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Stoichiometry

The combustio...

Question

The combustion of one mole of benzene takes place at $298 K$ and $1 a t m .$ After combustion, $C O_{2} (g)$ and $H_{2} O (l)$ are produced and $3267.0 k J$ of heat is liberated. Calculate the standard enthalpy of formation, $△_{f} H^{⊖}$ of benzene. Standard enthalpies of formation of $C O_{2} (g)$ and $H_{2} O (l)$ are $- 393.5 k J m o l^{- 1}$ and $- 285.83 k J m o l^{- 1}$ respectively.

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Solution

Step 1: Enthalpy of combustion of benzene

The enthalpy of combustion of $1 m o l$ of benzene is:

$C_{6} H_{6} (l) + \frac{15}{2} O_{2} ⟶ 6 C O_{2} (g) + 3 H_{2} O (l); △_{f} H^{⊖} = - 3267 k J m o l^{- 1} \dots (i)$

Step 2: Enthalpy of formation of $C O_{2}$

The enthalpy of formation of $1 m o l$ of $C O_{2} (g) :$

$C (g r a p h i t e) + O_{2} (g) ⟶ C O_{2} (g); △_{f} H^{⊖} = - 393.5 k J m o l^{- 1} \dots (i i)$

Step 3: Enthalpy of formation of $H_{2} O$

The enthalpy of formation of $1 m o l$ of $H_{2} O (l)$ is:

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{C} H^{⊖} = - 285.83 k J m o l^{- 1} \dots (i i i)$

Step 4: Enthalpy of formation of benzene

The formation reaction of benzene is given by:

$6 C (g r a p h i t e) + 3 H_{2} (g) \to C_{6} H_{6} (l); △_{f} H^{⊖} = ? \dots (i v)$

Multiplying equation $(i i)$ by $6$ and equation $(i i i)$ by $3$ we get,

$6 C (g r a p h i t e) + 6 O_{2} (g) \to 6 C O_{2} (g); △_{f} H^{⊖} = - 2361 k J m o l^{- 1}$

$3 H_{2} (g) + \frac{3}{2} O_{2} (g) \to 3 H_{2} O (l); △_{f} H^{⊖} = - 857.49 k J m o l^{- 1}$

Summing up the above two equations:

$6 C (g r a p h i t e) + 3 H_{2} (g) + \frac{15}{2} O_{2} (g) \to 6 C O_{2} (g) + 3 H_{2} O (l);$

$△_{f} H^{⊖} = - 3218.49 k J m o l^{- 1} \dots . (v)$

Reversing equation $(i) :$

$6 C O_{2} (g) + 3 H_{2} O (l) \to C_{6} H_{6} (l) + \frac{15}{2} O_{2}; △_{f} H^{⊖} = + 3267.0 k J m o l^{- 1} \dots . (v i)$

Adding equation $(v)$ and $(v i),$ we get

$6 C (g r a p h i t e) + 3 H_{2} (g) \to C_{6} H_{6} (l) \dots . (i v);$

$(△ H^{⊖})_{B e n z e n e} = - 3218.49 + 3267$

$(△_{f} H^{⊖})_{B e n z e n e} = 48.51 k J m o l^{- 1}$

Final Answer: $(△_{f} H^{⊖})_{B e n z e n e} = 48.51 k J m o l^{- 1} .$

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