Physical Chemistry Thermodynamics: The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard...

4.1 Ratings (831 Votes)

number of moles of benzoic acid = mass/ molar mass

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.9862 / 122.12

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â =8.076*10^-3 mol

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

Â Â Â Â Â Â Â Â Â Â Â Â Â = 26.06 KJ

Q = C*(Tf-Ti)

26.06 KJ = C* (25.67 â€“ 21.84) oC

C = 6.804 KJ/oC

Answer: 6.804 KJ/oC

Q = C*(Tf-Ti)

Â Â = 6.804*(22.28 â€“ 21.22)

Â Â =7.212 KJ

number of moles of glucose= mass/molar mass

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 0.4654 / 180

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.586*10^-3 mol

delta H = -Q/number of mol

Â Â Â Â = - 7.212 KJ /(2.586*10^-3)mol

Â Â Â Â = -2789.3 KJ/mol

Answer: -2789.3 KJ/mol

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