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Question: Reaction\(H_{2}(g) + I_{2}(g) \rightarrow 2HI(g);\Delta H = - 12.40kcal\). According to this, the h...

ReactionH2(g)+I2(g)→2HI(g);ΔH=−12.40kcalH_{2}(g) + I_{2}(g) \rightarrow 2HI(g);\Delta H = - 12.40kcal.

According to this, the heat of formation of HI will be

A

12.4 kcal

B

–12.4 kcal

C

–6.20 kcal

D

6.20 kcal

Answer

–6.20 kcal

Explanation

Solution

H2(g)+I2(g)→2HI(g)H_{2}(g) + I_{2}(g) \rightarrow 2HI(g)

For 2M,ΔH=−12.40kcal2M,\Delta H = - 12.40kcal

1M,−12.402=−6.20kcal1M,\frac{- 12.40}{2} = - 6.20kcal