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 70

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For the reaction SbCl₅(g) $For the reaction SbCl5(g) SbCl3(g)+ Cl2(g), \Delta G°f (SbCl5)= -334.34 kJ/mol \Delta G°f (SbCl3)= -301.25 kJ/mol \Delta H°f (SbCl5)= -394.34 kJ/mol \Delta H°f (SbCl3)= -313.80 kJ/mol Calculate the value of the equilibrium constant (Kp)at 800 K and 1 atm pressure.$ 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 the value of the equilibrium constant (K_p)at 800 K and 1 atm pressure.

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

Short Answer

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

Correct Answer:
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