Question

Physical Chemistry Thermodynamics:

The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its value has been accurately determined to be —3226.7 kJ mol-1, (a) When 0.9862 g of benzoic acid was oxidized, the temperature rose from 21.84°C to 25.67°C. What is the heat capacity of the calorimeter? (b) In a separate experiment, 0.4654 g of glucose (C6H1206) was oxidized in the same calorimeter, and the temperature rose from 21.22°C to 22.28°C. Calculate the enthalpy of combustion of glucose, the value of delta r U for the combustion, and the molar enthalpy of formation of glucose.

Answer 1

a)

number of moles of benzoic acid = mass/ molar mass

                                                   = 0.9862 / 122.12

                                               =8.076*10^-3 mol

Heat released = 3226.7 KJ/mol*8.076*10^-3 mol

              = 26.06 KJ

Q = C*(Tf-Ti)

26.06 KJ = C* (25.67 – 21.84) oC

C = 6.804 KJ/oC

Answer: 6.804 KJ/oC

b)

Q = C*(Tf-Ti)

   = 6.804*(22.28 – 21.22)

   =7.212 KJ

number of moles of glucose= mass/molar mass

                           = 0.4654 / 180

                                         = 2.586*10^-3 mol

delta H = -Q/number of mol

     = - 7.212 KJ /(2.586*10^-3)mol

     = -2789.3 KJ/mol

Answer: -2789.3 KJ/mol