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C2H6 + 7/2 O2 → 2CO2 +3H2O Finding ∆Hcombustion= ? ∆Hcombustion =∑Bonds energy of reactants - ∑Bonds energy of products= [6 (C-C) + 1(C-C) + 7/2 (OO) ] – [2(2(C=O)) + 3(2(O-H))] =[6(412kJ/mol) + 1 (347 kJ/mol) + 7/2 (496kJ/mol)] – [4(805kJ/mol) + 6(464kJ/mol)] = 4562kJ/mol – 6004kJ/mol = -1442kJ/mol Therefore, the enthalpy of combustion of ethane is -1442kJ/mol.
 * 3a)** **Given the following bond enthalpies (bond energies) in kJ/mol**
 * C-H 412; C-C 347; O-H 464; O=O 498; C=O 805; C-O358**
 * Enthalpy of combustion = ?**

1) C(s) +O2 (g) → CO2 (g) ∆Hcomb = -393kJ 2) H2 (g) + ½ O2 (g)→ H2O (g) ∆Hcomb = -285.6kJ 3) C2H6 (g) + 7/2 O2 (g) → 2CO2 (g) + 3H2O (l) ∆Hcomb = -1560kJ Formation equation for C2H6 (g) 2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = ? Using Hess’s Law: a) multiply 1) by 2 b) multiply 2) by 3 c) multiply 3) by -1 (reverse the equation) 1) 2C (s) + 2O2 (g) → 2CO2 (g) ∆Hcomb = -786kJ 2) 3H2 (g) + 3/2 O2 (g) → 3H2O (g) ∆Hcomb = -856.8kJ 3) 2CO2 (g) + 3H2O → C2H6 (g) + 7/2 O2 (g) ∆Hcomb = 1560kJ After cancelling out reactants and products that are same, we get: 2C (s) + 3H2(g) → C2H6 (g) ∆Hformation =-786kJ -856.8kJ +1560kJ -82.8kJ 2C (s) + 3H2(g) → C2H6 (g) ∆Hformation = -82.8kJ Therefore, the standard enthalpy of formation is -82.8kJ. 7) The Activation energy from A to B is: 350kJ - 250kJ =100kJ Therefore, the activation energy for reaction A to B is 100kJ
 * 4)Given the following standard enthalpies of combustion**
 * C (s) -393kJ/mol; H2(g) -285.6kJ/mol; C2H6 (g) -1560kJ/mol**
 * Calculate the standard enthalpies of formation of ethane.**

1 e 2 c 3d 4c 5a 6d 7e 8b 9e
 * Multiple Choice:**