Examlex
For the reaction SbCl5(g) SbCl3(g)+ Cl2(g),
G°f (SbCl5)= -334.34 kJ/mol
G°f (SbCl3)= -301.25 kJ/mol
H°f (SbCl5)= -394.34 kJ/mol
H°f (SbCl3)= -313.80 kJ/mol
Calculate the value of the equilibrium constant (Kp)for this reaction at 298 K.
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