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

Question 32

Short Answer

For the reaction SbCl5(g) For the reaction SbCl<sub>5</sub>(g) <sub> </sub> SbCl<sub>3</sub>(g)+ Cl<sub>2</sub>(g), \Delta G°<sub>f</sub> (SbCl<sub>5</sub>)= -334.34 kJ/mol \Delta G°<sub>f</sub> (SbCl<sub>3</sub>)= -301.25 kJ/mol \Delta H°<sub>f</sub> (SbCl<sub>5</sub>)= -394.34 kJ/mol \Delta H°<sub>f</sub> (SbCl<sub>3</sub>)= -313.80 kJ/mol Calculate the value of the equilibrium constant (K<sub>p</sub>)at 800 K and 1 atm pressure.
SbCl3(g)+ Cl2(g),
Δ\Delta G°f (SbCl5)= -334.34 kJ/mol
Δ\Delta G°f (SbCl3)= -301.25 kJ/mol
Δ\Delta H°f (SbCl5)= -394.34 kJ/mol
Δ\Delta H°f (SbCl3)= -313.80 kJ/mol
Calculate the value of the equilibrium constant (Kp)at 800 K and 1 atm pressure.

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