For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334

The Answer of For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334

For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334

The Answer of For the Reaction SbCl 5 (g) SbCl 3 (g)+ Cl 2 (g)$\Delta$ G° f (SbCl 5 )= -334

Solved

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

Question 83

Short Answer

For the reaction SbCl₅(g) $For the reaction SbCl5(g) SbCl3(g)+ Cl2(g), \Delta G°f (SbCl5)= -334.34 kJ/mol \Delta G°f (SbCl3)= -301.25 kJ/mol \Delta H°f (SbCl5)= -394.34 kJ/mol \Delta H°f (SbCl3)= -313.80 kJ/mol Calculate the value of the equilibrium constant (Kp)for this reaction at 298 K.$
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) SbCl3(g)+ Cl2(g) Δ\DeltaΔ G°f (SbCl5)= -334

Short Answer

Correct Answer:VerifiedUnlock this answer nowGet Access to more Verified Answers free of chargeAccess For Free

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

Correct Answer:
Verified
Unlock this answer now
Get Access to more Verified Answers free of charge