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Heat of Formation

Calculate the...

Question

Calculate the heat of the formation of anhydrous $A l_{2} C l_{6}$ from the following data:
a. $2 A l (s) + 6 H C l (a q) ⟶ A l_{2} C l_{6} (a q) + 3 H_{2} + 1004.2 k J$
b. $H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g) + 184.1 k J$
c. $H C l (g) + a q ⟶ H C l (a q) + 73.2 k J$
d. $A l_{2} C l_{6} (s) + a q ⟶ A l_{2} C l_{6} (a q) + 643.1 k J$

Δ_{f} H = - 1352.6 k J m o l^{- 1}

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Δ_{f} H = + 1352.6 k J m o l^{- 1}

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Δ_{f} H = - 2638.67 k J m o l^{- 1}

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None of these

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Solution

The correct option is A $Δ_{f} H = - 1352.6 k J m o l^{- 1}$
As given,
a. $2 A l (s) + 6 H C l (a q) ⟶ A l_{2} C l_{6} (a q) + 3 H_{2} + 1004.2 k J$
b. $H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g) + 184.1 k J$
c. $H C l (g) + a q ⟶ H C l (a q) + 73.2 k J$
d. $A l_{2} C l_{6} (s) + a q ⟶ A l_{2} C l_{6} (a q) + 643.1 k J$
So for
$2 A l (s) + 3 C l_{2} (g) \to A l_{2} C l_{6}; Δ H_{f} = ?$
From given data
$Δ H_{f} = - (1004.2 - 643.1 + 6 * 73.2 + 3 * 184.1) = - 1352.6 k J / m o l$

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