For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334$\Delta$

The Answer of For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334$\Delta$

For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334$\Delta$

The Answer of For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334$\Delta$

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For the Reaction SbCl₅(g) SbCl₃(g)+ Cl₂(g) $\Delta$ G°_f (SbCl₅)= -334 $\Delta$

Question 95

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For the reaction SbCl₅(g) $For the reaction SbCl5(g) SbCl3(g)+ Cl2(g), \Delta G°f (SbCl5)= -334.34 kJ/mol \Delta G°f (SbCl3)= -301.25 kJ/mol \Delta H°f (SbCl5)= -394.34 kJ/mol \Delta H°f (SbCl3)= -313.80 kJ/mol Calculate \Delta G at 800 K and 1 atm pressure (assume \Delta S° and \Delta H° do not change with temperature).$ SbCl₃(g)+ Cl₂(g),
$\Delta$ G°_f (SbCl₅)= -334.34 kJ/mol
$\Delta$ G°_f (SbCl₃)= -301.25 kJ/mol
$\Delta$ H°_f (SbCl₅)= -394.34 kJ/mol
$\Delta$ H°_f (SbCl₃)= -313.80 kJ/mol
Calculate $\Delta$ G at 800 K and 1 atm pressure (assume $\Delta$ S°and $\Delta$ H° do not change with temperature).

For the Reaction SbCl5(g) SbCl3(g)+ Cl2(g) Δ\DeltaΔ G°f (SbCl5)= -334 Δ\DeltaΔ

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For the Reaction SbCl₅(g) SbCl₃(g)+ Cl₂(g) $\Delta$ G°_f (SbCl₅)= -334 $\Delta$

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