Trinitroglycerin, C3H5N3O9 (usually referred to simply as nitroglycerin), has be

Question

Trinitroglycerin, C3H5N3O9 (usually referred to simply as nitroglycerin), has been widely used as an explosive. Alfred Nobel used it to make dynamite in 1866. Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart) by dilating the blood vessels. The enthalpy of decomposition at 1 atm pressure of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas at 25 °C is –1541.4 kJ/mol. (a) Write a balanced chemical equation for the decomposition of trinitroglycerin. (b) Calculate the standard heat of formation of trinitroglycerin. (c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. If the sample is eventually oxidized in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released?

Explanation / Answer

a) The reaction is  C3H5(ONO3)3-->CO2+H20+N2+O2

The balanced equation is

4C3H5(ONO3)3-->12CO2+10H20+6N2+O2 (1) delH =-1541.4 Kj/mol

Enthalpy of products

CO2= -393.5 Kj/mol N2= 0, O2= 0 and H2O= -241.8 Kj/mol

Enthalpy of products =12*-393.5+10*-241.8=-7166 Kj

Reactant enthalpy=-1541.4* 4= -6165.6 Kj

Standard heat of formation = -7166 +6165.6==950.4 Kj

Standard heat of formation/ mole = -950.4/4=237.6 Kj/mol

c) moles of trinitroglycerin= 0.6*10-3g/227= 2.643*10-6gmol

Enthalpy of decomposition = -1541.4 Kj/mol (exothermic reaction)

heat released = 1541.4* 2.643*10-6*103 j= 4.074 joules=4.074*0.00024*1000 cal=0.9776 cal

Standard heat of formation = Enthalpy of products - Enthalpy of reactants

=12*

b)

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