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 ArU for the combustion, and the molar enthalpy of formation of glucose.

Answer 1

Heat liberated by the burning of benzoic acid = heat absorbed by calorimeter

Moles of benzoic acid used = 0.9862gm/ 122.12 gmmol^-1 = 8.076*10^-3 moles

Heat liberated = 3226.7 kJ mol-1 * 8.076*10^-3 moles =

Heat absorbed = heat capacity * temperature change = heat capacity * (25.67-21.84)

3.83* heat capacity= 26058.83 J

heat capacity = 26058.83 J/3.83 = 14.099 J/oC

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heat absorbed by the calorimeter = heat capacity*(22.28-21.22) =14.099 J/oC *1.05 oC =14.95 J

Number of moles of glucose burned = 0.4654/180gmmol^-1 =2.59 *10^-3 moles

Enthalpy of combustion =dH= 14.95 J/2.59 *10^-3 moles =5.77 kJ/mol

dH = dU + PdV

as this is a constant volume calorimeter dV = 0, so, dU = dV

So, dU = 5.77 kJ/mol