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Bond Order and Enthalpy

For the react...

Question

For the reaction, $4 A l (s) + 3 O_{2} (g) + 6 H_{2} O + 4 O H^{-} ⇋ 4 [A l (O H)_{4}^{-}]$ ; $E_{c e l l}^{\circ} = 2.73 V$ .
If $Δ G_{f}^{\circ} (O H^{-}) = 157 k J m o l^{- 1}$ and $Δ G_{f}^{\circ} (H_{2} O) = - 237.2 k j m o l^{- 1}$ , determine $Δ G_{f}^{\circ} [A l (O H)_{4}]$ .

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Solution

Given reaction, $4 A l (s) + {3 O}_{2} (g) + 6 H_{2} O + 4 {O H}^{-} ⇌ 4 [{A l (O H)}_{4}^{-}]$
We have $G_{r e a c t i o n}^{0} = - n F E_{c e l l}^{0}$
$= - 4 \times 96500 \times 2.73$
$= 1053780$ $J {m o l}^{- 1}$
$= 1053.78$ $K J {m o l}^{-}$
and ${Δ G}_{r e a c t i o n}^{0} = {Δ G}_{p r o d u c t s}^{0} - {Δ G}_{r e a c t a n t s}^{0}$
$1053.78 = 4 \times {Δ G}_{f}^{0} [{A l ({O H}_{4})}^{-}] - [4 \times (0) + 3 \times (0) + 6 \times (- 237.2) + 4 \times (157)]$
$1053.78 = 4 \times {Δ G}_{f} [{A l ({O H}_{4})}^{-}] - [- 795.2]$
${Δ G}_{f}^{0} [{A l ({O H}_{4})}^{-}] = \frac{1053.78 - 795.2}{4}$
$∴$ ${Δ G}_{f}^{0} [{A l ({O H}_{4})}^{-}] = 64.645$ $K J {m o l}^{- 1}$

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Similar questions

Q. For the reaction,

4 A J (s) + {3 O}_{2} (g) + 6 H_{2} O (l) + 4 O H^{-} (a q) \to 4 [A J (O H)_{4}^{-} (a q)];

E_{c e l l} = 2.73 V . I f Δ G_{f} (O H^{-}) = - 157 k J m o l^{- 1}

Δ G_{f} (H_{2} O) = - 237.2 k J m o l^{- 1},

Δ G_{f} [A J (O H)_{4}^{-}]

Q. For the reaction :

4 A l (s) + 3 O_{2} (g) + 6 H_{2} O + 4 ⊖ O H \to 4 [A l (O H)_{4}]^{\oplus}

E_{c e l l}^{⊖} = 2.73 V

Δ_{f} G^{⊖} [\oplus O H] = - 157 k J m o l^{- 1}

;

Δ_{f} G^{⊖} (H_{2} O) = - 237.2 k J m o l^{- 1}

Δ_{f} G^{⊖} [A l (O H)_{4}]^{⊖}

is:
[free energy of formation of

(A l (O H)_{4}^{⊖}]

.

E_{c e l l}^{o}

4 A l (s) + 3 O_{2} (s) + 6 H_{2} O + 4 O H^{-} \to 4 [A l (O H)_{4}^{-}]

△ G_{f}^{o}

O H^{-}

H_{2} O

are - 157 kJ mol

^{- 1}

and - 237.2 kJ mol

^{- 1}

△ G_{f}^{o}

[A l (O H)_{4}]^{-}

x

- x

Δ G^{⊖}

for the following reaction is:

4 A 1 + 3 O_{2} + 6 H_{2} O + 4 ⊖ O H \to 4 A 1 (O H)_{4}^{⊖}

E_{c e l l}^{⊖} = 2.73 V

Δ_{f} G_{(⊖ O H)}^{⊖} = - 157 k J m o l^{- 1}

Δ_{f} G_{(H_{2} O)}^{⊖} = - 237 k J m o l^{- 1}

For the reaction at $25^{\circ} C N_{2} O_{4 (g)} ⇌ 2 N O_{2 (g)}$ , if $Δ G_{f}^{\circ}$ for $N_{2} O_{4}$ and $N O_{2}$ are 23.49 and 12.39 $K C a l . m o l^{- 1}$ , then $l o g (K_{p})$ for the reaction is:

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