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).

Question

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).

Quizplus · Accepted Answer

The Answer of 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).

For the Reaction SbCl5(g) →\rarr→ SbCl3(g)+ Cl2(g) Δ\DeltaΔ G°f (SbCl5)= -334 Δ\DeltaΔ G°f (SbCl3)= -301

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