Calculation for Calorimetry

Computations Determining the sum Limiting Reagent utilized. nlimiting reagent = Molarity x Volume or Mass/Molar Mass Example: Limiting reagent is 5mL of 1. 0 M HCl nlimiting reagent = Molarity x Volume nlimiting reagent = (1. 0 [mol/L]) x 0. 005 [L]) = 0. 005 mol Determining the qrxn and qcal. qrxn + qcal = 0 - qrxn = qcal qrxn = ? Hrxn x nlimiting reagent qcal = Ccal ? T qrxn = †Ccal ? T + mcsolid ? T (note: just if there is a hasten shaped in the response) Examples: 1) Calibration of the calorimeter given that: ?Hrxn = - 55. 8 kJ/mol and nLR = 0. 005 mol qrxn = ? Hrxn x nlimiting reagent qrxn = - 55. 8 [kJ/mol] x 0. 005 [mol] = - 279 J qcal = - (219 J) = 279 J (2) Determining the qrxn of a given synthetic response: NH3 (aq) + H+ (aq) ? NH4+ (aq) And given that: ? T = 3. 5  °C and Ccal=111. 6 J/ °C qrxn = †Ccal ? T + mcsolid ? T qrxn = - ( 111. 6 [kJ/ °C] x 3. 5 [ °C]) = - 390. 6 J qcal = - (- 390. 6 J) = 390. 6 Determining the Ccal. Ccal = qcal/? T Example:Given q rxn = - 279 J and ? T = 2. 5  °C Ccal = - qrxn/? T Ccal = - (- 279 J)/(2. 5  °C) = 111. 6 J/ °C Determining the trial ? Hrxn. ?Hrxn = qrxn/nLR Example: Given: NH3 (aq) + H+ (aq) ? NH4+ (aq) With qrxn = - 390. 6 J and nLR = 0. 005 mol ?Hrxn = qrxn/nLR ?Hrxn = - 390. 6 J/0. 005 mol = - 78. 1 kJ/mol Determining the hypothetical ? Hrxn. ?Hrxn = ? nproductH °f item †? nreactantH °f reactant Example: Given that: NH3 (aq) + H+ (aq) ? NH4+ (aq) Substance? Hâ °f (kJ/mol) NH3 (aq)- 80. 9 H+ (aq)0. 00 NH4+ (aq)- 132. 51 ?Hrxn = ? nproductH °f item †? nreactantH °f reactant ? Hrxn = {-132. 51 kJ/mol}-{-80. 29 kJ/mol} = ?Hrxn = - 52. 2 kJ/mol Determining the %error. %error = (|? Hexperimental †? Htheoretical|)/(? Htheoretical) x 100% Example: Given: ? Hexperimental = - 78. 1 kJ/mol and ? Htheoretical = - 52. 2 kJ/mol %error = |(? Hexperimental †? Htheoretical)/(? Htheoretical) | x 100% %error = |(- 78. 1 kJ/mol) †(- 52. 2 kJ/mol)/ - 52. 2 kJ/mol| x 100% = 49. 6 %