Nitroglycerin is a powerful explosive, giving four different gases when detonated.2 C3H5(NO3)3 (l) → 3 N2 (g) + 1/2 O2 (g) + 6 CO2 (g) + 5 H2O (g)Given that the enthalpy of formation of nitroglycerin, ΔHf°, is −364 kJ/mol, calculate the energy (heat at constant pressure) released by this reaction.Δ= ΣnrxnHfΔH(prod.) − ΣmfΔH(reac.)= (3 mol)(0) + (1/2 mol)(0) + (6 mol)(−393.5 kJ/mol) + (5 mol)(−241.8 kJ/mol) − (2 mol)(−364 kJ/mol) rxnΔH= −2842 kJ rxnΔHb) What is delta H when 4.65 mol of products is formed?
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I'm not able to read all of the "funny" symbols you printed but if I've translated your question properly, I think the reaction produces 14.5 mols for 2842 kJ heat produced. You want 4.65 mols products so 2842 kJ/mol x (4.65/14.5) = ?
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