Use Table 5.2 on slide 15 of the accompanying PowerPoint and calculate the energ
ID: 1054029 • Letter: U
Question
Use Table 5.2 on slide 15 of the accompanying PowerPoint and calculate the energy required to combust 1 mole of ethanol and 1 mole of H_2. Assume there is one mole of ethanol for live first equation and one mole of H_2 for the second equation. In order to figure this out, follow the example calculation as on slides 14-17 for each reaction using the bond energies in Table 5.2. You will need to draw the Lewis structure for each molecule because some of the bonds are double bonds. If that is the case, watch out and use the correct bond energy in table provided. Your answer should be in kJ/mol (since you arc dividing by 1 mol of the substance reacted) C_2H_5OH + O_2 rightarrow CO_2 + H_2 O(l) net energy change=? kJ Balance the reaction first! H_2 + 1/2 O_2 rightarrow H_2 O net energy change =? kJ Part II. Calculate the net energy change for each reaction in slide 19. Report your answer in kJ/g and show your work. ? C_2 H_5OH + O_2 rightarrow CO_2 + H_2 O(l) unbalanced net energy change = ? kJ/g H_2 + 1/2 O_2 rightarrow H_2 O net energy change = ?kJ/gExplanation / Answer
The balanced reaction is C2H5OH + 3O2---->2CO2+3H2O(l)
C2H5OH is CH3CH2OH. There are 5 C-H bonds, one C- C bond and one O-H bond
bond enthalpies ( Kj/mol): C-C =348 Kj/mol, C-H = 412, O-H = 463 and C-O = 360
enthalpy of C2H5OH= 5*412+348+463 = 2871 Kj
bond enthalpy of 3 O=O = 3*498=1494 Kj
enthalpy required to break the bonds = 2871+1494 = 4365 Kj
there are 2 O-H bonds in one mole of water, bond enthalpy =6*467 = 2802 Kj
there are 2 C=0 in CO2, hence bond enthalpy =4*799=3196
enthalpy required to form the products = 2802+3196=5998 Kj
enthalpy change = Bond energy of reactants - bond energy of products = 4365-5998=-1663 Kj
this is the bond energy per moles of C2H5OH ,i.e 46 gm of ethanol
enthalpy change -1663/46 Kj/gm =-35.5 Kj/gm
b) for the second reaction there are 1 H-H bond ( bond energy = 432 Kj/mole , 0.5 O=O bonds ( O=0 =494)
similalry = 432+498/2 - 2802=-2123 Kj
enthalpy change =-2123/2 Kj/gm of H2=-1061.5 Kj/gm of H2
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