dissolved in 250.0mL of water at 21.0EC, what is the final temperature of the solution?A: 30.00g NaoH x 1mol/39.9988g =0.75004molQrxn =-44.505kg/mol NaOH x0.75004 mol = -33.3804KJqH2O = -Qrxn =33380.4JΔH= q/Cs x m= 33380.4J/ 4.184J/gC x 250.0g = 31912CTf = Ti + ΔT = 21.0C + 31.9C =
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