Question

Using the equations 2 C₆H₆ (l) + 15 O₂ (g) → 12 CO₂ (g) + 6 H₂O (g)∆H° = -6271 kJ/mol C (s) + O₂ (g) → CO₂ (g) ∆H° = -393.5 kJ/mol 2 H₂ (g) + O₂ (g) → 2 H₂O (g) ∆H° = -483.6 kJ/mol Determine the enthalpy for the reaction 6 C (s) + 3 H₂ (g) → C₆H₆ (l).

Answer 1

Answer : The enthalpy for the reaction is 49.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of $C_6H_6$ will be,

$6C(s)+3H_2(g)\rightarrow C_6H_6(l)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $2C_6H_6(g)+15O_2(g)\rightarrow 12CO_2(g)+6H_2O(l)$    

$\Delta H_1=-6271kJ/mole$

(2) $C(s)+O_2(g)\rightarrow CO_2(g)$    

$\Delta H_2=-393.5kJ/mole$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$    

$\Delta H_3=-483.6kJ/mole$

Now we will reverse the reaction 1 and divide by 2, multiply reaction 2 by 6 and reaction 3 by 3/2 then adding all the equations, we get :

(1) $6CO_2(g)+3H_2O(l)\rightarrow C_6H_6(g)+\frac{15}{2}O_2(g)$    

$\Delta H_1=-\frac{-6271kJ/mole}{2}=3135.5kJ/mol$

(2) $6C(s)+6O_2(g)\rightarrow 6CO_2(g)$    

$\Delta H_2=6\times (-393.5kJ/mole)=-2361kJ/mol$

(3) $3H_2(g)+\frac{3}{2}O_2(g)\rightarrow 3H_2O(l)$    

$\Delta H_3=\farc{3}{2}\times (-483.6kJ/mole)=-725.4kJ/mol$

The expression for enthalpy of formation of $C_6H_6$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(3135.5kJ/mole)+(-2361kJ/mole)+(-725.4kJ/mole)$

$\Delta H=49.1kJ/mole$

Therefore, the enthalpy for the reaction is 49.1 kJ/mol