At 10000C- Zns-1-2- ZnOs- G0 - - 360 kJ mol-1 C1 -1-2 O2g- G0-460kJmol-1The correct statement is-

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Standard XII

Chemistry

Wurtz Halide Reaction

At 10000C, ...

Question

At $1000^{0} C$ ,
$Z n_{(} s) + \frac{1}{2} \to Z n O_{(s)}; △ G^{0} = - 360 k J m o l^{- 1}$
$C_{(1)} + \frac{1}{2} O_{2 (g)}; △ G^{0} = - 460 k J m o l^{- 1}$
The correct statement is:

zinc can be oxidised by carbon monoxide

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zinc blend is produced during the reaction

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zinc oxide can be reduced by graphite

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zinc can be oxidised by graphite

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Solution

The correct option is C zinc oxide can be reduced by graphite
Formation of $^{'} C O^{'}$ , is more spontaneous as it has more negative $△ G$ value. Then $Z n O$ changes to $Z n$ because of less negative $^{'} △ G^{'}$ value. So, $Z n O$ can be reduced by Graphite $Z n O + C \to Z n + C O$

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Similar questions

Q. 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

Q. Consider the following reactions at 1000

^{\circ} C

(i)

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

(ii)

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

Choose the correct statement at 1000

^{\circ} C

Q. Consider the following reactions at

1000^{o} C

;
(A)

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

;

Δ G^{o} = - 360 k J {m o l}^{- 1}

(B)

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}

Choose the correct statement at

1000^{o} C

Q. Anthracene can be electrolytically oxidised to anthraquinone.
What mass of anthraquinone can be produced by the passage of

1

ampere current for an hour at

100

% efficiency?

F e^{2 +}

as well as

H N O_{2}

can be used to estimate

M n O_{4}^{-}

in acidic medium. If 100 mL of 1 M

F e^{2 +}

can be oxidised by

V

mL of 1 M

M n O_{4}^{-} / H^{+}

solution, calculate volume of 1 M

H N O_{2}

that can be oxidises by

V

mL of 1 M

M n O_{4}^{-} / H^{+}

solution.

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At 10000C,Zn(s)+12→ZnO(s);△G0=−360kJmol−1C(1)+12O2(g);△G0=−460kJmol−1The correct statement is:

The correct option is C zinc oxide can be reduced by graphiteFormation of ′CO′, is more spontaneous as it has more negative △G value. Then ZnO changes to Zn because of less negative ′△G′ value. So, ZnO can be reduced by Graphite ZnO+C→Zn+CO

At $1000^{0} C$ ,
$Z n_{(} s) + \frac{1}{2} \to Z n O_{(s)}; △ G^{0} = - 360 k J m o l^{- 1}$
$C_{(1)} + \frac{1}{2} O_{2 (g)}; △ G^{0} = - 460 k J m o l^{- 1}$
The correct statement is:

The correct option is C zinc oxide can be reduced by graphite
Formation of $^{'} C O^{'}$ , is more spontaneous as it has more negative $△ G$ value. Then $Z n O$ changes to $Z n$ because of less negative $^{'} △ G^{'}$ value. So, $Z n O$ can be reduced by Graphite $Z n O + C \to Z n + C O$