CH 301 CH 9 calorim trickier example_answers

CH 301 CH 9 calorim trickier example_answers - CH301 Notes:...

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Unformatted text preview: CH301 Notes: A trickier example of getting a thermochemical equation from data! We will use a coffee-cup calorimeter of heat capacity 27.8 J/oC to find the thermochemical equation for the acid-base neutralization reaction of NaOH and CH3COOH. Data: Add 25.00 mL of NaOH at 23.00o C to 25.00 mL of CH3COOH which is already in the calorimeter and at the same temperature. After mixing, the resulting temperature is observed to be 25.947o C. Assume that the specific heat of the mixture is the same as that of water, 4.18 J/g0C and that the density of the mixture is 1.02 g/mL. temperature change: 25.947o C 23o C = 2.947o C heat absorbed by calorimeter body = q calorimeter = 2.947o C * 27.8 J/oC = 81.9J Now calculate the amount of heat given off in the reaction. We use the volume of the combined solutions and the density of the resulting solution to find its mass, then use its specific heat to find the heat given off in the reaction. Volume of solution in calorimeter = 25.00mL + 25.00mL = 50.00mL Mass of this solution = 50.00mL x 1.02g/mL = 51.0g Heat absorbed by solution = q solution = 51.0g * 4.18 J/g0C * 2.947o C = 628J Total heat absorbed = q calorimeter + q solution = 81.9J + 628J = 709.9J Since this was done in a coffee cup calorimeter, by finding q, we have found H reaction. = - 709.9J (heat is emitted exothermic reaction!) Now we find the specific amount to be quoted per mole of reaction: that is, determine the heat of reaction for this particular acid-base neutralization: CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l) We now must determine Hrxn for 1 mole of the reaction... This is a limiting reagent problem: We have H for the reaction (not necessarily using 1 mol). We must determine the number of moles of reactants consumed which requires a limiting reactant calculation. CH 3 COOH aq + NaOH aq 0.500 mmol NaOH 1 mL NaOH NaCH 3 COO aq + H 2 O l 25.00 mL NaOH 1 mmol NaCH 3 COO 1 mmol NaOH 12.5 mmol NaCH 3 COO 25.00 mL CH 3 COOH 0.600 mmol CH 3 COOH 1 mL CH 3 COOH 1 mmol NaCH 3 COO 1 mmol CH 3 COOH 15.0 mmol NaCH 3 COO The limiting reactant is NaOH. We now know that 709.9 J were evolved when 12.5 mmol of the salt NaCH3COO were made. Again, the equation for this reaction is: CH 3 COOH aq + NaOH aq NaCH 3 COO aq + H 2 O l This reaction is balanced and makes 1 mole of NaCH3COO (from looking at the stoichiometric coefficients) The amount of 709.9J was found for 12.5 mmoles of NaCH3COO. We calculate Hrxn for 1 mole of reaction, which in this case makes 1 mole of NaCH3COO. 12.5mmol = 0.0125mol (709.9kJ / 0.0125mol NaCH3COO) * (1 mol NaCH3COO /1 mol reaction ) = 56792J/mol reaction = 56.8 kJ/mol reaction Finally we can write the thermochemical equation: CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l) + 56.8kJ/nol rxn ...
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This note was uploaded on 02/12/2012 for the course CH 301 taught by Professor Fakhreddine/lyon during the Summer '07 term at University of Texas at Austin.

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