Select equation-s- having endothermic step-Na-g-Cl-g-rightarrow NaCl-s-N-g-rightarrow N-g-Al-2-g-rightarrow Al-3-g-O-g-rightarrow O-2-g-

-A-O-x2212-g-x2192-O2-x2212-g-x394-He-g-ve-B-Na-g-CI-x2212-g-x2192-Na-CI-x2212-s-x394-HI-E-x2212-ve-C-N-g-x2192-N-x2212-g-x394-He-g-ve-D-Al2-g-x2192-Al3-g-x394-HI-E-ve

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Standard XII

Chemistry

Question

Select equation(s) having endothermic step:
$N a^{+} (g) + C l^{-} (g) \to N a C l (s)$
$N (g) \to N^{-} (g)$
$A l^{2 +} (g) \to A l^{3 +} (g)$
$O^{-} (g) \to O^{2 -} (g)$

O^{-} (g) \to O^{2 -} (g)

A l^{2 +} (g) \to A l^{3 +} (g)

N a^{+} (g) + C l^{-} (g) \to N a C l (s)

N (g) \to N^{-} (g)

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Solution

Verified by Toppr

$(A) O^{-} (g) \to O^{2 -} (g); Δ H_{e . g .} = (+) v e$
$(B) N a^{+} (g) + C I^{-} (g) \to N a^{+} C I^{-} (s); Δ H_{I . E .} = (-) v e$
$(C) N (g) \to N^{-} (g); Δ H_{e . g .} = (+) v e$
$(D) A l^{2 +} (g) \to A l^{3 +} (g); Δ H_{I . E .} = (+) v e$

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Similar Questions

Select equations having exothermic step.
(I)

S_{(g)}^{-} \to S_{(g)}^{2 -}

(II)

N a_{(g)}^{+} + C l_{(g)}^{-} \to N a C l_{(s)}

(III)

N_{(g)} \to N_{(g)}^{-}

(IV)

A l_{(g)}^{2 +} \to A l_{(g)}^{3 +}

Choose the correct code(s).

View Solution

Select the equations that have an exothermic step.

I . S_{(g)}^{-} \to S_{(g)}^{2 -}

I I . N a_{(g)}^{+} + C l_{(g)}^{-} \to N a C l_{(s)}

I I I . N_{(g)} \to N_{(g)}^{-}

I V . A l_{(g)}^{2 +} \to A l_{(g)}^{3 +}

Choose the correct code:

View Solution

Select the equations that have an exothermic step.

I . S_{(g)}^{-} \to S_{(g)}^{2 -}

I I . N a_{(g)}^{+} + C l_{(g)}^{-} \to N a C l_{(s)}

I I I . N_{(g)} \to N_{(g)}^{-}

I V . A l_{(g)}^{2 +} \to A l_{(g)}^{3 +}

Choose the correct code:

View Solution

The formation of the oxide ion

O^{2 -} (g)

requires first an exothermic and then, an endothermic step as shown below,

O (g) + e ⟶ O^{-} (g)

;

Δ H = - 142 k J / m o l

O^{-} (g) + e ⟶ O^{2 -} (g)

;

Δ H = 844 k J / m o l

.
This is because:

View Solution

The formation of the oxide ion

O^{2 -} (g)

requires first an exothermic and then an endothermic step as shown below:

O (g) + e^{-} ⟶ O^{-} (g)

;

Δ H = - 142 k J / m o l

O^{-} (g) + e^{-} ⟶ O^{2 -} (g)

;

Δ H = 844 k J / m o l

This is because:

View Solution

Select equation(s) having endothermic step:Na+(g)+Cl−(g)→NaCl(s)N(g)→N−(g)Al2+(g)→Al3+(g)O−(g)→O2−(g)

(A)O−(g)→O2−(g);ΔHe.g.=(+)ve(B)Na+(g)+CI−(g)→Na+CI−(s);ΔHI.E.=(−)ve(C)N(g)→N−(g);ΔHe.g.=(+)ve(D)Al2+(g)→Al3+(g);ΔHI.E.=(+)ve

Select equation(s) having endothermic step:
$N a^{+} (g) + C l^{-} (g) \to N a C l (s)$
$N (g) \to N^{-} (g)$
$A l^{2 +} (g) \to A l^{3 +} (g)$
$O^{-} (g) \to O^{2 -} (g)$

$(A) O^{-} (g) \to O^{2 -} (g); Δ H_{e . g .} = (+) v e$
$(B) N a^{+} (g) + C I^{-} (g) \to N a^{+} C I^{-} (s); Δ H_{I . E .} = (-) v e$
$(C) N (g) \to N^{-} (g); Δ H_{e . g .} = (+) v e$
$(D) A l^{2 +} (g) \to A l^{3 +} (g); Δ H_{I . E .} = (+) v e$