Question

$2C_4{H}_{10}⟶C_8{H}_{18}+H_2$. Given that bond energy of $C-C,C-H$ are $347.3$ and $414.2$

kJ mol${}^{-1}$ and the heat of formation of hydrogen atom is $217.55$ kJ mol${}^{-1}$. Find $\Delta H$ for the above reaction in kJ mol${}^{-1}$.

(If $\Delta H=x$, input the answer to the nearest integer as $\frac{x}{9}$).

Answer 1

For $2C_4{H}_{10}⟶C_8{H}_{18}+H_2;\Delta H=?$

$\therefore \Delta H=$ bond energy data used for the formation of bond + bond energy data used for dissociation of bond

$\Delta H=-[7\times E_{(C--C)}+18\times E_{(C--H)}+E_{(H--H)}]+2[3\times E_{(C--C)}+10\times E_{(C--H)}]$

$\Delta H=-e_{(C--C)}+2\times e_{(C--H)}-e_{(H--H)}$

$\because$ Given, $\frac{1}{2}{H}_2⟶H;\Delta H=217.55$ kJ

$\therefore e_{(H--H)}=2\times 217.55=435.10$ kJ

$\therefore \Delta H=-347.3+2\times \left(414.2\right)-1\times \left(435.1\right)$

$\Delta H=46.0$ kJ mol${}^{-1}$

Now,

$\frac{46}{9}=5.11\equiv 5$ kJ mol${}^{-1}$