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

question 23

Short Answer

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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