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Question
Using the data (all value are in kilocalorie per mole at 25∘C ) given below, the bond energy of C−C and bond C−H bond is m kcal and n kcal, respectively. Calculate the value of m+n
ΔH∘combustion~of~ethane=−372.0ΔH∘Combustion of propane=−530.0ΔH∘forC(graphite)→C(g)=+172.0Bond energy of H - H bond=+104.0ΔH∘f of H2O(l)=−68.0ΔH∘f of CO2(g)=−94.0
ΔH∘combustion~of~ethane=−372.0ΔH∘Combustion of propane=−530.0ΔH∘forC(graphite)→C(g)=+172.0Bond energy of H - H bond=+104.0ΔH∘f of H2O(l)=−68.0ΔH∘f of CO2(g)=−94.0
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Solution
C2H6(g)+72O2→2CO2(g)+3H2O(l):ΔH=−372.0ΔH∘f(C2H6)=2×(−94.0)+3×(−68.0)+372.0=−20 kcalC3H8(g)+5O2→3CO2(g)+4H2O(l);ΔH=−530.0ΔH∘f(C3H8)=3×(−94.0)+4×(−68.0)+530.0=−24 kcal2C(s)+3H2(g)→C2H6(g); ΔH=−20.02C(g)→2C(s); ΔH=−344.06H(g)→3H2(g); ΔH=−312.0–––––––––––––––––––––––––––––––––––––Adding 2C(g)+6H(g)→C2H6(g);ΔH=−676 kcal
So, enthalpy of formation of 6C - H bond and one C - C bond is −676.0 kcal.
3C(s)+4H2(g)→C3H8(g); ΔH=−24.03C(g)→3C(s); ΔH=−516.08H(g)→4H2(g); ΔH=−416.0–––––––––––––––––––––––––––––––––––––Adding 3C(g)+8H(g)→C3H8(g);ΔH=−956.0 kcal
So, enthalpy of formation of 8C - H and 2C - C bonds in −956 k cal.
Let the bond energy of C - C be x and of C - H be y kcal
In ethanex+6y=676In propane2x+8y=956on solvingx=82 and y=99
Thus, bond energy of C−C=82 kcal and
bond energy of C−H=99 kcal
So, enthalpy of formation of 6C - H bond and one C - C bond is −676.0 kcal.
3C(s)+4H2(g)→C3H8(g); ΔH=−24.03C(g)→3C(s); ΔH=−516.08H(g)→4H2(g); ΔH=−416.0–––––––––––––––––––––––––––––––––––––Adding 3C(g)+8H(g)→C3H8(g);ΔH=−956.0 kcal
So, enthalpy of formation of 8C - H and 2C - C bonds in −956 k cal.
Let the bond energy of C - C be x and of C - H be y kcal
In ethanex+6y=676In propane2x+8y=956on solvingx=82 and y=99
Thus, bond energy of C−C=82 kcal and
bond energy of C−H=99 kcal
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