For the processes $K^{+} (g) I ⟶ K (g) I I ⟶ K^{-} (g)$

Energy is released in (I) and absorbed in (II)

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Energy is absorbed in (I) and released in (II)

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Energy is absorbed in both the processes

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Energy is released in both the processes

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Solution

The correct option is A
Energy is released in (I) and absorbed in (II)

(I) $K^{+} (g) + e^{-} ⟶ K (g) ⟶$ Negative electron gain enthalpy

(II) $K (g) + e^{-} ⟶ K^{-} (g) ⟶$ Positive electron gain enthalpy

In first process energy is released while in the second energy is absorbed.

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Similar questions

K^{+} (g) I \to K (g) I I - \to K (s)

For the processes $K^{+} (g) I ⟶ K (g) I I ⟶ K^{-} (g)$

K_{p}

K_{c}

(i) N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

(i i) 2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g)

K_{p_{1}} = K_{c} (R T)^{x}

K_{p_{2}} = K_{c} (R T)^{y}

2 N H_{3} (g) P t \to N_{2} (g) + 3 H_{2} (g)

K_{c}

(i) 2 N O C l (g) ⇌ 2 N O (g) + C l_{2} (g); K_{P} = 1.8 \times 10^{- 2} a t 500 K

(i i) C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g); K_{p} = 167 a t 1073 K

R = 0.0831 b a r L m o l^{- 1} K^{- 1}

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