Evaluate ∆U at 500 K for the processMgs+2HClg→ MgCl2s+H2g; ∆H=-109kcalThe volume change of solids to be ignored

∆H=∆U+∆ngRT -109=∆U+-1 × 2 × 5001000kcal -109=∆U-1 kcal ∆U=-109 + 1=-108kcal

Evaluate ∆U at 500 K for the processMgs+2HClg→ MgCl2s+H2g; ∆H=-109kcalThe volume change of solids to be ignored

∆H=∆U+∆ngRT -109=∆U+-1 × 2 × 5001000kcal -109=∆U-1 kcal ∆U=-109 + 1=-108kcal

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Evaluate $∆ U$ at 500 K for the process
$Mg (s) + 2 HCl (g) \to {MgCl}_{2} (s) + H_{2} (g); ∆ H = - 109 kcal$
The volume change of solids to be ignored

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-204 Kcal

-108 Kcal

-9 Kcal

+9 kcal

answer is B.

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$∆ H = ∆ U + ∆ n_{g} RT - 109 = ∆ U + (\frac{- 1 \times 2 \times 500}{1000}) kcal - 109 = ∆ U - 1 kcal ∆ U = - 109 + 1 = - 108 kcal$

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Evaluate ∆U at 500 K for the processMgs+2HClg→ MgCl2s+H2g; ∆H=-109kcalThe volume change of solids to be ignored

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Evaluate $∆ U$ at 500 K for the process
$Mg (s) + 2 HCl (g) \to {MgCl}_{2} (s) + H_{2} (g); ∆ H = - 109 kcal$
The volume change of solids to be ignored