ENTHALPY OF P4O10 (S) IS -2984 KJ/MOL, H2O (L) IS -187
KJ/MOL, AND H3PO4(AQ) IS -1288 KJ/MOL
1. Tetraphosphorus decaoxide and water react according to the following balanced equation: P4010 (8) + 6H20 (1) 4H3PO4 (aq) (a) Using Handout 9, look up the standard enthalpy of formations of P4010 (s), H20 (1), and H3PO4 (aq). (b) Write out the formation reactions for P4010 (s), H20 (1), and H3PO4 (aq). (c) Using the formation reactions and their corresponding standard enthalpy of formations, use Hess's Law to calculate the standard enthalpy change for the reaction above. (d) Using the standard enthalpy of formations from (a), use the short method to calculate the standard enthalpy change for the reaction above. 2. Using the standard enthalpy of formations (also called the standard heat of formations) on Handout 9, calculate AH° for the combustion of octane, in kJ/gram of octane, and AH° for the combustion of ethanol, in kJ/gram of ethanol. Determine which fuel provides the most energy per gram (the standard heat of formation for octane is -269 kJ/mol). (a) 2C8H18(1) + 2502(g) → 16CO2(g) + 18H2O(g) (b) C2H5OH(1) + 302(g) → 2CO2(g) + 3H2O(g) 3. Using the standard heat of formations on Handout 9, calculate AH° for each of the following reactions in kJ per gram of sodium: (a) 4Na(s) + O2(g) + 2Na2O(s) (b) 2Na(s) + 2H20(1) + 2NaOH(aq) + H2(g) (c) 2Na(s) + CO2(g) → Na2O(s) + CO(g) Explain why a water or carbon dioxide fire extinguisher might not be effective in putting out a sodium fire. 4. The standard heat of combustion of butane gas, C4H10, is -49.52 kJ/gram of butane. Using the standard heat of formations on Handout 9, calculate the standard heat of formation of butane gas.
ENTHALPY OF P4O10 (S) IS -2984 KJ/MOL, H2O (L) IS -187 KJ/MOL, AND H3PO4(AQ) IS -1288 KJ/MOL
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ENTHALPY OF P4O10 (S) IS -2984 KJ/MOL, H2O (L) IS -187 KJ/MOL, AND H3PO4(AQ) IS -1288 KJ/MOL
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