Question

What is the heat of formation of $C_6{H}_6$, give that the heat of combustion of Benzene, carbon and Hydrogen are $-782,-94$ and $-68K$ respectvely

• A. $+14K.Cal$
• B. $-14K.Cal$
• C. $+28K.Cal$
• D. $-28K.Cal$

Answer 1

The correct option is A $+14K.Cal$

$6C+3H_2\to C_6{H}_6⟶\left(i\right)$ ; ${\Delta H}_f=?$

$C_6{H}_6+\frac{15}{2}{O}_2⟶{6CO}_2+3H_{2}O⟶\left(ii\right)$ ; ${\Delta H}_h=-782K$ $Cal$

given, that heat of combustion of benzene $=-782K$ $Cal$

carbon $=-94K$ $Cal$

Hydrogen $=-68K$ $Cal$

$C+O_2\to {CO}_2⟶\left(iii\right)$ ${\Delta H}_{f_2}=-94K$ $Cal$

$H_2+\frac{1}{2}{O}_2\to H_{2}O⟶\left(iv\right)$ ${\Delta H}_{f_3}=-68K$ $Cal$

$\left(i\right)=6\left(ii\right)+3\left(iv\right)-ii$

$\therefore {\Delta H}_f=({6\Delta H}_2+3{\Delta H}_3)-{\Delta H}_1$

${\Delta H}_f=(6\times (-94)+3\times (-68))-(-782)$

${\Delta H}_f=14K.Cal$

$\Rightarrow$ Heat of formation of Benzene is $+14$ $K.Cal$