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 the value of the equilibrium constant (K_p)for this reaction at 298 K.

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 the value of the equilibrium constant (K_p)for this reaction at 298 K.

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 the value of the equilibrium constant (K_p)for this reaction at 298 K.

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