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Ellingham Diagram

Consider the ...

Question

Consider the following reaction at $1000^{\circ} C$ ,
$i . Z n (s) + \frac{1}{2} O_{2} (g) \to Z n O (s); Δ G = - 360 KJ/mol$
$i i . C (s) + \frac{1}{2} O_{2} (g) \to C O (s); Δ G = - 460 KJ/mol$
Then choose the correct statement from the following

Zinc can be oxidized by

C O

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Z n O

can be reduced by

C

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Z n O

can be reduced by

C O

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None of the above

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Solution

The correct option is B $Z n O$ can be reduced by $C$
In Ellingham diagram, the $Δ_{f} G^{o}$ of $C$ lies below that of $Z n O$ . If the reduction is carried out at very high temperatures, the C will act as a reductant and can reduce $Z n O$ .

$Z n O (s) \to Z n (s) + \frac{1}{2} O_{2} (g) Δ_{f} G^{o} = - 360$

$C (s) + \frac{1}{2} O_{2} (g) \to C O (s) Δ_{f} G^{o} = - 460$

Net reaction
$Z n O (s) + C (s) \to Z n (s) + C O Δ_{f} G^{o} = - 100$
As $Δ_{f} G^{o} = - 100 k J / m o l$ , the reaction is spontaneous.

Suggest Corrections

Similar questions

1000^{\circ} C

i . Z n (s) + \frac{1}{2} O_{2} (g) \to Z n O (s); Δ G = - 360 KJ/mol

i i . C (s) + \frac{1}{2} O_{2} (g) \to C O (s); Δ G = - 460 KJ/mol

10000^{\circ} C

(a) Z n (s) + \frac{1}{2} O_{2} (g) \to Z n O (s); Δ G^{\circ} = - 360 k J / m o l (b) C (s) + \frac{1}{2} O_{2} (g) \to C O (s); Δ G^{\circ} = - 460 k J / m o l

^{\circ} C

Z n_{(s)} + 1 / 2 O_{2 (g)} ⟶ Z n O_{(s)}, Δ G^{\circ} = - 360 k J m o l^{- 1}

C_{(s)} + 1 / 2 O_{2 (g)} ⟶ C O_{(g)}, Δ G^{\circ} = - 460 k J m o l^{- 1}

^{\circ} C

Which statement is correct about the following reaction?
ZnO + CO → Zn + $C O_{2}$

1000^{o} C

Z n (s) + \frac{1}{2} O_{2} (g) \to Z n O (s)

Δ G^{o} = - 360 k J {m o l}^{- 1}

C_{g r a p h i t e} + \frac{1}{2} O_{2} (g) \to C O (g) Δ G^{o} = - 460 k J {m o l}^{- 1}

1000^{o} C

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