Physical Chemistry Thermodynamics: The enthalpy of combustion of benzoic acid (C
ID: 1049417 • Letter: P
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.
Explanation / Answer
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