Benzoic acid 1.35 g, is reacted with oxygen in a constant volumecalorimeter to f

Question

Benzoic acid 1.35 g, is reacted with oxygen in a constant volumecalorimeter to form H2 O(l) and CO2 (g)at 298 K. the mass of the water volume in the inner bath is1.376*103 g. The temperature of the calorimeter and itsconstant rises 3.11 K as a result of this reaction. Calculate thecalorimeter constant.

Explanation / Answer

C7H6O2 + (15/2) O2 = >   3H2O + 7CO2 delH = stoichiometric coefficient* enthalpy of formation delH = (7*-393.509 kJ/mol) + (3*-285.83 kJ/mol) - (15/2 * 0) -(1*-385.2 kJ/mol) delH = -3226.853 kJ/mol benzoic acid delU = delH - del(PV) = delH - RT*deln deln = change number of gas moles = (7 - 15/2) = -1/2 = -0.5 delU = -3226.853 kJ/mol - (8.314 e-3 kJ/mol/K * 298 K* -0.5) delU = -3225.61421 kJ/mol benzoic acid molar mass benzoic acid = 122.12 g/mol Heat released, 1.35 g benzoic acid *( 1 mol/ 122.12 g) * (3225.61421 kJ/mol) =35.6581984 kJ released heat released = heat absorbed by water + heat absorbed bycalorimeter heat released = mass water*heat capacity*T + calorimeter constant*T calorimeter constant = (heat released - mass water*heatcapacity*T) / T calorimeter constant = (35.66 kJ - 1.376e3 g *4.18e-3 kJ/g/K *3.11K)/ 3.11 K calorimeter constant = 5.71397865 kJ/ K calorimeter constant = 5.71 kJ/K Clarification request, The temperature of the calorimeter and itsconstant rises 3.11 K as a result of thisreaction. Calculate the calorimeter constant. Do you mean "contents?"

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