For the Reaction SbCl₅(g) $\rarr$ SbCl₃(g)+ Cl₂(g) $\Delta$ G°_f (SbCl₅)= -334 $\Delta$ G°_f (SbCl₃)= -301

question 109

Short Answer

For the reaction SbCl₅(g) $\rarr$ 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 $\Delta$ G at 800 K and 1 atm pressure (assume $\Delta$ S° and $\Delta$ H° do not change with temperature).

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