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 (K_p)for this reaction at 298 K.

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The Answer of 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 (K_p)for this reaction at 298 K.

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

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