Question

Using the data (all values are in ${\text{kcal mol}}^{-1}$ at ${25}^o\text{C}$) given below, calculate bond energy of $C-C$ bond.
$\Delta H_{\text{combustion}}^o$ (ethane) $=-372.0$
$\Delta H_{\text{combustion}}^o$ (propane) $=-530.0$
$\Delta H_{\text{combustion}}^o$(graphite)$\to C\left(g\right)=-172.0$
Bond energy of $H-H=104.0$
$\Delta H_f^o$ of $H_{2}O\left(l\right)=68.0$
$\Delta H_f^o$ of $C{O}_2\left(g\right)=94.0$

Answer 1

For formation of $C_3{H}_8$
$3C+4H_2\to C_3{H}_8$, $\Delta H_1=?$
For formation of $C_2{H}_6$
$2C+3H_2\to C_2{H}_6$, $\Delta H_2=?$
Thus
$\left(i\right)\Delta H_1=-\left[3\right(C-C)+8(C-H\left)\right]+[3C_{s-g}+4(H-H\left)\right]$
$\left(ii\right)\Delta H_2=-\left[2\right(C-C)+6(C-H\left)\right]+[2C_{s-g}+3(H-H\left)\right]$
Let the bond energy of $C-C$ and $C-H$ bonds be $x\text{kcal}$ and $y\text{kcal}$ respectively.
Then, we have
$\left(iii\right)\Delta H_1=-(2x+8y)+[3\times 172+4\times 104]$
$\left(iv\right)\Delta H_2=-(x+6y)+[2\times 172+3\times 104]$
Given
$\left(v\right)C+O_2\to C{O}_2;\Delta H=-94.0\text{kcal}$
$\left(vi\right)H_2+\frac{1}{2}{O}_2\to H_{2}O;\Delta H=-68.0\text{kcal}$
$\left(vii\right)C_2{H}_6+\frac{7}{2}{O}_2\to 2C{O}_2+3H_{2}O;\Delta H=-372.0\text{kcal}$
$\left(viii\right)C_2{H}_8+5O_2\to 3C{O}_2+4H_{2}O;\Delta H=-530.0\text{kcal}$
$\therefore 2\times \left(v\right)+3\times \left(vi\right)-\left(vii\right)$, we get
$\left(ix\right)2C+3H_2\to C_2{H}_6;\Delta H_2=-20.0\text{kcal}$
Again $3\times \left(v\right)+4\times \left(vi\right)-\left(viii\right)$ gives,
$\left(x\right)3C+4H_2\to C_3{H}_8;\Delta H_1=-24.0\text{kcal}$
Solving equation $\left(iii\right),\left(iv\right),\left(ix\right)$ and $\left(x\right)$, we get
$x+6y=6762x+8y=956$
$\therefore x=82\text{kcal}$ and $y=99\text{kcal}$
Hence bond energy of $C-C$ bond $=82\text{kcal}$ and bond energy of $C-H$ bond $=99\text{kcal}$