According to the following reaction, heat of formation of $H I$ will be:

$H_{2} (g) + I_{2} (g) \to 2 H I$ ; $Δ H = 12.40 k c a l$

+ 12.40 k c a l

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- 12.4 k c a l

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- 6.20 k c a l

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+ 6.20 k c a l

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Solution

The correct option is D $+ 6.20 k c a l$

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Similar questions

$H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = 51.9$ kJ

According to the above reaction, the heat of formation of $H I$ (in kJ/mol) is:

H_{2} (g) + I_{2} (s) \to 2 H I (g); Δ H = 51.9 K J / m o l e

H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 9.2 K J / m o l e

The heat of reaction of $I_{2} (s) \to I_{2} (g)$ is:

N_{2} (g) + O_{2} (g) \to 2 N O (g)

Δ H = 181

2 C O_{2} (g) \to 2 C O (g) + O_{2} (g); Δ H = 566

H_{2} (g) + I_{2} (g) \to 2 H I (g); Δ H = - 94

H_{2} (g) + F_{2} (g) \to 2 H F (g); Δ H = - 541

H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)

Δ H = - 3000 c a l o r i e s

The heat of formation of HI from the equation

$H_{2} + I_{2} \to 2 H I, Δ H = + 52 K J$ is

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