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For the Reaction SbCl5(g)
SbCl3(g)+ Cl2(g) Δ\Deltaf (SbCl5)= -334

Question 56

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 Will this reaction proceed spontaneously at 298 K and 1 atm pressure?
SbCl3(g)+ Cl2(g),
Δ\Deltaf (SbCl5)= -334.34 kJ/mol
Δ\Deltaf (SbCl3)= -301.25 kJ/mol
Δ\Deltaf (SbCl5)= -394.34 kJ/mol
Δ\Deltaf (SbCl3)= -313.80 kJ/mol
Will this reaction proceed spontaneously at 298 K and 1 atm pressure?

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